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Precalculus—

Appendix A
A Review of Basic Concepts and Skills
APPENDIX OUTLINE
A.1 Algebraic Expressions and the Properties of Real Numbers A-1 A.2 Exponents, Scientific Notation, and a Review of Polynomials A-10 A.3 Solving Linear Equations and Inequalities A-24 A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring A-38 A.5 Rational Expressions and Equations A-52 A.6 Radicals, Rational Exponents, and Radical Equations A-64

A.1

Algebraic Expressions and the Properties of Real Numbers
To effectively use mathematics as a problem-solving tool, we must develop the ability to translate written or verbal information into a mathematical model. After obtaining a model, many applications require that you work effectively with algebraic terms and expressions. The basic ideas involved are reviewed here.

LEARNING OBJECTIVES
In Section A.1 you will review how to:

A. Identify terms,
coefficients, and expressions Create mathematical models Evaluate algebraic expressions Identify and use properties of real numbers Simplify algebraic expressions

B. C. D.

A. Terms, Coefficients, and Algebraic Expressions
An algebraic term is a collection of factors that may include numbers, variables, or expressions within parentheses. Here are some examples: (a) 3 (b) Ϫ6P (c) 5xy (d) Ϫ8n2 (e) n (f ) 2 1 x ϩ 3 2 If a term consists of a single nonvariable number, it is called a constant term. In (a), 3 is a constant term. Any term that contains a variable is called a variable term. We call the constant factor of a term the numerical coefficient or simply the coefficient. The coefficients for (a), (b), (c), and (d) are 3, Ϫ6, 5, and Ϫ8, respectively. In (e), the coefficient of n is 1, since 1 # n ϭ 1n ϭ n. The term in (f ) has two factors as written, 2 and 1 x ϩ 3 2 . The coefficient is 2. An algebraic expression can be a single term or a sum or difference of terms. To avoid confusion when identifying the coefficient of each term, the expression can be rewritten using algebraic addition if desired: A Ϫ B ϭ A ϩ 1 ϪB 2 . For instance, 4 Ϫ 3x ϭ 4 ϩ 1 Ϫ3x 2 shows the coefficient of x is Ϫ3. To identify the coefficient of a rational term, it sometimes helps to decompose the term, rewriting it using a unit 2 x 1 ϭ1 fraction as in n Ϫ 5 5 1 n Ϫ 2 2 and 2 ϭ 2 x.
A-1

E.

Precalculus—

A-2

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 1



Identifying Terms and Coefficients State the number of terms in each expression as given, then identify the coefficient of each term. xϩ3 Ϫ 2x a. 2x Ϫ 5y b. c. Ϫ 1 x Ϫ 12 2 d. Ϫ2x2 Ϫ x ϩ 5 7

Solution



We can begin by rewriting each subtraction using algebraic addition.
Rewritten: Number of terms: Coefficient(s):
2 a. 2x ϩ 1 Ϫ5y 2 b. 1 7 1 x ϩ 3 2 ϩ 1Ϫ2x 2 c. Ϫ1 1 x Ϫ 12 2 d. Ϫ2x ϩ 1 Ϫ1x 2 ϩ 5

two 2 and Ϫ5
1 7

two and Ϫ2

one Ϫ1

three Ϫ2, Ϫ1, and 5


A. You’ve just seen how we can identify terms, coefficients, and expressions

Now try Exercises 7 through 14

B. Translating Written or Verbal Information into a Mathematical Model
The key to solving many applied problems is finding an algebraic expression that accurately models relationships described in context. First, we assign a variable to represent an unknown quantity, then build related expressions using words from the English language that suggest mathematical operations. Variables that remind us of what they represent are often used in the modeling process, such as D ϭ RT for Distance equals Rate times Time. These are often called descriptive variables. Capital letters are also used due to their widespread appearance in other fields. EXAMPLE 2


Translating English Phrases into Algebraic Expressions Assign a variable to the unknown number, then translate each phrase into an algebraic expression. a. twice a number, increased by five b. eleven less than eight times the width c. ten less than triple the payment d. two hundred fifty dollars more than double the amount

Solution



a. Let n represent the number. Then 2n represents twice the number, and 2n ϩ 5 represents twice the number, increased by five. b. Let W represent the width. Then 8W represents eight times the width, and 8W Ϫ 11 represents 11 less than eight times the width. c. Let p represent the payment. Then 3p represents triple the payment, and 3p Ϫ 10 represents 10 less than triple the payment. d. Let A represent the amount in dollars. Then 2A represents double the amount, and 2A ϩ 250 represents 250 dollars more than double the amount. Now try Exercises 15 through 28


Identifying and translating such phrases when they occur in context is an important problem-solving skill. Note how this is done in Example 3. EXAMPLE 3


Creating a Mathematical Model The cost for a rental car is $35 plus 15 cents per mile. Express the cost of renting a car in terms of the number of miles driven.

Solution



B. You’ve just seen how we can create mathematical models

Let m represent the number of miles driven. Then 0.15m represents the cost for each mile and C ϭ 35 ϩ 0.15m represents the total cost for renting the car. Now try Exercises 29 through 40


Precalculus—

Section A.1 Algebraic Expressions and the Properties of Real Numbers

A-3

C. Evaluating Algebraic Expressions
We often need to evaluate expressions to investigate patterns and note relationships. Evaluating a Mathematical Expression 1. Replace each variable with open parentheses ( ). 2. Substitute the values given for each variable. 3. Simplify using the order of operations. In this process, it’s best to use a vertical format, with the original expression written first, the substitutions shown next, followed by the simplified forms and the final result. The numbers substituted or “plugged into” the expression are often called the input values, with the result called the output value. EXAMPLE 4


Evaluating an Algebraic Expression Evaluate the expression x3 Ϫ 2x2 ϩ 5 for x ϭ Ϫ3. For x ϭ Ϫ3: x3 Ϫ 2x2 ϩ 5 ϭ 1 2 3 Ϫ 2 1 2 2 ϩ 5 ϭ 1 Ϫ3 2 3 Ϫ 2 1 Ϫ3 2 2 ϩ 5 ϭ Ϫ27 Ϫ 2 1 9 2 ϩ 5 ϭ Ϫ27 Ϫ 18 ϩ 5 ϭ Ϫ40

Solution
WORTHY OF NOTE
In Example 4, note the importance of the first step in the evaluation process: replace each variable with open parentheses. Skipping this step could easily lead to confusion as we try to evaluate the squared term, since Ϫ32 ϭ Ϫ9, while 1 Ϫ3 2 2 ϭ 9. Also see Exercises 55 and 56.



replace variables with open parentheses simplify: 1 Ϫ3 2 3 ϭ Ϫ27, 1 Ϫ3 2 2 ϭ 9 simplify: 2 1 9 2 ϭ 18 result substitute Ϫ3 for x

When the input is Ϫ3, the output is Ϫ40. Now try Exercises 41 through 60


If the same expression is evaluated repeatedly, results are often collected and analyzed in a table of values, as shown in Example 5. As a practical matter, the substitutions and simplifications are often done mentally or on scratch paper, with the table showing only the input and output values. EXAMPLE 5


Evaluating an Algebraic Expression Evaluate x2 Ϫ 2x Ϫ 3 to complete the table shown. Which input value(s) of x cause the expression to have an output of 0?

Solution



Input x Ϫ2 Ϫ1 0 1 2 3 4

1 Ϫ2 2 2 Ϫ 2 1 Ϫ2 2 Ϫ 3 ϭ 5 0 Ϫ3 Ϫ4 Ϫ3 0 5

Output x2 ؊ 2x ؊ 3

The expression has an output of 0 when x ϭ Ϫ1 and x ϭ 3. Now try Exercises 61 through 66


Precalculus—

A-4

APPENDIX A A Review of Basic Concepts and Skills

Figure A.1

C. You’ve just seen how we can evaluate algebraic expressions

Graphing calculators provide an efficient means of evaluating many expressions. After entering the expression on the Y= screen (Figure A.1), we can set up the table using the keystrokes 2nd (TBLSET). For this exercise, we’ll put the table in the “Indpnt: Auto Ask” mode, which will have the calculator “automatically” generate the input and output values. In this mode, we can tell the calculator where to start the inputs (we chose TblStart ϭ Ϫ2), and have the calculator produce the input values using any increment desired (we choose ¢ Tbl ϭ 1), as shown in Figure A.2. We access the completed table using 2nd GRAPH (TABLE), and the result for Example 5 is shown in Figure A.3. For exercises that combine the skills from Examples 3 through 5, see Exercises 91 to 98.
WINDOW

Figure A.2

Figure A.3

D. Properties of Real Numbers
While the phrase, “an unknown number times five,” is accurately modeled by the expression n5 for some number n, in algebra we prefer to have numerical coefficients precede variable factors. When we reorder the factors as 5n, we are using the commutative property of multiplication. A reordering of terms involves the commutative property of addition. The Commutative Properties Given that a and b represent real numbers: ADDITION: a ϩ b ϭ b ϩ a Terms can be combined in any order without changing the sum. MULTIPLICATION: a # b ϭ b # a Factors can be multiplied in any order without changing the product.

Each property can be extended to include any number of terms or factors. While the commutative property implies a reordering or movement of terms (to commute implies back-and-forth movement), the associative property implies a regrouping 3 2 or reassociation of terms. For example, the sum 1 3 4 ϩ 5 2 ϩ 5 is easier to compute if we 3 2 regroup the addends as 3 4 ϩ 1 5 ϩ 5 2 . This illustrates the associative property of addition. Multiplication is also associative. The Associative Properties Given that a, b, and c represent real numbers: 1a ϩ b2 ϩ c ϭ a ϩ 1b ϩ c2 Terms can be regrouped. ADDITION: 1a # b2 # c ϭ a # 1b # c2 MULTIPLICATION:

Factors can be regrouped.

EXAMPLE 6



Simplifying Expressions Using Properties of Real Numbers Use the commutative and associative properties to simplify each calculation. a.
3 8 3 8

Ϫ 19 ϩ 5 8

Solution



a.

3 5 Ϫ 19 ϩ 5 8 ϭ Ϫ19 ϩ 8 ϩ 8 5 ϭ Ϫ19 ϩ 1 3 8 ϩ 82 ϭ Ϫ19 ϩ 1 ϭ Ϫ18

b. 3 Ϫ2.5 # 1 Ϫ1.2 2 4 # 10

commutative property (order changes) associative property (grouping changes) simplify result

Precalculus—

Section A.1 Algebraic Expressions and the Properties of Real Numbers

A-5

WORTHY OF NOTE
Is subtraction commutative? Consider a situation involving money. If you had $100, you could easily buy an item costing $20: $100 Ϫ $20 leaves you with $80. But if you had $20, could you buy an item costing $100? Obviously $100 Ϫ $20 is not the same as $20 Ϫ $100. Subtraction is not commutative. Likewise, 100 Ϭ 20 is not the same as 20 Ϭ 100, and division is not commutative.

b. 3 Ϫ2.5 # 1 Ϫ1.2 2 4 # 10 ϭ Ϫ2.5 # 3 1 Ϫ1.2 2 # 10 4 ϭ Ϫ2.5 # 1 Ϫ12 2 ϭ 30

associative property (grouping changes) simplify result

Now try Exercises 67 and 68



For any real number x, x ϩ 0 ϭ x and 0 is called the additive identity since the original number was returned or “identified.” Similarly, 1 is called the multiplicative identity since 1 # x ϭ x. The identity properties are used extensively in the process of solving equations. The Additive and Multiplicative Identities Given that x is a real number, xϩ0ϭx Zero is the identity for addition. 1#xϭx One is the identity for multiplication.

For any real number x, there is a real number Ϫx such that x ϩ 1 Ϫx 2 ϭ 0. The number Ϫx is called the additive inverse of x, since their sum results in the additive iden#1 tity. Similarly, the multiplicative inverse of any nonzero number x is 1 x , since x x ϭ 1 p q (the multiplicative identity). This property can also be stated as q # p ϭ 1 1 p, q 0 2 for q p p any rational number q. Note that q and p are reciprocals. The Additive and Multiplicative Inverses x ϩ 1 Ϫx 2 ϭ 0 Given that p, q, and x represent real numbers 1 p, q 02: p q # ϭ1 q p and p are multiplicative inverses.
p q q

x and Ϫx are additive inverses. EXAMPLE 7


Determining Additive and Multiplicative Inverses Replace the box to create a true statement: # Ϫ3 x ϭ 1 # x a. b. x ϩ 4.7 ϩ 5 ϭx ϭ Ϫ4.7, since 4.7 ϩ 1 Ϫ4.7 2 ϭ 0 Now try Exercises 69 and 70


Solution



a.

ϭ

5 5 Ϫ3 # , since ϭ1 Ϫ3 Ϫ3 5

b.

Note that if no coefficient is indicated, it is assumed to be 1, as in x ϭ 1x, 1 x2 ϩ 3x 2 ϭ 1 1 x2 ϩ 3x 2 , and Ϫ 1 x3 Ϫ 5x2 2 ϭ Ϫ1 1 x3 Ϫ 5x2 2 . The distributive property of multiplication over addition is widely used in a study of algebra, because it enables us to rewrite a product as an equivalent sum and vice versa. The Distributive Property of Multiplication over Addition Given that a, b, and c represent real numbers: a 1 b ϩ c 2 ϭ ab ϩ ac A factor outside a sum can be distributed to each addend in the sum. ab ϩ ac ϭ a 1 b ϩ c 2

A factor common to each addend in a sum can be “undistributed” and written outside a group.

Precalculus—

A-6

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 8



Simplifying Expressions Using the Distributive Property Apply the distributive property as appropriate. Simplify if possible. a. 7 1 p ϩ 5.2 2 a. 7 1 p ϩ 5.2 2 ϭ 7p ϩ 7 1 5.2 2 ϭ 7p ϩ 36.4 c. 7x3 Ϫ x3 ϭ 7x3 Ϫ 1x3 ϭ 1 7 Ϫ 1 2 x3 ϭ 6x3 b. Ϫ 1 2.5 Ϫ x 2 b. Ϫ 1 2.5 Ϫ x 2 ϭ Ϫ1 1 2.5 Ϫ x 2 ϭ Ϫ1 1 2.5 2 Ϫ 1 Ϫ1 2 1 x 2 ϭ Ϫ2.5 ϩ x d. 5 1 5 1 n ϩ n ϭ a ϩ bn 2 2 2 2 6 ϭ a bn 2 ϭ 3n Now try Exercises 71 through 78


c. 7x3 Ϫ x3

d.

Solution



5 1 nϩ n 2 2

WORTHY OF NOTE
From Example 8b we learn that a negative sign outside a group changes the sign of all terms within the group: Ϫ 1 2.5 Ϫ x 2 ϭ Ϫ2.5 ϩ x.

D. You’ve just seen how we can identify and use properties of real numbers

E. Simplifying Algebraic Expressions
Two terms are like terms only if they have the same variable factors (the coefficient is 3 2 not used to identify like terms). For instance, 3x2 and Ϫ1 7 x are like terms, while 5x 2 and 5x are not. We simplify expressions by combining like terms using the distributive property, along with the commutative and associative properties. Many times the distributive property is used to eliminate grouping symbols and combine like terms within the same expression. EXAMPLE 9


Simplifying an Algebraic Expression 7 1 2p2 ϩ 1 2 Ϫ 1 1 p2 ϩ 3 2

Simplify the expression completely: 7 1 2p2 ϩ 1 2 Ϫ 1 p2 ϩ 3 2 .
original expression; note coefficient of Ϫ1 distributive property commutative and associative properties (collect like terms) distributive property result

Solution



ϭ 14p 2 ϩ 7 Ϫ 1p 2 Ϫ 3 ϭ 1 14p2 Ϫ 1p2 2 ϩ 1 7 Ϫ 3 2 ϭ 1 14 Ϫ 1 2 p2 ϩ 4 ϭ 13p2 ϩ 4

Now try Exercises 79 through 88 The steps for simplifying an algebraic expression are summarized here: To Simplify an Expression 1. Eliminate parentheses by applying the distributive property. 2. Use the commutative and associative properties to group like terms. 3. Use the distributive property to combine like terms.
E. You’ve just seen how we can simplify algebraic expressions



As you practice with these ideas, many of the steps will become more automatic. At some point, the distributive property, the commutative and associative properties, as well as the use of algebraic addition will all be performed mentally.

Precalculus—

Section A.1 Algebraic Expressions and the Properties of Real Numbers

A-7

A.1 EXERCISES


CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. A term consisting of a single number is called a(n) term. 2. A term containing a variable is called a(n) term. 3. The constant factor in a variable term is called the .

#2# 4. When 3 # 14 # 2 3 is written as 3 3 14, the property has been used.
5. Discuss/Explain why the additive inverse of Ϫ5 is 5, while the multiplicative inverse of Ϫ5 is Ϫ1 5. 6. Discuss/Explain how we can rewrite the sum 3x ϩ 6y as a product, and the product 2 1 x ϩ 7 2 as a sum.



DEVELOPING YOUR SKILLS
Create a mathematical model using descriptive variables.

Identify the number of terms in each expression and the coefficient of each term.

7. 3x Ϫ 5y 9. 2x ϩ xϩ3 4

8. Ϫ2a Ϫ 3b 10. nϪ5 ϩ 7n 3

29. The length of the rectangle is three meters less than twice the width. 30. The height of the triangle is six centimeters less than three times the base. 31. The speed of the car was fifteen miles per hour more than the speed of the bus. 32. It took Romulus three minutes more time than Remus to finish the race. 33. Hovering altitude: The helicopter was hovering 150 ft above the top of the building. Express the altitude of the helicopter in terms of the building’s height.

11. Ϫ2x2 ϩ x Ϫ 5 13. Ϫ 1 x ϩ 5 2

12. 3n2 ϩ n Ϫ 7 14. Ϫ 1 n Ϫ 3 2

Translate each phrase into an algebraic expression.

15. seven fewer than a number 16. a number decreased by six 17. the sum of a number and four 18. a number increased by nine 19. the difference between a number and five is squared 20. the sum of a number and two is cubed 21. thirteen less than twice a number 22. five less than double a number 23. a number squared plus the number doubled 24. a number cubed less the number tripled 25. five fewer than two-thirds of a number 26. fourteen more than one-half of a number 27. three times the sum of a number and five, decreased by seven 28. five times the difference of a number and two, increased by six

34. Stacks on a cruise liner: The smoke stacks of the luxury liner cleared the bridge by 25 ft as it passed beneath it. Express the height of the stacks in terms of the bridge’s height. 35. Dimensions of a city park: The length of a rectangular city park is 20 m more than twice its width. Express the length of the park in terms of the width.

Precalculus—

A-8

APPENDIX A A Review of Basic Concepts and Skills

36. Dimensions of a parking lot: In order to meet the city code while using the available space, a contractor planned to construct a parking lot with a length that was 50 ft less than three times its width. Express the length of the lot in terms of the width. 37. Cost of milk: In 2010, a gallon of milk cost two and one-half times what it did in 1990. Express the cost of a gallon of milk in 2010 in terms of the 1990 cost. 38. Cost of gas: In 2010, a gallon of gasoline cost two and one-half times what it did in 1990. Express the cost of a gallon of gas in 2010 in terms of the 1990 cost. 39. Pest control: In her pest control business, Judy charges $50 per call plus $12.50 per gallon of insecticide for the control of spiders and certain insects. Express the total charge in terms of the number of gallons of insecticide used. 40. Computer repairs: As his reputation and referral business grew, Keith began to charge $75 per service call plus an hourly rate of $50 for the repair and maintenance of home computers. Express the cost of a service call in terms of the number of hours spent on the call.
Evaluate each algebraic expression given x ‫ ؍‬2 and y ‫ ؍‬؊3.

Rewrite each expression using the given property and simplify if possible.

67. Commutative property of addition a. Ϫ5 ϩ 7 b. Ϫ2 ϩ n c. Ϫ4.2 ϩ a ϩ 13.6 d. 7 ϩ x Ϫ 7 68. Associative property of multiplication a. 2 # 1 3 # 6 2 b. 3 # 1 4 # b 2 # c. Ϫ1.5 # 1 6 # a 2 d. Ϫ6 # 1 Ϫ5 6 x2
Replace the box so that a true statement results.

69. a. x ϩ 1 Ϫ3.2 2 ϩ b. n Ϫ 5 6ϩ

ϭx

ϭn

70. a. b.

#2 3 x ϭ 1x #
n ϭ 1n Ϫ3

Apply the distributive property and simplify if possible.

71. Ϫ5 1 x Ϫ 2.6 2
1 73. 2 3 1 Ϫ5 p ϩ 9 2

72. Ϫ12 1 v Ϫ 3.2 2
2 74. 5 6 1 Ϫ15 q ϩ 24 2

41. 4x Ϫ 2y 43. Ϫ2x2 ϩ 3y2 45. 2y ϩ 5y Ϫ 3
2

42. 5x Ϫ 3y 44. Ϫ5x2 ϩ 4y2 46. 3x ϩ 2x Ϫ 5
2

75. 3a ϩ 1 Ϫ5a 2
3 77. 2 3x ϩ 4x

76. 13m ϩ 1 Ϫ5m 2 78.
5 12 y

47. Ϫ2 1 3y ϩ 1 2 49. 3x y 53.
1 2x 2

51. 1 Ϫ3x 2 2 Ϫ 4xy Ϫ y2 52. 1 Ϫ2x 2 2 Ϫ 5xy Ϫ y2 55. 1 3x Ϫ 2y 2 2 Ϫ12y ϩ 5 Ϫ3x ϩ 1 Ϫ
1 3y

50. 6xy2 54.
2 3x

48. Ϫ3 1 2y ϩ 5 2

Ϫ3 8y

Simplify by removing all grouping symbols (as needed) and combining like terms.

56. 1 2x Ϫ 3y 2 2 58. Ϫ3y ϩ 1

Ϫ

1 2y

79. 3 1 a2 ϩ 3a 2 Ϫ 1 5a2 ϩ 7a 2 80. 2 1 b2 ϩ 5b 2 Ϫ 1 6b2 ϩ 9b 2 81. x2 Ϫ 1 3x Ϫ 5x2 2 82. n2 Ϫ 1 5n Ϫ 4n2 2

57.

12x ϩ 1 Ϫ3 2

59. 1Ϫ12y # 4

60. 7 # 1Ϫ27y

83. 1 3a ϩ 2b Ϫ 5c 2 Ϫ 1 a Ϫ b Ϫ 7c 2
5 85. 3 5 1 5n Ϫ 4 2 ϩ 8 1 n ϩ 16 2 3 86. 2 3 1 2x Ϫ 9 2 ϩ 4 1 x ϩ 12 2

Evaluate each expression for integers from ؊3 to 3 inclusive. Verify results using a graphing calculator. What input(s) give an output of zero?

84. 1 x Ϫ 4y ϩ 8z 2 Ϫ 1 8x Ϫ 5y Ϫ 2z 2

61. x2 Ϫ 3x Ϫ 4 65. x3 Ϫ 6x ϩ 4

63. Ϫ3 1 1 Ϫ x 2 Ϫ 6

62. x2 Ϫ 2x Ϫ 3

64. 5 1 3 Ϫ x 2 Ϫ 10 66. x3 ϩ 5x ϩ 18

87. 1 3a2 Ϫ 5a ϩ 7 2 ϩ 2 1 2a2 Ϫ 4a Ϫ 6 2

88. 2 1 3m2 ϩ 2m Ϫ 7 2 Ϫ 1 m2 Ϫ 5m ϩ 4 2

Precalculus—

Section A.1 Algebraic Expressions and the Properties of Real Numbers


A-9

WORKING WITH FORMULAS
kL d2 90. Volume and pressure: P ‫؍‬ k V If temperature remains constant, the pressure of a gas held in a closed container is related to the volume of gas by the formula shown, where P is the pressure in pounds per square inch, V is the volume of gas in cubic inches, and k is a constant that depends on given conditions. Find the pressure exerted by the gas if k ϭ 440,310 and V ϭ 22,580 in3.

89. Electrical resistance: R ‫؍‬

The electrical resistance in a wire depends on the length and diameter of the wire. This resistance can be modeled by the formula shown, where R is the resistance in ohms, L is the length in feet, and d is the diameter of the wire in inches. Find the resistance if k ϭ 0.000025, d ϭ 0.015 in., and L ϭ 90 ft



APPLICATIONS

Translate each key phrase into an algebraic expression, then evaluate as indicated.

91. Cruising speed: A turbo-prop airliner has a cruising speed that is one-half the cruising speed of a 767 jet aircraft. (a) Express the speed of the turbo-prop in terms of the speed of the jet, and (b) determine the speed of the airliner if the cruising speed of the jet is 550 mph. 92. Softball toss: Macklyn can throw a softball twothirds as far as her father. (a) Express the distance that Macklyn can throw a softball in terms of the distance her father can throw. (b) If her father can throw the ball 210 ft, how far can Macklyn throw the ball? 93. Dimensions of a lawn: The length of a rectangular lawn is 3 ft more than twice its width. (a) Express the length of the lawn in terms of the width. (b) If the width is 52 ft, what is the length? 94. Pitch of a roof: To obtain the proper pitch, the crossbeam for a roof truss must be 2 ft less than three-halves the rafter. (a) Express the length of the crossbeam in terms of the rafter. (b) If the rafter is 18 ft, how long is the crossbeam?


95. Postage costs: In 2009, a first class stamp cost 29¢ more than it did in 1978. Express the cost of a 2009 stamp in terms of the 1978 cost. If a stamp cost 15¢ in 1978, what was the cost in 2009? 96. Minimum wage: In 2009, the federal minimum wage was $4.95 per hour more than it was in 1976. Express the 2009 wage in terms of the 1976 wage. If the hourly wage in 1976 was $2.30, what was it in 2009? 97. Repair costs: The TV repair shop charges a flat fee of $43.50 to come to your house and $25 per hour for labor. Express the cost of repairing a TV in terms of the time it takes to repair it. If the repair took 1.5 hr, what was the total cost? 98. Repair costs: At the local car dealership, shop charges are $79.50 to diagnose the problem and $85 per shop hour for labor. Express the cost of a repair in terms of the labor involved. If a repair takes 3.5 hr, how much will it cost?

EXTENDING THE CONCEPT
100. Historically, several attempts have been made to create metric time using factors of 10, but our current system won out. If 1 day was 10 metric hours, 1 metric hour was 10 metric minutes, and 1 metric minute was 10 metric seconds, what time would it really be if a metric clock read 4:3:5? Assume that each new day starts at midnight.

99. If C must be a positive odd integer and D must be a negative even integer, then C2 ϩ D2 must be a: a. positive odd integer. b. positive even integer. c. negative odd integer. d. negative even integer. e. cannot be determined.

Precalculus—

A.2

Exponents, Scientific Notation, and a Review of Polynomials
In this section, we review basic exponential properties and operations on polynomials. Although there are five to eight exponential properties (depending on how you count them), all can be traced back to the basic definition involving repeated multiplication.

LEARNING OBJECTIVES
In Section A.2 you will review how to:

A. Apply properties of B. C. D. E. F.
exponents Perform operations in scientific notation Identify and classify polynomial expressions Add and subtract polynomials Compute the product of two polynomials Compute special products: binomial conjugates and binomial squares

A. The Properties of Exponents
An exponent is a superscript number or letter occurring to the upper right of a base number, and indicates how many times the base occurs as a factor. For b # b # b ϭ b3, we say b3 is written in exponential form. In some cases, we may refer to b3 as an exponential term. Exponential Notation For any positive integer n, bn ϭ b # b # b # . . . # b n times and b # b # b # . . . # b ϭ bn n times ⎞ ⎜ ⎜ ⎜ ⎬ ⎜ ⎜ ⎠ ⎞ ⎜ ⎜ ⎜ ⎬ ⎜ ⎜ ⎠

The Product and Power Properties
There are two properties that follow immediately from this definition. When b3 is multiplied by b2, we have an uninterrupted string of five factors: b3 # b2 ϭ 1 b # b # b 2 # 1 b # b 2 , which can then be written as b5. This is an example of the product property of exponents. Product Property of Exponents For any base b and positive integers m and n: bm # bn ϭ bm ϩ n In words, the property says, to multiply exponential terms with the same base, keep the common base and add the exponents. A special application of the product property uses repeated factors of the same exponential term, as in (x2)3. Using the product property, we have 1 x2 2 1 x2 21 x2 2 ϭ x6. Notice the same result can be found more quickly by # multiplying the inner exponent by the outer exponent: 1 x2 2 3 ϭ x2 3 ϭ x6. We generalize this idea to state the power property of exponents. In words the property says, to raise an exponential term to a power, keep the same base and multiply the exponents. Power Property of Exponents For any base b and positive integers m and n: 1 bm 2 n ϭ bm n
#

WORTHY OF NOTE
In this statement of the product property and the exponential properties that follow, it is assumed that for any expression of the form 0m, m 7 0 (hence 0m ϭ 0).

EXAMPLE 1



Multiplying Terms Using Exponential Properties Compute each product. 2 a. Ϫ4x3 # 1 b. 1 p3 2 2 # 1 p4 2 5 2x

Solution



a. Ϫ4x3

2 #1 # 1 3 # x2 2 2 x ϭ 1 Ϫ4 2 21 x

b. 1 p3 2 2

#

ϭ 1 Ϫ2 2 1 x3 ϩ 2 2 ϭ Ϫ2 x5 # # 4 5 1 p 2 ϭ p3 2 # p4 5 ϭ p6 # p20 ϭ p6 ϩ 20 ϭ p26

commutative and associative properties simplify; product property result power property simplify product property result

Now try Exercises 7 through 12 ᮣ
A-10

Precalculus—

Section A.2 Exponents, Scientific Notation, and a Review of Polynomials

A-11

The power property can easily be extended to include more than one factor within the parentheses. This application of the power property is sometimes called the product to a power property and can be extended to include any number of factors. We can also raise a quotient of exponential terms to a power. The result is called the quotient to a power property. In words the properties say, to raise a product or quotient of exponential terms to a power, multiply every exponent inside the parentheses by the exponent outside the parentheses. Product to a Power Property For any bases a and b, and positive integers m, n, and p: 1 ambn 2 p ϭ amp # bnp

Quotient to a Power Property For any bases a and b 0, and positive integers m, n, and p: a EXAMPLE 2


am p amp b ϭ np bn b

Simplifying Terms Using the Power Properties Simplify using the power property (if possible): Ϫ5a3 2 a. 1 Ϫ3a 2 2 b. Ϫ3a2 c. a b 2b

Solution
WORTHY OF NOTE
Regarding Examples 2a and 2b, note the difference between the expressions 1 Ϫ3a 2 2 ϭ 1 Ϫ3 # a 2 2 and Ϫ3a2 ϭ Ϫ3 # a2. In the first, the exponent acts on both the negative 3 and the a; in the second, the exponent acts on only the a and there is no “product to a power.”



a. 1 Ϫ3a 2 2 ϭ 1 Ϫ3 2 2 # 1 a1 2 2 ϭ 9a2 3 2 1 Ϫ5 2 2 1 a3 2 2 Ϫ5a b ϭ c. a 2b 22b2 25a6 ϭ 4b2

b. Ϫ3a2 is in simplified form

Now try Exercises 13 through 24 ᮣ Applications of exponents sometimes involve linking one exponential term with another using a substitution. The result is then simplified using exponential properties.

EXAMPLE 3



Applying the Power Property after a Substitution The formula for the volume of a cube is V ϭ S3, where S is the length of one edge. If the length of each edge is 2x2: a. Find a formula for volume in terms of x. b. Find the volume if x ϭ 2.

2x2 2x2

Solution



a. V ϭ S 3 ϭ 1 2x2 2 3 ϭ 8x6
substitute 2x 2 for S

b. For V ϭ 8x6, V ϭ 81226 substitute 2 for x ϭ 8 # 64 or 512 1 2 2 6 ϭ 64 The volume of the cube would be 512 units3. Now try Exercises 25 and 26 ᮣ

2x2

S

Precalculus—

A-12

APPENDIX A A Review of Basic Concepts and Skills

The Quotient Property of Exponents
x ϭ 1 for x 0, we note a x x5 x # x # x # x # x ϭ x3 , pattern that helps to simplify a quotient of exponential terms. For 2 ϭ x#x x the exponent of the final result appears to be the difference between the exponent in the numerator and the exponent in the denominator. This seems reasonable since the subtraction would indicate a removal of the factors that reduce to 1. Regardless of how many factors are used, we can generalize the idea and state the quotient property of exponents. In words the property says, to divide two exponential terms with the same base, keep the common base and subtract the exponent of the denominator from the exponent of the numerator. By combining exponential notation and the property Quotient Property of Exponents For any base b 0 and positive integers m and n: bm ϭ bm Ϫ n bn

Zero and Negative Numbers as Exponents
If the exponent of the denominator is greater than the exponent in the numerator, the x2 quotient property yields a negative exponent: 5 ϭ x2 Ϫ 5 ϭ xϪ3. To help understand x what a negative exponent means, let’s look at the expanded form of the expression: x2 x # x1 1 ϭ ϭ 3 . A negative exponent can literally be interpreted as “write x#x#x#x#x x5 x the factors as a reciprocal.” A good way to remember this is 2Ϫ3
three factors of 2 written as a reciprocal

2Ϫ3 1 1 ϭ 3ϭ 1 8 2

!

Since the result would be similar regardless of the base used, we can generalize this idea and state the property of negative exponents. Property of Negative Exponents For any base b 0 and integer n: bϪn 1 ϭ n 1 b
WORTHY OF NOTE
The use of zero as an exponent should not strike you as strange or odd; it’s simply a way of saying that no factors of the base remain, since all terms have been reduced to 1. 8 23 For 3 , we have ϭ 1, or 8 2 2#2#2 ϭ 1, or 23 Ϫ 3 ϭ 20 ϭ 1. 2#2#2
1 1 1

!

1 bn Ϫn ϭ b 1

a Ϫn b n a b ϭa b ;a a b

0

x3 x3 ϭ 1 ϭ x3 Ϫ 3 ϭ x0 using the by division, and x3 x3 quotient property, we conclude that x0 ϭ 1 as long as x 0. We can also generalize this observation and state the meaning of zero as an exponent. In words the property says, any nonzero quantity raised to an exponent of zero is equal to 1. Finally, when we consider that Zero Exponent Property For any base b 0: b0 ϭ 1

Precalculus—

Section A.2 Exponents, Scientific Notation, and a Review of Polynomials

A-13

EXAMPLE 4



Simplifying Expressions Using Exponential Properties Simplify using exponential properties. Answer using positive exponents only. 2a3 Ϫ2 a. a 2 b b. 1 3hkϪ2 2 3 1 6hϪ2kϪ3 2 Ϫ2 b 1 Ϫ2m2n3 2 5 c. 1 3x 2 0 ϩ 3x0 ϩ 3Ϫ2 d. 1 4mn2 2 3 3 Ϫ2 2 2 2a b a. a 2 b ϭ a 3 b property of negative exponents b 2a 1 b2 2 2 ϭ 2 3 2 power properties 2 1a 2 ϭ b. 1 3hkϪ2 2 3 1 6hϪ2kϪ3 2 Ϫ2 ϭ 33h3 1 kϪ2 2 3 # 6Ϫ2 1 hϪ2 2 Ϫ2 1 kϪ3 2 Ϫ2 ϭ3hk ϭ3 ϭ ϭ
3 3 3 Ϫ6

Solution



b4 4a6

result power property simplify product property simplify a 6Ϫ2 ϭ 1 62 ϭ 1 b 36

#6

Ϫ2

#6 hk # h3 ϩ 4 # kϪ6 ϩ 6

Ϫ2 4 6

27h k 36 3h7 4

7 0

result 1 k 0 ϭ 1 2

c. 1 3x 2 0 ϩ 3x0 ϩ 3Ϫ2 ϭ 1 ϩ 3 1 1 2 ϩ ϭ4ϩ 1 9

1 32

zero exponent property; property of negative exponents

simplify: (3x )0 ϭ 1, 3x 0 ϭ 3 и 1 ϭ 3

1 37 ϭ4 ϭ 9 9
WORTHY OF NOTE
Notice in Example 4(c), we have 1 3x 2 0 ϭ 1 3 # x 2 0 ϭ 1, while 3x0 ϭ 3 # x0 ϭ 3 1 1 2 . This is another example of operations and grouping symbols working together: 1 3x 2 0 ϭ 1 because any quantity to the zero power is 1. However, for 3x0 there are no grouping symbols, so the exponent 0 acts only on the x and not the 3: 3x0 ϭ 3 # x0 ϭ 3 1 1 2 ϭ 3.

result

d.

