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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution and Answer Guide GUSTAFSON/HUGHES, C OLLEGE ALGEBRA 2023, 9780357723654; C HAPTER R: A REVIEW OF BASIC ALGEBRA

TABLE OF CONTENTS End of Section Exercise Solutions ....................................................................................... 1 Exercises R.1 ..................................................................................................................................1 Exercises R.2 .............................................................................................................................. 19 Exercises R.3 ..............................................................................................................................43 Exercises R.4 ............................................................................................................................. 66 Exercises R.5 ..............................................................................................................................92 Exercises R.6 ............................................................................................................................. 115 Chapter Review Solutions................................................................................................ 144 Chapter Test Solutions .................................................................................................... 167 Group Activity Solutions ...................................................................................................175

END OF SECTION EXERCISE SOLUTIONS EXERCISES R.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Simplify. 10 – 20 Solution

10  20  10

2. Simplify. 5 – (–10) Solution

5   10   15

3. Division by what number is undefined?

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution Division by 0 is undefined 4. Given

5 0 0 , , and . Which one is equivalent to 0? 0 5 0

Solution 0 0 5 5. Write the inequality symbol for greater than. Solution The inequality symbol for greater than is > 6. Write the math symbol for infinity. Solution The math symbol for infinity is Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A ______ is a collection of objects. Solution set 8. If every member of one set B is also a member of a second set A, then B is called a ________ of A. Solution subset 9. If A and B are two sets, the set that contains all members that are in sets A and B or both is called the ________ of A and B. Solution union 10. If A and B are two sets, the set that contains all members that are in both sets is called the ________of A and B. Solution intersection 11. A real number is any number that can be expressed as a __________ . Solution decimal

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

12. A __________ is a letter that is used to represent a number. Solution variable 13. The smallest prime number is __________. Solution 2 14. All integers that are exactly divisible by 2 are called _________integers. Solution even 15. Natural numbers greater than 1 that are not prime are called __________numbers. Solution composite 16. Fractions such as 23 , 82 , and  97 are called _________ numbers. Solution rational 17. Irrational numbers are ________ that don’t terminate and don’t repeat. Solution decimals 18. The symbol ________is read as “is less than or equal to.” Solution  19. On a number line, the __________ numbers are to the left of 0. Solution negative 20. The only integer that is neither positive nor negative is _________. Solution 0 21. The Associative Property of Addition states that  x  y   z  _________. Solution

x   y  z

22. The Commutative Property of Multiplication states that xy = __________. Solution yx

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

23. Use the Distributive Property to complete the statement: 5(m + 2) = ___________. Solution

5m 5  2

24. The statement (m + n) p = p(m + n) illustrates the __________ Property of ________. Solution Commutative, Multiplication 25. The graph of an __________ is a portion of a number line. Solution interval 26. The graph of an open interval has _________ endpoints. Solution no 27. The graph of a closed interval has __________ endpoints. Solution two 28. The graph of a _________ interval has one endpoint. Solution half-open 29. Except for 0, the absolute value of every number is _________. Solution positive 30. The __________ between two distinct points on a number line is always positive. Solution distance Let N = the set of natural numbers W = the set of whole numbers Z = the set of integers Q = the set of rational numbers R = the set of real numbers Determine whether each statement is true or false. Read the symbol  , as “is a subset of.” 31. N  W Solution Every natural number is a whole number, so N  W. TRUE

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

32. Q  R Solution Every rational number is a real number, so Q  R. TRUE 33. Q  N Solution The rational number 21 is not a natural number, so Q | N. FALSE 34. Z  Q Solution Every integer is a rational number, so Z  Q. TRUE 35. W  Z Solution Every whole number is an integer, so W  Z. TRUE 36. R  Z Solution The real number

2 is not an integer, so R | Z FALSE

Practice

,

f , e , c , a =

, B 

d n a



g , f , e , d =

e , d , c , b , a =

t e L



 . Find each set.

37. A  B Solution A  B  {a, b, c, d, e, f, g} 38. A  B Solution A  B  {d, e}

39. A  C Solution A  C  {a, c, e}

40. B  C Solution B  C  {a, c, d, e, f, g}

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Determine whether the decimal form of each fraction terminates or repeats. 7 41. 16 Solution 7  0.4375; terminates 16 42.

5 8

Solution 5  0.625; terminates 8 43.

5 11 Solution

5 = 0.454545...; repeats 11 44.

7 12 Solution

7  0.583333...; repeats 12 Consider the following set:

7 , 6 , 5 7 . 2 , 2

, 2 , 1 , 0 , 23

, 4 , 5

  



45. Which numbers are natural numbers? Solution natural: 1, 2, 6, 7 46. Which numbers are whole numbers? Solution whole: 0, 1, 2, 6, 7 47. Which numbers are integers? Solution integers: –5, –4, 0, 1, 2, 6, 7 48. Which numbers are rational numbers? Solution rational: 5,  4,  23 , 0, 1, 2, 2.75, 6, 7

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

49. Which numbers are irrational numbers? Solution irrational: 2 50. Which numbers are prime numbers? Solution prime: 2,7 51. Which numbers are composite numbers? Solution composite: 6 52. Which numbers are even integers? Solution even: –4, 0, 2, 6 53. Which numbers are odd integers? Solution odd: –5, 1, 7 54. Which numbers are negative numbers? Solution negative: 5, 4,  23 Graph each subset of the real numbers on a number line. 55. The natural numbers between 1 and 5 Solution

56. The composite numbers less than 10 Solution

57. The prime numbers between 10 and 20 Solution

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

58. The integers from –2 to 4 Solution

59. The integers between –5 and 0 Solution

60. The even integers between –9 and –1 Solution

61. The odd integers between –6 and 4 Solution

62. 0.7, 1.75, and 3 87 Solution

Write each inequality in interval notation and graph the interval. 63. x  2 Solution

x  2   2,  

64. x  4 Solution

x  4   , 4 

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

65. 0  x  5 Solution

0  x  5   0,5

66. 2  x  3 Solution

2  x  3   2, 3

67. x  4 Solution

x  4   4,  

68. x  3 Solution

x  3    , 3

69. 2  x  2 Solution  2  x  2  [  2, 2]

70. 4  x  1 Solution  4  x  1  (  4, 1]

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

71. x  5 Solution x  5  ( ,5]

72. x  1 Solution x  1  [1, )

73. 5  x  0 Solution 5  x  0  ( 5, 0]

74. 3  x  4 Solution 3  x  4  [ 3, 4)

75. 2  x  3 Solution 2  x  3  [ 2, 3]

76. 4  x  4 Solution 4  x  4  [ 4, 4]

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

77. 6  x  2 Solution 6  x  2  2  x  6  [2, 6]

78. 3  x  2 Solution 3  x  2  2  x  3  [ 2, 3]

Write each pair of inequalities as the intersection of two intervals and graph the result. 79. x  5 and x  4 Solution

x  5 and x  4   5,      , 4 

 5,  

 , 4  ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

 5,     , 4  80. x  3 and x  6 Solution

x  3 and x  6  [3, )    , 6 

  3,  

 , 6 

  3,     , 6 

81. x  8 and x  3 Solution x  8 and x  3  [8, )  (, 3]

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

  8,  

 ,  3 __________________________________  8,     ,  3 

82. x  1 and x  7 Solution

x  1 and x  7   1,    ( , 7]

 1,    , 7  ________________________________

 1,     , 7  Write each inequality as the union of two intervals and graph the result. 83. x  2 or x  2 Solution

x  2 or x  2   , 2   2,  

84. x  5 or x  0 Solution

x  5 or x  0  ( , 5]   0,  

85. x  1 or x  3 Solution x  1 or x  3  ( , 1]  [3, )

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

86. x  3 or x  2 Solution

x  3 or x  2   , 3  [2,)

Write each expression without using absolute value symbols. 87. 13 Solution Since 13  0, 13  13. 88. 17 Solution





Since 17  0, 17   17  17. 89. 0 Solution Since 0  0, 0  0. 90.  63 Solution Since 63  0, 63  63.  63    63  63

91.  8 Solution

Since  8  0, 8    8   8.  8    8   8

92. 25 Solution





Since 25  0, 25   25  25.

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

93.  32 Solution Since 32  0, 32  32.  32    32   32

94.  6 Solution

Since  6  0, 6    6   6.  6    6   6

95.   5 Solution Since   5  0,

  5     5     5  5   .

96. 8   Solution

Since 8    0, 8    8   . 97.    Solution

   0  0 98. 2  Solution

Since 2  0, 2  2 . 99. x  1 and x  2 Solution

If x  2, then x  1  0. Then x  1  x  1. 100. x  1 and x  2 Solution

If x  2, then x  1  0. Then x  1    x  1 .

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

101. x  4 and x  0 Solution

If x  0, then x  4  0. Then x  4    x  4  .

102. x  7 and x  10 Solution

x  10, then x  7  0. Then x  7  x  7.