1 Ϫ2m2n3 2 5 1 4mn2 2 3

ϭ ϭ ϭ

1 Ϫ2 2 5 1 m2 2 5 1 n3 2 5 Ϫ32m10n15 64m3n6 Ϫ1m7n9 2 m7n9 2 43m3 1 n2 2 3

power property

simplify

quotient property

ϭϪ

result

Now try Exercises 27 through 66 ᮣ

Precalculus—

A-14

APPENDIX A A Review of Basic Concepts and Skills

Summary of Exponential Properties For real numbers a and b, and integers m, n, p (excluding 0 raised to a nonpositive power) bm # bn ϭ bm ϩ n Product property: # Power property: 1 bm 2 n ϭ bm n Product to a power: 1 ambn 2 p ϭ amp # bnp am p amp Quotient to a power: a n b ϭ np 1 b 0 2 b b bm Quotient property: ϭ bm Ϫ n 1 b 0 2 bn Zero exponents: b0 ϭ 1 1 b 0 2 bϪn 1 1 a Ϫn b n Negative exponents: ϭ n , Ϫn ϭ bn, a b ϭ a b 1 a, b 0 2 a 1 b b b

A. You’ve just seen how we can apply properties of exponents

B. Exponents and Scientific Notation
In many technical and scientific applications, we encounter numbers that are either extremely large or very, very small. For example, the mass of the Moon is over 73 quintillion kilograms (73 followed by 18 zeroes), while the constant for universal gravitation contains 10 zeroes before the first nonzero digit. When computing with numbers of this magnitude, scientific notation has a distinct advantage over the common decimal notation (base-10 place values).
WORTHY OF NOTE
Recall that multiplying by 10’s (or multiplying by 10k, k 7 0 2 shifts the decimal point to the right k places, making the number larger. Dividing by 10’s (or multiplying by 10Ϫk, k 7 0) shifts the decimal point to the left k places, making the number smaller.

Scientific Notation A non-zero number written in scientific notation has the form where 1 Յ 0 N 0 6 10 and k is an integer. To convert a number from decimal notation into scientific notation, we begin by placing the decimal point to the immediate right of the first nonzero digit (creating a number less than 10 but greater than or equal to 1) and multiplying by 10k. Then we determine the power of 10 (the value of k) needed to ensure that the two forms are equivalent. When writing large or small numbers in scientific notation, we sometimes round the value of N to two or three decimal places. N ϫ 10k

EXAMPLE 5



Converting from Decimal Notation to Scientific Notation The mass of the Moon is about 73,000,000,000,000,000,000 kg. Write this number in scientific notation.

Solution



Place decimal to the right of first nonzero digit (7) and multiply by 10k. 73,000,000,000,000,000,000 ϭ 7.3 ϫ 10k To return the decimal to its original position would require 19 shifts to the right, so k must be positive 19. 73,000,000,000,000,000,000 ϭ 7.3 ϫ 1019 The mass of the Moon is 7.3 ϫ 1019 kg. Now try Exercises 67 and 68 ᮣ

Precalculus—

Section A.2 Exponents, Scientific Notation, and a Review of Polynomials

A-15

Converting a number from scientific notation to decimal notation is simply an application of multiplication or division with powers of 10. EXAMPLE 6


Converting from Scientific Notation to Decimal Notation The constant of gravitation is 6.67 ϫ 10Ϫ11. Write this number in common decimal form.

Solution



Since the exponent is negative 11, shift the decimal 11 places to the left, using placeholder zeroes as needed to return the decimal to its original position: 6.67 ϫ 10Ϫ11 ϭ 0.000 000 000 066 7 Now try Exercises 69 and 70 ᮣ Computations that involve scientific notation typically use real number properties and the properties of exponents.

EXAMPLE 7



Storage Space on a Hard Drive A typical 320-gigabyte portable hard drive can hold about 340,000,000,000 bytes of information. A 2-hr DVD movie can take up as much as 8,000,000,000 bytes of storage space. Find the number of movies (to the nearest whole movie) that can be stored on this hard drive.

Solution



Using the ideas from Example 5, the hard drive holds 3.4 ϫ 1011 bytes, while the DVD requires 8.0 ϫ 109 bytes. Divide to find the number of DVDs the hard drive will hold. 3.4 ϫ 1011 3.4 1011 ϭ ϫ 8.0 8.0 ϫ 109 109 ϭ 0.425 ϫ 102 ϭ 42.5
rewrite the expression divide; subtract exponents result

The drive will hold approximately 42 DVD movies. A calculator check is shown in the figure.
B. You’ve just seen how we can perform operations in scientific notation

Now try Exercises 71 and 72 ᮣ

C. Identifying and Classifying Polynomial Expressions
A monomial is a term using only whole number exponents on variables, with no variables in the denominator. One important characteristic of a monomial is its degree. For a monomial in one variable, the degree is the same as the exponent on the variable. The degree of a monomial in two or more variables is the sum of exponents occurring on variable factors. A polynomial is a monomial or any sum or difference of monomial 2 Ϫ2 ϩ 2n Ϫ 7 is not (the terms. For instance, 1 2 x Ϫ 5x ϩ 6 is a polynomial, while 3n exponent Ϫ2 is not a whole number). Identifying polynomials is an important skill because they represent a very different kind of real-world model than nonpolynomials. In addition, there are different families of polynomials, with each family having different characteristics. We classify polynomials according to their degree and number of terms. The degree of a polynomial in one variable is the largest exponent occurring on the variable. The degree of a polynomial in more than one variable is the largest sum of exponents in any one term. A polynomial with two terms is called a binomial (bi means two) and a polynomial with three terms is called a trinomial (tri means three). There are special names for polynomials with four or more terms, but for these, we simply use the general name polynomial (poly means many).

Precalculus—

A-16

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 8



Classifying and Describing Polynomials For each expression: a. Classify as a monomial, binomial, trinomial, or polynomial. b. State the degree of the polynomial. c. Name the coefficient of each term.

Solution



Expression 5x y Ϫ 2xy
2

Classification binomial binomial polynomial (four terms) binomial trinomial

Degree three two three one two

Coefficients 5, Ϫ2 1, Ϫ0.81 1, Ϫ3, 9, Ϫ27
Ϫ3 4 ,

x2 Ϫ 0.81 z3 Ϫ 3z2 ϩ 9z Ϫ 27
Ϫ3 4 x 2

ϩ5

5

2x ϩ x Ϫ 3

2, 1, Ϫ3

Now try Exercises 73 through 78 ᮣ A polynomial expression is in standard form when the terms of the polynomial are written in descending order of degree, beginning with the highest-degree term. The coefficient of the highest-degree term is called the leading coefficient. EXAMPLE 9


Writing Polynomials in Standard Form Write each polynomial in standard form, then identify the leading coefficient.

Solution



Polynomial 9Ϫx
2 2 3

Standard Form Ϫx ϩ 9
2

Leading Coefficient Ϫ1 3
Ϫ3 4

5z ϩ 7z ϩ 3z Ϫ 27 Ϫ3 ϩ 2x2 ϩ x C. You’ve just seen how we can identify and classify polynomial expressions
3 2 ϩ 1Ϫ 4 2x

3z ϩ 7z ϩ 5z Ϫ 27
3 2

2x ϩ x Ϫ 3

Ϫ3 4 x 2

ϩ2

2

Now try Exercises 79 through 84 ᮣ

D. Adding and Subtracting Polynomials
Adding polynomials simply involves using the distributive, commutative, and associative properties to combine like terms (at this point, the properties are usually applied mentally). As with real numbers, the subtraction of polynomials involves adding the opposite of the second polynomial using algebraic addition. This can be viewed as distributing Ϫ1 to the second polynomial and combining like terms. EXAMPLE 10


Adding and Subtracting Polynomials Perform the indicated operations: 1 0.7n3 ϩ 4n2 ϩ 8 2 ϩ 1 0.5n3 Ϫ n2 Ϫ 6n 2 Ϫ 1 3n2 ϩ 7n Ϫ 10 2 .

Solution



0.7n3 ϩ 4n2 ϩ 8 ϩ 0.5n3 Ϫ n2 Ϫ 6n Ϫ 3n2 Ϫ 7n ϩ 10 ϭ 0.7n3 ϩ 0.5n3 ϩ 4n2 Ϫ 1n2 Ϫ 3n2 Ϫ 6n Ϫ 7n ϩ 8 ϩ 10 ϭ 1.2n3 Ϫ 13n ϩ 18

eliminate parentheses (distributive property) use properties to collect like terms combine like terms

Now try Exercises 85 through 90 ᮣ

Precalculus—

Section A.2 Exponents, Scientific Notation, and a Review of Polynomials

A-17

Sometimes it’s easier to add or subtract polynomials using a vertical format and aligning like terms. Note the use of a placeholder zero in Example 11. EXAMPLE 11


Subtracting Polynomials Using a Vertical Format Compute the difference of x3 Ϫ 5x ϩ 9 and x3 ϩ 3x2 ϩ 2x Ϫ 8 using a vertical format.

Solution



x3 ϩ 0x2 Ϫ 5x ϩ 9 x3 ϩ 0x2 Ϫ 5x ϩ 9 3 2 Ϫ 1 x ϩ 3x ϩ 2x Ϫ 8 2 ¡ Ϫx3 Ϫ 3x2 Ϫ 2x ϩ 8 Ϫ3x2 Ϫ 7x ϩ 17 The difference is Ϫ3x2 Ϫ 7x ϩ 17. Now try Exercises 91 and 92 ᮣ

D. You’ve just seen how we can add and subtract polynomials

E. The Product of Two Polynomials
Monomial Times Monomial
The simplest case of polynomial multiplication is the product of monomials shown in Example 1a. These were computed using exponential properties and the properties of real numbers.

Monomial Times Polynomial
To compute the product of a monomial and a polynomial, we use the distributive property. EXAMPLE 12


Multiplying a Monomial by a Polynomial Find the product: Ϫ2a2 1 a2 Ϫ 2a ϩ 1 2 . Ϫ2a2 1 a2 Ϫ 2a ϩ 1 2 ϭ Ϫ2a2 1 a2 2 Ϫ 1 Ϫ2a2 21 2a1 2 ϩ 1 Ϫ2a2 21 1 2 ϭ Ϫ2a4 ϩ 4a3 Ϫ 2a2

Solution



distribute simplify

Now try Exercises 93 and 94 ᮣ

Binomial Times Polynomial
For products involving binomials, we still use a version of the distributive property — this time to distribute one polynomial to each term of the other polynomial factor. Note the distribution can be performed either from the left or from the right. EXAMPLE 13


Multiplying a Binomial by a Polynomial Multiply as indicated: a. 1 2z ϩ 1 2 1 z Ϫ 2 2 b. 1 2v Ϫ 3 2 1 4v2 ϩ 6v ϩ 9 2

Solution



a. 1 2z ϩ 1 21 z Ϫ 2 2 ϭ 2z 1 z Ϫ 2 2 ϩ 1 1 z Ϫ 2 2 distribute to every term in the first binomial eliminate parentheses (distribute again) ϭ 2z2 Ϫ 4z ϩ 1z Ϫ 2 2 ϭ 2z Ϫ 3z Ϫ 2 simplify 2 2 b. 1 2v Ϫ 3 21 4v ϩ 6v ϩ 9 2 ϭ 2v 1 4v ϩ 6v ϩ 9 2 Ϫ3 1 4v2 ϩ 6v ϩ 9 2 distribute ϭ 8v3 ϩ 12v2 ϩ 18v Ϫ 12v2 Ϫ 18v Ϫ 27 simplify ϭ 8v3 Ϫ 27 combine like
terms

Now try Exercises 95 through 100 ᮣ

Precalculus—

A-18

APPENDIX A A Review of Basic Concepts and Skills

The F-O-I-L Method
By observing the product of two binomials in Example 13(a), we note a pattern that can make the process more efficient. The product of two binomials can quickly be computed using the First, Outer, Inner, Last (FOIL) method, an acronym giving the respective position of each term in a product of binomials in relation to the other terms. We illustrate here using the product 1 2x Ϫ 1 2 1 3x ϩ 2 2 . The F-O-I-L Method for Multiplying Binomials
Last First
S S

WORTHY OF NOTE

Consider the product 1 x ϩ 3 21 x ϩ 2 2 in the context of area. If we view x ϩ 3 as the length of a rectangle (an unknown length plus 3 units), and x ϩ 2 as its width (the same unknown length plus 2 units), a diagram of the total area would look like the following, with the result x2 ϩ 5x ϩ 6 clearly visible.

6x2 ϩ 4x Ϫ 3x Ϫ 2
First Outer Inner Last
S

x

3

1 2x Ϫ 1 2 1 3x ϩ 2 2
S S

Inner Outer

S

S

S

Combine like terms 6x2 ϩ x Ϫ 2

x

x2

3x

2

2x x2

6 ϩ 5x ϩ 6

(x ϩ 3)(x ϩ 2) ϭ

The first term of the result will always be the product of the first terms from each binomial, and the last term of the result is the product of their last terms. We also note that here, the middle term is found by adding the outermost product with the innermost product. As you practice with the F-O-I-L process, much of the work can be done mentally and you can often compute the entire product without writing anything down except the answer.


EXAMPLE 14

Multiplying Binomials Using F-O-I-L Compute each product mentally: a. 1 5n Ϫ 1 21 n ϩ 2 2 b. 1 2b ϩ 3 2 1 5b Ϫ 6 2

10n ϩ (Ϫ1n) ϭ 9n

S

product of first two terms

S

sum of outer and inner products

S

Solution



a. 1 5n Ϫ 1 2 1 n ϩ 2 2 :

5n2 ϩ 9n Ϫ 2
product of last two terms

Ϫ12b ϩ 15b ϭ 3b

b. 1 2b ϩ 3 21 5b Ϫ 6 2 : 10b2 ϩ 3b Ϫ 18
S product of first two terms S sum of outer and inner products S product of last two terms

E. You’ve just seen how we can compute the product of two polynomials

Now try Exercises 101 through 116 ᮣ

F. Special Polynomial Products
Certain polynomial products are considered “special” for two reasons: (1) the product follows a predictable pattern, and (2) the result can be used to simplify expressions, graph functions, solve equations, and/or develop other skills.

Binomial Conjugates
Expressions like x ϩ 7 and x Ϫ 7 are called binomial conjugates. For any given binomial, its conjugate is found by using the same two terms with the opposite sign

Precalculus—

Section A.2 Exponents, Scientific Notation, and a Review of Polynomials

A-19

between them. Example 15 shows that when we multiply a binomial and its conjugate, the “outers” and “inners” sum to zero and the result is a difference of two squares. EXAMPLE 15


Multiplying Binomial Conjugates Compute each product mentally: a. 1 x ϩ 7 21 x Ϫ 7 2 b. 1 2x Ϫ 5y 2 1 2x ϩ 5y 2
Ϫ7x ϩ 7x ϭ 0x

2 2 c. a x ϩ b a x Ϫ b 5 5

Solution



a. 1 x ϩ 7 2 1 x Ϫ 7 2 ϭ x2 Ϫ 49

difference of squares 1 x 2 2 Ϫ 1 7 2 2

10xy ϩ (Ϫ10xy) ϭ 0xy

b. 1 2x Ϫ 5y 21 2x ϩ 5y 2 ϭ 4x2 Ϫ 25y2

difference of squares: 1 2x 2 2 Ϫ 1 5y 2 2

2xϩ2x ϭ 0 Ϫ5 5

2 2 4 c. a x ϩ b a x Ϫ b ϭ x2 Ϫ 5 5 25

2 2 difference of squares: x2 Ϫ a b 5

Now try Exercises 117 through 124 ᮣ In summary, we have the following. The Product of a Binomial and Its Conjugate Given any expression that can be written in the form A ϩ B, the conjugate of the expression is A Ϫ B and their product is a difference of two squares: 1 A ϩ B 2 1 A Ϫ B 2 ϭ A 2 Ϫ B2

Binomial Squares

Expressions like 1 x ϩ 7 2 2 are called binomial squares and are useful for solving many equations and sketching a number of basic graphs. Note 1 x ϩ 7 2 2 ϭ 1 x ϩ 7 21 x ϩ 7 2 ϭ x2 ϩ 14x ϩ 49 using the F-O-I-L process. The expression x2 ϩ 14x ϩ 49 is called a perfect square trinomial because it is the result of expanding a binomial square. If we write a binomial square in the more general form 1 A ϩ B 2 2 ϭ 1 A ϩ B 21 A ϩ B 2 and compute the product, we notice a pattern that helps us write the expanded form more quickly.
LOOKING AHEAD
Although a binomial square can always be found using repeated factors and F-O-I-L, learning to expand them using the pattern is a valuable skill. Binomial squares occur often in a study of algebra and it helps to find the expanded form quickly.

1A ϩ B22 ϭ 1A ϩ B2 1A ϩ B2
2 2

repeated multiplication
2

ϭ A ϩ AB ϩ AB ϩ B ϭ A ϩ 2AB ϩ B
2

F-O-I-L simplify (perfect square trinomial)

The first and last terms of the trinomial are squares of the terms A and B. Also, the middle term of the trinomial is twice the product of these two terms: AB ϩ AB ϭ 2AB. The F-O-I-L process shows us why. Since the outer and inner products are identical, we always end up with two. A similar result holds for 1 A Ϫ B 2 2 and the process can be summarized for both cases using the Ϯ symbol.

Precalculus—

A-20

APPENDIX A A Review of Basic Concepts and Skills

The Square of a Binomial

Given any expression that can be written in the form 1 A Ϯ B 2 2, 1. 1 A ϩ B 2 2 ϭ A2 ϩ 2AB ϩ B2 2. 1 A Ϫ B 2 2 ϭ A2 Ϫ 2AB ϩ B2 CAUTION


Note the square of a binomial always results in a trinomial (three terms). In particular, 1 A ϩ B 2 2 A2 ϩ B2.

EXAMPLE 16



Find each binomial square without using F-O-I-L: a. 1 a ϩ 9 2 2 ϭ a2 ϩ 2 1 a # 9 2 ϩ 92 ϭ a2 ϩ 18a ϩ 81 2 b. 1 3x Ϫ 5 2 ϭ 1 3x 2 2 Ϫ 2 1 3x # 5 2 ϩ 52 ϭ 9x2 Ϫ 30x ϩ 25 2 c. 1 3 ϩ 1x 2 ϭ 9 ϩ 2 1 3 # 1x 2 ϩ 1 1x 2 2 ϭ 9 ϩ 6 1x ϩ x a. 1 a ϩ 9 2 2 b. 1 3x Ϫ 5 2 2

c. 1 3 ϩ 1x 2 2
simplify

Solution



1 A ϩ B 2 2 ϭ A 2 ϩ 2AB ϩ B 2

1 A Ϫ B 2 2 ϭ A 2 Ϫ 2AB ϩ B 2 simplify 1 A ϩ B 2 2 ϭ A 2 ϩ 2AB ϩ B 2 simplify

F. You’ve just seen how we can compute special products: binomial conjugates and binomial squares

Now try Exercises 125 through 136 ᮣ With practice, you will be able to go directly from the binomial square to the resulting trinomial.

A.2 EXERCISES


CONCEPTS AND VOCABULARY
1. The equation 1 x2 2 3 ϭ x6 is an example of the property of exponents. 2. The equation 1 x3 2 Ϫ2 ϭ property of

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

4. The expression 2x2 Ϫ 3x Ϫ 10 can be classified as a of degree , with a leading coefficient of . 5. Discuss/Explain why one of the following expressions can be simplified further, while the other cannot: (a) Ϫ7n4 ϩ 3n2; (b) Ϫ7n4 # 3n2. 6. Discuss/Explain why the degree of 2x2y3 is greater than the degree of 2x 2 ϩ y 3. Include additional examples for contrast and comparison.

1 is an example of the x6 exponents.

3. The sum of the “outers” and “inners” for 1 2x ϩ 5 2 2 is , while the sum of the outers and inners for 1 2x ϩ 5 2 1 2x Ϫ 5 2 is .


DEVELOPING YOUR SKILLS

Determine each product using the product and/or power properties.

7.

10. 1 Ϫ1.5vy2 2 1 Ϫ8v4y 2

2 2# n 21n5 3

8. 24g5

# 3 g9
8

9. 1 Ϫ6p2q 2 1 2p3q3 2 12. d 2 # d 4 # 1 c5 2 2 # 1 c3 2 2

11. 1 a2 2 4 # 1 a3 2 2 # b2 # b5

Precalculus—

Section A.2 Exponents, Scientific Notation, and a Review of Polynomials

A-21

Simplify using the product to a power property.

13. 1 6pq2 2 3 p 2 17. a b 2q 21.

15. 1 3.2hk2 2 3 19. 1 Ϫ0.7c 2 1 10c d 2
4 2 2 2 23. 1 Ϫ3 8 x 2 1 16xy 2 3 2 13 4x y 2

14. 1 Ϫ3p2q 2 2 b 3 18. a b 3a
3 2 22. 1 4 5x 2

43. a 45. a

16. 1 Ϫ2.5h5k 2 2 20. 1 Ϫ2.5a 2 1 3a b 2
3 2

5m2n3 2 b 2r4 5p2q3r4 Ϫ2pq2r b 4
2

44. a 46. a

4p3 3x2y

b

3

9p3q2r3 12p5qr

b 2

3

3 2 2

2 2 3

Use properties of exponents to simplify the following. Write the answer using positive exponents only.

2 2 # 1 1 2mn2 2 24. 1 2 3m n 2

47. 49.
3x2

9p6q4 Ϫ12p4q6 20hϪ2 12h5 1 a2 2 3 a4 # a5 aϪ3 # b Ϫ4 b cϪ2 Ϫ6 1 2xϪ3 2 2 10xϪ2 14aϪ3bc0 Ϫ7 1 3a2bϪ2c 2 3

48. 50. 52. 54. 56. 58.

5m5n2 10m5n 5k3 20kϪ2 1 53 2 4

25. Volume of a cube: The 3x2 formula for the volume of a cube is V ϭ S3, where S is the length of one edge. If the length of each edge is 3x2, 3x2 a. Find a formula for volume in terms of the variable x. b. Find the volume of the cube if x ϭ 2. 26. Area of a circle: The formula for the area of a circle is A ϭ ␲r2, where r is the length of the radius. If the radius is given as 5x3, a. Find a formula for area in terms of the variable x. b. Find the area of the circle if x ϭ 2.

51.

53. a 55.
5x3

59 1 pϪ4q8 2 2 p5qϪ2

57.

18nϪ3 Ϫ8 1 3nϪ2 2 3 Ϫ3 1 2x3yϪ4z 2 2 18xϪ2yz0

59. 40 ϩ 50

61. 2Ϫ1 ϩ 5Ϫ1 63. 30 ϩ 3Ϫ1 ϩ 3Ϫ2 65. Ϫ5x0 ϩ 1 Ϫ5x 2 0

62. 4Ϫ1 ϩ 8Ϫ1

60. 1 Ϫ3 2 0 ϩ 1 Ϫ7 2 0

64. 2Ϫ2 ϩ 2Ϫ1 ϩ 20 66. Ϫ2n0 ϩ 1 Ϫ2n 2 0

Simplify using the quotient property or the property of negative exponents. Write answers using positive exponents only.

Convert the following numbers to scientific notation.

27. 29.

Ϫ6w Ϫ2w2
5

28. 30.

8z 16z5 5m3n5 10mn2 3 mϪ2

7

67. In mid-2009, the U.S. Census Bureau estimated the world population at nearly 6,770,000,000 people. 68. The mass of a proton is generally given as 0.000 000 000 000 000 000 000 000 001 670 kg.
Convert the following numbers to decimal notation.

Ϫ3 31. 1 2 32

Ϫ12a3b5 4a2b4 2 hϪ3

Ϫ1 32. 1 5 62

69. The smallest microprocessors in common use measure 6.5 ϫ 10Ϫ9 m across. 70. In 2009, the estimated net worth of Bill Gates, the founder of Microsoft, was 5.8 ϫ 1010 dollars.
Compute using scientific notation. Show all work.

33.

34.

35. 1 Ϫ2 2 Ϫ3 37.
1 Ϫ3 1Ϫ 2 2

36. 1 Ϫ4 2 Ϫ2 38.
2 Ϫ2 1Ϫ 3 2

Simplify each expression using the quotient to a power property.

39. a 41. a

2p 4 q3

b

2

40. a 42. a

Ϫ5v4 2 b 7w3 Ϫ0.5a3 2 b 0.4b2

71. The average distance between the Earth and the planet Jupiter is 465,000,000 mi. How many hours would it take a satellite to reach the planet if it traveled an average speed of 17,500 mi per hour? How many days? Round to the nearest whole. 72. In fiscal terms, a nation’s debt-per-capita is the ratio of its total debt to its total population. In the year 2009, the total U.S. debt was estimated at $11,300,000,000,000, while the population was estimated at 305,000,000. What was the U.S. debtper-capita ratio for 2009? Round to the nearest whole dollar.

0.2x2 3 b 0.3y3

Precalculus—

A-22

APPENDIX A A Review of Basic Concepts and Skills

Identify each expression as a polynomial or nonpolynomial (if a nonpolynomial, state why); classify each as a monomial, binomial, trinomial, or none of these; and state the degree of the polynomial.

96. 1 s Ϫ 3 21 5s ϩ 4 2

97. 1 x Ϫ 3 2 1 x2 ϩ 3x ϩ 9 2

73. Ϫ35w3 ϩ 2w2 ϩ 1 Ϫ12w 2 ϩ 14
2 74. Ϫ2x3 ϩ 2 3 x Ϫ 12x ϩ 1.2

98. 1 z ϩ 5 21 z2 Ϫ 5z ϩ 25 2

4 75. 5nϪ2 ϩ 4n ϩ 117 76. 3 ϩ 2.7r2 ϩ r ϩ 1 r 2 3 77. p Ϫ 5 78. q3 ϩ 2qϪ2 Ϫ 5q
Write each polynomial in standard form and name the leading coefficient.

100. 1 2h2 Ϫ 3h ϩ 8 21 h Ϫ 1 2 101. 1 7v Ϫ 4 2 1 3v Ϫ 5 2 103. 1 3 Ϫ m 2 1 3 ϩ m 2
1 107. 1 x ϩ 1 2 21 x ϩ 4 2

99. 1 b2 Ϫ 3b Ϫ 28 21 b ϩ 2 2 102. 1 6w Ϫ 1 21 2w ϩ 5 2 104. 1 5 ϩ n 2 1 5 Ϫ n 2
5 108. 1 z ϩ 1 3 2 1z ϩ 6 2

105. 1 p Ϫ 2.5 2 1 p ϩ 3.6 2 106. 1 q Ϫ 4.9 2 1 q ϩ 1.2 2
3 109. 1 m ϩ 3 4 21 m Ϫ 4 2 2 110. 1 n Ϫ 2 5 21 n ϩ 5 2

79. 7w ϩ 8.2 Ϫ w3 Ϫ 3w2 80. Ϫ2k2 Ϫ 12 Ϫ k 81. c3 ϩ 6 ϩ 2c2 Ϫ 3c
2 83. 12 Ϫ 2 3x 2

111. 1 3x Ϫ 2y 2 1 2x ϩ 5y 2 112. 1 6a ϩ b 2 1 a ϩ 3b 2

82. Ϫ3v3 ϩ 14 ϩ 2v2 ϩ 1 Ϫ12v 2 84. 8 ϩ 2n ϩ 7n
Find the indicated sum or difference.

113. 1 4c ϩ d 2 1 3c ϩ 5d 2 114. 1 5x ϩ 3y 2 1 2x Ϫ 3y 2 115. 1 2x2 ϩ 5 21 x2 Ϫ 3 2 116. 1 3y2 Ϫ 2 21 2y2 ϩ 1 2
For each binomial, determine its conjugate and find the product of the binomial with its conjugate.

117. 4m Ϫ 3 119. 7x Ϫ 10 121. 6 ϩ 5k 123. x ϩ 16 125. 1 x ϩ 4 2 2

118. 6n ϩ 5 120. c ϩ 3 122. 11 Ϫ 3r 124. p Ϫ 12 126. 1 a Ϫ 3 2 2

85. 1 3p3 Ϫ 4p2 ϩ 2p Ϫ 7 2 ϩ 1 p2 Ϫ 2p Ϫ 5 2 86. 1 5q2 Ϫ 3q ϩ 4 2 ϩ 1 Ϫ3q2 ϩ 3q Ϫ 4 2 87. 1 5.75b2 ϩ 2.6b Ϫ 1.9 2 ϩ 1 2.1b2 Ϫ 3.2b 2
1 2 2 89. 1 3 4 x Ϫ 5x ϩ 2 2 Ϫ 1 2 x ϩ 3x Ϫ 4 2

88. 1 0.4n2 ϩ 5n Ϫ 0.5 2 ϩ 1 0.3n2 Ϫ 2n ϩ 0.75 2
1 2 2 3 2 90. 1 5 9 n ϩ 4n Ϫ 2 2 Ϫ 1 3 n Ϫ 2n ϩ 4 2

Find each binomial square.

127. 1 4g ϩ 3 2 2

91. Subtract q5 ϩ 2q4 ϩ q2 ϩ 2q from q6 ϩ 2q5 ϩ q4 ϩ 2q3 using a vertical format. 92. Find x4 ϩ 2x3 ϩ x2 ϩ 2x decreased by x4 Ϫ 3x3 ϩ 4x2 Ϫ 3x using a vertical format.
Compute each product.

129. 1 4p Ϫ 3q 2 2 131. 1 4 Ϫ 1x 2 2
Compute each product.

128. 1 5x Ϫ 3 2 2

130. 1 5c ϩ 6d 2 2 132. 1 1x ϩ 7 2 2

133. 1 x Ϫ 3 2 1 y ϩ 2 2

93. Ϫ3x 1 x2 Ϫ x Ϫ 6 2 95. 1 3r Ϫ 5 21 r Ϫ 2 2

134. 1 a ϩ 3 2 1 b Ϫ 5 2

94. Ϫ2v2 1 v2 ϩ 2v Ϫ 15 2


135. 1 k Ϫ 5 2 1 k ϩ 6 2 1 k ϩ 2 2

136. 1 a ϩ 6 2 1 a Ϫ 1 2 1 a ϩ 5 2

WORKING WITH FORMULAS

137. Medication in the bloodstream: M ‫ ؍‬0.5t 4 ؉ 3t 3 ؊ 97t 2 ؉ 348t If 400 mg of a pain medication are taken orally, the number of milligrams in the bloodstream is modeled by the formula shown, where M is the number of milligrams and t is the time in hours, 0 Յ t 6 5. Construct a table of values for t ϭ 1 through 5, then answer the following. a. How many milligrams are in the bloodstream after 2 hr? After 3 hr? b. Based on part a, would you expect the number of milligrams in the bloodstream after 4 hr to be less or more? Why? c. Approximately how many hours until the medication wears off (the number of milligrams in the bloodstream is 0)?

Precalculus—

Section A.2 Exponents, Scientific Notation, and a Review of Polynomials

A-23

138. Amount of a mortgage payment: M ‫؍‬

Aa

r r n b a1 ؉ b 12 12

r n b ؊1 12 The monthly mortgage payment required to pay off (or amortize) a loan is given by the formula shown, where M is the monthly payment, A is the original amount of the loan, r is the annual interest rate, and n is the term of the loan in months. Find the monthly payment (to the nearest cent) required to purchase a $198,000 home, if the interest rate is 6.5% and the home is financed over 30 yr. a1 ؉



APPLICATIONS
number of people in the pool at any time can be approximated by the formula S 1 t 2 ϭ Ϫt 2 ϩ 10t, where S is the number of swimmers and t is the number of hours the pool has been open (8 A.M.: t ϭ 0, 9 A.M.: t ϭ 1, 10 A.M.: t ϭ 2, etc.). a. How many swimmers are in the pool at 6 P.M.? Why? b. Between what times would you expect the largest number of swimmers? c. Approximately how many swimmers are in the pool at 3 P.M.? d. Create a table of values for t ϭ 1, 2, 3, 4, . . . and check your answer to part b. 143. Maximizing revenue: A sporting goods store finds that if they price their video games at $20, they make 200 sales per day. For each decrease of $1, 20 additional video games are sold. This means the store’s revenue can be modeled by the formula R ϭ 1 20 Ϫ 1x 2 1 200 ϩ 20x 2 , where x is the number of $1 decreases. Multiply out the binomials and use a table of values to determine what price will give the most revenue. 144. Maximizing revenue: Due to past experience, a jeweler knows that if they price jade rings at $60, they will sell 120 each day. For each decrease of $2, five additional sales will be made. This means the jeweler’s revenue can be modeled by the formula R ϭ 1 60 Ϫ 2x 2 1 120 ϩ 5x 2 , where x is the number of $2 decreases. Multiply out the binomials and use a table of values to determine what price will give the most revenue.

139. Attraction between particles: In electrical theory, the force of attraction between two particles P and kPQ Q with opposite charges is modeled by F ϭ 2 , d where d is the distance between them and k is a constant that depends on certain conditions. This is known as Coulomb’s law. Rewrite the formula using a negative exponent. 140. Intensity of light: The intensity of illumination from a light source depends on the distance from k the source according to I ϭ 2 , where I is the d intensity measured in footcandles, d is the distance from the source in feet, and k is a constant that depends on the conditions. Rewrite the formula using a negative exponent. 141. Rewriting an expression: In advanced mathematics, negative exponents are widely used because they are easier to work with than rational expressions. 5 3 2 Rewrite the expression 3 ϩ 2 ϩ 1 ϩ 4 using x x x negative exponents. 142. Swimming pool hours: A swimming pool opens at 8 A.M. and closes at 6 P.M. In summertime, the



EXTENDING THE CONCEPT
146. If a 2x ϩ 1 2 1 b ϭ 5, then the expression 4x2 ϩ 2 2x 4x is equal to what number?

145. If 1 3x2 ϩ kx ϩ 1 2 Ϫ 1 kx2 ϩ 5x Ϫ 7 2 ϩ 1 2x2 Ϫ 4x Ϫ k 2 ϭ Ϫx2 Ϫ 3x ϩ 2, what is the value of k?

Precalculus—

A.3

Solving Linear Equations and Inequalities
In a study of algebra, you will encounter many families of equations, or groups of equations that share common characteristics. Of interest to us here is the family of linear equations in one variable, a study that lays the foundation for understanding more advanced families. This section will also lay the foundation for solving a formula for a specified variable, a practice widely used in science, business, industry, and research.

LEARNING OBJECTIVES
In Section A.3 you will review how to:

A. Solve linear equations
using properties of equality Recognize equations that are identities or contradictions Solve linear inequalities Solve compound inequalities Solve basic applications of linear equations and inequalities Solve applications of basic geometry

B.

A. Solving Linear Equations Using Properties of Equality
An equation is a statement that two expressions are equal. From the expressions 3 1 x Ϫ 1 2 ϩ x and Ϫx ϩ 7, we can form the equation 3 1 x Ϫ 1 2 ϩ x ϭ Ϫx ϩ 7,
Table A.1
x Ϫ2 Ϫ1 31x ؊ 12 ؉ x Ϫ11 Ϫ7 Ϫ3 1 5 9 13 ؊x ؉ 7 9 8 7 6 5 4 3

C. D. E.

F.

which is a linear equation in one variable (the 0 exponent on any variable is a 1). To solve an equa1 tion, we attempt to find a specific input or x-value 2 that will make the equation true, meaning the left3 hand expression will be equal to the right. Using 4 Table A.1, we find that 3 1 x Ϫ 1 2 ϩ x ϭ Ϫx ϩ 7 is a true equation when x is replaced by 2, and is a false equation otherwise. Replacement values that make the equation true are called solutions or roots of the equation.


CAUTION

From Appendix A.1, an algebraic expression is a sum or difference of algebraic terms. Algebraic expressions can be simplified, evaluated or written in an equivalent form, but cannot be “solved,” since we’re not seeking a specific value of the unknown.