103.

x 7 x 7

and x >7

Solution Since x  7 is positive, x  7 is its own absolute value. x 7 x 7

104.

x 8 x 8



x 7 1 x 7

and x  8

Solution

Since x  8 is negative, x  8    x  8

x 8 x 8



x 8

  x  8

 1

Find the distance between each pair of points on the number line. 105. 3 and 8

Solution

distance  8  3  5  5 106. –5 and 12

Solution

distance  12   5   17  17

107. –8 and –3

Solution

distance  3   8   5  5

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

108. 6 and –20

Solution

distance  20  6  26  26 109. –100 and 50

Solution d  ba d  20  6  26 110. –200 and –50

Solution d  ba

d  50   100  50  100  150

Fix It In Exercises 111 and 112, identify the error made and fix it. 111. Let A = {s, n, i, c, k, e, r, s} and B = {t, w, i, x}, find A ∪ B

Solution A  {s, n, i , c, k, e, r , s} and B  {t, w, i, x} A  B  {s, n, i , c, k, e, r , s, t, w, x} 112. Graph the inequality x < 1 on a number line.

Solution Graph x  1

Applications 113. What subset of the real numbers would you use to describe the populations of Memphis and Miami?

Solution Since population must be positive and never has a fractional part, the set of natural numbers should be used. 114. What subset of the real numbers would you use to describe the subdivisions of an inch on a ruler?

Solution Since the subdivisions on a ruler are measured in fractions of an inch, the set of rational numbers should be used.

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

115. What subset of the real numbers would you use to report temperatures in London and Lisbon?

Solution Since temperatures are usually reported without fractional parts and may be either positive or negative (or zero), the set of integers should be used. 116. What subset of the real numbers would you use to describe the prices of hoodies at Old Navy?

Solution Since the financial condition of a business is usually described in terms of dollars and cents (fractional parts of a dollar), the set of rational numbers should be used. 117. Temperature The average low temperature in International Falls, Minnesota, in January is –7°F. The average high temperature is 15°F. Determine the degrees difference between the average high and the average low.

Solution change  7  15  22  22 The change is 22° F. 118. Temperature Harbin, China, is one of the world’s coldest cities and known for its ice and snow festivals. In February, the average nightly low temperature is –20°C and the average daily high temperature is –7°C. What is the temperature drop from day to night?

Solution

change  20   7   13  13 The change is 13° C.

Discovery and Writing 119. Explain why – x could be positive.

Solution –x will represent a positive number if x itself is negative. For instance, if x = – 3, then –x = – (–3) = 3, which is a positive number. 120. Explain why every integer is a rational number.

Solution Every integer is a rational number because every integer is equal to itself over 1. 121. Is the statement ab  a  b always true? Explain.

Solution The statement is always true. 122. Is the statement

a a   b  0 always true? Explain. b b

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution The statement is always true. 123. Is the statement a  b  a  b always true? Explain.

Solution The statement is not always true. (For example, let a  5 and b  2.) 124. Explain why it is incorrect to write a  b  c if a  b and b  c.

Solution The statement a  b  c could be interpreted to mean that a  c, when this is not necessarily true. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 125. There are six integers between –3 and 3.

Solution False. There are 5 integers: –2, –1, 0, 1, and 2. 126.

725 is a rational number because 725 and 0 are integers. 0 Solution 725 False. is not a rational number because the denominator cannot equal 0. 0

127. ∞ is a real number.

Solution False. ∞ is not a number at all. 128. a  b  b  a

Solution True.  13  129.   5,  , 3,  4  Solution True. (You cannot find an element in the 1st set that is not in the 2nd set.) 130.   

Solution True. (You cannot find an element in the 1st set that is not in the 2nd set.)

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra





131. There are six subsets of 11, 22, 33 .

Solution False. There are eight subsets. 132. A set is always a subset of itself.

Solution True.

EXERCISES R.2 Getting Ready

Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Match each expression with the proper description given below. 5

x7 3 4 5 2 7  x4  3 8 x y x  2  x  x x2 x 



a. b. c. d. e.

 

Product of exponential expressions with the same base Quotient of exponential expressions with the same base Power of an exponential expression Power of a product Power of a quotient

Solution a. Product of exponential expressions with the same base is x 3  x 8 x7 b. Quotient of exponential expressions with the same base is 2 x

 

c. Power of an exponential expression is x 2



d. Power of a product is x 3 y 4  x4  e. Power of a quotient is  2  x 



2. Simplify the expression. x  x  x  x  x  x

Solution x  x  x  x  x  x  x6 3. Simplify the expression.

yyyyyy yyy

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution yyyyyy y6 63  3  y   y3 yyy y

    x  x  x   x

4. Fill in the box. x 2

Solution

 x    x  x  x   x 2

5. If x = –5 what is x3 + x?

Solution

 5   5  125   5  130 3

  x x

6. These look alike. Match each with its correct simplification x 5  x 5 x 5 a. b. c.

2x 5 x 10 x 25

Solution a. 2x 5  x 5  x 5 b.

x 10  x 5  x 5

x 25  x 5

 

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. Each quantity in a product is called a __________ of the product.

Solution factor 8. A _________ number exponent tells how many times a base is used as a factor.

Solution natural 9. In the expression (2x)3, ___________ is the exponent and _________ is the base.

Solution 3, 2x 10. The expression xn is called an __________ expression.

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution exponential 11. A number is in _________ notation when it is written in the form N  10n , where 1  N  10 and n is an __________ .

Solution scientific, integer 12. Unless __________ indicate otherwise, _________are performed before additions.

Complete each exponent rule. Assume x 

. 0

Solution Answers may vary.

13. x m x n  ________

Solution x m x n  x mn 14.

 x   ________ m

Solution

x   x m

15.

 xy   _________ n

Solution

 xy   x y n

16.

xm  __________ xn

Solution xm  x m n n x 17. x 0  _________

Solution x0  1

18. x  n  __________

Solution 1 x n  n x

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Practice Write each number or expression without using exponents. 19. 132

Solution 132  13  13  169

20. 103

Solution

103  10  10  10  1,000 21.  5 2

Solution  52   1  5  5   25

22.  5 

Solution

 5   5 5  25 2

23. 4 x 3

Solution 4x3  4  x  x  x 24.  4 x 

Solution

 4 x    4 x  4 x  4 x  3

25.  5x 

Solution

 5 x    5 x  5 x  5 x  5 x  4

26. 6x 2

Solution 6 x 2  6  x  x

27. 8x 4

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 8 x 4   8  x  x  x  x

28.  8x 

Solution

 8x    8 x  8x  8 x  8 x  4

Write each expression using exponents. 29. 7 xxx

Solution 7 xxx  7 x 3

30. 8 yyyy

Solution

8 yyyy  8 y 4 31.

  x   x  Solution

  x   x    1 1 x  x 2

   

32. 2a 2a 2a

Solution

 2a  2a  2a   2  2  2  a  8a 3

  

33. 3t 3t 3t



Solution

 3t  3t  3t    3 3 3 t  27t 3

    

34.  2b 2b 2b 2b

Solution

  2b  2b  2b  2b   1  2  2  2  2  b4  16b4

35. xxxyy

Solution

xxxyy  x3 y 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

36. aaabbbb

Solution aaabbbb  a 3 b 4

Use a calculator to simplify each expression. 37. 2.23

Solution 2.23  10.648

38. 7.14

Solution 7.14  2541.1681

39.  0.54

Solution 0.54  0.0625

40.  0.2

Solution

 0.2   0.0016 4

Simplify each expression. Write all answers without using negative exponents. Assume that all variables are restricted to those numbers for which the expression is defined. 41. x 2 x 3

Solution x 2 x 3  x 23  x 5 3

42. y y

Solution

y 3 y 4  y 3 4  y 7

 

43. z 2

Solution

z   z 44.  t  2

23

 z6

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

t   t 6

45.

y y  5

67

 t 42

Solution

y y   y   y 5





46. a3a6 a 4

Solution

a a  a  a a  a 3

  z 

47. z 2

Solution

z  z   z z  z 2

  t 

48. t 3

Solution

t  t   t t  t 3

  a 

49. a2

12 10

Solution

a  a   a a  a 2

  a 

50. a2

Solution

a  a   a a  a 2

51.

3 x 

Solution

 3 x   3 x  27 x 3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

52.  2 y 

Solution

 2 y    2 y  16 y 4



53. x 2 y



Solution

x y  x  y  x y 3



54. x 3 z 4



Solution

x z   x  z   x z 3

 a2  55.    b

Solution

  a

3 a2  a2     b3  b 

 x  56.  3  y 

Solution 4

 x  x4 x4   3  4 y 12 y  y3

 

57.   x 

Solution

x   1 0

58. 4 x 0

Solution 4x0  4  1  4

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

59.  4 x 

Solution

4x   1 0

60. 2x 0

Solution 2 x 0  2  1  2

61. z 4

Solution 1 z 4  4 z 62.

1 t 2

Solution 1  t2 2 t 2 3 63. y y

Solution

y 2 y 3  y 5 

64.  m 2 m3

Solution  m 2 m 3   m 1   m



65. x 3 x 4



2

Solution

x x   x   x 3

4



66. y 2 y 3

2



1

2

4

Solution

y y   y   y 2

4



1 y4

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

67.

x7 x3

Solution x7  x 7 3  x 4 x3 68.

r5 r2

Solution r5  r 52  r 3 r2 69.

a 21 a 17

Solution a 21  a 21 17  a 4 a 17 70.

t 13 t4

Solution t 13  t 13  4  t 9 t4

x  71. 2

x2 x

Solution

x   x  x 2

x x

72.