Solving equations using a table is too time consuming to be practical. Instead we attempt to write a sequence of equivalent equations, each one simpler than the one before, until we reach a point where the solution is obvious. Equivalent equations are those that have the same solution set, and can be obtained by using the distributive property to simplify the expressions on each side of the equation. The additive and multiplicative properties of equality are then used to obtain an equation of the form x ϭ constant. The Additive Property of Equality If A, B, and C represent algebraic expressions and A ϭ B, then A ϩ C ϭ B ϩ C The Multiplicative Property of Equality If A, B, and C represent algebraic expressions and A ϭ B, then AC ϭ BC and A B ϭ , 1C C C 02

In words, the additive property says that like quantities, numbers, or terms can be added to both sides of an equation. A similar statement can be made for the multiplicative property. These properties are combined into a general guide for solving linear equations, which you’ve likely encountered in your previous studies. Note that not all steps in the guide are required to solve every equation.

A-24

Precalculus—

Section A.3 Solving Linear Equations and Inequalities

A-25

Guide to Solving Linear Equations in One Variable • Eliminate parentheses using the distributive property, then combine any like terms. • Use the additive property of equality to write the equation with all variable terms on one side, and all constants on the other. Simplify each side. • Use the multiplicative property of equality to obtain an equation of the form x ϭ constant. • For applications, answer in a complete sentence and include any units of measure indicated. For our first example, we’ll use the equation 3 1 x Ϫ 1 2 ϩ x ϭ Ϫx ϩ 7 from our initial discussion.

EXAMPLE 1 Solution



Solving a Linear Equation Using Properties of Equality Solve for x: 3 1 x Ϫ 1 2 ϩ x ϭ Ϫx ϩ 7. 3 1 x Ϫ 1 2 ϩ x ϭ Ϫx ϩ 7 3x Ϫ 3 ϩ x ϭ Ϫx ϩ 7 4x Ϫ 3 ϭ Ϫx ϩ 7 5x Ϫ 3 ϭ 7 5x ϭ 10 xϭ2



original equation distributive property combine like terms add x to both sides (additive property of equality) add 3 to both sides (additive property of equality) multiply both sides by 1 5 or divide both sides by 5 (multiplicative property of equality)

As we noted in Table R.1, the solution is x ϭ 2. Now try Exercises 7 through 12


To check a solution by substitution means we substitute the solution back into the original equation (this is sometimes called back-substitution), and verify the lefthand side is equal to the right. For Example 1 we have: 3 1 2 Ϫ 1 2 ϩ 2 ϭ Ϫ2 ϩ 7 3112 ϩ 2 ϭ 5 5 ϭ 5✓ 3 1 x Ϫ 1 2 ϩ x ϭ Ϫx ϩ 7
original equation substitute 2 for x simplify solution checks

If any coefficients in an equation are fractional, multiply both sides by the least common denominator (LCD) to clear the fractions. Since any decimal number can be written in fraction form, the same idea can be applied to decimal coefficients.

EXAMPLE 2 Solution



Solving a Linear Equation with Fractional Coefficients
1 4 1n ϩ 82 1 4n ϩ 2 1 Solve for n: 1 4 1n ϩ 82 Ϫ 2 ϭ 2 1n Ϫ 62.



A. You’ve just seen how we can solve linear equations using properties of equality

Ϫ2ϭ1 2 1n Ϫ 62 Ϫ2ϭ1 2n Ϫ 3 1 1 4n ϭ 2n Ϫ 3 1 4 1 4n 2 ϭ 4 1 1 2n Ϫ 3 2 n ϭ 2n Ϫ 12 Ϫn ϭ Ϫ12 n ϭ 12

original equation distributive property combine like terms multiply both sides by LCD ϭ 4 distributive property subtract 2n multiply by Ϫ1

Verify the solution is n ϭ 12 using back-substitution. Now try Exercises 13 through 30


Precalculus—

A-26

APPENDIX A A Review of Basic Concepts and Skills

B. Identities and Contradictions
Example 1 illustrates what is called a conditional equation, since the equation is true for x ϭ 2, but false for all other values of x. The equation in Example 2 is also conditional. An identity is an equation that is always true, no matter what value is substituted for the variable. For instance, 2 1 x ϩ 3 2 ϭ 2x ϩ 6 is an identity with a solution set of all real numbers, written as 5 x|x ʦ ‫ ޒ‬6, or x ʦ 1 Ϫ q , q 2 in interval notation. Contradictions are equations that are never true, no matter what real number is substituted for the variable. The equations x Ϫ 3 ϭ x ϩ 1 and Ϫ3 ϭ 1 are contradictions. To state the solution set for a contradiction, we use the symbol “л” (the null set) or “{ }” (the empty set). Recognizing these special equations will prevent some surprise and indecision in later chapters.

EXAMPLE 3



Solving Equations (Special Cases) Solve each equation and state the solution set. a. 2 1 x Ϫ 4 2 ϩ 10x ϭ 8 ϩ 4 1 3x ϩ 1 2 b. 8x Ϫ 1 6 Ϫ 10x 2 ϭ 24 ϩ 6 1 3x Ϫ 5 2

Solution



a. 2 1 x Ϫ 4 2 ϩ 10x ϭ 8 ϩ 4 1 3x ϩ 1 2 2x Ϫ 8 ϩ 10x ϭ 8 ϩ 12x ϩ 4 12x Ϫ 8 ϭ 12x ϩ 12 Ϫ8 ϭ 12

original equation distributive property combine like terms subtract 12x; contradiction

b. 8x Ϫ 1 6 Ϫ 10x 2 ϭ 24 ϩ 6 1 3x Ϫ 5 2 8x Ϫ 6 ϩ 10x ϭ 24 ϩ 18x Ϫ 30 18x Ϫ 6 ϭ 18x Ϫ 6 Ϫ6 ϭ Ϫ6

Since Ϫ8 is never equal to 12, the original equation is a contradiction. The solution set is empty: { }
original equation distributive property combine like terms subtract 18x; identity

The result shows that the original equation is an identity, with an infinite number of solutions: 5 x | x ʦ ‫ ޒ‬6 . You may recall this notation is read, “the set of all numbers x, such that x is a real number.” Now try Exercises 31 through 36


B. You’ve just seen how we can recognize equations that are identities or contradictions

In Example 3(a), our attempt to solve for x ended with all variables being eliminated, leaving an equation that is always false —a contradiction (Ϫ8 is never equal to 12). There is nothing wrong with the solution process, the result is simply telling us the original equation has no solution. In Example 3(b), all variables were again eliminated but the end result was always true —an identity (Ϫ6 is always equal to Ϫ6). Once again we’ve done nothing wrong mathematically, the result is just telling us that the original equation will be true no matter what value of x we use for an input.

C. Solving Linear Inequalities
A linear inequality resembles a linear equality in many respects: Linear Inequality (1) (2) x 6 3 3 p Ϫ 2 Ն Ϫ12 8 Related Linear Equation xϭ3 3 p Ϫ 2 ϭ Ϫ12 8

A linear inequality in one variable is one that can be written in the form ax ϩ b 6 c, where a, b, and c ʦ ‫ ޒ‬and a 0. This definition and the following properties also apply when other inequality symbols are used. Solutions to simple inequalities are easy to spot. For instance, x ϭ Ϫ2 is a solution to x 6 3 since Ϫ2 6 3. For more involved inequalities we use the additive property of inequality and the multiplicative property of

Precalculus—

Section A.3 Solving Linear Equations and Inequalities

A-27

inequality. Similar to solving equations, we solve inequalities by isolating the variable on one side to obtain a solution form such as variable 6 number. The Additive Property of Inequality If A, B, and C represent algebraic expressions and A 6 B, then A ϩ C 6 B ϩ C Like quantities (numbers or terms) can be added to both sides of an inequality. While there is little difference between the additive property of equality and the additive property of inequality, there is an important difference between the multiplicative property of equality and the multiplicative property of inequality. To illustrate, we begin with Ϫ2 6 5. Multiplying both sides by positive three yields Ϫ6 6 15, a true inequality. But notice what happens when we multiply both sides by negative three: Ϫ2 1 Ϫ3 2 6 5 1 Ϫ3 2 6 6 Ϫ15 Ϫ2 6 5
original inequality multiply by negative three false

The result is a false inequality, because 6 is to the right of Ϫ15 on the number line. Multiplying (or dividing) an inequality by a negative quantity reverses the order relationship between two quantities (we say it changes the sense of the inequality). We must compensate for this by reversing the inequality symbol. 6 7 Ϫ15
change direction of symbol to maintain a true statement

For this reason, the multiplicative property of inequality is stated in two parts. The Multiplicative Property of Inequality If A, B, and C represent algebraic expressions and A 6 B, then AC 6 BC if C is a positive quantity (inequality symbol remains the same).


If A, B, and C represent algebraic expressions and A 6 B, then AC 7 BC if C is a negative quantity (inequality symbol must be reversed).

EXAMPLE 4

Solving an Inequality Solve the inequality, then graph the solution set and write it in interval notation: Ϫ2 1 5 3 x ϩ 2 Յ 6.

Solution



WORTHY OF NOTE
As an alternative to multiplying or dividing by a negative value, the additive property of inequality can be used to ensure the variable term will be positive. From Example 4, the inequality Ϫ4x Յ 2 can be written as Ϫ2 Յ 4x by adding 4x to both sides and subtracting 2 from both sides. This gives the solution Ϫ1 2 Յ x, which is equivalent to x Ն Ϫ1 2.

1 5 Ϫ2 xϩ Յ 3 2 6 Ϫ2 1 5 6 a x ϩ b Յ 162 3 2 6 Ϫ4x ϩ 3 Յ 5 Ϫ4x Յ 2 Ϫ4x 2 Ն Ϫ4 Ϫ4 1 xՆϪ 2
Ϫ1 2

original inequality

clear fractions (multiply by LCD) simplify subtract 3 divide by Ϫ4, reverse inequality sign result

• Graph:

• Interval notation: x ʦ 3 Ϫ1 2, q 2

Ϫ3 Ϫ2 Ϫ1

[

0

1

2

3

4

Now try Exercises 37 through 46



Precalculus—

A-28

APPENDIX A A Review of Basic Concepts and Skills

To check a linear inequality, you often have an infinite number of choices— any number from the solution set/interval. If a test value from the solution interval results in a true inequality, all numbers in the interval are solutions. For Example 4, using 5 x ϭ 0 results in the true statement 1 2 Յ 6 ✓. Some inequalities have all real numbers as the solution set: 5 x| x ʦ ‫ ޒ‬6, while other inequalities have no solutions, with the answer given as the empty set: { }.

EXAMPLE 5



Solving Inequalities Solve the inequality and write the solution in set notation: a. 7 Ϫ 1 3x ϩ 5 2 Ն 2 1 x Ϫ 4 2 Ϫ 5x b. 3 1 x ϩ 4 2 Ϫ 5 6 2 1 x Ϫ 3 2 ϩ x

Solution



a. 7 Ϫ 1 3x ϩ 5 2 Ն 2 1 x Ϫ 4 2 Ϫ 5x 7 Ϫ 3x Ϫ 5 Ն 2x Ϫ 8 Ϫ 5x Ϫ3x ϩ 2 Ն Ϫ3x Ϫ 8 2 Ն Ϫ8

original inequality distributive property combine like terms add 3x

b. 3 1 x ϩ 4 2 Ϫ 5 3x ϩ 12 Ϫ 5 3x ϩ 7 7

Since the resulting statement is always true, the original inequality is true for all real numbers. The solution is all real numbers ‫ޒ‬. 6 6 6 6 21x Ϫ 32 ϩ x 2x Ϫ 6 ϩ x 3x Ϫ 6 Ϫ6
original inequality distribute combine like terms subtract 3x

Since the resulting statement is always false, the original inequality is false for all real numbers. The solution is { }.
C. You’ve just seen how we can solve linear inequalities

Now try Exercises 47 through 52



D. Solving Compound Inequalities
In some applications of inequalities, we must consider more than one solution interval. These are called compound inequalities, and these require us to take a close look at the operations of union “ ´ ” and intersection “ ¨”. The intersection of two sets A and B, written A ¨ B, is the set of all elements common to both sets. The union of two sets A and B, written A ´ B, is the set of all elements that are in either set. When stating the union of two sets, repetitions are unnecessary.

EXAMPLE 6



Finding the Union and Intersection of Two Sets

For set A ϭ 5 Ϫ2, Ϫ1, 0, 1, 2, 3 6 and set B ϭ 5 1, 2, 3, 4, 5 6 , determine A ¨ B and A ´ B. A ¨ B is the set of all elements in both A and B: A ʝ B ϭ 5 1, 2, 3 6 . A ´ B is the set of all elements in either A or B: A ´ B ϭ 5 Ϫ2, Ϫ1, 0, 1, 2, 3, 4, 5 6. Now try Exercises 53 through 58


Solution



WORTHY OF NOTE
For the long term, it may help to rephrase the distinction as follows. The intersection is a selection of elements that are common to two sets, while the union is a collection of the elements from two sets (with no repetitions).

Notice the intersection of two sets is described using the word “and,” while the union of two sets is described using the word “or.” When compound inequalities are formed using these words, the solution is modeled after the ideas from Example 6. If “and” is used, the solutions must satisfy both inequalities. If “or” is used, the solutions can satisfy either inequality.

Precalculus—

Section A.3 Solving Linear Equations and Inequalities

A-29

EXAMPLE 7



Solving a Compound Inequality Solve the compound inequality, then write the solution in interval notation: Ϫ3x Ϫ 1 6 Ϫ4 or 4x ϩ 3 6 Ϫ6.

Solution



Begin with the statement as given: Ϫ3x Ϫ 1 6 Ϫ4 Ϫ3x 6 Ϫ3 x 7 1 or or or 4x ϩ 3 6 Ϫ6 4x 6 Ϫ9 9 x 6 Ϫ 4
original statement isolate variable term solve for x, reverse first inequality symbol

The solution x 7 1 or x 6 Ϫ9 4 is better understood by graphing each interval separately, then selecting both intervals (the union).
WORTHY OF NOTE
The graphs from Example 7 clearly show the solution consists of two disjoint (disconnected) intervals. This is reflected in the “or” statement: x 6 Ϫ9 4 or x 7 1, and in the interval notation. Also, note the 9 solution x 6 Ϫ4 or x 7 1 is not equivalent to Ϫ9 4 7 x 7 1, as there is no single number that is both greater than 1 and less than Ϫ9 4 at the same time.

x Ͼ 1:

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

0

1

)

2

3

4

5

6

x Ͻ Ϫ9 : 4

Ϫ9 4

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

)

0

1

2

3

4

5

6

x Ͻ Ϫ 9 or x Ͼ 1: 4

Ϫ9 4

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

)

0

1

)

2

3

4

5

6

9 Interval notation: x ʦ a Ϫ q , Ϫ b ´ 1 1, q 2 . 4 Now try Exercises 59 and 60


EXAMPLE 8



Solving a Compound Inequality Solve the compound inequality, then write the solution in interval notation: 3x ϩ 5 7 Ϫ13 and 3x ϩ 5 6 Ϫ1.

Solution



Begin with the statement as given: 3x ϩ 5 7 Ϫ13 3x 7 Ϫ18 x 7 Ϫ6 and and and 3x ϩ 5 6 Ϫ1 3x 6 Ϫ6 x 6 Ϫ2
original statement subtract five divide by 3

The solution x 7 Ϫ6 and x 6 Ϫ2 can best be understood by graphing each interval separately, then noting where they intersect.
WORTHY OF NOTE
The inequality a 6 b (a is less than b) can equivalently be written as b 7 a (b is greater than a). In Example 8, the solution is read, “x 7 Ϫ6 and x 6 Ϫ2,” but if we rewrite the first inequality as Ϫ6 6 x (with the “arrowhead” still pointing at Ϫ6 2 , we have Ϫ6 6 x and x 6 Ϫ2 and can clearly see that x must be in the single interval between Ϫ6 and Ϫ2.

x Ͼ Ϫ6: x Ͻ Ϫ2: x Ͼ Ϫ6 and x Ͻ Ϫ2:

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

)

0

1

2

3

4

5

6

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

)

0

1

2

3

4

5

6

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

)

Interval notation: x ʦ 1 Ϫ6, Ϫ2 2 . Now try Exercises 61 through 72


)

0

1

2

3

4

5

6

Precalculus—

A-30

APPENDIX A A Review of Basic Concepts and Skills

The solution from Example 8 consists of the single interval 1 Ϫ6, Ϫ2 2 , indicating the original inequality could actually be joined and written as Ϫ6 6 x 6 Ϫ2, called a joint inequality. We solve joint inequalities in much the same way as linear inequalities, but must remember they have three parts (left, middle, and right). This means operations must be applied to all three parts in each step of the solution process, to obtain a solution form such as smaller number 6 x 6 larger number. The same ideas apply when other inequality symbols are used.

EXAMPLE 9



Solving a Joint Inequality Solve the joint inequality, then graph the solution set and write it in interval 2x ϩ 5 Ն Ϫ6. notation: 1 7 Ϫ3

Solution



2x ϩ 5 Ն Ϫ6 Ϫ3 Ϫ3 6 2x ϩ 5 Յ 18 Ϫ8 6 2x Յ 13 13 Ϫ4 6 x Յ 2 1 7 • Graph:
)

original inequality multiply all parts by Ϫ3; reverse the inequality symbols subtract 5 from all parts divide all parts by 2
13 2 0
13 2 4

D. You’ve just seen how we can solve compound inequalities

• Interval notation: x ʦ 1 Ϫ4,

Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

1

2

3

4

5

6

[

7

8

Now try Exercises 73 through 78



E. Solving Basic Applications of Linear Equations and Inequalities
Applications of linear equations and inequalities come in many forms. In most cases, you are asked to translate written relationships or information given verbally into an equation using words or phrases that indicate mathematical operations or relationships. Here, we’ll practice this skill using ideas that were introduced in Appendix A.1, where we translated English phrases into mathematical expressions. Very soon these skills will be applied in much more significant ways.

EXAMPLE 10



Translating Written Information into an Equation Translate the following relationships into equations, then solve: In an effort to lower the outstanding balance on her credit card, Laura paid $10 less than triple her normal payment. If she sent the credit card company $350.75, how much was her normal payment? (See Section A.1, Examples 2 and 3.)

Solution



Let p represent her normal payment. Then “triple her normal payment” would be 3p, and “ten less than triple” would be 3p Ϫ 10. Since “she sent the company $350.75,” we have

Precalculus—

Section A.3 Solving Linear Equations and Inequalities

A-31

3p Ϫ 10 ϭ 350.75 3p ϭ 360.75 p ϭ 120.25

equation form add 10 divide by 3

Laura’s normal payment is $120.25 per month. A calculator check is shown in the figure.

Now try Exercises 83 through 90



Inequalities are widely used to help gather information, and to make comparisons that will lead to informed decisions.

EXAMPLE 11



Using an Inequality to Compute Desired Test Scores Justin earned scores of 78, 72, and 86 on the first three out of four exams. What score must he earn on the fourth exam to have an average of at least 80?

Solution



The current scores are 78, 72, and 86. An average of at least 80 means A Ն 80. In organized form:
Test 1 78 Test 2 72 Test 3 86 Test 4 x Computed Average 78 ϩ 72 ϩ 86 ϩ x 4 Minimum 80

Let x represent Justin’s score on the fourth exam, then represents his average score. 78 ϩ 72 ϩ 86 ϩ x Ն 80 4 78 ϩ 72 ϩ 86 ϩ x Ն 320 236 ϩ x Ն 320 x Ն 84
E. You’ve just seen how we can solve applications of linear equations and inequalities

78 ϩ 72 ϩ 86 ϩ x 4

average must be greater than or equal to 80 multiply by 4 simplify solve for x (subtract 236)

Justin must score at least an 84 on the last exam to earn an 80 average. Now try Exercises 91 through 100


F. Solving Applications of Basic Geometry
As your translation skills grow, your ability to solve a wider range of more significant applications will grow as well. In many cases, the applications will involve some basic geometry and the most often used figures and formulas appear here. For a more complete review of geometry, see Appendix II Geometry Review with Unit Conversions, which is posted online at www.mhhe.com/coburn.

Perimeter and Area
Perimeter is a measure of the distance around a two dimensional figure. As this is a linear measure, results are stated in linear units as in centimeters (cm), feet (ft), kilometers (km), miles (mi), and so on. If no unit is specified, simply write the result as x units. Area is a measure of the surface of a two dimensional figure, with results stated in square units as in x units2. Some of the most common formulas involving perimeter and area are given in Table A.2A.

Precalculus—

A-32

APPENDIX A A Review of Basic Concepts and Skills

Table A.2A
Definition and Diagram a three-sided polygon Triangle s1 s3 a quadrilateral with four right angles and opposite sides parallel Rectangle W L a rectangle with four equal sides Square S P ϭ 4S A ϭ S2 P ϭ 2L ϩ 2W A ϭ LW s2 b h P ϭ s1 ϩ s2 ϩ s3 Aϭ 1 bh 2 Perimeter Formula (linear units or units) Area Formula (square units or units2)

a quadrilateral with one pair of parallel sides (called bases b1 and b2) Trapezoid s1 s4 the set of all points lying in a plane that are an equal distance (called the radius r) from a given point (called the center C). s2 s3 h b2 b1 sum of all sides P ϭ s1 ϩ s2 ϩ s3 ϩ s4 Aϭ h 1 b1 ϩ b2 2 2

Circle

r C

C ϭ 2␲r or C ϭ ␲d

A ϭ ␲r2

If an exercise or application uses a formula, begin by stating the formula first. Using the formula as a template for the values substituted will help to prevent many careless errors.

EXAMPLE 12A



Computing the Area of a Trapezoidal Window A basement window is shaped like an isosceles trapezoid (base angles equal, nonparallel sides equal in length), with a height of 10 in. and bases of 1.5 ft and 2 ft. What is the area of the glass in the window?
1.5 ft

10 in.

2 ft

Precalculus—

Section A.3 Solving Linear Equations and Inequalities

A-33

Solution



Before applying the area formula, all measures must use the same unit. In inches, we have 1.5 ft ϭ 18 in. and 2 ft ϭ 24 in. h 1 b1 ϩ b2 2 2 10 in. 1 18 in. ϩ 24 in. 2 Aϭ 2 A ϭ 1 5 in. 2 1 42 in. 2 A ϭ 210 in2 Aϭ
given formula substitute 10 for h, 18 for b1, and 24 for b2 simplify result

The area of the glass in the window is 210 in2. Now try Exercises 101 and 102


Volume
Volume is a measure of the amount of space occupied by a three dimensional object and is measured in cubic units. Some of the more common formulas are given in Table A.2B.
Table A.2B
Definition and Diagram Rectangular solid a six-sided, solid figure with opposite faces congruent and adjacent faces meeting at right angles a rectangular solid with six congruent, square faces S the set of all points in space, an equal distance (called the radius) from a given point (called the center) union of all line segments connecting two congruent circles in parallel planes, meeting each at a right angle union of all line segments connecting a given point (vertex) to a given circle (base) and whose altitude meets the center of the base at a right angle union of all line segments connecting a given point (vertex) to a given square (base) and whose altitude meets the center of the base at a right angle 4 V ϭ ␲r3 3 Volume Formula (cubic units or units3) V ϭ LWH

H L W

Cube

V ϭ S3

Sphere

C

r

Right circular cylinder

h r

V ϭ ␲r2h

Right circular cone

h r

1 V ϭ ␲r2h 3

Right pyramid

h s

1 V ϭ s2h 3

Precalculus—

A-34

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 12B



Computing the Volume of a Composite Figure Sand at a cement factory is being dumped from a conveyor belt into a pile shaped like a right circular cone atop a right circular cylinder (see figure). How many cubic feet of sand are there at the moment the cone is 6 ft high with a diameter of 10 ft?
6 ft 3 ft 10 ft

Solution



Total Volume ϭ volume of cylinder ϩ volume of cone 1 V ϭ ␲r2h1 ϩ ␲r2h2 3 1 ϭ ␲1522132 ϩ ␲1522162 3 ϭ 75␲ ϩ 50␲ ϭ 125␲ There are about 392.7 ft of sand in the pile.
3

verbal model formula model (note h1 h2 2

substitute 5 for r, 3 for h1, and 6 for h2 simplify result (exact form)

F. You’ve just seen how we can solve applications of basic geometry

Now try Exercises 103 and 104



A.3 EXERCISES


CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. A(n) is an equation that is always true, regardless of the value while a(n) is an equation that is always false, regardless of the value. 2. For inequalities, the three ways of writing a solution set are notation, a number line graph, and notation. 3. The mathematical sentence 3x ϩ 5 6 7 is a(n) inequality, while Ϫ2 6 3x ϩ 5 6 7 is a(n) inequality.


4. The The

of sets A and B is written A ʝ B. of sets A and B is written A ´ B.

5. Discuss/Explain the similarities and differences between the properties of equality for equations and those for inequalities. 6. Discuss/Explain the use of the words “and” and “or” in the statement of compound inequalities. Include a few examples to illustrate.

DEVELOPING YOUR SKILLS
7. 4x ϩ 3 1 x Ϫ 2 2 ϭ 18 Ϫ x
Solve each equation.
1 13. 1 5 1 b ϩ 10 2 Ϫ 7 ϭ 3 1 b Ϫ 9 2 Ϫ1 15. 2 3 1m ϩ 62 ϭ 2

Solve each equation. Check your answer by substitution.

8. 15 Ϫ 2x ϭ Ϫ4 1 x ϩ 1 2 ϩ 9

10. Ϫ12 Ϫ 5w ϭ Ϫ9 Ϫ 1 6w ϩ 7 2

9. 21 Ϫ 1 2v ϩ 17 2 ϭ Ϫ7 Ϫ 3v

1 14. 1 6 1 n Ϫ 12 2 ϭ 4 1 n ϩ 8 2 Ϫ 2 Ϫ8 16. 4 5 1 n Ϫ 10 2 ϭ 9

11. 8 Ϫ 1 3b ϩ 5 2 ϭ Ϫ5 ϩ 2 1 b ϩ 1 2 12. 2a ϩ 4 1 a Ϫ 1 2 ϭ 3 Ϫ 1 2a ϩ 1 2

1 17. 1 2x ϩ 5 ϭ 3x ϩ 7

1 18. Ϫ4 ϩ 2 3y ϭ 2y Ϫ 5

19.

xϩ3 x ϩ ϭ7 5 3

20.

zϪ4 z Ϫ2ϭ 6 2

Precalculus—

Section A.3 Solving Linear Equations and Inequalities

A-35

21. 15 ϭ Ϫ6 Ϫ

23. 0.2 1 24 Ϫ 7.5a 2 Ϫ 6.1 ϭ 4.1

3p 8

22. Ϫ15 Ϫ

2q ϭ Ϫ21 9

Solve each inequality and write the solution in set notation.

24. 0.4 1 17 Ϫ 4.25b 2 Ϫ 3.15 ϭ 4.16

47. 7 Ϫ 2 1 x ϩ 3 2 Ն 4x Ϫ 6 1 x Ϫ 3 2

25. 6.2v Ϫ 1 2.1v Ϫ 5 2 ϭ 1.1 Ϫ 3.7v n n 2 ϩ ϭ 2 5 3 m 2 m Ϫ ϭ 28. 3 5 4 p p 29. 3p Ϫ Ϫ 5 ϭ Ϫ 2p ϩ 6 4 6 q q 30. ϩ 1 Ϫ 3q ϭ 2 Ϫ 4q ϩ 6 8 27.

48. Ϫ3 Ϫ 6 1 x Ϫ 5 2 Յ 2 1 7 Ϫ 3x 2 ϩ 1

26. 7.9 Ϫ 2.6w ϭ 1.5w Ϫ 1 9.1 ϩ 2.1w 2

49. 4 1 3x Ϫ 5 2 ϩ 18 6 2 1 5x ϩ 1 2 ϩ 2x 51. Ϫ6 1 p Ϫ 1 2 ϩ 2p Յ Ϫ2 1 2p Ϫ 3 2

50. 8 Ϫ 1 6 ϩ 5m 2 7 Ϫ9m Ϫ 1 3 Ϫ 4m 2 52. 9 1 w Ϫ 1 2 Ϫ 3w Ն Ϫ2 1 5 Ϫ 3w 2 ϩ 1
Determine the intersection and union of sets A, B, C, and D as indicated, given A ‫ ؍‬5؊3, ؊2, ؊1, 0, 1, 2, 3 6 , B ‫ ؍‬5 2, 4, 6, 8 6, C ‫ ؍‬5؊ 4, ؊2, 0, 2, 4 6 , and D ‫ ؍‬5 4, 5, 6, 7 6 .

53. A ʝ B and A ´ B 54. A ʝ C and A ´ C 55. A ʝ D and A ´ D 56. B ʝ C and B ´ C

Identify the following equations as an identity, a contradiction, or a conditional equation, then state the solution.

57. B ʝ D and B ´ D 58. C ʝ D and C ´ D
Express the compound inequalities graphically and in interval notation.

31. Ϫ3 1 4z ϩ 5 2 ϭ Ϫ15z Ϫ 20 ϩ 3z 32. 5x Ϫ 9 Ϫ 2 ϭ Ϫ5 1 2 Ϫ x 2 Ϫ 1

33. 8 Ϫ 8 1 3n ϩ 5 2 ϭ Ϫ5 ϩ 6 1 1 ϩ n 2 34. 2a ϩ 4 1 a Ϫ 1 2 ϭ 1 ϩ 3 1 2a ϩ 1 2 35. Ϫ4 1 4x ϩ 5 2 ϭ Ϫ6 Ϫ 2 1 8x ϩ 7 2

59. x 6 Ϫ2 or x 7 1 60. x 6 Ϫ5 or x 7 5 61. x 6 5 and x Ն Ϫ2 62. x Ն Ϫ4 and x 6 3 63. x Ն 3 and x Յ 1 64. x Ն Ϫ5 and x Յ Ϫ7

36. Ϫ 1 5x Ϫ 3 2 ϩ 2x ϭ 11 Ϫ 4 1 x ϩ 2 2
Write the solution set illustrated on each graph in set notation and interval notation.

Solve the compound inequalities and graph the solution set.

65. 4 1 x Ϫ 1 2 Յ 20 or x ϩ 6 7 9 67. Ϫ2x Ϫ 7 Յ 3 and 2x Յ 0

37. 38. 39. 40.

Ϫ3 Ϫ2 Ϫ1

[

66. Ϫ3 1 x ϩ 2 2 7 15 or x Ϫ 3 Յ Ϫ1 68. Ϫ3x ϩ 5 Յ 17 and 5x Յ 0
1 69. 3 5x ϩ 2 7 3 10

0

1

2

3

Ϫ3 Ϫ2 Ϫ1

0

) [

1

2

3

Ϫ3 Ϫ2 Ϫ1

[

and Ϫ4x 7 1

0

1

2

3

70.
4

2 3x

Ϫ Յ 0 and Ϫ3x 6 Ϫ2
5 6

Ϫ3 Ϫ2 Ϫ1

[

0

1

2

)

3

71. 72.

3x x ϩ 6 Ϫ3 or x ϩ 1 7 Ϫ5 8 4 2x x ϩ 6 Ϫ2 or x Ϫ 3 7 2 5 10

Solve the inequality and write the solution in set notation. Then graph the solution and write it in interval notation.

41. 5a Ϫ 11 Ն 2a Ϫ 5 42. Ϫ8n ϩ 5 7 Ϫ2n Ϫ 12 43. 2 1 n ϩ 3 2 Ϫ 4 Յ 5n Ϫ 1 3x x ϩ 6 Ϫ4 45. 8 4

73. Ϫ3 Յ 2x ϩ 5 6 7 74. 2 6 3x Ϫ 4 Յ 19 75. Ϫ0.5 Յ 0.3 Ϫ x Յ 1.7 76. Ϫ8.2 6 1.4 Ϫ x 6 Ϫ0.9 77. Ϫ7 6 Ϫ3 4 x Ϫ 1 Յ 11 78. Ϫ21 Յ Ϫ2 3x ϩ 9 6 7

44. Ϫ5 1 x ϩ 2 2 Ϫ 3 6 3x ϩ 11 y 2y ϩ 6 Ϫ2 46. 5 10

Precalculus—

A-36


APPENDIX A A Review of Basic Concepts and Skills

WORKING WITH FORMULAS
81. Body mass index: B ‫؍‬ 704W H2

79. Euler’s Polyhedron Formula: V ؉ F ؊ E ‫ ؍‬2 Discovered by Leonhard Euler in 1752, this simple but powerful formula states that in any regular polyhedron, the number of vertices V and faces F is always two more than the number of edges E. (a) Verify the formula for a simple cube. (b) Verify the formula for the octahedron shown in the figure. (c) If a dodecahedron has 12 faces and 30 edges, how many vertices does it have? 80. Area of a Regular Polygon: A ‫؍‬ 1 ap 2

The U.S. government publishes a body mass index formula to help people consider the risk of heart disease. An index “B” of 27 or more means that a person is at risk. Here W represents weight in pounds and H represents height in inches. If your height is 5 ¿ 8 – what range of weights will help ensure you remain safe from the risk of heart disease?
Source: www.surgeongeneral.gov/topics.

82. Lift capacity: 75S ؉ 125B Յ 750 The capacity in pounds of the lift used by a roofing company to place roofing shingles and buckets of roofing nails on rooftops is modeled by the formula shown, where S represents packs of shingles and B represents buckets of nails. Use the formula to find (a) the largest number of shingle packs that can be lifted, (b) the largest number of nail buckets that can be lifted, and (c) the largest number of shingle packs that can be lifted along with three nail buckets.

The area of any regular polygon can be found a using the formula shown, where a is the apothem of the polygon (perpendicular distance from center to any edge), and p is the perimeter. (a) Verify the formula using a square with sides of length 6 cm. (b) If the hexagon shown has an area of 259.8 cm2 with sides 10 cm in length, what is the length a of the apothem?


APPLICATIONS

Write an equation to model the given information and solve.

83. Celebrity Travel: To avoid paparazzi and overzealous fans, the arrival gates of planes carrying celebrities are often kept secret until the last possible moment. While awaiting the arrival of Angelina Jolie, a large crowd of fans and photographers had gathered at Terminal A, Gate 18. However, the number of fans waiting at Gate 32 was twice that number increased by 5. If there were 73 fans at Gate 32, how many were waiting at Gate 18? (See Section A.1, Example 2a.) 84. Famous Architecture: The Hall of Mirrors is the central gallery of the Palace of Versailles and is one of the most famous rooms in the world. The length of this hall is 11 m less than 8 times the width. If the hall is 73 m long, what is its width? (See Section A.1, Example 2b.) 85. Dietary Goals: At the picnic, Mike abandoned his diet and consumed 13 calories more than twice the number of calories he normally allots for lunch. If he consumed 1467 calories, how many calories are normally allotted for lunch?

86. Marathon Training: While training for the Chicago marathon, Christina’s longest run of the week was 5 mi less than double the shortest. If the longest run was 11.2 mi, how long was the shortest? 87. Actor’s Ages: At the time of this writing, actor Will Smith (Enemy of the State, Seven Pounds, others), was 1 yr older than two-thirds the age of Samuel Jackson (The Negotiator, Die Hard III, others). If Will Smith was 41 at this time, how old was Samuel Jackson? 88. Football versus Fútbol: The area of a regulation field for American football is about 410 square meters (m2) less than three-fifths of an Olympicsized soccer field. If an American football field covers 5350 m2, what is the area of an Olympic soccer field?

Precalculus—

Section A.3 Solving Linear Equations and Inequalities

A-37

89. Forensic Studies: In forensic studies, skeletal remains are analyzed to determine the height, gender, race, age, and other characteristics of the decedent. For instance, the height of a male individual is approximated as 34 in. more than three and one-third times the length of the radial bone. If a live individual is 74 in. tall, how long is his radial bone?
Write an inequality to model the given information and solve.

90. Famous Waterways: The Suez Canal and the Panama Canal are two of the most important waterways in the world, saving ships thousands of miles as they journey from port to destination. The length of the Suez Canal is 39 kilometers (km) less than three times the length of the Panama Canal. If the Egyptian canal is 192 km long, how long is the Central American canal? 96. Area of a triangle: Using the triangle shown, find the height that will guarantee an area equal to or greater than 48 in2.

91. Exam scores: Jacques is going to college on an academic scholarship that requires him to maintain at least a 75% average in all of his classes. So far he has scored 82%, 76%, 65%, and 71% on four exams. What scores are possible on his last exam that will enable him to keep his scholarship? 92. Timed trials: In the first three trials of the 100-m butterfly, Johann had times of 50.2, 49.8, and 50.9 sec. How fast must he swim the final timed trial to have an average time of at most 50 sec? 93. Checking account balance: If the average daily balance in a certain checking account drops below $1000, the bank charges the customer a $7.50 service fee. The table Weekday Balance gives the daily balance Monday $1125 for one customer. What $850 must the daily balance be Tuesday Wednesday $625 for Friday to avoid a service charge? Thursday $400 94. Average weight: In the Lineman Weight National Football Left tackle 318 lb League, many consider an offensive line to be Left guard 322 lb “small” if the average Center 326 lb weight of the five down Right guard 315 lb linemen is less than Right tackle ? 325 lb. Using the table, what must the weight of the right tackle be so that the line will not be considered small? 95. Area of a rectangle: Given the rectangle shown, what is the range of values for the width, in order to keep the area less than 150m2?
20 m

h

12 in.

97. Heating and cooling subsidies: As long as the outside temperature is over 45°F and less than 85°F 1 45 6 F 6 85 2 , the city does not issue heating or cooling subsidies for low-income families. What is the corresponding range of Celsius temperatures C? Recall that F ϭ 9 5 C ϩ 32. 98. U.S. and European shoe sizes: To convert a European male shoe size “E” to an American male shoe size “A,” the formula A ϭ 0.76E Ϫ 23 can be used. Lillian has five sons in the U.S. military, with shoe sizes ranging from size 9 to size 14 1 9 Յ A Յ 14 2 . What is the corresponding range of European sizes? Round to the nearest half-size. 99. Power tool rentals: Sunshine Equipment Co. rents its power tools for a $20 fee, plus $4.50/hr. Kealoha’s Rentals offers the same tools for an $11 fee plus $6.00/hr. How many hours h must a tool be rented to make the cost at Sunshine a better deal? 100. Moving van rentals: Stringer Truck Rentals will rent a moving van for $15.75/day plus $0.35 per mile. Bertz Van Rentals will rent the same van for $25/day plus $0.30 per mile. How many miles m must the van be driven to make the cost at Bertz a better deal?

w

Precalculus—

A-38

APPENDIX A A Review of Basic Concepts and Skills 5 in. 103. Trophy bases: The base of a new trophy 7 in. has the form of a cylinder sitting atop a rectangular solid. 2 in. If the base is to be 10 in. cast in a special 10 in. aluminum, determine the volume of aluminum to be used.

101. Cost of drywall: After the studs are up, the 3 ft wall shown in the figure 15 ft must be covered in 10 ft 7 ft drywall. (a) How many square feet of drywall 19 ft are needed? (b) If drywall is sold only in 4-ft by 8-ft sheets, approximately how many sheets are required for this job?

102. Paving a walkway: Current plans 104. Grain storage: The dimensions of call for building a circular fountain a grain silo are shown in the figure. 6m 6 m in diameter with a circular If the maximum storage capacity of walkway around it that is 1.5 m the silo is 95% of the total volume wide. (a) What is the approximate of the silo, how many cubic meters area of the walkway? (b) If the 1.5 m of corn can be stored? concrete for the walkway is to be 6 cm deep, what volume of cement must be used 1 1 cm ϭ 0.01 m 2 ?


16 m

6m

EXTENDING THE CONCEPT
Place the correct inequality symbol in the blank to make the statement true.

105. Solve for x: Ϫ3 1 4x2 ϩ 5x Ϫ 2 2 ϩ 7x ϭ 6 1 4 Ϫ x Ϫ 2x2 2 Ϫ 19

106. Solve for n: 55 3 Ϫ 3 4 Ϫ 2 1 5 Ϫ 9n 2 4 6 ϩ 15 ϭ Ϫ65 5 ϩ 2 3 n Ϫ 10 1 9 ϩ n 2 4 6 107. Use your local library, the Internet, or another resource to find the highest and lowest point on each of the seven continents. Express the range of altitudes for each continent as a joint inequality. Which continent has the greatest range? 108. The sum of two consecutive even integers is greater than or equal to 12 and less than or equal to 22. List all possible values for the two integers.

109. If m 7 0 and n 6 0, then mn 110. If m 7 n and p 7 0, then mp 111. If m 6 n and p 7 0, then mp 112. If m Յ n and p 6 0, then mp 113. If m 7 n, then Ϫm
1 114. If 0 6 m 6 n, then m

0. np. np. np. Ϫn.
1 n.

115. If m 7 0 and n 6 0, then m2 116. If m 6 0, then m3 0.

n.

A.4

Factoring Polynomials and Solving Polynomial Equations by Factoring
It is often said that knowing which tool to use is just as important as knowing how to use the tool. In this section, we review the tools needed to factor an expression, an important part of solving polynomial equations. This section will also help us decide which factoring tool is appropriate when many different factorable expressions are presented.

LEARNING OBJECTIVES
In Section A.4 you will review:

A. Factoring out the greatest B. C. D. E.
common factor Common binomial factors and factoring by grouping Factoring quadratic polynomials Factoring special forms and quadratic forms Solving Polynomial Equations by Factoring

A. The Greatest Common Factor
To factor an expression means to rewrite the expression as an equivalent product. The distributive property is an example of factoring in action. To factor 2x2 ϩ 6x, we might first rewrite each term using the common factor 2x: 2x2 ϩ 6x ϭ 2x # x ϩ 2x # 3, then apply the distributive property to obtain 2x 1 x ϩ 3 2 . We commonly say that we have factored out 2x. The greatest common factor (or GCF) is the largest factor common to all terms in the polynomial.

Precalculus—

Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

A-39

EXAMPLE 1



Factoring Polynomials Factor each polynomial: a. 12x2 ϩ 18xy Ϫ 30y b. x5 ϩ x2
mentally: 6 # 2x2 ϩ 6 # 3xy Ϫ 6 # 5y

Solution



a. 6 is common to all three terms: 12x2 ϩ 18xy Ϫ 30y ϭ 6 1 2x2 ϩ 3xy Ϫ 5y 2 x5 ϩ x2 ϭ x2 1 x3 ϩ 1 2

b. x2 is common to both terms:
mentally: x 2 # x 3 ϩ x 2 # 1

A. You’ve just seen how we can factor out the greatest common factor

Now try Exercises 7 and 8



B. Common Binomial Factors and Factoring by Grouping
If the terms of a polynomial have a common binomial factor, it can also be factored out using the distributive property. EXAMPLE 2


Factoring Out a Common Binomial Factor Factor: a. 1 x ϩ 3 2 x2 ϩ 1 x ϩ 3 2 5 b. x2 1 x Ϫ 2 2 Ϫ 3 1 x Ϫ 2 2 b. x2 1 x Ϫ 2 2 Ϫ 3 1 x Ϫ 2 2 ϭ 1 x Ϫ 2 2 1 x2 Ϫ 3 2 Now try Exercises 9 and 10


Solution



a. 1 x ϩ 3 2 x2 ϩ 1 x ϩ 3 2 5 ϭ 1 x ϩ 3 2 1 x2 ϩ 5 2

One application of removing a binomial factor involves factoring by grouping. At first glance, the expression x3 ϩ 2x2 ϩ 3x ϩ 6 appears unfactorable. But by grouping the terms (applying the associative property), we can remove a monomial factor from each subgroup, which then reveals a common binomial factor. x3 ϩ 2x2 ϩ 3x ϩ 6 ϭ x2 1 x ϩ 2 2 ϩ 3 1 x ϩ 2 2 ϭ 1 x ϩ 2 2 1 x2 ϩ 3 2

This grouping of terms must take into account any sign changes and common factors, as seen in Example 3. Also, it will be helpful to note that a general four-term polynomial A ϩ B ϩ C ϩ D is factorable by grouping only if AD ϭ BC. EXAMPLE 3


Factoring by Grouping Factor 3t3 ϩ 15t2 Ϫ 6t Ϫ 30. Notice that all four terms have a common factor of 3. Begin by factoring it out. 3t3 ϩ 15t2 Ϫ 6t Ϫ 30 ϭ 3 1 t3 ϩ 5t2 Ϫ 2t Ϫ 10 2 ϭ 3 1 t3 ϩ 5t2 Ϫ 2t Ϫ 10 2 ϭ 3 3 t2 1 t ϩ 5 2 Ϫ 2 1 t ϩ 5 2 4 ϭ 3 1 t ϩ 5 21 t2 Ϫ 2 2
original polynomial factor out 3 group remaining terms factor common monomial factor common binomial

Solution



Now try Exercises 11 and 12



Precalculus—

A-40

APPENDIX A A Review of Basic Concepts and Skills

B. You’ve just seen how we can factor common binomial factors and factor by grouping

When asked to factor an expression, first look for common factors. The resulting expression will be easier to work with and help ensure the final answer is written in completely factored form. If a four-term polynomial cannot be factored as written, try rearranging the terms to find a combination that enables factoring by grouping.

C. Factoring Quadratic Polynomials
A quadratic polynomial is one that can be written in the form ax2 ϩ bx ϩ c, where a, b, c ʦ ‫ ޒ‬and a 0. One common form of factoring involves quadratic trinomials such as x2 ϩ 7x ϩ 10 and 2x2 Ϫ 13x ϩ 15. While we know 1 x ϩ 5 21 x ϩ 2 2 ϭ x2 ϩ 7x ϩ 10 and 1 2x Ϫ 3 21 x Ϫ 5 2 ϭ 2x2 Ϫ 13x ϩ 15 using F-O-I-L, how can we factor these trinomials without seeing the original expression in advance? First, it helps to place the trinomials in two families—those with a leading coefficient of 1 and those with a leading coefficient other than 1. When a ϭ 1, the only factor pair for x2 (other than 1 # x2 2 is x # x and the first term in each binomial will be x: (x )(x ). The following observation helps guide us to the complete factorization. Consider the product 1 x ϩ b 2 1 x ϩ a 2 : 1 x ϩ b 21 x ϩ a 2 ϭ x2 ϩ ax ϩ bx ϩ ab
F-O-I-L distributive property

ax 2 ؉ bx ؉ c, where a ‫ ؍‬1

ϭ x2 ϩ 1 a ϩ b 2 x ϩ ab

Note the last term is the product ab (the lasts), while the coefficient of the middle term is a ϩ b (the sum of the outers and inners). Since the last term of x2 Ϫ 8x ϩ 7 is 7 and the coefficient of the middle term is Ϫ8, we are seeking two numbers with a product of positive 7 and a sum of negative 8. The numbers are Ϫ7 and Ϫ1, so the factored form is 1 x Ϫ 7 21 x Ϫ 1 2 . It is also helpful to note that if the constant term is positive, the binomials will have like signs, since only the product of like signs is positive. If the constant term is negative, the binomials will have unlike signs, since only the product of unlike signs is negative. This means we can use the sign of the linear term (the term with degree 1) to guide our choice of factors. Factoring Trinomials with a Leading Coefficient of 1 If the constant term is positive, the binomials will have like signs: 1 x ϩ 2 1 x ϩ 2 or 1 x Ϫ 2 1 x Ϫ 2 , to match the sign of the linear (middle) term. If the constant term is negative, the binomials will have unlike signs: 1x ϩ 2 1x Ϫ 2, with the larger factor placed in the binomial whose sign matches the linear (middle) term.

EXAMPLE 4



Factoring Trinomials Factor these expressions: a. Ϫx2 ϩ 11x Ϫ 24 b. x2 Ϫ 10 Ϫ 3x

Solution



a. First rewrite the trinomial in standard form as Ϫ1 1 x2 Ϫ 11x ϩ 24 2 . For x2 Ϫ 11x ϩ 24, the constant term is positive so the binomials will have like signs. Since the linear term is negative, Ϫ1 1 x2 Ϫ 11x ϩ 24 2 ϭ Ϫ1 1 x Ϫ 21 x Ϫ 2 ϭ Ϫ1 1 x Ϫ 8 21 x Ϫ 3 2
like signs, both negative 1 Ϫ8 2 1 Ϫ3 2 ϭ 24; Ϫ8 ϩ 1 Ϫ3 2 ϭ Ϫ11

Precalculus—

Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

A-41

b. First rewrite the trinomial in standard form as x2 Ϫ 3x Ϫ 10. The constant term is negative so the binomials will have unlike signs. Since the linear term is negative, x2 Ϫ 3x Ϫ 10 ϭ 1 x ϩ 2 1 x Ϫ 2 ϭ 1x ϩ 22 1x Ϫ 52
unlike signs, one positive and one negative 5 7 2, 5 is placed in the second binomial; 1 2 2 1 Ϫ5 2 ϭ Ϫ10; 2 ϩ 1 Ϫ5 2 ϭ Ϫ3

Now try Exercises 13 and 14



Sometimes we encounter prime polynomials, or polynomials that cannot be factored. For x2 ϩ 9x ϩ 15, the factor pairs of 15 are 1 # 15 and 3 # 5, with neither pair having a sum of ϩ9. We conclude that x2 ϩ 9x ϩ 15 is prime. ax 2 ؉ bx ؉ c, where a 1

If the leading coefficient is not one, the possible combinations of outers and inners are more numerous. Furthermore, the sum of the outer and inner products will change depending on the position of the possible factors. Note that 1 2x ϩ 3 21 x ϩ 9 2 ϭ 2x2 ϩ 21x ϩ 27 and 1 2x ϩ 9 21 x ϩ 3 2 ϭ 2x2 ϩ 15x ϩ 27 result in a different middle term, even though identical numbers were used. To factor 2x2 Ϫ 13x ϩ 15, note the constant term is positive so the binomials must have like signs. The negative linear term indicates these signs will be negative. We then list possible factors for the first and last terms of each binomial, then sum the outer and inner products.
Possible First and Last Terms for 2x2 and 15 1. 1 2x Ϫ 1 21 x Ϫ 15 2 2. 1 2x Ϫ 15 21 x Ϫ 1 2 3. 1 2x Ϫ 3 21 x Ϫ 5 2 4. 1 2x Ϫ 5 21 x Ϫ 3 2
WORTHY OF NOTE
The number of trials needed to factor a polynomial can also be reduced by noting that the two terms in any binomial cannot share a common factor (all common factors are removed in a preliminary step).

Sum of Outers and Inners Ϫ30x Ϫ 1x ϭ Ϫ31x Ϫ2x Ϫ 15x ϭ Ϫ17x Ϫ10x Ϫ 3x ϭ Ϫ13x Ϫ6x Ϫ 5x ϭ Ϫ11x d

As you can see, only possibility 3 yields a linear term of Ϫ13x, and the correct factorization is then 1 2x Ϫ 3 21 x Ϫ 5 2 . With practice, this trial-and-error process can be completed very quickly. If the constant term is negative, the number of possibilities can be reduced by finding a factor pair with a sum or difference equal to the absolute value of the linear coefficient, as we can then arrange the sign in each binomial to obtain the needed result as shown in Example 5. Factoring a Trinomial Using Trial and Error Factor 6z2 Ϫ 11z Ϫ 35. Note the constant term is negative (binomials will have unlike signs) and ͿϪ11Ϳ ϭ 11. The factors of 35 are 1 # 35 and 5 # 7. Two possible first terms are: (6z )(z ) and (3z )(2z ), and we begin with 5 and 7 as factors of 35.
Outer and Inner Products Sum 1. (6z 2. (6z 5)(z 7)(z 7) 5) 42z ϩ 5z 47z 30z ϩ 7z 37z Difference 42z Ϫ 5z 37z 30z Ϫ 7z 23z 3. (3z 4. (3z 5)(2z 7)(2z 7) 5) Outer and Inner Products Sum 21z ϩ 10z 31z 15z ϩ 14z 29z Difference 21z Ϫ 10z 11z 15z Ϫ 14z 1z

EXAMPLE 5



Solution



(6z

)(z

)

(3z

)(2z

)

Precalculus—

A-42

APPENDIX A A Review of Basic Concepts and Skills

Since possibility 3 yields a linear term of 11z, we need not consider other factors of 35 and write the factored form as 6z2 Ϫ 11z Ϫ 35 ϭ 1 3z 5 21 2z 7 2 . The signs can then be arranged to obtain a middle term of Ϫ11z: 1 3z ϩ 5 2 1 2z Ϫ 7 2 , Ϫ21z ϩ 10z ϭ Ϫ11z ✓.
C. You’ve just seen how we can factor quadratic polynomials

Now try Exercises 15 and 16



D. Factoring Special Forms and Quadratic Forms
Next we consider methods to factor each of the special products we encountered in Appendix A.2.

The Difference of Two Squares
WORTHY OF NOTE
In an attempt to factor a sum of two perfect squares, say v2 ϩ 49, let’s list all possible binomial factors. These are (1) 1 v ϩ 7 21 v ϩ 7 2 , (2) 1 v Ϫ 7 21 v Ϫ 7 2 , and (3) 1 v ϩ 7 21 v Ϫ 7 2 . Note that (1) and (2) are the binomial squares 1 v ϩ 7 2 2 and 1 v Ϫ 7 2 2, with each product resulting in a “middle” term, whereas (3) is a binomial times its conjugate, resulting in a difference of squares: v2 Ϫ 49. With all possibilities exhausted, we conclude that the sum of two squares is prime!

Multiplying and factoring are inverse processes. Since 1 x Ϫ 7 21 x ϩ 7 2 ϭ x2 Ϫ 49, we know that x2 Ϫ 49 ϭ 1 x Ϫ 7 21 x ϩ 7 2 . In words, the difference of two squares will factor into a binomial and its conjugate. To find the terms of the factored form, rewrite each term in the original expression as a square: 1 2 2. Factoring the Difference of Two Perfect Squares Given any expression that can be written in the form A2 Ϫ B2, A2 Ϫ B2 ϭ 1 A ϩ B 21 A Ϫ B 2

Note that the sum of two perfect squares A2 ϩ B2 cannot be factored using real numbers (the expression is prime). As a reminder, always check for a common factor first and be sure to write all results in completely factored form. See Example 6(c). Factoring the Difference of Two Perfect Squares Factor each expression completely. a. 4w2 Ϫ 81 b. v2 ϩ 49 c. Ϫ3n2 ϩ 48
1 d. z4 Ϫ 81

EXAMPLE 6



Solution



a. 4w2 Ϫ 81 ϭ 1 2w 2 2 Ϫ 92 ϭ 1 2w ϩ 9 2 1 2w Ϫ 9 2 b. v2 ϩ 49 is prime. c. Ϫ3n2 ϩ 48 ϭ Ϫ3 1 n2 Ϫ 16 2 ϭ Ϫ3 3 n2 Ϫ 1 4 2 2 4 ϭ Ϫ3 1 n ϩ 4 2 1 n Ϫ 4 2 1 2 d. z4 Ϫ 81 ϭ 1 z2 2 2 Ϫ 1 1 92 1 2 2 ϭ 1z ϩ 9 2 1z Ϫ 1 92 2 2 ϭ 1 z2 ϩ 1 2 3 z Ϫ 11 9 32 4 1 1 2 ϭ 1z ϩ 9 2 1z ϩ 3 2 1z Ϫ 1 32 e. x2 Ϫ 7 ϭ 1 x 2 2 Ϫ 1 17 2 2 ϭ 1 x ϩ 17 21 x Ϫ 17 2

e. x2 Ϫ 7

write as a difference of squares A 2 Ϫ B 2 ϭ 1 A ϩ B 21 A Ϫ B 2 factor out Ϫ3 write as a difference of squares A 2 Ϫ B 2 ϭ 1 A ϩ B 21 A Ϫ B 2 write as a difference of squares write as a difference of squares (z 2 ϩ 1 9 is prime) result write as a difference of squares A 2 Ϫ B 2 ϭ 1 A ϩ B 21 A Ϫ B 2 A 2 Ϫ B 2 ϭ 1 A ϩ B 21 A Ϫ B 2

Now try Exercises 17 and 18



Perfect Square Trinomials

Since 1 x ϩ 7 2 2 ϭ x2 ϩ 14x ϩ 49, we know that x2 ϩ 14x ϩ 49 ϭ 1 x ϩ 7 2 2. In words, a perfect square trinomial will factor into a binomial square. To use this idea effectively, we must learn to identify perfect square trinomials. Note that the first and last terms of x2 ϩ 14x ϩ 49 are the squares of x and 7, and the middle term is twice the product of these two terms: 2 1 7x 2 ϭ 14x. These are the characteristics of a perfect square trinomial.

Precalculus—

Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

A-43

Factoring Perfect Square Trinomials Given any expression that can be written in the form A2 Ϯ 2AB ϩ B2, 1. A2 ϩ 2AB ϩ B2 ϭ 1 A ϩ B 2 2 2. A2 Ϫ 2AB ϩ B2 ϭ 1 A Ϫ B 2 2

EXAMPLE 7



Factoring a Perfect Square Trinomial Factor 12m3 Ϫ 12m2 ϩ 3m. 12m3 Ϫ 12m2 ϩ 3m ϭ 3m 1 4m2 Ϫ 4m ϩ 1 2
check for common factors: GCF ϭ 3m factor out 3m

Solution



For the remaining trinomial 4m2 Ϫ 4m ϩ 1 p 1. Are the first and last terms perfect squares?

4m2 ϭ 1 2m 2 2 and 1 ϭ 1 1 2 2 ✓ Yes. 2 # 2m # 1 ϭ 4m ✓ Yes.

2. Is the linear term twice the product of 2m and 1? Factor as a binomial square: 4m2 Ϫ 4m ϩ 1 ϭ 1 2m Ϫ 1 2 2 This shows 12m3 Ϫ 12m2 ϩ 3m ϭ 3m 1 2m Ϫ 1 2 2. Now try Exercises 19 and 20


CAUTION



As shown in Example 7, be sure to include the GCF in your final answer. It is a common error to “leave the GCF behind.”

In actual practice, these calculations can be performed mentally, making the process much more efficient.

Sum or Difference of Two Perfect Cubes
Recall that the difference of two perfect squares is factorable, but the sum of two perfect squares is prime. In contrast, both the sum and difference of two perfect cubes are factorable. For either A3 ϩ B3 or A3 Ϫ B3 we have the following: 1. Each will factor into the product of a binomial and a trinomial: 2. The terms of the binomial are the quantities being cubed: 3. The terms of the trinomial are the square of A, the product AB, and the square of B, respectively: 4. The binomial takes the same sign as the original expression 5. The middle term of the trinomial takes the opposite sign of the original expression (the last term is always positive): ( (A (A )(
trinomial

) ) AB AB B 2) B 2)

binomial

B)( B)(A2

(A Ϯ B)(A2

(A Ϯ B)(A2 ϯ AB ϩ B 2)

Factoring the Sum or Difference of Two Perfect Cubes: A3 Ϯ B3 1. A3 ϩ B3 ϭ 1 A ϩ B 2 1 A2 Ϫ AB ϩ B2 2 2. A3 Ϫ B3 ϭ 1 A Ϫ B 21 A2 ϩ AB ϩ B2 2

Precalculus—

A-44

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 8



Factoring the Sum and Difference of Two Perfect Cubes Factor completely: a. x3 ϩ 125 b. Ϫ5m3n ϩ 40n4
write terms as perfect cubes factoring template A S x and B S 5 check for common factors 1 GCF ϭ Ϫ5n 2 write terms as perfect cubes factoring template A S m and B S 2n simplify factored form

Solution



a.

x3 ϩ 125 ϭ x3 ϩ 53 Use A3 ϩ B3 ϭ 1 A ϩ B 21 A2 Ϫ AB ϩ B2 2 x3 ϩ 53 ϭ 1 x ϩ 5 2 1 x2 Ϫ 5x ϩ 25 2

b.

Ϫ5m3n ϩ 40n4 ϭ Ϫ5n 1 m3 Ϫ 8n3 2 ϭ Ϫ5n 3 m3 Ϫ 1 2n 2 3 4 Use A3 Ϫ B3 ϭ 1 A Ϫ B 21 A2 ϩ AB ϩ B2 2 m3 Ϫ 1 2n 2 3 ϭ 1 m Ϫ 2n 2 3 m2 ϩ m 1 2n 2 ϩ 1 2n 2 2 4 ϭ 1 m Ϫ 2n 21 m2 ϩ 2mn ϩ 4n2 2 3 4 1 Ϫ5m n ϩ 40n ϭ Ϫ5n 1 m Ϫ 2n 21 m2 ϩ 2mn ϩ 4n2 2 .

The results for parts (a) and (b) can be checked using multiplication. Now try Exercises 21 and 22


Quadratic Forms and u-Substitution
For any quadratic expression ax2 ϩ bx ϩ c in standard form, the degree of the leading term is twice the degree of the middle term. Generally, a trinomial is in quadratic form if it can be written as a 1 __ 2 2 ϩ b 1 __ 2 ϩ c, where the parentheses “hold” the same factors. The equation x4 Ϫ 13x2 ϩ 36 ϭ 0 is in quadratic form since 1 x2 2 2 Ϫ 13 1 x2 2 ϩ 36 ϭ 0. In many cases, we can factor these expressions using a placeholder substitution that transforms them into a more recognizable form. In a study of algebra, the letter “u” often plays this role. If we let u represent x2, the expression 1 x2 2 2 Ϫ 13 1 x2 2 ϩ 36 becomes u2 Ϫ 13u ϩ 36, which can be factored into 1 u Ϫ 9 2 1 u Ϫ 4 2 . After “unsubstituting” (replace u with x2), we have 1 x2 Ϫ 9 2 1 x2 Ϫ 4 2 ϭ 1 x ϩ 3 21 x Ϫ 3 21 x ϩ 2 2 1 x Ϫ 2 2 . EXAMPLE 9


Factoring a Quadratic Form

Write in completely factored form: 1 x2 Ϫ 2x 2 2 Ϫ 2 1 x2 Ϫ 2x 2 Ϫ 3. Expanding the binomials would produce a fourth-degree polynomial that would be very difficult to factor. Instead we note the expression is in quadratic form. Letting u represent x2 Ϫ 2x (the variable part of the “middle” term), 1 x2 Ϫ 2x 2 2 Ϫ 2 1 x2 Ϫ 2x 2 Ϫ 3 becomes u2 Ϫ 2u Ϫ 3. To finish up, write the expression in terms of x, substituting x2 Ϫ 2x for u. ϭ 1 x2 Ϫ 2x Ϫ 3 2 1 x2 Ϫ 2x ϩ 1 2 ϭ 1x Ϫ 32 1x ϩ 12 1x Ϫ 122
substitute x2 Ϫ 2x for u

Solution



u2 Ϫ 2u Ϫ 3 ϭ 1 u Ϫ 3 2 1 u ϩ 1 2

factor

The resulting trinomials can be further factored.
x2 Ϫ 2x ϩ 1 ϭ 1 x Ϫ 1 2 2

Now try Exercises 23 and 24



D. You’ve just seen how we can factor special forms and quadratic forms

It is well known that information is retained longer and used more effectively when it’s placed in an organized form. The “factoring flowchart” provided in Figure A.4 offers a streamlined and systematic approach to factoring and the concepts involved. However, with some practice the process tends to “flow” more naturally than following a chart, with many of the decisions becoming automatic.

Precalculus—

Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

A-45

Factoring Polynomials

Standard Form: decreasing order of degree

Greatest Common Factor (positive leading coefficient)

Number of Terms

Two

Three

Four

Difference of squares

Difference of cubes

Sum of cubes

Trinomials (a ϭ 1)

Trinomials (a 1)

Factor by grouping

Advanced methods (Section 4.2)

• Can any result be factored further?

• Polynomials that cannot be factored are said to be prime.

Figure A.4

For additional practice with these ideas, see Exercises 25 through 52.

E. Polynomial Equations and the Zero Product Property
The ability to solve linear and quadratic equations is the foundation on which a large percentage of our future studies are built. Both are closely linked to the solution of other equation types, as well as to the graphs of these equations. In standard form, linear and quadratic equations have a known number of terms, so we commonly represent their coefficients using the early letters of the alphabet, as in ax2 ϩ bx ϩ c ϭ 0. However, these equations belong to the larger family of polynomial equations. To write a general polynomial, where the number of terms is unknown, we often represent the coefficients using subscripts on a single variable, such as a1, a2, a3, and so on. Polynomial Equations A polynomial equation of degree n is one of the form anxn ϩ an Ϫ 1xnϪ1 ϩ p ϩ a1x1 ϩ a0 ϭ 0 where an, an Ϫ 1, p , a1, a0 are real numbers and an 0.

As a prelude to solving polynomial equations of higher degree, we’ll first look at quadratic equations and the zero product property. As before, a quadratic equation is one that can be written in the form ax2 ϩ bx ϩ c ϭ 0, where a, b, and c are real numbers and a 0. As written, the equation is in standard form, meaning the terms are in decreasing order of degree and the equation is set equal to zero.

Precalculus—

A-46

APPENDIX A A Review of Basic Concepts and Skills

Quadratic Equations A quadratic equation can be written in the form ax2 ϩ bx ϩ c ϭ 0, with a, b, c ʦ ‫ޒ‬, and a 0.

Notice that a is the leading coefficient, b is the coefficient of the linear (first degree) term, and c is a constant. All quadratic equations have degree two, but can have one, two, or three terms. The equation n2 Ϫ 81 ϭ 0 is a quadratic equation with two terms, where a ϭ 1, b ϭ 0, and c ϭ Ϫ81. EXAMPLE 10


Determining Whether an Equation Is Quadratic State whether the given equation is quadratic. If yes, identify coefficients a, b, and c. Ϫ3 a. 2x2 Ϫ 18 ϭ 0 b. z Ϫ 12 Ϫ 3z2 ϭ 0 c. xϩ5ϭ0 4 d. z3 Ϫ 2z2 ϩ 7z ϭ 8 e. 0.8x2 ϭ 0

Solution
WORTHY OF NOTE
The word quadratic comes from the Latin word quadratum, meaning square. The word historically refers to the “four sidedness” of a square, but mathematically to the area of a square. Hence its application to polynomials of the form ax2 ϩ bx ϩ c, where the variable of the leading term is squared.



Standard Form a. b. c. d. e. 2x Ϫ 18 ϭ 0
2

Quadratic yes, deg 2 yes, deg 2 no, deg 1 no, deg 3 yes, deg 2 aϭ2

Coefficients bϭ0 bϭ1 c ϭ Ϫ18 c ϭ Ϫ12 a ϭ Ϫ3

Ϫ3z2 ϩ z Ϫ 12 ϭ 0 Ϫ3 xϩ5ϭ0 4 z3 Ϫ 2z2 ϩ 7z Ϫ 8 ϭ 0 0.8x2 ϭ 0

(linear equation) (cubic equation) a ϭ 0.8 bϭ0 cϭ0


Now try Exercises 53 through 64

With quadratic and other polynomial equations, we generally cannot isolate the variable on one side using only properties of equality, because the variable is raised to different powers. Instead we attempt to solve the equation by factoring and applying the zero product property. Zero Product Property If A and B represent real numbers or real-valued expressions and A # B ϭ 0, then A ϭ 0 or B ϭ 0. In words, the property says, If the product of any two (or more) factors is equal to zero, then at least one of the factors must be equal to zero. We can use this property to solve higher degree equations after rewriting them in terms of equations with lesser degree. As with linear equations, values that make the original equation true are called solutions or roots of the equation.

Precalculus—

Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

A-47

EXAMPLE 11



Solving Equations Using the Zero Product Property Solve by writing the equations in factored form and applying the zero product property. a. 3x2 ϭ 5x b. Ϫ5x ϩ 2x2 ϭ 3 c. 4x2 ϭ 12x Ϫ 9

Solution



3x2 ϭ 5x 3x2 Ϫ 5x ϭ 0 x 1 3x Ϫ 5 2 ϭ 0 x ϭ 0 or 3x Ϫ 5 ϭ 0 5 x ϭ 0 or xϭ 3 2 b. Ϫ5x ϩ 2x ϭ 3 2x2 Ϫ 5x Ϫ 3 ϭ 0 1 2x ϩ 1 21 x Ϫ 3 2 ϭ 0 2x ϩ 1 ϭ 0 or x Ϫ 3 ϭ 0 1 xϭϪ or xϭ3 2 c. 4x2 ϭ 12x Ϫ 9 2 4x Ϫ 12x ϩ 9 ϭ 0 1 2x Ϫ 3 21 2x Ϫ 3 2 ϭ 0 2x Ϫ 3 ϭ 0 or 2x Ϫ 3 ϭ 0 3 3 or xϭ xϭ 2 2 a.

given equation standard form factor set factors equal to zero (zero product property) result given equation standard form factor set factors equal to zero (zero product property) result given equation standard form factor set factors equal to zero (zero product property) result

3 This equation has only the solution x ϭ , which we call a repeated root. 2 Now try Exercises 65 through 88


CAUTION



Consider the equation x2 Ϫ 2x Ϫ 3 ϭ 12. While the left-hand side is factorable, the result is 1 x Ϫ 3 21 x ϩ 1 2 ϭ 12 and finding a solution becomes a “guessing game” because the equation is not set equal to zero. If you misapply the zero factor property and say that x Ϫ 3 ϭ 12 or x ϩ 1 ϭ 12, the “solutions” are x ϭ 15 or x ϭ 11, which are both incorrect! After subtracting 12 from both sides, x2 Ϫ 2x Ϫ 3 ϭ 12 becomes x2 Ϫ 2x Ϫ 15 ϭ 0 giving 1 x Ϫ 5 21 x ϩ 3 2 ϭ 0 with solutions x ϭ 5 or x ϭ Ϫ3.

EXAMPLE 12



Solving Polynomials by Factoring Solve by factoring: 4x3 Ϫ 40x ϭ 6x2. 4x3 Ϫ 40x ϭ 6x2 4x Ϫ 6x2 Ϫ 40x ϭ 0 2x 1 2x2 Ϫ 3x Ϫ 20 2 ϭ 0 2x 1 2x ϩ 5 21 x Ϫ 4 2 ϭ 0 2x ϭ 0 or 2x ϩ 5 ϭ 0 or x Ϫ 4 ϭ 0 5 x ϭ 0 or xϭϪ or xϭ4 2
3

Solution



given equation standard form common factor is 2x factored form zero product property result — solve for x

Substituting these values into the original equation verifies they are solutions. Now try Exercises 89 through 92


Example 12 reminds us that in the process of factoring polynomials, there may be a common monomial factor. This factor is also set equal to zero in the solution process (if the monomial is a constant, no solution is generated).

Precalculus—

A-48

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 13



Solving Higher Degree Equations Solve each equation by factoring. a. x3 Ϫ 4x ϩ 20 ϭ 5x2 b. x4 Ϫ 10x2 ϩ 9 ϭ 0
original equation standard form; factor by grouping remove common factors from each group factor common binomial factored form

Solution



a.

x 1x Ϫ 52 Ϫ 41x Ϫ 52 ϭ 0 1 x Ϫ 5 21 x2 Ϫ 4 2 ϭ 0 1 x Ϫ 5 21 x ϩ 2 21 x Ϫ 2 2 ϭ 0 x Ϫ 5 ϭ 0 or x ϩ 2 ϭ 0 x ϭ 5 or x ϭ Ϫ2
2

x3 Ϫ 4x ϩ 20 ϭ 5x2 x3 Ϫ 5x2 Ϫ 4x ϩ 20 ϭ 0

or x Ϫ 2 ϭ 0 or xϭ2

zero product property solve

The solutions are x ϭ 5, x ϭ Ϫ2, and x ϭ 2. b. The equation appears to be in quadratic form and we begin by substituting u for x2 and u2 for x4. original equation x4 Ϫ 10x2 ϩ 9 ϭ 0 2 substitute u for x 2 and u 2 for x 4 u Ϫ 10u ϩ 9 ϭ 0 factored form 1 u Ϫ 9 21 u Ϫ 1 2 ϭ 0 zero product property u Ϫ 9 ϭ 0 or u Ϫ 1 ϭ 0 substitute x 2 for u x2 Ϫ 9 ϭ 0 or x2 Ϫ 1 ϭ 0 1 x ϩ 3 21 x Ϫ 3 2 ϭ 0 or 1 x ϩ 1 2 1 x Ϫ 1 2 ϭ 0 factor x ϭ Ϫ3 or x ϭ 3 or x ϭ Ϫ1 or x ϭ 1 zero product property The solutions are x ϭ Ϫ3, x ϭ 3, x ϭ Ϫ1, and x ϭ 1. Now try Exercises 93 through 100
E. You’ve just seen how we can solve polynomial equations by factoring


In Examples 12 and 13, we were able to solve higher degree polynomial equations by “breaking them down” into linear and quadratic forms. This basic idea can be applied to other kinds of equations as well.