4 3

 x1  x

s9 s 3

s  2

Solution s9 s 3 s 12   s 12 4  s8 2 4 2 s s

 

 m3  73.  2  n 

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

   

3 m3  m3  m9   2   3 n6 n  n2

 t4  74.  3  t 

Solution 3

3 3  t4  4 3  t1  t3  3   t t 

a  75. 3



  

2

aa2

Solution

a   a 3

76.

2

6

 a 6  3  a 9 

1 a9

r 9r 3

r  2

Solution r 9 r 3 r6 6   6   r  r 12 3 6 2 r r

 

 a 3  77.  1  b 

4

Solution

 a3   1  b   t 4  78.  3  t 

4

a   a  b  b 3

1

4

2

Solution

 t 4   3  t 

2

t   t  t  t  t 4 3

2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 r 4 r 6  79.  3 3  r r 

Solution 2

2  r 4 r 6   r 2  2  r 4  r 3 r 3    r 0   r     1  4 r

x x  80. x x  3

5

 

3

Solution

x x   x   x  x x x  x  x 3

5

1

3

 x 5 y 2  81.  3 2  x y 

3

2

3

11



1 x 11

Solution 4

 x 5 y 2   x5 x3   x8  x 32  3 2    2 2    4   16 y x y  y y  y   x 7 y 5  82.  7 4  x y 

Solution 3

 x 7 y 5   y5 y4   y9  y 27  7 4    7 7    14   42 x x y  x x  x   5 x 3 y 2  83.  2 3    3x y 

2

Solution  5 x 3 y 2   2 3    3x y 

2

 3 x 2 y 3   3x 2 x 3 y 2   3x5  9 x 10   3 2         3   5y  25 y 2  5x y   5y   

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 3 x 2 y 5  84.  2 6   2x y 

3

Solution  3 x 2 y 5   2 x 2 y 6     3 x 5 y 3  85.  5 3   6x y 

3

 2 x 2 y 6   2y5   2  8      2 2 6    4   2 5  12 3 3 x y 3 x x y 3 x y 27 x y      

2

Solution  3 x 5 y 3   5 3   6x y 

2

 6 x 5 y 3   2y3 y3   2y6  4 y 12   5 3    5 5    10   20 x  3x y   1x x   x 

 12 x 4 y 3 z 5  86.  4 3 5    4x y z 

Solution 3

 12 x 4 y 3 z 5   3y3 y3   3y6  27 y 18          4 3 5   1x 4 x 4 z 5 z 5   x 8 z 10  x 24 z 30  4x y z     

8 z y  87. 5 y z  5 yz  2

2

3

1

2

1

Solution

8 z y  8 z y 64z y 64z z 64z     5 y z 5 y z 5 y z 25 y y 25 y  5 y z  5 yz  2

2

3

1

2

1

 m n p   mn p  88.  mn p   mn p 2

2

4

2

6

1

4

1

Solution

 m n p   mn p   m n p  m n p  m n p  m m p p  m p nn n  mn p   mn p m n p  m n p m n p 2

2

4

2

1

6

8

14

4

12

1

2

1

5

13

Simplify each expression. 5 62   9  5   89.   2 4 2  3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution



5[62   9  5 ] 4  2  3



5[36  4] 4  1



5[40] 4  1



200  50 4

6[3   4  7  ] 2

90.



5 2  4 2



Solution 6[3   4  7  ] 2



5 2  4



5  2  16 

and z 



6[3  9]

5  14 



6[ 6] 36 18   70 70 35





, 0

, 2

Let x  

6[3   3  ]

and evaluate each expression.

91. x 2

Solution x 2   2   4 2

92.  x 2

Solution  x 2    2   1  4  4 2

93. x 3

Solution x 3   2   8 3

94.  x 3

Solution  x 3    2   1   8   8 3

95.   xz 

Solution

  xz   [1   2  3]  6  216 3

96.  xz

Solution

 xz 3  1   2  33  2  27  54

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

97.



 x2 z3 z y 2



Solution



 x2z3

  [ 2  3 ]  [4  27]  108  12 2

z2  y 2

98.

90

32  02



z2 x2  y 2



x z

Solution



z2 x2  y 2

  3 [ 2  0 ]  9 4  0  9  4  36   3 2

 8 3

 2  3

x3z

24

2 3 99. 5x  3 y z

Solution 5 x 2  3 y 3 z  5  2   3  0   3   5  4   3  0  3  20  0  20 2

100. 3  x  z   2  y  z  2

Solution 3  x  z   2  y  z   3  2  3   2  0  3   3  5   2  3   3  25   2  27  2

 75   54   21

101.