A.4 EXERCISES


CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. To factor an expression means to rewrite the expression as an equivalent . 3. The difference of two perfect squares always factors into the product of a(n) and its . 5. Discuss/Explain why 4x2 Ϫ 36 ϭ 1 2x Ϫ 6 2 1 2x ϩ 6 2 is not written in completely factored form, then rewrite it so it is factored completely.

2. If a polynomial will not factor, it is said to be a(n) polynomial. 4. The expression x2 ϩ 6x ϩ 9 is said to be a(n) trinomial, since its factored form is a perfect (binomial) square. 6. Discuss/Explain why a3 ϩ b3 is factorable, but a2 ϩ b2 is not. Demonstrate by writing x3 ϩ 64 in factored form, and by exhausting all possibilities for x2 ϩ 64 to show it is prime.

Precalculus—

Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring


A-49

DEVELOPING YOUR SKILLS
Perfect Square Trinomials

Factor each expression using the method indicated. Greatest Common Factor

7. a. Ϫ17x2 ϩ 51 c. Ϫ3a4 ϩ 9a2 Ϫ 6a3

b. 21b3 Ϫ 14b2 ϩ 56b

19. a. a2 Ϫ 6a ϩ 9 c. 4m2 Ϫ 20m ϩ 25 20. a. x2 ϩ 12x ϩ 36 c. 25p2 Ϫ 60p ϩ 36

b. b2 ϩ 10b ϩ 25 d. 9n2 Ϫ 42n ϩ 49 b. z2 Ϫ 18z ϩ 81 d. 16q2 ϩ 40q ϩ 25

8. a. Ϫ13n2 Ϫ 52 b. 9p2 ϩ 27p3 Ϫ 18p4 c. Ϫ6g5 ϩ 12g4 Ϫ 9g3
Common Binomial Factor

Sum/Difference of Perfect Cubes

10. a. 5x 1 x Ϫ 3 2 Ϫ 2 1 x Ϫ 3 2 b. 1 v Ϫ 5 2 2v ϩ 1 v Ϫ 5 2 3 c. 3p 1 q2 ϩ 5 2 ϩ 7 1 q2 ϩ 5 2
Grouping

9. a. 2a 1 a ϩ 2 2 ϩ 3 1 a ϩ 2 2 b. 1 b2 ϩ 3 2 3b ϩ 1 b2 ϩ 3 2 2 c. 4m 1 n ϩ 7 2 Ϫ 11 1 n ϩ 7 2

21. a. 8p3 Ϫ 27 c. g3 Ϫ 0.027 22. a. 27q3 Ϫ 125 c. b3 Ϫ 0.125
u-Substitution

b. m3 ϩ 1 8 4 d. Ϫ2t ϩ 54t
8 b. n3 ϩ 27 d. 3r4 Ϫ 24r

23. a. x4 Ϫ 10x2 ϩ 9 c. x6 Ϫ 7x3 Ϫ 8

b. x4 ϩ 13x2 ϩ 36

11. a. 9q3 ϩ 6q2 ϩ 15q ϩ 10 b. h5 Ϫ 12h4 Ϫ 3h ϩ 36 c. k5 Ϫ 7k3 Ϫ 5k2 ϩ 35 12. a. 6h3 Ϫ 9h2 Ϫ 2h ϩ 3 b. 4k3 ϩ 6k2 Ϫ 2k Ϫ 3 c. 3x2 Ϫ xy Ϫ 6x ϩ 2y
Trinomial Factoring where ͦaͦ ‫ ؍‬1

24. a. x6 Ϫ 26x3 Ϫ 27 b. 3 1 n ϩ 5 2 2 ϩ 2 1 n ϩ 5 2 Ϫ 21 c. 2 1 z ϩ 3 2 2 ϩ 3 1 z ϩ 3 2 Ϫ 54 25. Completely factor each of the following (recall that “1” is its own perfect square and perfect cube). a. n2 Ϫ 1 b. n3 Ϫ 1 c. n3 ϩ 1 d. 28x3 Ϫ 7x 26. Carefully factor each of the following trinomials, if possible. Note differences and similarities. b. x2 ϩ x Ϫ 6 a. x2 Ϫ x ϩ 6 d. x2 Ϫ x Ϫ 6 c. x2 ϩ x ϩ 6 e. x2 Ϫ 5x ϩ 6 f. x2 ϩ 5x Ϫ 6
Factor each expression completely, if possible. Rewrite the expression in standard form (factor out “Ϫ1” if needed) and factor out the GCF if one exists. If you believe the expression will not factor, write “prime.”

13. a. Ϫp2 ϩ 5p ϩ 14 c. n2 ϩ 20 Ϫ 9n 14. a. Ϫm2 ϩ 13m Ϫ 42 c. v2 ϩ 10v ϩ 15
Trinomial Factoring where a

b. q2 Ϫ 4q ϩ 12 b. x2 ϩ 12 ϩ 13x

1

15. a. 3p2 Ϫ 13p Ϫ 10 c. 10u2 Ϫ 19u Ϫ 15 16. a. 6v2 ϩ v Ϫ 35 c. 15z2 Ϫ 22z Ϫ 48
Difference of Perfect Squares

b. 4q2 ϩ 7q Ϫ 15 b. 20x2 ϩ 53x ϩ 18

27. a2 ϩ 7a ϩ 10 29. 2x2 Ϫ 24x ϩ 40 31. 64 Ϫ 9m2 33. Ϫ9r ϩ r2 ϩ 18 b. 9x2 Ϫ 49 d. 121h2 Ϫ 144 b. 25w Ϫ d. 16z4 Ϫ 81
2 1 49

28. b2 ϩ 9b ϩ 20 30. 10z2 Ϫ 140z ϩ 450 32. 25 Ϫ 16n2 34. 28 ϩ s2 Ϫ 11s 36. 3k2 ϩ 10k ϩ 8 38. 4p2 Ϫ 20p ϩ 25 40. Ϫ28z3 ϩ 16z2 ϩ 80z 42. Ϫ30n Ϫ 4n2 ϩ 2n3 44. b2 Ϫ 9b Ϫ 36 46. 27r3 ϩ 64

17. a. 4s2 Ϫ 25 c. 50x2 Ϫ 72 e. b2 Ϫ 5 18. a. 9v Ϫ c. v4 Ϫ 1 e. x2 Ϫ 17
2 1 25

35. 2h2 ϩ 7h ϩ 6 37. 9k2 Ϫ 24k ϩ 16 39. Ϫ6x3 ϩ 39x2 Ϫ 63x 41. 12m2 Ϫ 40m ϩ 4m3 43. a2 Ϫ 7a Ϫ 60 45. 8x3 Ϫ 125

Precalculus—

A-50

APPENDIX A A Review of Basic Concepts and Skills

47. m2 ϩ 9m Ϫ 24 49. x3 Ϫ 5x2 Ϫ 9x ϩ 45

48. n2 Ϫ 14n Ϫ 36 50. x3 ϩ 3x2 Ϫ 4x Ϫ 12

62. z2 Ϫ 6z ϩ 9 Ϫ z3 ϭ 0

63. 1 x Ϫ 1 2 2 ϩ 1 x Ϫ 1 2 ϩ 4 ϭ 9

51. Match each expression with the description that fits best. a. prime polynomial b. standard trinomial a ϭ 1 c. perfect square trinomial d. difference of cubes e. binomial square f. sum of cubes g. binomial conjugates h. difference of squares i. standard trinomial a 1 3 A. x ϩ 27 B. 1 x ϩ 3 2 2 C. x2 Ϫ 10x ϩ 25 D. x2 Ϫ 144 E. x2 Ϫ 3x Ϫ 10 F. 8s3 Ϫ 125t3 G. 2x2 Ϫ x Ϫ 3 H. x2 ϩ 9 I. 1 x Ϫ 7 2 and 1 x ϩ 7 2 52. Match each polynomial to its factored form. Two of them are prime. a. 4x2 Ϫ 9 b. 4x2 Ϫ 28x ϩ 49 c. x3 Ϫ 125 d. 8x3 ϩ 27 e. x2 Ϫ 3x Ϫ 10 f. x2 ϩ 3x ϩ 10 g. 2x2 Ϫ x Ϫ 3 h. 2x2 ϩ x Ϫ 3 i. x2 ϩ 25 A. 1 x Ϫ 5 21 x2 ϩ 5x ϩ 25 2 B. 1 2x Ϫ 3 21 x ϩ 1 2 C. 1 2x ϩ 3 21 2x Ϫ 3 2 2 D. 1 2x Ϫ 7 2 E. prime trinomial F. prime binomial G. 1 2x ϩ 3 21 x Ϫ 1 2 2 H. 1 2x ϩ 3 21 4x Ϫ 6x ϩ 9 2 I. 1 x Ϫ 5 21 x ϩ 2 2
Determine whether each equation is quadratic. If so, identify the coefficients a, b, and c. If not, discuss why.

64. 1 x ϩ 5 2 2 Ϫ 1 x ϩ 5 2 ϩ 4 ϭ 17
Solve using the zero factor property. Be sure each equation is in standard form and factor out any common factors before attempting to solve. Check all answers in the original equation.

65. x2 Ϫ 15 ϭ 2x 67. m2 ϭ 8m Ϫ 16 69. 5p2 Ϫ 10p ϭ 0 71. Ϫ14h2 ϭ 7h 73. a2 Ϫ 17 ϭ Ϫ8 75. g2 ϩ 18g ϩ 70 ϭ Ϫ11 76. h2 ϩ 14h Ϫ 2 ϭ Ϫ51

66. z2 Ϫ 10z ϭ Ϫ21 68. Ϫ10n ϭ n2 ϩ 25 70. 6q2 Ϫ 18q ϭ 0 72. 9w ϭ Ϫ6w2 74. b2 ϩ 8 ϭ 12

77. m3 ϩ 5m2 Ϫ 9m Ϫ 45 ϭ 0 78. n3 Ϫ 3n2 Ϫ 4n ϩ 12 ϭ 0 79. 1 c Ϫ 12 2 c Ϫ 15 ϭ 30 81. 9 ϩ 1 r Ϫ 5 2 r ϭ 33 82. 7 ϩ 1 s Ϫ 4 2 s ϭ 28 80. 1 d Ϫ 10 2 d ϩ 10 ϭ Ϫ6

83. 1 t ϩ 4 21 t ϩ 7 2 ϭ 54 85. 2x2 Ϫ 4x Ϫ 30 ϭ 0 87. 2w2 Ϫ 5w ϭ 3 88. Ϫ3v2 ϭ Ϫv Ϫ 2 89. 22x ϭ x3 Ϫ 9x2 90. x3 ϭ 13x2 Ϫ 42x 91. 3x3 ϭ Ϫ7x2 ϩ 6x 92. 7x2 ϩ 15x ϭ 2x3

84. 1 g ϩ 17 21 g Ϫ 2 2 ϭ 20 86. Ϫ3z2 ϩ 12z ϩ 36 ϭ 0

93. p3 ϩ 7p2 Ϫ 63 ϭ 9p 94. q3 Ϫ 4q ϩ 24 ϭ 6q2 95. x3 Ϫ 25x ϭ 2x2 Ϫ 50 96. 3c2 ϩ c ϭ c3 ϩ 3 97. x4 Ϫ 29x2 ϩ 100 ϭ 0 98. z4 Ϫ 20z2 ϩ 64 ϭ 0 99. 1 b2 Ϫ 3b 2 2 Ϫ 14 1 b2 Ϫ 3b 2 ϩ 40 ϭ 0

53. 2x Ϫ 15 Ϫ x2 ϭ 0 2 55. x Ϫ 7 ϭ 0 3 1 57. x2 ϭ 6x 4 59. 2x2 ϩ 7 ϭ 0

54. 21 ϩ x2 Ϫ 4x ϭ 0 56. 12 Ϫ 4x ϭ 9 58. 0.5x ϭ 0.25x 60. 5 ϭ Ϫ4x2
2

61. Ϫ3x2 ϩ 9x Ϫ 5 ϩ 2x3 ϭ 0

100. 1 d2 Ϫ d 2 2 Ϫ 8 1 d2 Ϫ d 2 ϩ 12 ϭ 0

Precalculus—

Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring


A-51

WORKING WITH FORMULAS
102. Volume of a cylindrical shell: ␲R2h ؊ ␲r2h
r The volume of a cylindrical shell (a larger cylinder with a smaller cylinder removed) can be found using the formula shown, where R is the radius of the larger cylinder and r is the radius of the smaller. Factor the expression R completely and use the result to find the volume of a shell where R ϭ 9 cm, r ϭ 3 cm, and h ϭ 10 cm. Answer in exact form and in approximate form rounded to the nearest whole number.

101. Surface area of a cylinder: 2␲ r2 ؉ 2␲ rh
The surface area of a cylinder is given by the formula shown, where h is the height of the cylinder and r is the radius. Factor out the GCF and use the result to find the surface area of a cylinder where r ϭ 35 cm and h ϭ 65 cm. Answer in exact form and in approximate form rounded to the nearest whole number.



APPLICATIONS

In many cases, factoring an expression can make it easier to evaluate as in the following applications.

103. Conical shells: The volume of a conical shell (like the shell of an ice cream cone) is given by the 1 1 formula V ϭ ␲R2h Ϫ ␲r2h, where R is the outer 3 3 radius and r is the inner radius of the cone. Write the formula in completely factored form, then find the volume of a shell when R ϭ 5.1 cm, r ϭ 4.9 cm, and h ϭ 9 cm. Answer in exact form and in approximate form rounded to the nearest tenth. 104. Spherical shells: The volume of a spherical shell (like the outer r shell of a cherry cordial) is given R 4 4 3 3 by the formula V ϭ 3␲R Ϫ 3␲r , where R is the outer radius and r is the inner radius of the shell. Write the right-hand side in completely factored form, then find the volume of a shell where R ϭ 1.8 cm and r ϭ 1.5 cm. Answer in exact form and in approximate form rounded to the nearest tenth. 105. Volume of a box: The volume of a rectangular box x inches in height is given by the relationship V ϭ x3 ϩ 8x2 ϩ 15x. Factor the right-hand side to determine: (a) The number of inches that the width exceeds the height, (b) the number of inches the length exceeds the height, and (c) the volume given the height is 2 ft. 106. Shipping textbooks: A publisher ships paperback books stacked x copies high in a box. The total number of books shipped per box is given by the relationship B ϭ x3 Ϫ 13x2 ϩ 42x. Factor the

right-hand side to determine (a) how many more or fewer books fit the width of the box (than the height), (b) how many more or fewer books fit the length of the box (than the height), and (c) the number of books shipped per box if they are stacked 10 high in the box. 107. Space-Time relationships: Due to the work of Albert Einstein and other physicists who labored on space-time relationships, it is known that the faster an object moves the shorter it appears to become. This phenomenon is modeled by the v 2 Lorentz transformation L ϭ L0 1 Ϫ a b , c B where L0 is the length of the object at rest, L is the relative length when the object is moving at velocity v, and c is the speed of light. Factor the radicand and use the result to determine the relative length of a 12-in. ruler if it is shot past a stationary observer at 0.75 times the speed of light 1 v ϭ 0.75c 2 . 108. Tubular fluid flow: As a fluid flows through a tube, it is flowing faster at the center of the tube than at the sides, where the tube exerts a backward drag. Poiseuille’s law gives the velocity of the flow G 2 1 R Ϫ r2 2 , at any point of the cross section: v ϭ 4␩ where R is the inner radius of the tube, r is the distance from the center of the tube to a point in the flow, G represents what is called the pressure gradient, and ␩ is a constant that depends on the viscosity of the fluid. Factor the right-hand side and find v given R ϭ 0.5 cm, r ϭ 0.3 cm, G ϭ 15, and ␩ ϭ 0.25.

Precalculus—

A-52

APPENDIX A A Review of Basic Concepts and Skills

Solve by factoring.

109. Envelope sizes: Large mailing envelopes often come in standard sizes, with 5- by 7-in. and 9- by 12-in. envelopes being the most common. The next larger size envelope has an area of 143 in2, with a length that is 2 in. longer than the width. What are the dimensions of the larger envelope? 110. Paper sizes: Letter size paper is 8.5 in. by 11 in. Legal size paper is 81 2 in. by 14 in. The next larger (common) size of paper has an area of 187 in2, with

a length that is 6 in. longer than the width. What are the dimensions of the Ledger size paper?

Letter

Legal Ledger



EXTENDING THE CONCEPT
x 3 x2 ϩ 3x ϩ 1 4 ϩ 5. Then we group the inner terms with x and factor again: P ϭ x 3 x2 ϩ 3x ϩ 1 4 ϩ 5 ϭ x 3 x 1 x ϩ 3 2 ϩ 1 4 ϩ 5. The expression can now be evaluated using any input and the order of operations. If x ϭ 2, we quickly find that P ϭ 27. Use this method to evaluate H ϭ x3 ϩ 2x2 ϩ 5x Ϫ 9 for x ϭ Ϫ3.
Factor each expression completely.

111. Factor out a constant that leaves integer coefficients for each term: 1 3 3 2 4 a. 1 2x ϩ 8x Ϫ 4x ϩ 4 1 3 4 2 5 b. 2 3b Ϫ 6b ϩ 9b Ϫ 1 112. If x ϭ 2 is substituted into 2x3 ϩ hx ϩ 8, the result is zero. What is the value of h? 113. Factor the expression: 192x3 Ϫ 164x2 Ϫ 270x. 114. As an alternative to evaluating polynomials by direct substitution, nested factoring can be used. The method has the advantage of using only products and sums — no powers. For P ϭ x3 ϩ 3x2 ϩ 1x ϩ 5, we begin by grouping all variable terms and factoring x: P ϭ 3 x3 ϩ 3x2 ϩ 1x 4 ϩ 5 ϭ

115. x4 Ϫ 81 117. p6 Ϫ 1 119. q4 Ϫ 28q2 ϩ 75

116. 16n4 Ϫ 1 118. m6 Ϫ 64 120. a4 Ϫ 18a2 ϩ 32

A.5

Rational Expressions and Equations
A rational number is one that can be written as the quotient of two integers. Similarly, a rational expression is one that can be written as the quotient of two polynomials. We can apply the skills developed in a study of fractions (how to reduce, add, subtract, multiply, and divide) to rational expressions, sometimes called algebraic fractions.

LEARNING OBJECTIVES
In Section A.5 you will review how to:

A. Write a rational
expression in simplest form Multiply and divide rational expressions Add and subtract rational expressions Simplify compound fractions Solve rational equations

B. C. D. E.

A. Writing a Rational Expression in Simplest Form
A rational expression is in simplest form when the numerator and denominator have no common factors (other than 1). After factoring the numerator and denominator, we apply the fundamental property of rational expressions. Fundamental Property of Rational Expressions If P, Q, and R are polynomials, with Q, R 0, P#R P P P#R 112 and 1 2 2 ϭ ϭ Q#R Q Q Q#R

Precalculus—

Section A.5 Rational Expressions and Equations

A-53

In words, the property says (1) a rational expression can be simplified by canceling common factors in the numerator and denominator, and (2) an equivalent expression can be formed by multiplying numerator and denominator by the same nonzero polynomial. EXAMPLE 1


Simplifying a Rational Expression Write the expression in simplest form: x2 Ϫ 1 . x2 Ϫ 3x ϩ 2

Solution



1 x Ϫ 1 21 x ϩ 1 2 x2 Ϫ 1 ϭ 2 1 x Ϫ 1 21 x Ϫ 2 2 x Ϫ 3x ϩ 2 1 x Ϫ 1 21 x Ϫ 2 2 xϩ1 ϭ xϪ2 ϭ 1 x Ϫ 1 21 x ϩ 1 2
1

factor numerator and denominator

common factors reduce to 1

simplest form

WORTHY OF NOTE
If we view a and b as two points on the number line, we note that they are the same distance apart, regardless of the order they are subtracted. This tells us the numerator and denominator will have the same absolute value but be opposite in sign, giving a value of Ϫ1 (check using a few test values).

Now try Exercises 7 through 10



When simplifying rational expressions, we sometimes encounter expressions of aϪb aϪb . If we factor Ϫ1 from the numerator, we see that ϭ the form bϪa bϪa Ϫ1 1 b Ϫ a 2 ϭ Ϫ1. bϪa

CAUTION



When reducing rational numbers or expressions, only common factors can be reduced. It is incorrect to reduce (or divide out) individual terms: xϩ1 xϩ2 1 (except for x ϭ 0) 2 Ϫ6 ϩ 4 23 2 Ϫ3 ϩ 4 23, and

Note that after simplifying an expression, we are actually saying the resulting (simpler) expression is equivalent to the original expression for all values where both are defined. The first expression is not defined when x ϭ 1 or x ϭ 2, the second when x ϭ 2 (since the denominators would be zero). The calculator screens shown in Figure A.5 help to illustrate this fact, and it appears that we would very much prefer to be working with the simpler expression!

Figure A.5

Precalculus—

A-54

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 2



Simplifying a Rational Expression Write the expression in simplest form: 6 Ϫ 2x . x2 Ϫ 9

Solution



213 Ϫ x2 6 Ϫ 2x ϭ 2 1 x Ϫ 3 21 x ϩ 3 2 x Ϫ9 1 2 21 Ϫ1 2 ϭ xϩ3 Ϫ2 ϭ xϩ3

factor numerator and denominator 3Ϫx ϭ Ϫ1 xϪ3

reduce:

simplest form

A. You’ve just seen how we can write a rational expression in simplest form

Now try Exercises 11 through 16



B. Multiplication and Division of Rational Expressions
Operations on rational expressions use the factoring skills reviewed earlier, along with much of what we know about rational numbers. Multiplying Rational Expressions Given that P, Q, R, and S are polynomials with Q, S P R # ϭ PR Q S QS 1. Factor all numerators and denominators completely. 2. Reduce common factors. 3. Multiply numerator ϫ numerator and denominator ϫ denominator. 0,

EXAMPLE 3



Multiplying Rational Expressions Compute the product: 2a ϩ 2 # 3a2 Ϫ a Ϫ 2 . 3a Ϫ 3a2 9a2 Ϫ 4

Solution



2 1 a ϩ 1 2 1 3a ϩ 2 21 a Ϫ 1 2 2a ϩ 2 # 3a2 Ϫ a Ϫ 2 # ϭ 2 2 3a 1 1 Ϫ a 2 1 3a Ϫ 2 21 3a ϩ 2 2 3a Ϫ 3a 9a Ϫ 4 ϭ ϭ 3a 1 1 Ϫ a 2 1 3a Ϫ 2 21 3a ϩ 2 2 Ϫ2 1 a ϩ 1 2 3a 1 3a Ϫ 2 2 21a ϩ 12

factor

#

1 3a ϩ 2 21 a Ϫ 1 2
1

1Ϫ12

reduce:

aϪ1 ϭ Ϫ1 1Ϫa

simplest form

Now try Exercises 17 through 20



To divide fractions, we multiply the first expression by the reciprocal of the second (we sometimes say, “invert the divisor and multiply”). The quotient of two rational expressions is computed in the same way.

Precalculus—

Section A.5 Rational Expressions and Equations

A-55

Dividing Rational Expressions Given that P, Q, R, and S are polynomials with Q, R, S P R P S PS Ϭ ϭ # ϭ Q S Q R QR Invert the divisor and multiply. 0,

EXAMPLE 4



Dividing Rational Expressions Compute the quotient 4m3 Ϫ 12m2 ϩ 9m 10m2 Ϫ 15m Ϭ . m2 Ϫ 49 m2 ϩ 4m Ϫ 21

Solution



4m3 Ϫ 12m2 ϩ 9m 10m2 Ϫ 15m Ϭ m2 Ϫ 49 m2 ϩ 4m Ϫ 21 3 2 4m Ϫ 12m ϩ 9m m2 ϩ 4m Ϫ 21 # ϭ m2 Ϫ 49 10m2 Ϫ 15m m 1 4m2 Ϫ 12m ϩ 9 2 1 m ϩ 7 2 1 m Ϫ 3 2 # ϭ 1 m ϩ 7 21 m Ϫ 7 2 5m 1 2m Ϫ 3 2
1 1 1

invert and multiply

factor

ϭ ϭ

m 1 2m Ϫ 3 21 2m Ϫ 3 2 1 m ϩ 7 21 m Ϫ 3 2 1 m ϩ 7 21 m Ϫ 7 2 1 2m Ϫ 3 2 1 m Ϫ 3 2 51m Ϫ 72

#

5m 1 2m Ϫ 3 2

factor and reduce

lowest terms

Note that we sometimes refer to simplest form as lowest terms. Now try Exercises 21 through 42


CAUTION



1 w ϩ 7 21 w Ϫ 7 2 w Ϫ 2 # , it is a common mistake to think that all factors 1 w Ϫ 7 21 w Ϫ 2 2 w ϩ 7 “cancel,” leaving an answer of zero. Actually, all factors reduce to 1, and the result is a value of 1 for all inputs where the product is defined. For products like 1w ϩ 72 1w Ϫ 72 w Ϫ 2 # ϭ1 1 w Ϫ 7 21 w Ϫ 2 2 w ϩ 7
1 1 1

B. You’ve just seen how we can multiply and divide rational expressions

C. Addition and Subtraction of Rational Expressions
Recall that the addition and subtraction of fractions requires finding the lowest common denominator (LCD) and building equivalent fractions. The sum or difference of the numerators is then placed over this denominator. The procedure for the addition and subtraction of rational expressions is very much the same. Note that the LCD can also be described as the least common multiple (LCM) of all denominators. Addition and Subtraction of Rational Expressions 1. 2. 3. 4. Find the LCD of all rational expressions. Build equivalent expressions using the LCD. Add or subtract numerators as indicated. Write the result in lowest terms.

Precalculus—

A-56

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 5



Adding and Subtracting Rational Expressions Compute as indicated: 7 3 a. ϩ 10x 25x2 10x 5 Ϫ xϪ3 x Ϫ9
2

b.

Solution



a. The LCM for 10x and 25x2 is 50x2. 7 3 7 # 5x 3 #2 ϩ ϭ ϩ 2 10x 10x 5x 25x 25x2 2 35x 6 ϭ ϩ 2 50x 50x2 35x ϩ 6 ϭ 50x2

find the LCD write equivalent expressions

simplify

add the numerators and write the result over the LCD

The result is in simplest form. b. The LCM for x2 Ϫ 9 and x Ϫ 3 is 1 x Ϫ 3 2 1 x ϩ 3 2 . 5 10x 10x 5 #xϩ3 Ϫ ϭ Ϫ 2 xϪ3 xϪ3 xϩ3 1 x Ϫ 3 21 x ϩ 3 2 x Ϫ9 10x Ϫ 5 1 x ϩ 3 2 ϭ 1 x Ϫ 3 21 x ϩ 3 2 10x Ϫ 5x Ϫ 15 ϭ 1 x Ϫ 3 21 x ϩ 3 2 5x Ϫ 15 ϭ 1x Ϫ 32 1x ϩ 32 ϭ 1 x Ϫ 3 21 x ϩ 3 2 51x Ϫ 32
1

find the LCD write equivalent expressions

subtract numerators, write the result over the LCD distribute

combine like terms

ϭ

5 xϩ3

factor and reduce

Now try Exercises 43 through 48



EXAMPLE 6



Adding and Subtracting Rational Expressions Perform the operations indicated: b2 nϪ3 c 5 a. b. Ϫ 2 Ϫ 2 a nϩ2 n Ϫ4 4a

Solution



a. The LCM for n ϩ 2 and n2 Ϫ 4 is 1 n ϩ 2 2 1 n Ϫ 2 2 . 5 nϪ3 nϪ3 5 #nϪ2Ϫ Ϫ 2 ϭ nϩ2 1n ϩ 22 n Ϫ 2 1n ϩ 22 1n Ϫ 22 n Ϫ4 51n Ϫ 22 Ϫ 1n Ϫ 32 ϭ 1 n ϩ 2 21 n Ϫ 2 2 5n Ϫ 10 Ϫ n ϩ 3 ϭ 1 n ϩ 2 21 n Ϫ 2 2 4n Ϫ 7 ϭ 1 n ϩ 2 21 n Ϫ 2 2

write equivalent expressions subtract numerators, write the result over the LCD distribute

result

Precalculus—

Section A.5 Rational Expressions and Equations

A-57

b. The LCM for a and 4a2 is 4a2:

b2 c c 4a b2 Ϫ Ϫ # ϭ 2 2 a a 4a 4a 4a b2 4ac ϭ 2Ϫ 2 4a 4a b2 Ϫ 4ac ϭ 4a2

write equivalent expressions simplify subtract numerators, write the result over the LCD


Now try Exercises 49 through 64

CAUTION



When the second term in a subtraction has a binomial numerator as in Example 6a, be sure the subtraction is applied to both terms. It is a common error to write 51n Ϫ 22 nϪ3 5n Ϫ 10 Ϫ n Ϫ 3 X in which the subtraction is applied Ϫ ϭ 1 n ϩ 2 21 n Ϫ 2 2 1 n ϩ 2 21 n Ϫ 2 2 1 n ϩ 2 21 n Ϫ 2 2 to the first term only. This is incorrect!

C. You’ve just seen how we can add and subtract rational expressions

D. Simplifying Compound Fractions
Rational expressions whose numerator or denominator contain a fraction are called 2 3 Ϫ 3m 2 compound fractions. The expression is a compound fraction with a 3 1 Ϫ 4m 3m2 2 3 3 1 Ϫ and a denominator of Ϫ . The two methods commonly numerator of 3m 2 4m 3m2 used to simplify compound fractions are summarized in the following boxes. Simplifying Compound Fractions (Method I) 1. Add/subtract fractions in the numerator, writing them as a single expression. 2. Add/subtract fractions in the denominator, also writing them as a single expression. 3. Multiply the numerator by the reciprocal of the denominator and simplify if possible. Simplifying Compound Fractions (Method II) 1. Find the LCD of all fractions in the numerator and denominator. 2. Multiply the numerator and denominator by this LCD and simplify. 3. Simplify further if possible. Method II is illustrated in Example 7.

EXAMPLE 7



Simplifying a Compound Fraction Simplify the compound fraction: 2 3 Ϫ 3m 2 3 1 Ϫ 4m 3m2

Precalculus—

A-58

APPENDIX A A Review of Basic Concepts and Skills

Solution



The LCD for all fractions is 12m2. 2 2 3 3 12m2 a Ϫ Ϫ ba b 3m 2 3m 2 1 ϭ 3 1 3 1 12m2 Ϫ a Ϫ ba b 4m 4m 1 3m2 3m2 2 12m2 3 12m2 a ba b Ϫ a ba b 3m 1 2 1 ϭ 3 12m2 1 12m2 a ba b Ϫ a 2 ba b 4m 1 1 3m 8m Ϫ 18m2 ϭ 9m Ϫ 4 ϭ 2m 1 4 Ϫ 9m 2
Ϫ1

multiply numerator and denominator by 12m 2 ϭ

12m 2 1

distribute

simplify

9m Ϫ 4

ϭ Ϫ2m

factor and write in lowest terms

D. You’ve just seen how we can simplify compound fractions

Now try Exercises 65 through 74



E. Solving Rational Equations
In Appendix A.3 we solved linear equations using basic properties of equality. If any equation contained fractional terms, we “cleared the fractions” using the least common denominator (LCD). We can also use this idea to solve rational equations, or equations that contain rational expressions. Solving Rational Equations 1. Identify and exclude any values that cause a zero denominator. 2. Multiply both sides by the LCD and simplify (this will eliminate all denominators). 3. Solve the resulting equation. 4. Check all solutions in the original equation.

EXAMPLE 8



Solving a Rational Equation Solve for m: 2 4 1 ϭ 2 . Ϫ m mϪ1 m Ϫm

Solution



Since m2 Ϫ m ϭ m 1 m Ϫ 1 2 , the LCD is m 1 m Ϫ 1 2 , where m 0 and m 1. 4 1 2 d multiply by LCD b ϭ m1m Ϫ 12 c m 1 m Ϫ 1 2a Ϫ m mϪ1 m1m Ϫ 12 m1m Ϫ 12 m1m Ϫ 12 m1m Ϫ 12 2 1 4 a bϪ a bϭ a b distribute and simplify m 1 1 mϪ1 1 m1m Ϫ 12 denominators are eliminated 21m Ϫ 12 Ϫ m ϭ 4 2m Ϫ 2 Ϫ m ϭ 4 distribute mϭ6 solve for m

Precalculus—

Section A.5 Rational Expressions and Equations

A-59

Checking by substitution we have: 4 1 2 ϭ 2 original equation Ϫ m mϪ1 m Ϫm 1 4 2 Ϫ ϭ substitute 6 for m 2 162 162 Ϫ 1 162 Ϫ 162 1 4 1 Ϫ ϭ simplify 3 5 30 3 2 5 Ϫ ϭ common denominator 15 15 15 2 2 ϭ ✓ result 15 15 A calculator check is shown in the figure. Now try Exercises 75 through 80 ᮣ Multiplying both sides of an equation by a variable sometimes introduces a solution that satisfies the resulting equation, but not the original equation — the one we’re trying to solve. Such “solutions” are called extraneous roots and illustrate the need to check all apparent solutions in the original equation. In the case of rational equations, we are particularly aware that any value that causes a zero denominator is outside the domain and cannot be a solution. EXAMPLE 9


Solving a Rational Equation Solve: x ϩ 4x 12 ϭ1ϩ . xϪ3 xϪ3

Solution



The LCD is x Ϫ 3, where x 3. multiply both 4x 12 b ϭ 1 x Ϫ 3 2a 1 ϩ b 1 x Ϫ 3 2a x ϩ sides by LCD xϪ3 xϪ3 4x 12 xϪ3 xϪ3 distribute and ba b simplify ba b ϭ 1x Ϫ 32 112 ϩ a 1x Ϫ 32x ϩ a 1 xϪ3 1 xϪ3 x2 Ϫ 3x ϩ 12 ϭ x Ϫ 3 ϩ 4x denominators are eliminated set equation equal to zero x2 Ϫ 8x ϩ 15 ϭ 0 factor 1 x Ϫ 3 21 x Ϫ 5 2 ϭ 0 x ϭ 3 or x ϭ 5 zero factor property Checking shows x ϭ 3 is an extraneous root, and x ϭ 5 is the only valid solution.

E. You’ve just seen how we can solve rational equations

Now try Exercises 81 through 86 ᮣ

Precalculus—

A-60

APPENDIX A A Review of Basic Concepts and Skills

A.5 EXERCISES


CONCEPTS AND VOCABULARY
1. In simplest form, 1 a Ϫ b 2 / 1 a Ϫ b 2 is equal to , while 1 a Ϫ b 2 / 1 b Ϫ a 2 is equal to 3. As with numeric fractions, algebraic fractions require a for addition and subtraction.

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

.

2. A rational expression is in when the numerator and denominator have no common factors, other than . 4. Since x2 ϩ 9 is prime, the expression 1 x2 ϩ 9 2 / 1 x ϩ 3 2 is already written in . 1 x ϩ 3 21 x Ϫ 2 2 1 x Ϫ 2 21 x ϩ 3 2

State T or F and discuss/explain your response.

5.


x xϩ1 1 Ϫ ϭ xϩ3 xϩ3 xϩ3

6.

ϭ0

DEVELOPING YOUR SKILLS
16. a. c. x3 ϩ 4x2 Ϫ 5x x3 Ϫ x 12y2 Ϫ 13y ϩ 3 27y3 Ϫ 1 b. d. 5p2 Ϫ 14p Ϫ 3 5p2 ϩ 11p ϩ 2 ax2 Ϫ 5x2 Ϫ 3a ϩ 15 ax Ϫ 5x ϩ 5a Ϫ 25

Reduce to lowest terms.

7. a. 8. a.

aϪ7 Ϫ3a ϩ 21 xϪ4 Ϫ7x ϩ 28

b. b.

2x ϩ 6 4x2 Ϫ 8x 3x Ϫ 18 6x2 Ϫ 12x

x2 Ϫ 5x Ϫ 14 9. a. 2 x ϩ 6x Ϫ 7 10. a. 11. a. 12. a. 13. a. r2 ϩ 3r Ϫ 10 r2 ϩ r Ϫ 6 xϪ7 7Ϫx v2 Ϫ 3v Ϫ 28 49 Ϫ v2 Ϫ12a3b5 4a2bϪ4

a2 ϩ 3a Ϫ 28 b. a2 Ϫ 49 b. b. b. b. m2 ϩ 3m Ϫ 4 m2 Ϫ 4m 5Ϫx xϪ5 u2 Ϫ 10u ϩ 25 25 Ϫ u2 7x ϩ 21 63

Compute as indicated. Write final results in lowest terms.

17. 18. 19. 20. 21. 22. 23. 24. 25.

a2 Ϫ 4a ϩ 4 a2 Ϫ 2a Ϫ 3 # a2 Ϫ 9 a2 Ϫ 4 b2 ϩ 5b Ϫ 24 # b b2 Ϫ 6b ϩ 9 b2 Ϫ 64 x2 Ϫ 7x Ϫ 18 2x2 ϩ 7x ϩ 3 # x2 Ϫ 6x Ϫ 27 2x2 ϩ 5x ϩ 2 6v2 ϩ 23v ϩ 21 # 4v2 Ϫ 25 3v ϩ 7 4v2 Ϫ 4v Ϫ 15 p3 Ϫ 64 p3 Ϫ p2 Ϭ p2 ϩ 4p ϩ 16 p2 Ϫ 5p ϩ 4

y2 Ϫ 9 c. 3Ϫy 14. a. c. 15. a. c. 5mϪ3n5 Ϫ10mn2 n2 Ϫ 4 2Ϫn 2n3 ϩ n2 Ϫ 3n n3 Ϫ n2 x3 ϩ 8 x Ϫ 2x ϩ 4
2

m3n Ϫ m3 d. 4 m Ϫ m4n b. d. b. d. Ϫ5v ϩ 20 25 w4 Ϫ w4v w3v Ϫ w3 6x2 ϩ x Ϫ 15 4x2 Ϫ 9 mn2 ϩ n2 Ϫ 4m Ϫ 4 mn ϩ n ϩ 2m ϩ 2

a2 ϩ 3a Ϫ 28 a3 Ϫ 4a2 Ϭ a2 ϩ 5a Ϫ 14 a3 Ϫ 8 3Ϫx 3x Ϫ 9 Ϭ 4x ϩ 12 5x ϩ 15 5b Ϫ 10 2Ϫb Ϭ 7b Ϫ 28 5b Ϫ 20 a2 ϩ a # 3a Ϫ 9 a2 Ϫ 3a 2a ϩ 2

Precalculus—

Section A.5 Rational Expressions and Equations
2 p2 Ϫ 36 # 2 4p 2p 2p ϩ 12p

A-61

26. 27.

Compute as indicated. Write answers in lowest terms [recall that a ؊ b ‫ ؍‬؊1(b ؊ a)].

8 # 1 a2 Ϫ 2a Ϫ 35 2 2 a Ϫ 25

43. 45. 47. 49. 51. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62.

3 5 ϩ 2 2x 8x 7 1 Ϫ 2 3 4x y 8xy4 4p p Ϫ 36
2 2

44. 46. 48. 50. 52. Ϫ 2 yϩ6

15 7 Ϫ 2 16y 2y 3 5 ϩ 3 6a b 9ab3 3q q Ϫ 49
2

28. 1 m2 Ϫ 16 2 29.

#

m2 Ϫ 5m m2 Ϫ m Ϫ 20 xy Ϫ 3x Ϭ xy Ϫ 5y

xy Ϫ 3x ϩ 2y Ϫ 6 x2 Ϫ 3x Ϫ 10

Ϫ

2 pϪ6

Ϫ

3 2q Ϫ 14

2a Ϫ ab ϩ 7b Ϫ 14 ab Ϫ 2a Ϭ 30. 2 ab Ϫ 7a b Ϫ 14b ϩ 49 31. 32. 33. 34. m2 ϩ 2m Ϫ 8 m2 Ϫ 16 Ϭ m2 Ϫ 2m m2 18 Ϫ 6x 2x2 Ϫ 18 Ϭ x2 Ϫ 25 x3 Ϫ 2x2 Ϫ 25x ϩ 50 yϩ3 3y2 ϩ 9y

m 4 ϩ 4Ϫm m Ϫ 16 2 Ϫ5 mϪ7 yϩ1 y ϩ y Ϫ 30
2 2

p 2 ϩ 2 pϪ2 4Ϫp 4 Ϫ9 xϪ1

#y

2

ϩ 7y ϩ 12 y2 Ϫ 16

Ϭ

y2 ϩ 4y y2 Ϫ 4y

4n 3 Ϫ 4n Ϫ 20 n Ϫ 5n 1 a ϩ 2 aϩ4 a Ϫ a Ϫ 20 2x Ϫ 1 xϪ5 Ϫ 2 x ϩ 3x Ϫ 4 x ϩ 3x Ϫ 4
2

x2 ϩ 4x Ϫ 5 x2 Ϫ 1 # x ϩ 1 Ϭ x2 Ϫ 5x Ϫ 14 x2 Ϫ 4 x ϩ 5

x2 Ϫ 0.49 x2 Ϫ x ϩ 0.21 Ϭ 35. 2 x ϩ 0.5x Ϫ 0.14 x2 Ϫ 0.09 x2 Ϫ 0.25 x2 Ϫ 0.8x ϩ 0.15 Ϭ 36. 2 x ϩ 0.1x Ϫ 0.2 x2 Ϫ 0.16 4 4 n2 ϩ n ϩ 3 9 Ϭ 37. 13 2 1 n2 Ϫ n ϩ n2 Ϫ 15 15 25 n2 Ϫ q2 Ϫ 38.
2

3y Ϫ 4 y ϩ 2y ϩ 1
2

Ϫ

2y Ϫ 5 y ϩ 2y ϩ 1
2

4 9

Ϫ2 7 Ϫ 2 3a ϩ 12 a ϩ 4a 2 mϪ5 ϩ 2 m Ϫ9 m ϩ 6m ϩ 9
2

9 25

q2 ϩ Ϭ

3 1 q Ϫ qϪ 10 10
3 2

17 3 qϩ 20 20 1 q2 Ϫ 16
2

mϩ2 mϩ6 Ϫ 2 2 m Ϫ 25 m Ϫ 10m ϩ 25 yϩ2 5y ϩ 11y ϩ 2
2 2

ϩ

5 y ϩyϪ6
2

39. 40. 41.

3a Ϫ 24a Ϫ 12a ϩ 96 6a Ϫ 24 Ϭ 3 2 a Ϫ 11a ϩ 24 3a Ϫ 81 p3 ϩ p2 Ϫ 49p Ϫ 49 p2 ϩ 6p Ϫ 7 Ϭ p2 ϩ p ϩ 1 p3 Ϫ 1

mϪ4 m ϩ 2 3m Ϫ 11m ϩ 6 2m Ϫ m Ϫ 15

Write each term as a rational expression. Then compute the sum or difference indicated.

6n2 ϩ 5n ϩ 1 # 12n2 Ϫ 17n ϩ 6 4n 2 Ϫ 1 # 12n2 Ϫ 5n Ϫ 3 2n2 ϩ n 6n2 Ϫ 7n ϩ 2 4x Ϫ 25 2x Ϫ x Ϫ 15 # 4x ϩ 25x Ϫ 21 Ϭ 2 b x Ϫ 11x ϩ 30 x Ϫ 9x ϩ 18 12x2 Ϫ 5x Ϫ 3
2 2 2 2

63. a. pϪ2 Ϫ 5pϪ1

42. a

64. a. 3aϪ1 ϩ 1 2a 2 Ϫ1

b. xϪ2 ϩ 2xϪ3

b. 2yϪ1 Ϫ 1 3y 2 Ϫ1

Precalculus—

A-62

APPENDIX A A Review of Basic Concepts and Skills

Simplify each compound fraction. Use either method.

Solve each equation. Identify any extraneous roots.

5 1 Ϫ a 4 65. 25 1 Ϫ 2 16 a 1 pϪ2 67. 1 1ϩ pϪ2 pϩ 3 2 ϩ 3Ϫx xϪ3 69. 4 5 ϩ x xϪ3 2 y2 Ϫ y Ϫ 20 71. 3 4 Ϫ yϩ4 yϪ5

8 1 Ϫ 3 27 x 66. 2 1 Ϫ x 3 3 yϪ6 68. 9 yϩ yϪ6 1ϩ 1 2 Ϫ yϪ5 5Ϫy 70. 3 2 Ϫ y yϪ5 2 x2 Ϫ 3x Ϫ 10 72. 6 4 Ϫ xϩ2 xϪ5

75. 76. 77. 78. 79.

2 5 1 ϭ 2 ϩ x xϩ1 x ϩx 3 5 1 Ϫ 2 ϭ m mϩ3 m ϩ 3m 21 3 ϭ aϩ2 aϪ1 4 7 ϭ 2y Ϫ 3 3y Ϫ 5 1 1 1 Ϫ ϭ 2 3y 4y y 80. 3 1 1 Ϫ ϭ 2 5x 2x x

81. x ϩ 82. 83. 84. 85. 86.

14 2x ϭ1ϩ xϪ7 xϪ7

10 2x ϩxϭ1ϩ xϪ5 xϪ5 6 20 5 ϩ 2 ϭ nϩ3 nϪ2 n ϩnϪ6 7 1 2 Ϫ 2 ϭϪ pϩ2 pϩ3 p ϩ 5p ϩ 6 a 2a2 ϩ 5 3 Ϫ 2 ϭ 2a ϩ 1 aϪ3 2a Ϫ 5a Ϫ 3 Ϫ18 4n 3n ϭ ϩ 2 n Ϫ 1 3 n ϩ1 6n Ϫ n Ϫ 1
2

Rewrite each expression as a compound fraction. Then simplify using either method.

73. a. 74. a.

1 ϩ 3mϪ1 1 Ϫ 3mϪ1 4 Ϫ 9aϪ2 3aϪ2

b. b.

1 ϩ 2xϪ2 1 Ϫ 2xϪ2 3 ϩ 2nϪ1 5nϪ2



WORKING WITH FORMULAS
450P 100 ؊ P 88. Chemicals in the bloodstream: C ‫؍‬ 200H2 H3 ؉ 40

87. Cost to seize illegal drugs: C ‫؍‬

The cost C, in millions of 450P P dollars, for a government to find 100 ؊ P and seize P% 1 0 Յ P 6 100 2 of 40 a certain illegal drug is modeled 60 by the rational equation shown. 80 Complete the table (round to the nearest dollar) and answer the 90 following questions. 93 a. What is the cost of seizing 95 40% of the drugs? Estimate 98 the cost at 85%. 100 b. Why does cost increase dramatically the closer you get to 100%? c. Will 100% of the drugs ever be seized?

Rational equations are often used 200H2 to model chemical concentrations H H3 ؉ 40 in the bloodstream. The percent 0 concentration C of a certain drug 1 H hours after injection into muscle tissue can be modeled by 2 the equation shown (H Ն 0). 3 Complete the table (round to the 4 nearest tenth of a percent) and 5 answer the following questions. 6 a. What is the percent 7 concentration of the drug 3 hr after injection? b. Why is the concentration virtually equal at H ϭ 4 and H ϭ 5? c. Why does the concentration begin to decrease? d. How long will it take for the concentration to become less than 10%?

Precalculus—

Section A.5 Rational Expressions and Equations


A-63

APPLICATIONS
of words per minute after t weeks, 3 6 t 6 12. Use a table to determine how many weeks it takes for a student to be typing an average of forty-five words per minute. 92. Memory retention: A group of students is asked to memorize 50 Russian words that are unfamiliar to them. The number N of these words that the average student remembers D days later is modeled 5D ϩ 35 1 D Ն 1 2 . How many by the equation N ϭ D words are remembered after (a) 1 day? (b) 5 days? (c) 12 days? (d) 35 days? (e) 100 days? According to this model, is there a certain number of words that the average student never forgets? How many? 93. Pollution removal: For a steel mill, the cost C (in millions of dollars) to remove toxins from the 22P , where resulting sludge is given by C ϭ 100 Ϫ P P is the percent of the toxins removed. What percent can be removed if the mill spends $88,000,000 on the cleanup? Round to tenths of a percent.

89. Stock prices: When a hot new stock hits the market, its price will often rise dramatically and then taper 50 1 7d 2 ϩ 10 2 off over time. The equation P ϭ d3 ϩ 50 models the price of stock XYZ d days after it has “hit the market.” (a) Create a table of values showing the price of the stock for the first 10 days (rounded to the nearest dollar) and comment on what you notice. (b) Find the opening price of the stock. (c) Does the stock ever return to its original price? 90. Population growth: The Department of Wildlife introduces 60 elk into a new game reserve. It is projected that the size of the herd will grow 10 1 6 ϩ 3t 2 , where according to the equation N ϭ 1 ϩ 0.05t N is the number of elk and t is the time in years. (a) Approximate the population of elk after 14 yr. (b) If recent counts find 225 elk, approximately how many years have passed? 91. Typing speed: The number of words per minute that a beginner can type is approximated by the 60t Ϫ 120 , where N is the number equation N ϭ t



EXTENDING THE CONCEPT
96. Given the rational numbers 2 3 and , what is the 5 4 reciprocal of the sum of their reciprocals? Given a c that and are any two numbers—what is the b d reciprocal of the sum of their reciprocals?

94. One of these expressions is not equal to the others. Identify which and explain why. 20n a. b. 20 # n Ϭ 10 # n 10n 1 20 # n c. 20n # d. 10n 10 n 95. The average of A and B is x. The average of C, D, and E is y. The average of A, B, C, D, and E is: 3x ϩ 2y 2x ϩ 3y a. b. 5 5 21x ϩ y2 31x ϩ y2 c. d. 5 5

Precalculus—

A-64

APPENDIX A A Review of Basic Concepts and Skills

A.6

Radicals, Rational Exponents, and Radical Equations
Square roots and cube roots come from a much larger family called radical expressions. Expressions containing radicals can be found in virtually every field of mathematical study, and are an invaluable tool for modeling many real-world phenomena.

LEARNING OBJECTIVES
In Section A.6 you will review how to:

A. Simplify radical
expressions of the form n 1a n Rewrite and simplify radical expressions using rational exponents Use properties of radicals to simplify radical expressions Add and subtract radical expressions Multiply and divide radical expressions; write a radical expression in simplest form Solve equations and use formulas involving radicals

A. Simplifying Radical Expressions of the Form 1an
In previous coursework, you likely noted that 1a ϭ b only if b2 ϭ a. This definition cannot be applied to expressions like 1Ϫ16, since there is no number b such that b2 ϭ Ϫ16. In other words, the expression 1a represents a real number only if a Ն 0 (for a full review of the real numbers and other sets of numbers, see Appendix I at www.mhhe.com/coburn). Of particular interest to us now is an inverse operation for a2. In other words, what operation can be applied to a2 to return a? Consider the following.

n

B.

C.

D. E.

EXAMPLE 1



Evaluating a Radical Expression

Evaluate 2a2 for the values given: a. a ϭ 3 b. a ϭ 5 c. a ϭ Ϫ6

F.

Solution



a. 232 ϭ 19 ϭ3

b. 252 ϭ 125 ϭ5

c. 2 1 Ϫ6 2 2 ϭ 136 ϭ6


Now try Exercises 7 and 8

The pattern seemed to indicate that 2a2 ϭ a and that our search for an inverse operation was complete—until Example 1(c), where we found that 2 1 Ϫ6 2 2 Ϫ6. Using the absolute value concept, we can “repair” this apparent discrepancy and state a general rule for simplifying these expressions: 2a2 ϭ ͿaͿ. For expressions like 249x2 and 2y6, the radicands can be rewritten as perfect squares and simplified in the same manner: 249x2 ϭ 2 1 7x 2 2 ϭ 7ͿxͿ and 2y6 ϭ 2 1 y3 2 2 ϭ Ϳy3Ϳ. The Square Root of a2: 2a2 For any real number a, 2a2 ‫ ͦ ؍‬a ͦ .

EXAMPLE 2



Simplifying Square Root Expressions Simplify each expression. a. 2169x2 b. 2x2 Ϫ 10x ϩ 25
since x could be negative
2

Solution



b. 2x Ϫ 10x ϩ 25 ϭ 2 1 x Ϫ 5 2 ϭ Ϳx Ϫ 5Ϳ
2

a. 2169x2 ϭ Ϳ13xͿ ϭ 13ͿxͿ

since x ؊ 5 could be negative

Now try Exercises 9 and 10



A-64

Precalculus—

Section A.6 Radicals, Rational Exponents, and Radical Equations

A-65

CAUTION



In Appendix A.2, we noted that 1 A ϩ B 2 2 A2 ϩ B2, indicating that you cannot square the individual terms in a sum (the square of a binomial results in a perfect square trinomial). In a similar way, 2A2 ϩ B2 A ϩ B, and you cannot take the square root of individual terms. There is a big difference between the expressions 2A2 ϩ B2 and 2 1 A ϩ B 2 2 ϭ ͿA ϩ BͿ. Try evaluating each when A ϭ 3 and B ϭ 4.

To investigate expressions like 2x3, note the radicand in both 18 and 1Ϫ64 can be written as a perfect cube. From our earlier definition of cube roots we know 3 3 3 3 18 ϭ 2 1 2 2 3 ϭ 2, 1Ϫ64 ϭ 2 1 Ϫ4 2 3 ϭ Ϫ4, and that every real number has only one real cube root. For this reason, absolute value notation is not used or needed when taking cube roots.
3 3 a The Cube Root of a3: 2

3

3

3

For any real number a, 3 3 2 a ‫ ؍‬a.

EXAMPLE 3



Simplifying Cube Root Expressions Simplify each expression. 3 3 a. 2 b. 2 Ϫ27x3 Ϫ64n6

Solution



3 3 a. 2Ϫ27x3 ϭ 2 1 Ϫ3x 2 3 ϭ Ϫ3x

3 3 b. 2Ϫ64n6 ϭ 2 1 Ϫ4n2 2 3 ϭ Ϫ4n2

Now try Exercises 11 and 12



We can extend these ideas to fourth roots, fifth roots, and so on. For example, the 5 fifth root of a is b only if b5 ϭ a. In symbols, 1a ϭ b implies b5 ϭ a. Since an odd number of negative factors is always negative: 1 Ϫ2 2 5 ϭ Ϫ32, and an even number of negative factors is always positive: 1 Ϫ2 2 4 ϭ 16, we must take the index into account n when evaluating expressions like 1an. If n is even and the radicand is unknown, absolute value notation must be used.
WORTHY OF NOTE
Just as 1 Ϫ16 is not a real number, 6 4 1 Ϫ16 and 1 Ϫ16 do not represent real numbers. An even number of repeated factors is always positive!
2

The nth Root of an: 2an For any real number a, n 1. 1an ϭ ͿaͿ when n is even. Simplifying Radical Expressions Simplify each expression. 4 4 a. 1 b. 1 81 Ϫ81 4 5 4 e. 216m f. 232p5
5 c. 1 32 6 g. 2 1 m ϩ 5 2 6 5 d. 1 Ϫ32 7 h. 2 1 x Ϫ 2 2 7

n

2. 1an ϭ a when n is odd.

n

EXAMPLE 4



Solution



a. c. e.

A. You’ve just seen how we can simplify radical n expressions of the form 1a n

4 1 81 ϭ 3 5 132 ϭ 2 4 4 2 16m4 ϭ 2 1 2m 2 4 ϭ Ϳ2mͿ or 2ͿmͿ 6 6 g. 2 1 m ϩ 5 2 ϭ Ϳm ϩ 5Ϳ

4 b. 1Ϫ81 is not a real number 5 d. 1 Ϫ32 ϭ Ϫ2 5 5 f. 232p5 ϭ 2 1 2p 2 5 ϭ 2p 7 h. 2 1x Ϫ 227 ϭ x Ϫ 2

Now try Exercises 13 and 14



Precalculus—

A-66

APPENDIX A A Review of Basic Concepts and Skills

B. Radical Expressions and Rational Exponents
As an alternative to radical notation, a rational (fractional) exponent can be used, 3 along with the power property of exponents. For 1a3 ϭ a, notice that an exponent of one-third1 can 3 replace the cube root notation and produce the same result: 3 3 2 a ϭ 1 a3 2 3 ϭ a3 ϭ a. In the same way, an exponent of one-half can replace the 1 2 square root notation: 2a2 ϭ 1 a2 2 2 ϭ a2 ϭ ͿaͿ. In general, we have the following: Rational Exponents If a is a real number and n is an integer greater than 1, 1 n n then 1a ϭ 2a1 ϭ an n provided 1a represents a real number.

EXAMPLE 5



Simplifying Radical Expressions Using Rational Exponents Simplify by rewriting each radicand as a perfect nth power and converting to rational exponent notation. 3 3 4 4 3 8w a. 2Ϫ125 b. Ϫ 216x20 c. 2Ϫ81 d. B 27

Solution



a. 2Ϫ125 ϭ 2 1 Ϫ5 2 3 1 ϭ 3 1 Ϫ5 2 3 4 3 3 ϭ 1 Ϫ5 2 3 ϭ Ϫ5
3 3

b. Ϫ 216x20 ϭ Ϫ 2 1 2x5 2 4 1 ϭ Ϫ 3 1 2x5 2 4 4 4 ϭ Ϫ 0 2x5 0 ϭ Ϫ2ͿxͿ5
4 4

c. 2Ϫ81 ϭ 1 Ϫ81 2 4 is not a real number
4
1

d.

3 8w3 3 2w ϭ a b B 27 B 3 1 2w 3 3 2w ϭ ca b d ϭ 3 3
3

Now try Exercises 15 and 16
n n
1



WORTHY OF NOTE
Any rational number can be decomposed into the product of a unit fraction and an m 1 ϭ # m. integer: n n

Figure A.6 When a rational exponent is used, as in 1a ϭ 2a1 ϭ an, the denominator of the exponent represents the index number, while the m numerator of the exponent represents the original power on a. This an n m is true even when the exponent on a is something other than one! In (͙ a ) 4 3 other words, the radical expression 216 can be rewritten as 1 3 1 3 1 163 2 4 ϭ 1 161 2 4 or 164. This is further illustrated in Figure A.6 where we see the rational exponent has the form, “power over root.” To evaluate this expression without the aid of a calculator, we use the commutative property to rewrite 3 1 1 3 1 3 1 161 2 4 as 1 164 2 1 and begin with the fourth root of 16: 1 164 2 1 ϭ 23 ϭ 8. m In general, if m and n have no common factors (other than 1) the expression a n can be interpreted in the following two ways.

Rational Exponents If m n is a rational number expressed in lowest terms with n Ն 2, then
n

(compute 1a, then take the mth power), (compute am, then take the nth root), n provided 1a represents a real number.

1 1 2 a n ϭ 1 2a 2 m
m

n

or

1 2 2 a n ϭ 2am
m

n

Precalculus—

Section A.6 Radicals, Rational Exponents, and Radical Equations

A-67

Expressions with rational exponents are generally easier to evaluate if we compute the root first, then apply the exponent. Computing the root first also helps us determine whether or not an expression represents a real number.

EXAMPLE 6



Simplifying Expressions with Rational Exponents Simplify each expression, if possible. a. Ϫ492
3

b. 1 Ϫ49 2 2
3 1

c. 1 Ϫ8 2 3
2 3

d. Ϫ8Ϫ3
2 1

Solution
WORTHY OF NOTE
Ϫ8 While the expression 1 Ϫ8 2 3 ϭ 1 represents the real number Ϫ2, the 2 6 expression 1 Ϫ8 2 6 ϭ 1 1Ϫ8 2 2 is not a 1 2 real number, even though ϭ . 3 6 Note that the second exponent is not in lowest terms.
1 3



a. Ϫ492 ϭ Ϫ 1 492 2 3 ϭ Ϫ 1 149 2 3 ϭ Ϫ 1 7 2 3 or Ϫ343
3

c. 1 Ϫ8 2 3 ϭ 3 1 Ϫ8 2 34 2 3 ϭ 1 1Ϫ8 2 2
2 1

b. 1 Ϫ49 2 2 ϭ 3 1 Ϫ49 2 2 4 3, ϭ 1 1Ϫ49 2 3 not a real number d. Ϫ8Ϫ3 ϭ Ϫ 1 83 2 Ϫ2 3 ϭ Ϫ1 1 8 2 Ϫ2
2 1

ϭ 1 Ϫ2 2 2 or 4

ϭ Ϫ2Ϫ2 or Ϫ

B. You’ve just seen how we can rewrite and simplify radical expressions using rational exponents

1 4


Now try Exercises 17 through 22

C. Using Properties of Radicals to Simplify Radical Expressions
The properties used to simplify radical expressions are closely connected to the properties of exponents. For instance, the product to a power property holds even when n 1 1 1 1 1 1 is a rational number. This means 1 xy 2 2 ϭ x2y2 and 1 4 # 25 2 2 ϭ 42 # 252. When the second statement is expressed in radical form, we have 14 # 25 ϭ 14 # 225, with both forms having a value of 10. This suggests the product property of radicals, which can be extended to include cube roots, fourth roots, and so on. Product Property of Radicals If 1A and 1B represent real-valued expressions, then 1AB ϭ 1A # 1B
n n n n n

and

1A # 1B ϭ 1AB.

n

n

n

CAUTION



Note that this property applies only to a product of two terms, not to a sum or difference. In other words, while 29x2 ϭ Ϳ3xͿ, 29 ϩ x2 |3 ϩ x|!

One application of the product property is to simplify radical expressions. In genn eral, the expression 1a is in simplified form if a has no factors (other than 1) that are perfect nth roots.

EXAMPLE 7



Simplifying Radical Expressions Write each expression in simplest form using the product property. a. 118
3 b. 5 2 125x4

c.

Ϫ4 ϩ 220 2

3 3 d. 1.2 2 16n4 2 4n5

Solution



a. 118 ϭ 19 # 2 ϭ 19 12 ϭ 3 12

3 3 b. 5 2 125x4 ϭ 5 # 2 125 # x4 3 3 3 # 3 1 These steps can ϭ5# 2 125 # 2 x 2x 3 be done mentally e ϭ 5 # 5 # x # 1x 3 ϭ 25x 1 x

Precalculus—

A-68

APPENDIX A A Review of Basic Concepts and Skills

WORTHY OF NOTE
For expressions like those in Example 7(c), students must resist the “temptation” to reduce individual terms as in Ϫ4 ϩ 120 ϭ Ϫ2 ϩ 120. 2 Remember, only factors can be reduced.

c.

Ϫ4 ϩ 120 Ϫ4 ϩ 14 # 5 ϭ 2 2 Ϫ4 ϩ 2 15 ϭ 2 1 2 1 Ϫ2 ϩ 15 2 ϭ 2 ϭ Ϫ2 ϩ 15

3 3 3 d. 1.2 2 16n4 2 4n5 ϭ 1.2 2 64 # n9 3 3 9 ϭ 1.2 264 2 n 3 3 ϭ 1.2 264 2 1 n3 2 3 ϭ 1.2 1 4 2 n3 ϭ 4.8n3

Now try Exercises 23 through 26



When radicals are combined using the product property, the result may contain a perfect nth root, which should be simplified. Note that the index numbers must be the same in order to use this property. The quotient property of radicals can also be established using exponential 100 1100 ϭ ϭ 2 suggests the following: properties. The fact that A 25 125 Quotient Property of Radicals If 1A and 1B represent real-valued expressions with B 0, then n n 1A 1A n A n A ϭ n and n ϭ . AB A B 1B 1B Many times the product and quotient properties must work together to simplify a radical expression, as shown in Example 8A.
n n

EXAMPLE 8A



Simplifying Radical Expressions Simplify each expression: 218a5 a. 22a 81 3 A 125x3
3 81 1 81 ϭ 3 3 A 125x 2125x3 3 1 27 # 3 ϭ 5x 3 3 13 ϭ 5x 3

b. b.

Solution



a.

218a5 22a

ϭ

18a5 B 2a

ϭ 29a4 ϭ 3a2

Radical expressions can also be simplified using rational exponents.

EXAMPLE 8B



Using Rational Exponents to Simplify Radical Expressions Simplify using rational exponents: 3 4 a. 236p4q5 b. v 2 v
3 c. 1 m 1m 3 b. v 2v4 ϭ v1 # v3 3 4 ϭ v3 # v3 7 ϭ v3 6 1 ϭ v3v3 3 ϭ v2 1 v
4

Solution



C. You’ve just seen how we can use properties of radicals to simplify radical expressions

a. 236p4q5 ϭ 1 36p4q5 2 2 1 4 5 ϭ 362p2q2 4 1 ϭ 6p2q12 ϩ 2 2 1 ϭ 6p2q2q2 ϭ 6p2q2 1q
1

3 c. 1m 1m ϭ m3m2 1 1 ϭ m3 ϩ 2 5 ϭ m6 6 ϭ2 m5

1

1

Now try Exercises 27 through 30



Precalculus—

Section A.6 Radicals, Rational Exponents, and Radical Equations

A-69

D. Addition and Subtraction of Radical Expressions
3 Since 3x and 5x are like terms, we know 3x ϩ 5x ϭ 8x. If x ϭ 17, the sum becomes 3 3 3 3 17 ϩ 5 17 ϭ 8 17, illustrating how like radical expressions can be combined. Like radicals are those that have the same index and radicand. In some cases, we can identify like radicals only after radical terms have been simplified.

EXAMPLE 9



Adding and Subtracting Radical Expressions Simplify and combine (if possible). 3 3 a. 145 ϩ 2 120 b. 2 16x5 Ϫ x 2 54x2

Solution



a. 145 ϩ 2 120 ϭ 3 15 ϩ 2 1 2 15 2 ϭ 3 15 ϩ 4 15 ϭ 7 15
3

simplify radicals: 145 ϭ 19 # 5; 120 ϭ 14 # 5 like radicals result
2

3 b. 216x Ϫ x 254x ϭ 28 # 2 # x # x Ϫ x 2 27 # 2 # x2 3 3 ϭ 2x 22x2 Ϫ 3x 22x2 3 ϭ Ϫx 2 2x2

5

3

2

3

3

simplify radicals result


D. You’ve just seen how we can add and subtract radical expressions

Now try Exercises 31 through 34

E. Multiplication and Division of Radical Expressions; Radical Expressions in Simplest Form
Multiplying radical expressions is simply an extension of our earlier work. The multiplication can take various forms, from the distributive property to any of the special products reviewed in Appendix A.2. For instance, 1 A Ϯ B 2 2 ϭ A2 Ϯ 2AB ϩ B2, even if A or B is a radical term.

EXAMPLE 10



Multiplying Radical Expressions Compute each product and simplify. a. 5 13 1 16 Ϫ 4 13 2 b. 1 2 12 ϩ 6 13 2 1 3 110 ϩ 115 2 c. 1 x ϩ 17 2 1 x Ϫ 17 2 d. 1 3 Ϫ 12 2 2

Solution



b. 12 12 ϩ 6 13 213 110 ϩ 115 2 ϭ 6 120 ϩ 2 130 ϩ 18 130 ϩ 6 145 F-O-I-L ϭ 12 15 ϩ 20 130 ϩ 18 15 extract roots and simplify ϭ 30 15 ϩ 20 130 result c. 1 x ϩ 17 2 1 x Ϫ 17 2 ϭ x2 Ϫ 1 17 2 2 ϭ x2 Ϫ 7
2 2

a. 5 13 1 16 Ϫ 4 13 2 ϭ 5 118 Ϫ 20 1 13 2 2 ϭ 5 1 3 2 1 2 Ϫ 1 20 21 3 2 ϭ 15 12 Ϫ 60

distribute simplify: 1 18 ϭ 3 1 2, 1 13 2 2 ϭ 3 result

1A ϩ B2 1A Ϫ B2 ϭ A2 Ϫ B 2 result
2

LOOKING AHEAD
Notice that the answer for Example 10(c) contains no radical terms, since the outer and inner products sum to zero. This result will be used to simplify certain radical expressions in this section and later in Chapter 2.

d. 1 3 Ϫ 12 2 ϭ 1 3 2 Ϫ 2 1 3 2 1 12 2 ϩ 1 12 2 ϭ 9 Ϫ 6 12 ϩ 2 ϭ 11 Ϫ 6 12

1 A Ϫ B 2 2 ϭ A 2 Ϫ 2AB ϩ B 2 simplify each term result

Now try Exercises 35 through 38



One application of products and powers of radical expressions is to evaluate certain quadratic expressions, as illustrated in Example 11.

Precalculus—

A-70

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 11 Solution



Evaluating a Quadratic Expression Show that when x2 Ϫ 4x ϩ 1 is evaluated at x ϭ 2 ϩ 13, the result is zero. x2 Ϫ 4x ϩ 1 original expression 1 2 ϩ 13 2 Ϫ 4 1 2 ϩ 13 2 ϩ 1 substitute 2 ϩ 13 for x 4 ϩ 4 13 ϩ 3 Ϫ 8 Ϫ 4 13 ϩ 1 square binomial; distribute 1 4 ϩ 3 Ϫ 8 ϩ 1 2 ϩ 1 4 13 Ϫ 4 13 2 commutative and associative properties 0✓
2



A calculator check is shown in the figure.

Now try Exercises 39 through 42



When we applied the quotient property in Example 8A, we obtained a denominator free of radicals. Sometimes the denominator is not automatically free of radicals, and the need to write radical expressions in simplest form comes into play. This process is called rationalizing the denominator. Radical Expressions in Simplest Form A radical expression is in simplest form if: 1. The radicand has no perfect nth root factors. 2. The radicand contains no fractions. 3. No radicals occur in a denominator. As with other types of simplification, the desired form can be achieved in various ways. If the denominator is a single radical term, we multiply the numerator and denominator by the factors required to eliminate the radical in the denominator [see Examples 12(a) and 12(b)]. If the radicand is a rational expression, it is generally easier to build an equivalent fraction within the radical having perfect nth root factors in the denominator [see Example 12(c)].

EXAMPLE 12



Simplifying Radical Expressions Simplify by rationalizing the denominator. Assume a, x 2 Ϫ7 a. b. 3 1x 5 23 0.

Solution



a.

2 2 # 13 ϭ 5 13 5 13 13 2 13 2 13 ϭ ϭ 2 15 5 1 13 2 Ϫ7
3 1 x

multiply numerator and denominator by 13

simplify—denominator is now rational

b.

ϭ ϭ

3 3 3 1 x1 1 x 211 x2 3 2 Ϫ7 2x

3 3 Ϫ7 1 1 x 211 x2

3 multiply using two additional factors of 1x

3 3 2 x 3 2 Ϫ7 2 x ϭ x

product property

3 3 2 x ϭx

Now try Exercises 43 and 44



Precalculus—

Section A.6 Radicals, Rational Exponents, and Radical Equations

A-71

In some applications, the denominator may be a sum or difference containing a radical term. In this case, the methods from Example 12 are ineffective, and instead we multiply by a conjugate since 1 A ϩ B 21 A Ϫ B 2 ϭ A2 Ϫ B2. If either A or B is a square root, the result will be a denominator free of radicals.


EXAMPLE 13

Simplifying Radical Expressions Using a Conjugate Simplify the expression by rationalizing the denominator. 2 ϩ 13 Write the answer in exact form and approximate form 16 Ϫ 12 rounded to three decimal places.

Solution



2 ϩ 13 2 ϩ 13 # 16 ϩ 12 ϭ 16 Ϫ 12 16 Ϫ 12 16 ϩ 12 2 16 ϩ 2 12 ϩ 118 ϩ 16 ϭ 1 16 2 2 Ϫ 1 12 2 2 3 16 ϩ 2 12 ϩ 3 12 6Ϫ2 3 16 ϩ 5 12 ϭ 4 Ϸ 3.605 ϭ

multiply by the conjugate of the denominator FOIL difference of squares simplify

exact form approximate form

E. You’ve just seen how we can multiply and divide radical expressions and write a radical expression in simplest form

Now try Exercises 45 through 48



F. Equations and Formulas Involving Radicals
A radical equation is any equation that contains terms with a variable in the radicand. To solve a radical equation, we attempt to isolate a radical term on one side, then apply the appropriate nth power to free up the radicand and solve for the unknown. This is an application of the power property of equality. The Power Property of Equality If 1u and v are real-valued expressions and 1u ϭ v, then 1 1u 2 n ϭ vn u ϭ vn for n an integer, n Ն 2.
n n n

Raising both sides of an equation to an even power can also introduce a false solution (extraneous root). Note that by inspection, the equation x Ϫ 2 ϭ 1x has only the solution x ϭ 4. But the equation 1 x Ϫ 2 2 2 ϭ x (obtained by squaring both sides) has both x ϭ 4 and x ϭ 1 as solutions, yet x ϭ 1 does not satisfy the original equation. This means we should check all solutions of an equation where an even power is applied.