 3 x 3 z 2 6 x 2 z 3

Solution  3 x 3 z 2 1z 3 z 3 3 3 3       2 3 2 3 2 5 5 6x z 2x x z 2x 2  32  64 64 2  2 

 5x z  102. 2

3

5 xz 2

Solution

 5x z   25x z 2

3

5 xz

2

5 xz

6

2



5x 4 2

xz z



5x 3 z



5  2  4



5  8  81



40 40  81 81

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Express each number in scientific notation. 103. 372,000

Solution

372,000  3.72  105 104. 89,500

Solution

89,500  8.95  104 105. –177,000,000

Solution

177,000,000  1.77  108 106. –23,470,000,000

Solution

23,470,000,000  2.347  1010 107. 0.007

Solution 0.007  7  10 3

108. 0.00052

Solution 0.00052  5.2  10 4

109. –0.000000693

Solution  0.000000693  6.93  10 7

110. –0.000000089

Solution 0.000000089  8.9  10 8

111. one trillion

Solution

1,000,000,000,000  1  1012 112. one millionth

Solution 0.000001  1  10 6

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Express each number in standard notation. 113. 9.37  105

Solution

9.37  105  937,000 114. 4.26  109

Solution

4.26  109  4, 260,000,000 115. 2.21 × 10-5

Solution 2.21  10 5  0.0000221

116. 2.774  10 2

Solution 2.774  10 2  0.02774

117. 0.00032  104

Solution 0.00032  104  3.2

118. 9, 300.0  104

Solution

9,300.0  104  0.93 119.  3.2  10 3

Solution 3.2  10 3  0.0032

120. 7.25  103

Solution

7.25  103  7,250 Use the method of Example 9 to do each calculation. Write all answers in scientific notation. 121.

65, 000  45, 000  250, 000

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

65,000 45,000  6.5  10  4.5  10   6.5 4.5  10 4

4  4 5

250,000

2.5  105

2.5

 11.7  103  1.17  101  103  1.17  104

122.

0.000000045 0.00000012  45, 000, 000

Solution

0.0000000450.00000012   4.5  10  1.2  10    4.5 1.2  10    8

7

8  7  7

4.5  107

45, 000, 000

123.

4.5  1.2  1022

0.00000035 170, 000  0.00000085

Solution

0.00000035 170,000   3.5  10  1.7  10    3.5 1.7   10  7

0.00000085

8.5  10

7

7  5   7 

8.5  0.7  105

 7  101  105  7  104 124.

0.0000000144  12, 000  600, 000

Solution

0.0000000144  12,000   1.44  10  1.2  10    1.44  1.2  10  8

8  4  5

600,000

6  10

6  0.288  109

 2.88  101  109  2.88  1010 125.

 45,000, 000,000 212, 000 0.00018

Solution

 45, 000,000,000 212, 000    4.5  10  2.12  10    4.5 2.12  10 10

0.00018

126.

1.8  10

4

10  5   4 

1.8  5.3  1019

0.00000000275  4750  500, 000, 000, 000

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

0.00000000275 4, 750    2.75  10  4.75  10    2.75 4.75  10  9

9  3  11

500, 000, 000, 000

5  1011

5  2.6125  1017

Fix It In Exercise 127, identify the step the first error is made and fix it. 3

 3 x 3 y 2  127. Use the properties of exponents to simplify the expression   . 2  x y 

Solution Step 3 was incorrect  x2 y  Step 1:  3 2   3x y 

x y 2

Step 2:

Step 3:

Step 4:

 3x y  3

2

x6 y 3

27 x 9 y 6 y9 27 x 3

In Exercise 128, identify the error made and fix it. 128. 123,456,789 written in scientific notation is 1.23456789

10–8.

Solution

123, 456, 789  1.23456789  108 Applications Use scientific notation to compute each answer. Write all answers in scientific notation. 129. Speed of sound The speed of sound in air is 3.31 104 centimeters per second. Compute the speed of sound in meters per minute.

Solution

3.31  104 cm/sec 







3.31  104 6  101 3.31  104 cm 1m 60 sec    m/min 1 sec 100 cm 1 min 1  102  3.316  104 12 m/min = 1 = 19.86  103 m/min = 1.986  104 m/min

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

130. Volume of a box Calculate the volume of a box that has dimensions of 6000 by 9700 by 4700 millimeters.

Solution





V  lwh   6, 000 mm 9, 700 mm 4, 700 mm  6  103 9.7  103

4.7  10  mm 3

  6  9.7  4.7   103  3  3 mm3  273.54  109 mm3  2.7354  1011 mm3

131. Mass of a proton The mass of one proton is 0.00000000000000000000000167248 gram. Find the mass of one billion protons.

Solution

mass  1, 000, 000,000  0.00000000000000000000000167248 g 



 1  109

 1.67248  10 g   1.67248  10 g 24

15

132. Speed of light The speed of light in a vacuum is approximately 30,000,000,000 centimeters per second. Find the speed of light in miles per hour. (160,934.4 cm = 1 mile.)

Solution 30, 000, 000, 00 cm

1 mile 60 sec 60min   1 sec 160, 934.4 cm 1 min 1 hr 3  1010 6  101 6  101  mile/hr 1.609344  105  366  1010 1 15 mile/hr  1.609344  67.11  107 mile/hr  6.711  108 mile/hr

30, 000,000, 000 cm/sec 









133. License plates License plates come in various forms. The number of different license plates of the form three digits followed by three letters, is 10  10  10  26  26  26 . Write this expression using exponents. Then evaluate it and express the result in scientific notation.

Solution

10  10  10  26  26  26  103  263 ; 103  263  17,576,000  1.7576  107 134. Astronomy

The distance d, in miles, of the nth planet from the sun is given by the





formula d  9, 275, 200 3 2n  2  4  To the nearest million miles, find the distance of   Earth and the distance of Mars from the sun. Give each answer in scientific notation.

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution Earth: n  3





d  9, 275, 200[3 2n 2  4] 3 2

 9, 275, 200[3(2

 

)  4]

 9, 275, 200[3 2  4] 1

 9, 275, 200[10]  92, 752,000  93,000,000 9.3  107 mi Mars: n  4

   9, 275, 200[3  2   4]  9, 275, 200[3  2   4]

d  9, 275, 200[3 2n 2  4] 4 2 2

 9, 275, 200[16]  148, 403, 200  148,000,000 1.48  108 mi 135. New way to the center of the Earth The spectacular “blue marble” image is the most detailed true-color image of the entire Earth to date. A new NASA-developed technique estimates Earth’s center of mass within 1 millimeter (0.04 inch) a year by using a combination of four space-based techniques.

NASA Images

The distance from the Earth’s center to the North Pole (the polar radius) measures approximately 6356.750 km, and the distance from the center to the equator (the

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

equatorial radius) measures approximately 6378.135 km. Express each distance using scientific notation. Solution polar radius  6.356750  103 km equatorial radius  6.378135  103 km

136. Refer to Exercise 127. Given that 1 km is approximately equal to 0.62 miles, use scientific notation to express each distance in miles.

Solution polar radius  3.941185  10 mi 3

equatorial radius  3.9544437  10 mi 3

Discovery and Writing Write each expression with a single base. 137. x n x 2

Solution x n x 2  x n2 138.

xm x3

Solution xm  x m3 x3 139.

xm x2 x3

Solution xm x2 x m 2 m 23 x   x m 1 3 3 x x 140.

x 3m 5 x2

Solution x 3m5  x 3m52  x 3m 3 x2 141. x m  1 x 3

Solution x m  1 x 3  x m  1 3  x m  4

142. an3a3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution a n  3a 3  a n  3  3  a n

143. Explain why –x4 and (–x)4 represent different numbers.

Solution In the expression  x 4 , the base of the exponent is x, while in the expression   x  , 4

the base of the exponent is  x. 144. Explain why –x55 and (–x)55 represent equal numbers.

Solution

  x    1  x 55

  x 55

145. Explain how to write a number in scientific notation.

Solution Answers will vary. 102 is not in scientific notation.

146. Explain why 32

Solution 32  102 is not in scientific notation because 32 is not a number between 1 and 10. 147. Explain why x 11  x 11  x 121 .

Solution x 11  x 11  x 11 11  x 22 148. Explain why 112  113  1215 .

Solution 112  113  112  3  115

149. Explain why

y 50 y 10

 y5 .

Solution y 50  y 50  10  y 40 y 10 150. Explain why  6 xyz   6 x 6 y 6 z 6 . 6

Solution

6 xyz   6 x y z 6

Critical Thinking In Exercises 151–158, determine if the statement is true or false. If the statement is false, then correct it and make it true. 151. 00  1

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution False. 00 is undefined. 152.  0  1

Solution True 153. x  n  

1 xn

Solution 1 xn

False. x  n  154.  x  y 

n



1 xn



1 yn

Solution False.  x  y 

n



x  y

155. 2 1  2 2

Solution



True. 21  21 , 22  41 156.  2    2  1



2

Solution True.  2    21 ,  2   1

2

 41  

157. Young adults between the ages of 18 and 24 send an average of 110 text messages per day. If there are approximately 31.5 million young adults in the USA in this age group, how many text messages are sent in one year? Write the answer using scientific notation.