Precalculus—

A-72

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 14



Solving Radical Equations Solve each radical equation: a. 13x Ϫ 2 ϩ 12 ϭ x ϩ 10
3 b. 2 1 xϪ5ϩ4ϭ0

Solution



a. 13x Ϫ 2 ϩ 12 ϭ x ϩ 10 13x Ϫ 2 ϭ x Ϫ 2 1 13x Ϫ 2 2 2 ϭ 1 x Ϫ 2 2 2 3x Ϫ 2 ϭ x2 Ϫ 4x ϩ 4 0 ϭ x2 Ϫ 7x ϩ 6 0 ϭ 1x Ϫ 62 1x Ϫ 12 x Ϫ 6 ϭ 0 or x Ϫ 1 ϭ 0 x ϭ 6 or x ϭ 1 x ϭ 6:

original equation isolate radical term (subtract 12) apply power property, power is even simplify, square binomial set equal to zero factor apply zero product property result, check for extraneous roots

Check



13 1 6 2 Ϫ 2 ϩ 12 ϭ 1 6 2 ϩ 10 116 ϩ 12 ϭ 16 16 ϭ 16 ✓ 13 1 1 2 Ϫ 2 ϩ 12 ϭ 1 1 2 ϩ 10 11 ϩ 12 ϭ 11 13 ϭ 11x

Check



x ϭ 1:

The only solution is x ϭ 6; x ϭ 1 is extraneous. A calculator check is shown in the figures.
3 b. 2 1x Ϫ 5 ϩ 4 ϭ 0 3 1 x Ϫ 5 ϭ Ϫ2 3 1 1x Ϫ 5 2 3 ϭ 1 Ϫ2 2 3 x Ϫ 5 ϭ Ϫ8 x ϭ Ϫ3

original equation isolate radical term (subtract 4, divide by 2) apply power property, power is odd
3 simplify: 1x Ϫ 5 2 3 ϭ x Ϫ 5

solve

Substituting Ϫ3 for x in the original equation verifies it is a solution.

Now try Exercises 49 through 52



Sometimes squaring both sides of an equation still results in an equation with a radical term, but often there is one fewer than before. In this case, we simply repeat the process, as indicated by the flowchart in Figure A.7.

EXAMPLE 15 Solution



Solving Radical Equations Solve the equation: 1x ϩ 15 Ϫ 1x ϩ 3 ϭ 2. 1x ϩ 15 Ϫ 1x ϩ 3 ϭ 2 1x ϩ 15 ϭ 1x ϩ 3 ϩ 2 1 1x ϩ 15 2 2 ϭ 1 1x ϩ 3 ϩ 2 2 2 x ϩ 15 ϭ 1 x ϩ 3 2 ϩ 4 1x ϩ 3 ϩ 4 x ϩ 15 ϭ x ϩ 4 1x ϩ 3 ϩ 7 8 ϭ 4 1x ϩ 3 2 ϭ 1x ϩ 3 4ϭxϩ3 1ϭx



original equation isolate one radical power property 1 A ϩ B 2 2; A ϭ 1x ϩ 3, B ϭ 2 simplify isolate radical divide by four power property possible solution

Precalculus—

Section A.6 Radicals, Rational Exponents, and Radical Equations

A-73

Figure A.7
Radical Equations

Check



1x ϩ 15 Ϫ 1x ϩ 3 ϭ 2 1 1 1 2 ϩ 15 Ϫ 1 1 1 2 ϩ 3 ϭ 2 116 Ϫ 14 ϭ 2 4Ϫ2ϭ2 2 ϭ 2✓

original equation substitute 1 for x simplify solution checks

Isolate radical term

Now try Exercises 53 and 54



Apply power property

Since rational exponents are so closely related to radicals, the solution process for each is very similar. The goal is still to “undo” the radical (rational exponent) and solve for the unknown. Power Property of Equality
YES

Does the result contain a radical?

For real-valued expressions u and v, with positive integers m, n, and m n in lowest terms: If m is odd m and u n ϭ v,
m n

NO Solve using properties of equality

then 1 u n 2 m ϭ vm
n n

u ϭ vm

then 1 u n 2 m ϭ Ϯv m
m n n

If m is even m and u n ϭ v 1 v 7 0 2 , u ϭ Ϯvm
n

Check results in original equation

The power property of equality basically says that if certain conditions are satisfied, both sides of an equation can be raised to any needed power.

EXAMPLE 16



Solving Equations with Rational Exponents Solve each equation: 3 a. 3 1 x ϩ 1 2 4 Ϫ 9 ϭ 15 b. 1 x Ϫ 3 2 3 ϭ 4
2

Solution



a. 3 1 x ϩ 1 2 4 Ϫ 9 ϭ 15
3

Check



3 1 15 ϩ 1 2 Ϫ 9 ϭ 15
3 4 1 4

3 1x ϩ 12 4 ϭ 8 x ϩ 1 ϭ 16 x ϭ 15
3 4 4 3 4 3

1x ϩ 12 ϭ 8
3 4

3 original equation; m n ϭ 4

isolate variable term (add 9, divide by 3) apply power property, note m is odd simplify 3 83 ϭ 1 83 2 4 ϭ 16 4
4 1

result substitute 15 for x in the original equation simplify, rewrite exponent 116 ϭ 2 23 ϭ 8 solution checks
2 original equation; m n ϭ 3
4

3 1 16 2 Ϫ 9 ϭ 15 3 1 2 2 3 Ϫ 9 ϭ 15 3 1 8 2 Ϫ 9 ϭ 15 15 ϭ 15 ✓
3

b.

3 1x Ϫ 32 4 ϭ Ϯ 4 xϪ3ϭ Ϯ8 xϭ3Ϯ8
2 3 3 2 3 2

1x Ϫ 32 ϭ 4
2 3

apply power property, note m is even simplify 3 42 ϭ 1 42 2 3 ϭ 8 4
3 1

result

The solutions are 3 ϩ 8 ϭ 11 and 3 Ϫ 8 ϭ Ϫ5. Verify by checking both in the original equation. Now try Exercises 55 through 58


Precalculus—

A-74

APPENDIX A A Review of Basic Concepts and Skills

CAUTION



As you continue solving equations with radicals and rational exponents, be careful not to arbitrarily place the “Ϯ” sign in front of terms given in radical form. The expression 118 indicates the positive square root of 18, where 118 ϭ 3 12. The equation x2 ϭ 18 becomes x ϭ Ϯ 118 after applying the power property, with solutions x ϭ Ϯ3 12 1 x ϭ Ϫ3 12, x ϭ 3 12 2 , since the square of either number produces 18.

In Appendix A.4, we used a technique called u-substitution to factor expressions in quadratic form. The following equations are also in quadratic form since the de2 1 gree of the leading term is twice the degree of the middle term: x3 Ϫ 3x3 Ϫ 10 ϭ 0 and 1 x2 ϩ x 2 2 Ϫ 8 1 x2 ϩ x 2 ϩ 12 ϭ 0. The first equation and its solution appear in Example 17.

EXAMPLE 17 Solution



Solving Equations in Quadratic Form Solve using a u-substitution: x3 Ϫ 3x3 Ϫ 10 ϭ 0.
2 1



This equation is in quadratic form since it can be rewritten as: 1 1 1 x3 2 2 Ϫ 3 1 x3 2 1 Ϫ 10 ϭ degree of leading term is twice that of second 0, where the 1 2 term. If we let u ϭ x3, then u2 ϭ x3 and the equation becomes u2 Ϫ 3u1 Ϫ 10 ϭ 0, which is factorable. 1 u Ϫ 5 21 u ϩ 2 2 uϭ5 1 x3 ϭ 5 1 1 x3 2 3 ϭ 53 x ϭ 125 ϭ0 or u ϭ Ϫ2 1 or x3 ϭ Ϫ2 1 or 1 x3 2 3 ϭ 1 Ϫ2 2 3 or x ϭ Ϫ8
factor solution in terms of u resubstitute x 3 for u cube both sides: 1 3 132 ϭ 1 solve for x
1

Both solutions check. Now try Exercises 59 and 60
Figure A.8
Hypotenuse 90Њ Leg Leg


A right triangle is one that has a 90° angle (see Figure A.8). The longest side (opposite the right angle) is called the hypotenuse, while the other two sides are simply called “legs.” The Pythagorean theorem is a formula that says if you add the square of each leg, the result will be equal to the square of the hypotenuse. Furthermore, we note the converse of this theorem is also true. Pythagorean Theorem 1. For any right triangle with legs a and b and hypotenuse c, a2 ϩ b2 ϭ c2 2. For any triangle with sides a, b, and c, if a2 ϩ b2 ϭ c2, then the triangle is a right triangle.

Figure A.9
2

A geometric interpretation of the theorem is given in Figure A.9, which shows 3 ϩ 42 ϭ 52.
ea Ar in2 25

Area 16 in2

4

5

13 12 52 ϩ 122 ϭ 132 25 ϩ 144 ϭ 169

25 5 24 72 ϩ 242 ϭ 252 49 ϩ 576 ϭ 625

7

c

a

Area 9 in2

3

b a2 ϩ b2 ϭ c2 general case

Precalculus—

Section A.6 Radicals, Rational Exponents, and Radical Equations

A-75

EXAMPLE 18



Applying the Pythagorean Theorem An extension ladder is placed 9 ft from the base of a building in an effort to reach a third-story window that is 27 ft high. What is the minimum length of the ladder required? Answer in exact form using radicals, and in approximate form by rounding to one decimal place.

Solution



We can assume the building makes a 90° angle with the ground, and use the Pythagorean theorem to find the required length. Let c represent this length. c2 c2 c2 c2 c c c ϭ a2 ϩ b2 ϭ 1 9 2 2 ϩ 1 27 2 2 ϭ 81 ϩ 729 ϭ 810 ϭ 1810 ϭ 9 110 Ϸ 28.5 ft
Pythagorean theorem substitute 9 for a and 27 for b 92 ϭ 81, 272 ϭ 729 add definition of square root; c 7 0 exact form: 1810 ϭ 181 # 10 ϭ 9 110 approximate form

c

27 ft

The ladder must be at least 28.5 ft tall.
F. You’ve just seen how we can solve equations and use formulas involving radicals

9 ft

Now try Exercises 63 and 64



A.6 EXERCISES


CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. 1an ϭ ͿaͿ if n 7 0 is a(n)

n

integer.

2. The conjugate of 2 Ϫ 13 is 4. 1 x2 2 3 ϭ x2 3 ϭ x1 is an example of the property of exponents.
3 2 3 2

.

5. Discuss/Explain what it means when we say an expression like 1A has been written in simplest form.


3. By decomposing the rational exponent, we can 3 ? rewrite 16 4 as 1 16 ? 2 ?.

#

6. Discuss/Explain why it would be easier x1 2 to simplify the expression given using 1 rational exponents rather than radicals. x3

DEVELOPING YOUR SKILLS
10. a. 225n2 c. 2v10
3 11. a. 1 64 3 c. 2 216z12 3 12. a. 1 Ϫ8 3 c. 2 27q9

Evaluate the expression 2x2 for the values given.

7. a. x ϭ 9 8. a. x ϭ 7

b. x ϭ Ϫ10 b. x ϭ Ϫ8

d. 24a2 ϩ 12a ϩ 9
3 b. 2 Ϫ216x3

b. 2 1 y ϩ 2 2 2

Simplify each expression, assuming that variables can represent any real number.

d.

v3 B Ϫ8
3

9. a. 249p2 c. 281m4

b. 2 1 x Ϫ 3 2 2

3 b. 2 Ϫ125p3

d. 2x2 Ϫ 6x ϩ 9

d.

w3 B Ϫ64
3

Precalculus—

A-76

APPENDIX A A Review of Basic Concepts and Skills
6 b. 1 Ϫ64 5 d. 2Ϫ243x5 6 f. 2 1 h ϩ 2 2 6

4 14. a. 1 81 5 c. 2 1024z15 5 e. 2 1 q Ϫ 9 2 5 3 15. a. 1 Ϫ125

6 13. a. 1 64 5 c. 2243x10 5 e. 2 1 k Ϫ 3 2 5

4 b. Ϫ 281n12

4 b. 1 Ϫ81 5 d. 2Ϫ1024z20 6 f. 2 1 p ϩ 4 2 6

24. a. 28x6 2 3 c. 227a2b6 9 12 Ϫ 248 e. 8 25. a. 2.5 218a 22a3 c. x3y 4x5y B 3 B 12y

3 b. 3 2128a4b2

d. 254m6n8 Ϫ20 ϩ 232 4 2 b. Ϫ 23b 212b2 3 f.
3 3 d. 29v2u 23u5v2

c. 2Ϫ36
3 16. a. 1 Ϫ216

49v10 d. B 36 4 b. Ϫ 216m24 25x6 d. B 4 16 2 b. a b 25
3

26. a. 5.1 22p 232p5 c. 27. a. ab2 25ab4 B 3 B 27 28m5

4 b. Ϫ 25q 220q3 5 d. 25cd2 125cd b.
3 2 108n4 3 3

c. 2Ϫ121 17. a. 8
2 3

4 Ϫ2 c. a b 25
3

d. a

Ϫ27p6 8q3
3 2

b

2 3

22m 45 c. B 16x2 227y7 23y 20 B 4x4

3 2 4n 3 81 d. 12 A 8z9

18. a. 92 16 Ϫ4 c. a b 81
3

3

4 b. a b 9

28. a.
2

b.

3 2 72b5

23b2

3

Ϫ125v9 3 d. a b 27w6 b. a Ϫ
2

c.

125 d. Ϫ9 3 A 27x6
4 5 b. x 2 x 3 1 6

5 29. a. 2 32x10y15 4 3 c. 3 1b 4 e. 2b 1 b 4 30. a. 281a12b16 4 c. 32 a 3 4 e. 1 c 1 c

19. a. Ϫ1442 c. 1 Ϫ27 2 Ϫ3 20. a. Ϫ1002 c. 1 Ϫ125 2
Ϫ2 3
3

3

4 b 25

3 2

d.

26

d. Ϫa b. a Ϫ

27x b 64 49 b 36
3 2

3 Ϫ4 3

5 b. a 2a6

d.

3 1 3 4 2 3

x9 Ϫ3 d. Ϫa b 8
4

Simplify and add (if possible).

Use properties of exponents to simplify. Answer in exponential form without negative exponents.

21. a. 1 2n p 2 22. a. a 24x8 4x
1 2 3

2 Ϫ2 5 5

b. a

8y4 64y2
Ϫ1 4
3

3

b

1 3

31. a. b. c. d. 32. a. b. c. d.

12 272 Ϫ 9 298 8 248 Ϫ 3 2108 7 218m Ϫ 250m 2 228p Ϫ 3 263p Ϫ3 280 ϩ 2 2125 5 212 ϩ 2 227 3 212x Ϫ 5 275x 3 240q ϩ 9 210q

b

2

b. 1 2x y 2 4
3 4

Simplify each expression. Assume all variables represent nonnegative real numbers.

23. a. 218m2 3 3 c. 264m3n5 8 Ϫ6 ϩ 228 e. 2

3 b. Ϫ2 2 Ϫ125p3q7

d. 232p3q6 f. 27 Ϫ 272 6

3 3 33. a. 3x 1 54x Ϫ 5 216x4 b. 14 ϩ 13x Ϫ 112x ϩ 145 c. 272x3 ϩ 150 Ϫ 17x ϩ 127 3 3 34. a. 5 254m3 Ϫ 2m 216m3 b. 110b ϩ 1200b Ϫ 120 ϩ 140

c. 275r3 ϩ 132 Ϫ 127r ϩ 138

Precalculus—

Section A.6 Radicals, Rational Exponents, and Radical Equations

A-77

Compute each product and simplify the result.
2

b. 15 1 16 Ϫ 12 2 36. a. 1 0.3 15 2 c. 1 4 ϩ 13 21 4 Ϫ 13 2 d. 1 2 ϩ 15 2 2
2

35. a. 1 7 12 2 b. 13 1 15 ϩ 17 2 c. 1 n ϩ 15 21 n Ϫ 15 2 d. 1 6 Ϫ 13 2 2

46. a. 47. a. 48. a.

7 17 ϩ 3 110 Ϫ 3 13 ϩ 12 1 ϩ 12 16 ϩ 114

b. b. b.

12 1x ϩ 13 7 ϩ 16 3 Ϫ 3 12 1 ϩ 16 5 ϩ 2 13

37. a. 1 3 ϩ 2 17 21 3 Ϫ 2 17 2 b. 1 15 Ϫ 114 2 1 12 ϩ 113 2 c. 1 2 12 ϩ 6 16 2 1 3 110 ϩ 17 2 38. a. 1 5 ϩ 4 110 21 1 Ϫ 2 110 2 b. 1 13 ϩ 12 2 1 110 ϩ 111 2 c. 1 3 15 ϩ 4 12 2 1 115 ϩ 16 2

Solve each equation and check your solutions by substitution. Identify any extraneous roots.

49. a. Ϫ3 13x Ϫ 5 ϭ Ϫ9 b. x ϭ 13x ϩ 1 ϩ 3 50. a. Ϫ2 14x Ϫ 1 ϭ Ϫ10 b. Ϫ5 ϭ 15x Ϫ 1 Ϫ x
3 51. a. 2 ϭ 13m Ϫ 1 3 b. 2 1 7 Ϫ 3x Ϫ 3 ϭ Ϫ7 3 1 2m ϩ 3 ϩ2ϭ3 c. Ϫ5 3 3 d. 1 2x Ϫ 9 ϭ 1 3x ϩ 7 3 52. a. Ϫ3 ϭ 1 5p ϩ 2 3 b. 3 1 3 Ϫ 4x Ϫ 7 ϭ Ϫ4 3 1 6x Ϫ 7 Ϫ 5 ϭ Ϫ6 c. 4 3 3 d. 3 1 x ϩ 3 ϭ 2 1 2x ϩ 17

Use a substitution to verify the solutions to the quadratic equation given. Verify results using a calculator.

39. x2 Ϫ 4x ϩ 1 ϭ 0 a. x ϭ 2 ϩ 13 40. x2 Ϫ 10x ϩ 18 ϭ 0 a. x ϭ 5 Ϫ 17 41. x2 ϩ 2x Ϫ 9 ϭ 0 a. x ϭ Ϫ1 ϩ 110 42. x Ϫ 14x ϩ 29 ϭ 0 a. x ϭ 7 Ϫ 2 15
2

b. x ϭ 2 Ϫ 13 b. x ϭ 5 ϩ 17 b. x ϭ Ϫ1 Ϫ 110 b. x ϭ 7 ϩ 2 15

Rationalize each expression by building perfect nth root factors for each denominator. Assume all variables represent positive quantities.

43. a.

3 112 27 c. B 50b 5 e. 3 1a

b.

20 B 27x3 1 d. 3 A 4p

53. a. b. c. d. 54. a. b. c. d.

1x Ϫ 9 ϩ 1x ϭ 9 x ϭ 3 ϩ 223 Ϫ x 1x Ϫ 2 Ϫ 12x ϭ Ϫ2 112x ϩ 9 Ϫ 124x ϭ Ϫ3 1x ϩ 7 Ϫ 1x ϭ 1 12x ϩ 31 ϩ x ϭ 2 13x ϭ 1x Ϫ 3 ϩ 3 13x ϩ 4 Ϫ 17x ϭ Ϫ2

Ϫ4 125 44. a. b. B 12n3 120 5 3 c. d. 3 B 12x A 2m2 Ϫ8 e. 3 3 15 Simplify the following expressions by rationalizing the denominators. Where possible, state results in exact form and approximate form, rounded to hundredths. 45. a. 8 3 ϩ 111 b. 6 1x Ϫ 12

Write the equation in simplified form, then solve. Check all answers by substitution.

55. a. x5 ϩ 17 ϭ 9 3 b. Ϫ2x4 ϩ 47 ϭ Ϫ7 56. a. 0.3x2 Ϫ 39 ϭ 42 5 b. 0.5x3 ϩ 92 ϭ Ϫ43
2 5

3

57. a. 2 1 x ϩ 5 2 3 Ϫ 11 ϭ 7 4 b. Ϫ3 1 x Ϫ 2 2 5 ϩ 29 ϭ Ϫ19
3 58. a. 3x3 Ϫ 10 ϭ 1x 2 5 2 b. 2 1x Ϫ 4 ϭ x5
1

59. x3 Ϫ 2x3 Ϫ 15 ϭ 0 60. x3 Ϫ 9x2 ϭ Ϫ8
3

2

1

Precalculus—

A-78


APPENDIX A A Review of Basic Concepts and Skills

WORKING WITH FORMULAS
1

61. Fish length to weight relationship: L ‫ ؍‬1.13 1 W 2 3 The length to weight relationship of a female Pacific halibut can be approximated by the formula shown, where W is the weight in pounds and L is the length in feet. A fisherman lands a halibut that weighs 400 lb. Approximate the length of the fish (round to two decimal places).


62. Timing a falling object: t ‫؍‬

1s 4 The time it takes an object to fall a certain distance is given by the formula shown, where t is the time in seconds and s is the distance the object has fallen. Approximate the time it takes an object to hit the ground, if it is dropped from the top of a building that is 80 ft in height (round to hundredths).

APPLICATIONS
24 m

63. Length of a cable: A radio tower is secured by cables that are anchored in the ground 8 m from its base. If the cables are attached to the tower 24 m above the ground, what is the length of each cable? Answer in (a) exact form using radicals, and (b) approximate form by rounding to one decimal place.

65. a. Earth: 93 million mi b. Mars: 142 million mi c. Mercury: 36 million mi 66. a. Venus: 67 million mi b. Jupiter: 480 million mi c. Saturn: 890 million mi 67. Accident investigation: After an accident, police officers will try to determine the approximate velocity V that a car was traveling using the formula V ϭ 2 26L, where L is the length of the skid marks in feet and V is the velocity in miles per hour. (a) If the skid marks were 54 ft long, how fast was the car traveling? (b) Approximate the speed of the car if the skid marks were 90 ft long. 68. Wind-powered energy: If a wind-powered generator is delivering P units of power, the velocity V of the wind (in miles per hour) can be 3 P , where k is a constant determined using V ϭ Ak that depends on the size and efficiency of the generator. Rationalize the radical expression and use the new version to find the velocity of the wind if k ϭ 0.004 and the generator is putting out 13.5 units of power.

c

8m 64. Height of a kite: Benjamin Franklin is flying his kite in a storm once again. John Adams has walked to a position directly under the kite and is 75 ft from Ben. If the kite is 50 ft above John Adams’ head, how much string S has Ben let out? Answer in (a) exact form using radicals, and (b) approximate form by rounding to one decimal place.

S

50 ft

75 ft

The time T (in days) required for a planet to make one revolution around the sun is modeled 3 by the function T ‫ ؍‬0.407R2, where R is the maximum radius of the planet’s orbit (in millions of miles). This is known as Kepler’s third law of planetary motion. Use the equation given to approximate the number of days required for one complete orbit of each planet, given its maximum orbital radius.

Precalculus—

Section A.6 Radicals, Rational Exponents, and Radical Equations

A-79

69. Surface area: The lateral surface area (surface area excluding the base) h S of a cone is given by the formula 2 2 S ϭ ␲r 2r ϩ h , where r is the r radius of the base and h is the height of the cone. Find the lateral surface area of a cone that has a radius of 6 m and a height of 10 m. Answer in simplest form. 70. Surface area: The lateral surface a area S of a frustum (a truncated cone) is given by the formula h S ϭ ␲ 1 a ϩ b 2 2h2 ϩ 1 b Ϫ a 2 2, b where a is the radius of the upper base, b is the radius of the lower base, and h is the height. Find the surface area of a frustum where a ϭ 6 m, b ϭ 8 m, and h ϭ 10 m. Answer in simplest form. 71. Planetary motion: The time T (in days) for a planet to make one revolution (elliptical orbit) 3 around the sun is modeled by T ϭ 0.407R2, where R is the maximum radius of the planet’s orbit in

millions of miles (Kepler’s third law of planetary motion). Use the equation to approximate the maximum radius of each orbit, given the number of days it takes for one revolution. (See Exercises 65 and 66.) a. Mercury: 88 days b. Venus: 225 days c. Earth: 365 days d. Mars: 687 days e. Jupiter: 4333 days f. Saturn: 10,759 days 72. Wind-powered energy: If a wind-powered generator is delivering P units of power, the velocity V of the wind (in miles per hour) can be 3 P , where k is a constant determined using V ϭ Ak that depends on the size and efficiency of the generator. Given k ϭ 0.004, approximately how many units of power are being delivered if the wind is blowing at 27 miles per hour? (See Exercise 68.)



EXTENDING THE CONCEPT
76. Find a quick way to simplify the expression without the aid of a calculator.
5 6 3 2

73. a. x2 Ϫ 5 b. n2 Ϫ 19 74. a. 4v2 Ϫ 11 b. 9w2 Ϫ 17 75. The following terms form a pattern that continues until the sixth term is found: 23x ϩ 29x ϩ 227x ϩ p (a) Compute the sum of all six terms; (b) develop a system (investigate the pattern further) that will enable you to find the sum of 12 such terms without actually writing out the terms.

1 1 1 1 9 77. If 1 x2 ϩ xϪ2 2 2 ϭ , find the value of x2 ϩ xϪ2. 2

78. Rewrite by rationalizing the numerator: 1x ϩ h Ϫ 1x h Determine the values of x for which each expression represents a real number. 79. 80. 1x Ϫ 1 x2 Ϫ 4 x2 Ϫ 4 1x Ϫ 1

aaaa

The expression x2 ؊ 7 is not factorable using integer values. But the expression can be written in the form x2 ؊ 1 27 2 2, enabling us to factor it as a “binomial” and its conjugate: 1 x ؉ 27 2 1 x ؊ 27 2 . Use this idea to factor the following expressions.

a a a a a3 b

4 5

3 4

2 5

10 3

Precalculus—

A-80

APPENDIX A A Review of Basic Concepts and Skills

OVERVIEW OF APPENDIX A
Prerequisite Definitions, Properties, Formulas, and Relationships
Notation and Relations
concept notation {members} description braces enclose the members of a set example set of even whole numbers A ϭ 5 0, 2, 4, 6, 8, p 6 14 ʦ A odd numbers in A S ϭ 5 0, 6, 12, 18, 24, p 6 S ( A S ϭ 5 x | x ϭ 6n for n ʦ ‫ ޗ‬6

• Set notation:

ʦ • Is an element of Empty set or {} л • ( • Is a proper subset of

• Defining a set

5x | x p6

indicates membership in a set a set having no elements indicates the elements of one set are entirely contained in another the set of all x, such that x . . .

Sets of Numbers • Natural: ‫ ގ‬ϭ 5 1, 2, 3, 4, p 6

• Integers: ‫ ޚ‬ϭ 5 p , Ϫ3, Ϫ2, Ϫ1, 0, 1, 2, 3, p 6 • Irrational: ‫ ވ‬ϭ {numbers with a nonterminating,
nonrepeating decimal form}

• Whole: ‫ ޗ‬ϭ 5 0, 1, 2, 3, p 6
p

• Rational: ‫ ޑ‬ϭ e q , where p, q ʦ ‫ ;ޚ‬q 0 f • Real: ‫ ޒ‬ϭ {all rational and irrational numbers} Distance between numbers a and b on the number line

Absolute Value of a Number • The distance between a number n

Ϳa Ϫ bͿ or Ϳb Ϫ aͿ and zero (always positive) n if n Ն 0 0n 0 ϭ e Ϫn if n 6 0 For a complete review of these ideas, go to www.mhhe.com/coburn.

A.1 Properties of Real Numbers: For real numbers a, b, and c,
Commutative Property

• Addition: a ϩ b ϭ b ϩ a • Multiplication: a # b ϭ b # a
Identities

• Addition: 1 a ϩ b 2 ϩ c ϭ a ϩ 1 b ϩ c 2 • Multiplication: 1 a # b 2 # c ϭ a # 1 b # c 2 • Additive: a ϩ 1 Ϫa 2 ϭ 0
Inverses

Associative Property

• Additive: 0 ϩ a ϭ a • Multiplicative: 1 # a ϭ a

p q • Multiplicative: q # p ϭ 1; p, q

0

A.2 Properties of Exponents: For real numbers a and b, and integers m, n, and p • • • •
(excluding 0 raised to a nonpositive power), Product property: b # b ϭ b • Power property: 1 bm 2 n ϭ bmn am p amp Product to a power: 1 ambn 2 p ϭ amp # bnp • Quotient to a power: a n b ϭ np 1 b 0 2 m b b b Quotient property: n ϭ bm Ϫ n 1 b 0 2 0 b • Zero exponents: b ϭ 1 1 b 0 2 Ϫn n 1 a b • Scientific notation: N ϫ 10k; 1 Յ ͿNͿ 6 10, k ʦ ‫ޚ‬ Negative exponents: bϪn ϭ n ; a b ϭ a b a b b 1 a, b 0 2
m n mϩn

Precalculus—

Overview of Appendix A

A-81

A.2 Polynomials • A polynomial is a sum or difference of monomial terms • Polynomials are classified as a monomial, binomial, trinomial, or polynomial, depending on the number of terms • The degree of a polynomial in one variable is the same as the largest exponent occuring on the variable in any term • A polynomial expression is in standard form when written with the terms in descending order of degree A.3 Solving Linear Equations and Inequalities • Properties of Equality:
multiplicative property If A and B are algebraic expressions, where A ϭ B, then A # C ϭ B # C A B [C can be positive or negative] and ϭ ; C 0 C C A linear equation in one variable is one that can be written in the form ax ϩ b ϭ c, where the exponent on the variable is a 1. To solve a linear equation, we attempt to isolate the term containing the variable using the additive property, then solve for the variable using the multiplicative property. An equation can be an identity (always true), a contradiction (never true) or conditional (true or false depending on the input value[s]). If an equation contains fractions, multiplying both sides by the least common denominator of all fractions will “clear the denominators” and reduce the amount of work required to solve. Inequalities are solved using properties similar to those used for solving equations. The one exception is when multiplying or dividing by a negative quantity, as the inequality symbol must then be reversed to maintain the truth of the resulting statement. Solutions to an inequality can be given using a simple inequality, graphed on a number line, stated in set notation, or stated using interval notation. Given two sets A and B: A intersect B 1 A ¨ B 2 is the set of elements shared by both A and B (elements common to both sets). A union B 1 A ´ B 2 is the set of elements in either A or B (elements are combined to form a larger set). Compound inequalities are formed using the conjunction “and” or the conjunction “or.” The result can be either a joint inequality as in Ϫ3 6 x Յ 5, or a disjoint inequality, x 6 Ϫ2 or x 7 7. additive property If A and B are algebraic expressions, where A ϭ B, then A ϩ C ϭ B ϩ C.

• • • • • • • •

A.4 Special Factorizations • A2 Ϫ B2 ϭ 1 A ϩ B 21 A Ϫ B 2 • A2 Ϯ 2AB ϩ B2 ϭ 1 A Ϯ B 2 2 3 3 2 2 • A Ϫ B ϭ 1 A Ϫ B 21 A ϩ AB ϩ B 2 • A3 ϩ B3 ϭ 1 A ϩ B 21 A2 Ϫ AB ϩ B2 2 • Certain equations of higher degree can be solved using factoring skills and the zero product property. A.5 Rational Expressions: For polynomials P, Q, R, and S with no denominator of zero, • Lowest terms: • • •
P#R P P#R P ϭ Equivalence: ϭ • Q#R Q Q Q#R # P R P R PR R P S PS P Multiplication: # ϭ ϭ • Division: Ϭ ϭ # ϭ Q S Q#S QS Q S Q R QR Q PϩQ Q PϪQ P P Addition: ϩ ϭ • Subtraction: Ϫ ϭ R R R R R R Addition/subtraction with unlike denominators: 1. Find the LCD of all rational expressions. 2. Build equivalent expressions using LCD. 3. Add/subtract numerators as indicated. 4. Write the result in lowest terms. To solve rational equations, first clear denominators using the LCD, noting values that must be excluded. Multiplying an equation by a variable quantity sometimes introduces extraneous solutions. Check all results in the original equation.

• •

Precalculus—

A-82

APPENDIX A A Review of Basic Concepts and Skills

A.6 Properties of Radicals • 1a is a real number only for a Ն 0 n • 1a ϭ b, only if bn ϭ a • •
n





• •

if a Ն 0 n For any real number a, 1a ϭ ͿaͿ when n is even • For any real number a, 1an ϭ a when n is odd m If is a rational number written in lowest terms If a is a real number and n is an integer greater • 1 n n n m m than 1, then 1 a ϭ an provided 1 a represents a n n n Ն 2, then a n ϭ 11a 2 m and a n ϭ 1am with real number n provided 2a represents a real number. n n n n If 1A and 1B represent real numbers, • If 1A and 1B represent real numbers and B 0, n n n n 1AB ϭ 1 A # 1B 1A n A ϭ n BB 1B A radical expression is in simplest form when: 1. the radicand has no factors that are perfect nth roots, 2. the radicand contains no fractions, and 3. no radicals occur in a denominator. To solve radical equations, isolate the radical on one side, then apply the appropriate “nth power” to free up the radicand. Repeat the process if needed. See flowchart on page 74. n For equations with a rational exponent m n , isolate the variable term and raise both sides to the m power. If m is even, there will be two real solutions.
n

• 1a ϭ b, only if b2 ϭ a n • If n is even, 1a represents a real number only

A.6 Pythagorean Theorem • For any right triangle with legs a and b and
hypotenuse c: a2 ϩ b2 ϭ c2.

• For any triangle with sides a, b, and c, if

a2 ϩ b2 ϭ c2, then the triangle is a right triangle.

PRACTICE TEST
1. State true or false. If false, state why. 7 a. 1 3 ϩ 4 2 2 ϭ 25 b. ϭ 0 0 c. x Ϫ 3 ϭ Ϫ3 ϩ x d. Ϫ2 1 x Ϫ 3 2 ϭ Ϫ2x Ϫ 3 7. State the number of terms in each expression and identify the coefficient of each. cϩ2 ϩc a. Ϫ2v2 ϩ 6v ϩ 5 b. 3 8. Evaluate each expression given x ϭ Ϫ0.5 and y ϭ Ϫ2. Round to hundredths as needed. y a. 2x Ϫ 3y2 b. 12 Ϫ x 1 4 Ϫ x2 2 ϩ x 9. Translate each phrase into an algebraic expression. a. Nine less than twice a number is subtracted from the number cubed. b. Three times the square of half a number is subtracted from twice the number. 10. Create a mathematical model using descriptive variables. a. The radius of the planet Jupiter is approximately 119 mi less than 11 times the radius of the Earth. Express the radius of Jupiter in terms of the Earth’s radius. b. Last year, Video Venue Inc. earned $1.2 million more than four times what it earned this year. Express last year’s earnings of Video Venue Inc. in terms of this year’s earnings.

2. State the value of each expression. 3 a. 1 121 b. 1 Ϫ125 c. 1Ϫ36 d. 1400

4. Evaluate each expression: 1 a. 1 Ϫ4 2a Ϫ2 b b. 1 Ϫ0.6 21 Ϫ1.5 2 3 Ϫ2.8 c. d. 4.2 Ϭ 1 Ϫ0.6 2 Ϫ0.7 # 0.08 12 10 5. Evaluate using a calculator: 2000 a 1 ϩ b 12 6. State the value of each expression, if possible. a. 0 Ϭ 6 b. 6 Ϭ 0

3. Evaluate each expression: 7 1 1 5 a. Ϫ aϪ b b. Ϫ Ϫ 8 4 3 6 c. Ϫ0.7 ϩ 1.2 d. 1.3 ϩ 1 Ϫ5.9 2

Precalculus—

Practice Test

A-83

11. Simplify by combining like terms. a. 8v2 ϩ 4v Ϫ 7 ϩ v2 Ϫ v b. Ϫ4 1 3b Ϫ 2 2 ϩ 5b c. 4x Ϫ 1 x Ϫ 2x2 2 ϩ x 1 3 Ϫ x 2

12. Factor each expression completely. a. 9x2 Ϫ 16 b. 4v3 Ϫ 12v2 ϩ 9v c. x3 ϩ 5x2 Ϫ 9x Ϫ 45 13. Simplify using the properties of exponents. 5 a. Ϫ3 b. 1 Ϫ2a3 2 2 1 a2b4 2 3 b 5p2q3r4 2 m2 3 c. a b d. a b 2n Ϫ2pq2r4 14. Simplify using the properties of exponents. Ϫ12a3b5 a. 3a2b4 b. 1 3.2 ϫ 10Ϫ17 2 ϫ 1 2.0 ϫ 1015 2 aϪ3 # b Ϫ4 c. a Ϫ2 b d. Ϫ7x0 ϩ 1 Ϫ7x 2 0 c 15. Compute each product. a. 1 3x2 ϩ 5y 2 1 3x2 Ϫ 5y 2 b. 1 2a ϩ 3b 2 2

19. Maximizing revenue: Due to past experience, the manager of a video store knows that if a popular video game is priced at $30, the store will sell 40 each day. For each decrease of $0.50, one additional sale will be made. The formula for the store’s revenue is then R ϭ 1 30 Ϫ 0.5x 2 1 40 ϩ x 2 , where x represents the number of times the price is decreased. Multiply the binomials and use a table of values to determine (a) the number of 50¢ decreases that will give the most revenue and (b) the maximum amount of revenue. 20. Diagonal of a rectangular prism: Use the Pythagorean theorem to determine the length of the diagonal of the rectangular prism shown in the figure. (Hint: First find the diagonal of the base.)
42 cm

32 cm

24 cm

16. Add or subtract as indicated. a. 1 Ϫ5a3 ϩ 4a2 Ϫ 3 2 ϩ 1 7a4 ϩ 4a2 Ϫ 3a Ϫ 15 2 b. 1 2x2 ϩ 4x Ϫ 9 2 Ϫ 1 7x4 Ϫ 2x2 Ϫ x Ϫ 9 2 xϪ5 4 Ϫ n2 b. 2 5Ϫx n Ϫ 4n ϩ 4 x3 Ϫ 27 3x2 Ϫ 13x Ϫ 10 c. 2 d. x ϩ 3x ϩ 9 9x2 Ϫ 4 x2 Ϫ 25 x2 ϩ x Ϫ 20 e. 2 Ϭ 2 3x Ϫ 11x Ϫ 4 x Ϫ 8x ϩ 16 mϩ3 2 f. 2 Ϫ 51m ϩ 42 m ϩ m Ϫ 12 Ϫ8 A 27v3 Ϫ4 ϩ 132 d. 8 f. 1 x ϩ 15 2 1 x Ϫ 15 2 8 h. 16 Ϫ 12 b.
3

21. Solve each linear equation. a. Ϫ2b ϩ 7 ϭ Ϫ5 b. 3 1 2n Ϫ 6 2 ϩ 1 ϭ 7 1 2 3 c. 4m Ϫ 5 ϭ 11m ϩ 2 d. x ϩ ϭ 2 3 4 e. Ϫ8 1 3p ϩ 5 2 Ϫ 9 ϭ 6 1 3 Ϫ 4p 2 g 5g 1 f. Ϫ ϭ 3 Ϫ Ϫ 6 2 12 22. If one-fourth of the sum of a number and twelve is added to three, the result is sixteen. Find an equation model for this statement, then use it to find the number. 23. Solve each polynomial equation by factoring. a. x3 Ϫ 7x2 ϭ 4x Ϫ 28 b. Ϫ7r3 ϩ 21r2 ϩ 28r ϭ 0 c. g4 Ϫ 10g2 ϩ 9 ϭ 0 24. Solve each rational equation. 7 1 3 a. ϩ ϭ 5x 10 4x 3h 7 1 b. Ϫ 2 ϭ hϩ3 h h ϩ 3h 3 4n ϩ 20 n Ϫ ϭ 2 c. nϩ2 nϪ4 n Ϫ 2n Ϫ 8 25. Solve each radical equation. 2x2 ϩ 7 a. ϩ3ϭ5 2 b. 3 1x ϩ 4 ϭ x ϩ 4 c. 13x ϩ 4 ϭ 2 Ϫ 1x ϩ 2

Simplify or compute as indicated. 17. a.

18. a. 2 1 x ϩ 11 2 2 25 2 c. a b 16 e. 7 140 Ϫ 190 2 g. B 5x
Ϫ3

Precalculus—

Appendix B
Proof Positive — A Selection of Proofs from Precalculus
The Remainder Theorem
If a polynomial p(x) is divided by (x Ϫ c) using synthetic division, the remainder is equal to p(c). Proof of the Remainder Theorem From our previous work, any number c used in synthetic division will occur as the factor (x Ϫ c) when written as 1 quotient 21 divisor 2 ϩ remainder: p 1 x 2 ϭ 1 x Ϫ c 2 q 1 x 2 ϩ r. Here, q(x) represents the quotient polynomial and r is a constant. Evaluating p(c) gives p1c2 ϭ 1c Ϫ c2q1c2 ϩ r ϭ 0 # q1c2 ϩ r ϭr✓ p1x2 ϭ 1x Ϫ c2q1x2 ϩ r

The Factor Theorem
Given a polynomial p(x), (1) if p(c) ϭ 0, then x Ϫ c is a factor of p(x), and (2) if x Ϫ c is a factor of p(x), then p(c) ϭ 0. Proof of the Factor Theorem 1. Consider a polynomial p written in the form p 1 x 2 ϭ 1 x Ϫ c 2 q 1 x 2 ϩ r. From the remainder theorem we know p 1 c 2 ϭ r, and substituting p(c) for r in the equation shown gives: and x Ϫ c is a factor of p(x), if p 1 c 2 ϭ 0 p1x2 ϭ 1x Ϫ c2q1x2 ϩ p1c2

2. The steps from part 1 can be reversed, since any factor (x Ϫ c) of p(x), can be written in the form p 1 x 2 ϭ 1 x Ϫ c 2 q 1 x 2 . Evaluating at x ϭ c produces a result of zero: p1c2 ϭ 1c Ϫ c2q1x2 ϭ0✓

p1x2 ϭ 1x Ϫ c2q1x2 ✓

Complex Conjugates Theorem
Given p(x) is a polynomial with real number coefficients, complex solutions must occur in conjugate pairs. If a ϩ bi, b 0 is a solution, then a Ϫ bi must also be a solution. To prove this for polynomials of degree n 7 2, we let z1 ϭ a ϩ bi and z2 ϭ c ϩ di be complex numbers, and let z1 ϭ a Ϫ bi, and z2 ϭ c Ϫ di represent their conjugates, and observe the following properties:

A-84

Precalculus—

APPENDIX B Proof Positive — A Selection of Proofs from Precalculus

A-85

1. The conjugate of a sum is equal to the sum of the conjugates. sum: z1 ϩ z2 sum of conjugates: z1 ϩ z2 1a ϩ c2 ϩ 1b ϩ d2i product: z1 # z2 1 a ϩ bi 2 ϩ 1 c ϩ di 2
→ conjugate of sum →

1a ϩ c2 Ϫ 1b ϩ d2i ✓ product of conjugates: z1 # z2

1 a Ϫ bi 2 ϩ 1 c Ϫ di 2

2. The conjugate of a product is equal to the product of the conjugates. 1 a ϩ bi 2 # 1 c ϩ di 2 1 a Ϫ bi 2 # 1 c Ϫ di 2

1 ac Ϫ bd 2 ϩ 1 ad ϩ bc 2 i

ac ϩ adi ϩ bci ϩ bdi

2

→ conjugate of product →

1 ac Ϫ bd 2 Ϫ 1 ad ϩ bc 2 i ✓

ac Ϫ adi Ϫ cbi ϩ bdi2

Since polynomials involve only sums and products, and the complex conjugate of any real number is the number itself, we have the following: Proof of the Complex Conjugates Theorem Given polynomial p 1 x 2 ϭ anxn ϩ an Ϫ 1xn Ϫ 1 ϩ p ϩ a1x1 ϩ a0, where an, an Ϫ 1, p , a1, a0 are real numbers and z ϭ a ϩ bi is a zero of p, we must show that z ϭ a Ϫ bi is also a zero. evaluate p (x ) at z anzn ϩ an Ϫ 1zn Ϫ 1 ϩ p ϩ a1z1 ϩ a0 ϭ p 1 z 2 anzn ϩ an Ϫ 1zn Ϫ 1 ϩ p ϩ a1z1 ϩ a0 ϭ 0 anzn ϩ an Ϫ 1zn Ϫ 1 ϩ p ϩ a1z1 ϩ a0 ϭ 0 anzn ϩ an Ϫ 1zn Ϫ 1 ϩ p ϩ a1z1 ϩ a0 ϭ 0
nϪ1

p 1 z 2 ϭ 0 given

conjugate both sides property 1 property 2 conjugate of a real number is the number

1 2 ϩ p ϩ a1 1 z 2 ϩ a0 ϭ 0 an 1 z 2 ϩ an Ϫ 1 1 z n nϪ1 1 2 ϩ p ϩ a1 1 z 2 ϩ a0 ϭ 0 an 1 z 2 ϩ an Ϫ 1 1 z n

p1z2 ϭ 0



result

An immediate and useful result of this theorem is that any polynomial of odd degree must have at least one real root.

Linear Factorization Theorem
If p(x) is a complex polynomial of degree n Ն 1, then p has exactly n linear factors and can be written in the form p(x) ϭ an(x Ϫ c1)(x Ϫ c2) # . . . # (x Ϫ cn), where an 0 and c1, c2, . . . , cn are complex numbers. Some factors may have multiplicities greater than 1 (c1, c2, . . . , cn are not necessarily distinct). Proof of the Linear Factorization Theorem Given p(x) ϭ anxn ϩ anϪ1xnϪ1 ϩ . . . ϩ a1x ϩ a0 is a complex polynomial, the Fundamental Theorem of Algebra establishes that p(x) has a least one complex zero, call it c1. The factor theorem stipulates (x Ϫ c1) must be a factor of P, giving where q1(x) is a complex polynomial of degree n Ϫ 1. Since q1(x) is a complex polynomial in its own right, it too must also have a complex zero, call it c2. Then (x Ϫ c2) must be a factor of q1(x), giving where q2(x) is a complex polynomial of degree n Ϫ 2. Repeating this rationale n times will cause p(x) to be rewritten in the form where qn(x) has a degree of n Ϫ n ϭ 0, a nonzero constant typically called an. The result is p 1 x 2 ϭ an 1 x Ϫ c1 21 x Ϫ c2 2 # . . . # 1 x Ϫ cn 2 , and the proof is complete. p 1 x 2 ϭ 1 x Ϫ c1 21 x Ϫ c2 2 p 1 x 2 ϭ 1 x Ϫ c1 21 x Ϫ c2 2 q2 1 x 2 p 1 x 2 ϭ 1 x Ϫ c1 2 q1 1 x 2

#

...

#

1 x Ϫ cn 2 qn 1 x 2

Precalculus—

A-86

APPENDIX B Proof Positive — A Selection of Proofs from Precalculus

The Product Property of Logarithms
Given M, N, and b 1 are positive real numbers, logb(MN) ϭ logbM ϩ logbN. Proof of the Product Property For P ϭ logb M and Q ϭ logb N, we have bP ϭ M and bQ ϭ N in exponential form. It follows that logb 1 MN 2 ϭ logb 1 bPbQ 2 ϭPϩQ ϭ logb 1 bP ϩ Q 2 ϭ logb M ϩ logb N
substitute b P for M and b Q for N properties of exponents log property 3 substitute logb M for P and logb N for Q

The Quotient Property of Logarithms
Given M, N, and b 1 are positive real numbers, M logb a b ϭ logb M Ϫ logb N. N Proof of the Quotient Property For P ϭ logb M and Q ϭ logb N, we have bP ϭ M and bQ ϭ N in exponential form. It follows that logb a M bP b ϭ logb a Q b N b ϭPϪQ
substitute b P for M and b Q for N properties of exponents log property 3 substitute logb M for P and logb N for Q

ϭ logb 1 bP Ϫ Q 2 ϭ logb M Ϫ logb N

The Power Property of Logarithms
Given M, N, and b logb Mx ϭ x logb M. 1 are positive real numbers and any real number x,

Proof of the Power Property For P ϭ logb M, we have bP ϭ M in exponential form. It follows that logb 1 M 2 x ϭ logb 1 bP 2 x ϭ logb 1 bPx 2 ϭ Px ϭ 1 logb M 2 x ϭ x logb M
substitute b P for M properties of exponents log property 3 substitute logb M for P rewrite factors

Proof of the Determinant Formula for the Area of a Parallelogram
1 Since the area of a triangle is A ϭ ab sin ␪, the area of the corresponding parallelogram 2 is twice as large: A ϭ ab sin ␪. In terms of the vectors u ϭ Ha, bI and v ϭ Hc, dI, we have A ϭ ͿuͿͿvͿ sin ␪, and it follows that

Precalculus—

APPENDIX B Proof Positive — A Selection of Proofs from Precalculus

A-87

Area ϭ ͿuͿͿvͿ 21 Ϫ cos2␪,
u ␪ v

ϭ ͿuͿͿvͿ ϭ ͿuͿͿvͿ

B B
2



1u # v22 ͿuͿ2ͿvͿ2

Pythagorean identity substitute for cos ␪

ͿuͿ2ͿvͿ2 Ϫ 1 u # v 2 2 ͿuͿ2ͿvͿ2
2

common denominator simplify

ϭ 2ͿuͿ2ͿvͿ2 Ϫ 1 u # v 2 2
2 2 2 2 2

ϭ 2 1 a ϩ b 2 1 c ϩ d 2 Ϫ 1 ac ϩ bd 2
2 2 2 2 2

2 2 2 2 2

substitute expand simplify factor result

ϭ 2a c ϩ a d ϩ b c ϩ b d Ϫ 1 a c ϩ 2acbd ϩ b d 2 ϭ 2a d Ϫ 2acbd ϩ b c
2 2 2 2

ϭ 2 1 ad Ϫ bc 2 ϭ Ϳad Ϫ bcͿ

2

Proof of Heron’s Formula Using Algebra
Note that 2a2 Ϫ d2 ϭ h ϭ 2c2 Ϫ e2. It follows that
a d b h e c

2a2 Ϫ d2 ϭ 2c2 Ϫ e2 a2 Ϫ 1 b Ϫ e 2 2 ϭ c2 Ϫ e2 a2 Ϫ b2 Ϫ c2 ϭ Ϫ2be b2 ϩ c2 Ϫ a2 ϭe 2b This shows: 1 A ϭ bh 2 1 ϭ b 2c2 Ϫ e2 2 1 b2 ϩ c2 Ϫ a2 2 ϭ b c2 Ϫ a b 2 B 2b 1 b4 ϩ 2b2c2 Ϫ 2a2b2 Ϫ 2a2c2 ϩ a4 ϩ c4 ϭ b c2 Ϫ a b 2 B 4b2 1 4b2c2 b4 ϩ 2b2c2 Ϫ 2a2b2 Ϫ 2a2c2 ϩ a4 ϩ c4 ϭ b Ϫ a b 2 B 4b2 4b2 1 4b2c2 Ϫ b4 Ϫ 2b2c2 ϩ 2a2b2 ϩ 2a2c2 Ϫ a4 Ϫ c4 ϭ b 2 B 4b2 1 ϭ 22a2b2 ϩ 2a2c2 ϩ 2b2c2 Ϫ a4 Ϫ b4 Ϫ c4 4 1 ϭ 2 3 1 a ϩ b 2 2 Ϫ c2 4 3 c2 Ϫ 1 a Ϫ b 2 2 4 4 1 ϭ 2 1 a ϩ b ϩ c 21 a ϩ b Ϫ c 2 1 c ϩ a Ϫ b 2 1 c Ϫ a ϩ b 2 4 From this point, the conclusion of the proof is the same as the trigonometric development found on page 765. a2 Ϫ d2 ϭ c2 Ϫ e2

a2 Ϫ b2 ϩ 2be Ϫ e2 ϭ c2 Ϫ e2

Precalculus—

A-88

APPENDIX B Proof Positive — A Selection of Proofs from Precalculus

Proof that u ؒ v ‫ ؍‬compv u ؋ |v|
z u–v u y

Consider the vectors in the figure shown, which form a triangle. Applying the Law of Cosines to this triangle yields:
v ␪

0 u Ϫ v 0 2 ϭ 0 u 0 2 ϩ 0 v 0 2 Ϫ 2 0 u 0 0 v 0 cos ␪

x

Using properties of the dot product (page 680), we can rewrite the left-hand side as follows: 0 u Ϫ v 0 2 ϭ 1u Ϫ v2 # 1u Ϫ v2 ϭu#uϪu#vϪv#uϩv#v ϭ 0u 0 2 Ϫ 2 u # v ϩ 0v 0 2

Substituting the last expression for 0 u Ϫ v 0 2 from the Law of Cosines gives 0 u 0 2 Ϫ 2 u # v ϩ 0 v 0 2 ϭ 0 u 0 2 ϩ 0 v 0 2 Ϫ 2 0 u 0 0 v 0 cos ␪ Ϫ2 u # v ϭ Ϫ2 0 u 0 0 v 0 cos ␪ u # v ϭ 0 u 0 0 v 0 cos ␪ ϭ 0 u 0 cos ␪ ϫ 0 v 0 .

Substituting compv u for 0 u 0 cos ␪ completes the proof: u # v ϭ compv u ϫ 0 v 0

Proof of DeMoivre’s Theorem: (cos x ؉ i sin x)n ‫ ؍‬cos(nx) ؉ i sin(nx)
For n 7 0, we proceed using mathematical induction. 1. Show the statement is true for n ϭ 1 (base case): 1 cos x ϩ i sin x 2 1 ϭ cos 1 1x 2 ϩ i sin 1 1x 2 cos x ϩ i sin x ϭ cos x ϩ i sin x ✓ 1 cos x ϩ i sin x 2 k ϭ cos 1 kx 2 ϩ i sin 1 kx 2

2. Assume the statement is true for n ϭ k (induction hypothesis): 3. Show the statement is true for n ϭ k ϩ 1:

1 cos x ϩ i sin x 2 k ϩ 1 ϭ 1 cos x ϩ i sin x 2 k 1 cos x ϩ i sin x 2 1 ϭ 3 cos 1 kx 2 ϩ i sin 1 kx 2 4 1 cos x ϩ i sin x 2 induction hypothesis ϭ cos 1 kx 2 cos x Ϫ sin 1 kx 2 sin x ϩ i 3 cos 1 kx 2 sin x ϩ sin 1 kx 2 cos x 4 F-O-I-L ϭ cos 3 1 k ϩ 1 2 x 4 ϩ i sin 3 1 k ϩ 1 2 x 4 ✓ sum/difference identities By the principle of mathematical induction, the statement is true for all positive integers. For n 6 0 (the theorem is obviously true for n ϭ 0), consider a positive integer m, where n ϭ Ϫm. 1 cos x ϩ i sin x 2 n ϭ 1 cos x ϩ i sin x 2 Ϫm 1 ϭ 1 cos x ϩ i sin x 2 m 1 ϭ cos 1 mx 2 ϩ i sin 1 mx 2 ϭ cos 1 mx 2 Ϫ i sin 1 mx 2

negative exponent property

DeMoivre’s theorem for n 7 0 multiply numerator and denom by cos 1 mx 2 Ϫ i sin 1 mx 2 and simplify even/odd identities n ϭ Ϫm

ϭ cos 1 Ϫmx 2 ϩ i sin 1 Ϫmx 2 ϭ cos 1 nx 2 ϩ i sin 1 nx 2

Precalculus—

Appendix C
More on Synthetic Division
As the name implies, synthetic division simulates the long division process, but in a condensed and more efficient form. It’s based on a few simple observations of long division, as noted in the division (x3 Ϫ 2x2 Ϫ 13x Ϫ 17 2 Ϭ 1 x Ϫ 5 2 shown in Figure AI.1.
Figure AI.1 Figure AI.2

x ϩ 3x ϩ 2 x Ϫ 5ͤ x3 Ϫ 2x2 Ϫ 13x Ϫ 17
2

Ϫ 1 x3 Ϫ 5x2 2
2 2

x Ϫ 5ͤ 1

1 Ϫ2 5 3

3 Ϫ13

2 Ϫ17

Ϫ 1 3x Ϫ 15x 2

3x Ϫ 13x



Ϫ 1 2x Ϫ 10 2 Ϫ7
remainder

2x Ϫ 17



15 2 10 Ϫ7
remainder

A careful observation reveals a great deal of repetition, as any term in red is a duplicate of the term above it. In addition, since the dividend and divisor must be written in decreasing order of degree, the variable part of each term is unnecessary as we can let the position of each coefficient indicate the degree of the term. In other words, we’ll agree that 1 Ϫ2 Ϫ13 Ϫ17 represents the polynomial 1x3 Ϫ 2x2 Ϫ 13x Ϫ 17. Finally, we know in advance that we’ll be subtracting each partial product, so we can “distribute the negative,” shown at each stage. Removing the repeated terms and variable factors, then distributing the negative to the remaining terms produces Figure AI.2. The entire process can now be condensed by vertically compressing the rows of the division so that a minimum of space is used (Figure AI.3).
Figure AI.3 1 3 Figure AI.4 2 Ϫ17
quotient dividend products sums

x Ϫ 5ͤ 1

Ϫ2 5 3

Ϫ13 15 2

x Ϫ 5ͤ 1


1 Ϫ2 5 3

3 Ϫ13 15 2

2 Ϫ17 10 Ϫ7

dividend products remainder

10 7

1

quotient

Further, if we include the lead coefficient in the bottom row (Figure AI.4), the coefficients in the top row (in blue) are duplicated and no longer necessary, since the quotient and remainder now appear in the last row. Finally, note all entries in the product row (in red) are five times the sum from the prior column. There is a direct connection between this multiplication by 5 and the divisor x Ϫ 5, and in fact, it is the zero of the divisor that is used in synthetic division (x ϭ 5 from x Ϫ 5 ϭ 0). A simple change in format makes this method of division easier to use, and highlights the location of the divisor and remainder (the blue brackets in Figure AI.5). Note the process begins by “dropping the lead coefficient into place” (shown in bold). The full process of synthetic division is shown in Figure AI.6 for the same exercise. A-89

Precalculus—

A-90

APPENDIX C More on Synthetic Division

5 1 ↓ drop lead coefficient into place → 1

divisor (use 5, not Ϫ5) →

Figure AI. 5 Ϫ2 Ϫ13

Ϫ17

← coefficients of dividend ← quotient and remainder appear in this row

We then multiply this coefficient by the “divisor,” place the result in the next column and add. In a sense, we “multiply in the diagonal direction,” and “add in the vertical direction.” Continue the process until the division is complete.
Figure AI. 6 Ϫ2 Ϫ13 Ϫ17 5 15 10 3 2 Ϫ7

5 1
multiply by 5

← coefficients of dividend ← quotient and remainder appear in this row

↓ 1

The result is x2 ϩ 3x ϩ 2 ϩ

Ϫ7 , read from the last row. xϪ5



Precalculus—

Appendix D
Reduced Row-Echelon Form and More on Matrices
Reduced Row-Echelon Form
A matrix is in reduced row-echelon form if it satisfies the following conditions: 1. All null rows (zeroes for all entries) occur at the bottom of the matrix. 2. The first non-zero entry of any row must be a 1. 3. For any two consecutive, nonzero rows, the leading 1 in the higher row is to the left of the 1 in the lower row. 4. Every column with a leading 1 has zeroes for all other entries in the column. Matrices A through D are in reduced row-echelon form. 0 A ϭ £0 0 1 0 0 0 0 0 0 1 0 0 0 1 5 3§ 2 1 B ϭ £0 0 0 1 0 0 0 0 5 3§ 0 1 C ϭ £0 0 0 1 0 0 3 0 5 Ϫ2 § 0 1 D ϭ £0 0 0 1 0 5 2 0 0 0§ 1

Where Gaussian elimination places a matrix in row-echelon form (satisfying the first three conditions), Gauss-Jordan elimination places a matrix in reduced row-echelon form. To obtain this form, continue applying row operations to the matrix until the fourth condition above is also satisfied. For a 3 ϫ 3 system having a unique solution, the diagonal entries of the coefficient matrix will be 1’s, with 0’s for all other entries. To illustrate, we’ll extend Example 3 from Section 9.2 until reduced row-echelon form is obtained. EXAMPLE 4B


Solving Systems Using the Augmented Matrix

2x ϩ y Ϫ 2z ϭ Ϫ7 Solve using Gauss-Jordan elimination: • x ϩ y ϩ z ϭ Ϫ1 Ϫ2y Ϫ z ϭ Ϫ3 2x ϩ y Ϫ 2z ϭ Ϫ7 • x ϩ y ϩ z ϭ Ϫ1 Ϫ2y Ϫ z ϭ Ϫ3
1 1 1 Ϫ1 £2 1 Ϫ2 Ϫ7 § 0 Ϫ2 Ϫ1 Ϫ3 1 1 1 Ϫ1 £0 1 4 5§ 0 Ϫ2 Ϫ1 Ϫ3 1 £0 0 1 1 0 1 Ϫ1 4 5§ 1 1
R 1 4 R2

x ϩ y ϩ z ϭ Ϫ1 • 2x ϩ y Ϫ 2z ϭ Ϫ7 Ϫ2y Ϫ z ϭ Ϫ3 1 1 1 Ϫ1 £ 0 Ϫ1 Ϫ4 Ϫ5 § 0 Ϫ2 Ϫ1 Ϫ3 1 £0 0 1 £0 0 1 1 0 1 Ϫ1 4 5§ 7 7

matrix form →

1 1 1 Ϫ1 £2 1 Ϫ2 Ϫ7 § 0 Ϫ2 Ϫ1 Ϫ3 1 1 1 Ϫ1 £0 1 4 5§ 0 Ϫ2 Ϫ1 Ϫ3 1 £0 0 1 £0 0 1 1 0 0 1 0 1 Ϫ1 4 5§ 1 1 0 Ϫ3 0 1§ 1 1

Ϫ2R1 ϩ R2 S R2

Ϫ1R2

2R2 ϩ R3 S R3

R3 S R3 7

ϪR2 ϩ R1 S R1

0 Ϫ3 Ϫ6 1 4 5§ 0 1 1

3R3 ϩ R1 S R1 Ϫ4R3 ϩ R2 S R2

The final matrix is in reduced row-echelon form with solution (Ϫ3, 1, 1) just as in Section 6.1.
A-91

Precalculus—

A-92

APPENDIX D Reduced Row-Echelon Form and More on Matric

The Determinant of a General Matrix
To compute the determinant of a general square matrix, we introduce the idea of a cofactor. For an n ϫ n matrix A, Aij ϭ 1 Ϫ1 2 i ϩ j Ϳ Mij Ϳ is the cofactor of matrix element aij, where ͿMijͿ represents the determinant of the corresponding minor matrix. Note that i ϩ j is the sum of the row and column of the entry, and if this sum is even, 1 Ϫ1 2 i ϩ j ϭ 1, while if the sum is odd, 1 Ϫ1 2 i ϩ j ϭ Ϫ1 (this is how the sign table for a 3 ϫ 3 determinant was generated). To compute the determinant of an n ϫ n matrix, multiply each element in any row or column by its cofactor and add. The result is a tierlike process in which the determinant of a larger matrix requires computing the determinant of smaller matrices. In the case of a 4 ϫ 4 matrix, each of the minor matrices will be size 3 ϫ 3, whose determinant then requires the computation of other 2 ϫ 2 determinants. In the following illustration, two of the entries in the first row are zero for convenience. For Ϫ2 1 Aϭ ≥ 3 0 we have: det 1 A 2 ϭ Ϫ2 # 1 Ϫ1 2
1ϩ1

0 2 Ϫ1 Ϫ3 0 4 2

3 0 4 2

0 Ϫ2 ¥, 1 1 2 Ϫ1 Ϫ3 Ϫ2 1† 1

2 † Ϫ1 Ϫ3

Ϫ2 1 1 ϩ 3 1 † ϩ 1 3 2 # 1 Ϫ1 2 †3 1 0

Computing the first 3 ϫ 3 determinant gives Ϫ16, the second 3 ϫ 3 determinant is 14. This gives: det 1 A 2 ϭ Ϫ2 1 Ϫ16 2 ϩ 3 1 14 2 ϭ 74

Precalculus—

Appendix E
The Equation of a Conic
The Equation of an Ellipse
In Section 10.2, the equation 2 1 x ϩ c 2 2 ϩ y2 ϩ 2 1 x Ϫ c 2 2 ϩ y2 ϭ 2a was developed using the distance formula and the definition of an ellipse. To find the standard form of the equation, we treat this result as a radical equation, isolating one of the radicals and squaring both sides. 2 1 x ϩ c 2 2 ϩ y2 ϭ 2a Ϫ 2 1 x Ϫ c 2 2 ϩ y2
2 2 2 2

1 x ϩ c 2 ϩ y ϭ 4a Ϫ 4a 2 1 x Ϫ c 2 ϩ y ϩ 1 x Ϫ c 2 ϩ y
2 2

isolate one radical square both sides

2

We continue by simplifying the equation, isolating the remaining radical, and squaring again. x2 ϩ 2cx ϩ c2 ϩ y2 ϭ 4a2 Ϫ 4a 2 1 x Ϫ c 2 2 ϩ y2 ϩ x2 Ϫ 2cx ϩ c2 ϩ y2
2

a 2 1 x Ϫ c 2 ϩ y ϭ a Ϫ cx a2 3 1 x Ϫ c 2 2 ϩ y2 4 ϭ a4 Ϫ 2a2cx ϩ c2x2 a2x2 Ϫ 2a2cx ϩ a2c2 ϩ a2y2 ϭ a4 Ϫ 2a2cx ϩ c2x2 a2x2 Ϫ c2x2 ϩ a2y2 ϭ a4 Ϫ a2c2 x2 1 a2 Ϫ c2 2 ϩ a2y2 ϭ a2 1 a2 Ϫ c2 2 y2 x2 ϩ ϭ1 a2 a2 Ϫ c2
2 2

4cx ϭ 4a2 Ϫ 4a 2 1 x Ϫ c 2 2 ϩ y2

simplify isolate radical; divide by 4 square both sides expand and distribute a 2 on left add 2a 2cx and rewrite equation factor divide by a 2 1 a 2 Ϫ c 2 2

Since a 7 c, we know a2 7 c2 and a2 Ϫ c2 7 0. For convenience, we let b2 ϭ a2 Ϫ c2 and it also follows that a2 7 b2 and a 7 b (since c 7 0). Substituting b2 for a2 Ϫ c2 we obtain the standard form of the equation of an ellipse (major axis y2 x2 horizontal, since we stipulated a 7 b): 2 ϩ 2 ϭ 1. Note once again the x-intercepts a b are (Ϯa, 0), while the y-intercepts are (0, Ϯb).

The Equation of a Hyperbola

In Section 10.3, the equation 2 1 x ϩ c 2 2 ϩ y2 Ϫ 2 1 x Ϫ c 2 2 ϩ y2 ϭ 2a was developed using the distance formula and the definition of a hyperbola. To find the standard form of this equation, we apply the same procedures as before.

isolate one radical 2 1 x ϩ c 2 2 ϩ y2 ϭ 2a ϩ 2 1 x Ϫ c 2 2 ϩ y2 1 x ϩ c 2 2 ϩ y2 ϭ 4a2 ϩ 4a 2 1 x Ϫ c 2 2 ϩ y2 ϩ 1 x Ϫ c 2 2 ϩ y2 square both sides expand x2 ϩ 2cx ϩ c2 ϩ y2 ϭ 4a2 ϩ 4a 2 1 x Ϫ c 2 2 ϩ y2 ϩ x2 Ϫ 2cx ϩ c2 ϩ y2 binomials simplify 4cx ϭ 4a2 ϩ 4a 2 1 x Ϫ c 2 2 ϩ y2 cx Ϫ a2 ϭ a 2 1 x Ϫ c 2 2 ϩ y2 isolate radical; divide by 4 square both sides c2x2 Ϫ 2a2cx ϩ a4 ϭ a2 3 1 x Ϫ c 2 2 ϩ y2 4 2 2 2 4 2 2 2 2 2 2 2 c x Ϫ 2a cx ϩ a ϭ a x Ϫ 2a cx ϩ a c ϩ a y expand and distribute a 2 on the right add 2a 2cx and rewrite equation c2x2 Ϫ a2x2 Ϫ a2y2 ϭ a2c2 Ϫ a4 2 2 2 2 2 2 2 2 factor x 1c Ϫ a 2 Ϫ a y ϭ a 1c Ϫ a 2 2 2 y x divide by a 2 1 c 2 Ϫ a 2 2 Ϫ 2 ϭ1 A-93 a2 c Ϫ a2

Precalculus—

A-94

APPENDIX E The Equation of a Conic

From the definition of a hyperbola we have 0 6 a 6 c, showing c2 7 a2 and c2 Ϫ a2 7 0. For convenience, we let b2 ϭ c2 Ϫ a2 and substitute to obtain the stany2 x2 dard form of the equation of a hyperbola (transverse axis horizontal): 2 Ϫ 2 ϭ 1. a b Note the x-intercepts are (0, Ϯa) and there are no y-intercepts.

The Asymptotes of a Central Hyperbola
From our work in Section 10.3, a central hyperbola with a horizontal axis will have b asymptotes at y ϭ Ϯ x. To understand why, recall that for asymptotic behavior we a investigate what happens to the relation for large values of x, meaning as ͿxͿ S q . y2 x2 Starting with 2 Ϫ 2 ϭ 1, we have a b b2x2 Ϫ a2y2 ϭ a2b2 ay ϭbx Ϫab
2 2 2 2 2 2 2

clear denominators isolate term with y factor out b 2x 2 from right side

a2y2 ϭ b2x2 a 1 Ϫ

a b x2 b2 a2 y2 ϭ 2 x2 a 1 Ϫ 2 b a x b a2 yϭϮ x 1Ϫ 2 a B x

divide by a 2

square root both sides

As ͿxͿ S q ,

b a2 2 S 0, and we find that for large values of x, y Ϸ Ϯ a x. x

Precalculus—

Appendix F
Families of Polar Curves
Circles and Spiral Curves
r a a

Circle rϭa 2

Circle Circle r ϭ a cos ␪ r ϭ a sin ␪ a represents the diameter of the circle

Spiral r ϭ k␪

Roses: r ‫ ؍‬a sin(n␪) (illustrated here) and r ‫ ؍‬a cos(n␪)
a a a a

Four-petal rose r ϭ a sin(2␪)

Five-petal rose Eight-petal rose Three-petal rose r ϭ a sin(5␪) r ϭ a sin(4␪) r ϭ a sin(3␪) If n is odd S there are n petals, if n is even S there are 2n petals. |a| represents the maximum distance from the origin (the radius of a circumscribed circle)

Limaçons: r ‫ ؍‬a ؉ b sin ␪ (illustrated here) and r ‫ ؍‬a ؉ b cos ␪
|a| ϩ |b| |b| Ϫ |a| a |a| ϩ |b| a |a| Ϫ |b| |a| ϩ |b| a |a| Ϫ |b| a |a| ϩ |b|

Cardioid Apple Limaçon Eye-ball (limaçon where |a| ϭ |b|) (limaçon where |a| Ͼ |b|) (limaçon where |a| Ն 2|b|) (inner loop if |a| Ͻ |b|) r ϭ a ϩ b sin ␪ r ϭ a ϩ b sin ␪ r ϭ a ϩ b sin ␪ r ϭ a ϩ b sin ␪ |a| ϩ |b| represents the maximum distance from the origin (along the axis of symmetry)

Lemniscates: r 2 ‫ ؍‬a2sin(2␪) and r 2 ‫ ؍‬a2cos (2␪)
a a

Lemniscate Lemniscate r2 ϭ a2 cos(2␪) r2 ϭ a2 sin(2␪) a represents the maximum distance from the origin (the radius of a circumscribed circle)

A-95

College Algebra G&M—

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