Solution

110  365  31,500,000  1.1  102  3.65  102  3.15  107  12.64725  1011  1.264725  101  1011  1.264725  1012 158. Health authorities recommend that we drink eight 8-ounce glasses of water each day. How many glasses of water would you drink over a lifetime of 80 years? Write the answer using scientific notation.

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

8  365  80  8  3.65  102  8  101  233.6  103  2.336  102  103  2.336  105

EXERCISES R.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Write the expressions in radical form. a. 125 1/5 b.

 64 

2/3

Solution a.

1251/5  5 125

 64

2/3



 64  or  64 3

2. Write the expressions in exponential form. 9 64



 16 

Solution

  9 64

1/2



 16   16 . 4

5/4

3. Simplify

a. b. c. d.

Solution

x2  x

a. b.

x3  x

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

x4  x

x5  x

4. a. Determine

16 9 and

144.

b. Based on your answers for a. what can you conclude?

Solution a.

16 9  4  3   12

16 9 = 144

144  12

and

5. Simplify. a.

5 2  10 2  15 2

5 x  10 x  15 x

Solution a.

5 2  10 2  15 2  (5  10  15) 2  10 2

5 x  10 x  15 x  (5  10  15) x  10 x

6. a. What can you multiply

b. What can you multiply

2x by so that the radicand is a perfect square? 2x by so that the radicand is a perfect cube?

Solution a. When you multiply

2x by itself, the radicand will be a perfect square.

2x  2x  4x2 , where 4 x 2 is a perfect square b. When you multiply

2 x by

4 x 2 , the radicand will be a perfect cube.

2x  4x2  8x3 , where 8x 3 is a perfect cube Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If a = 0 and n is a natural number, then a 1/ n  __________.

Solution 0 8. If a > 0 and n is a natural number, then a 1/ n is a __________ number.

Solution positive

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

9. If a < 0 and n is an even number, then a 1/n is _________ a real number.

Solution not 10. 6 2/3 can be written as _________ or _________.

Solution

6  , 6  1/3

11.

1/3

a  _________.

Solution a 1/n

a2  _________.

12.

Solution a

13.

a n b  ________.

Solution n

14.

ab a  _________. b

Solution n

x  y ________

15.

x

Solution



16.

m n

x or

n m

x can be written as _______.

Solution mn

Practice Simplify each expression. 17. 9 1/ 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

 

91/2  32

1/2

3

18. 8 1/ 3

Solution

 

81/3  23  1  19.    25 

1/3

2

1/2

Solution

 1     25 

1/2

 16  20.    625 

 1 2       5    

1/2

1 5



1/4

Solution

 16     625 

 2 4       5    

1/4



2 5

21.  811/4

Solution

 

811/4   34  8  22.     27 

1/4

 3

1/3

Solution

 8     27 

 2  3        3    

1/3

23.  10, 000 

1/3



2 3

1/4

Solution

 10,000

1/4

 

 104

1/4

 10

24. 1024 1/5

Solution

 1,024

1/5

 

 45

1/5

4

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 27  25.     8 

1/3

Solution

 27     8 

1/3

 3 3        2    

1/3



3 2

26.  64 1/3

Solution

 

641/3   43 27.  64 

1/3

 4

1/2

Solution

 64

1/2

28.  125 

 not a real number

1/3

Solution

 125

1/3

   5     3

1/3

 5

Simplify each expression. Use absolute value symbols when necessary.



29. 16a2



1/2

Solution

 16a 

1/2



1/2

30. 25a4



  4a     2

1/2

4a

Solution

25a  4

31.

 16a  4

1/2

 

2   5a2   

1/2

 5 a2  5a2

1/4

Solution

 16a  4

1/4

  2a     4

1/4

2a

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



32. 64a3



1/3

Solution

 64a  3



33. 32a5



  4a    

1/3

 4a

1/5

Solution

 32a  5



34. 64a6



1/5

  2a     5

1/5

 2a

1/6

Solution

64a  6

  2a    

1/6



35. 216b6



1/6

2a

1/3

Solution

 216b  6



36. 256t 8



1/3





3   6b2   

1/3

 6b2

1/4

Solution

256t  8

 16a4  37.  2    25b 

  

  4t 2 

1/4

 4 t 2  4t 2

1/2

Solution

 16a4   2    25b 

 4a2 2       5b    

1/2

 a5  38.   10    32b 

1/2



4a2 4a2  5b 5b

1/5

Solution

 a5    10    32b 

1/5

 a 5     2    2b    

1/5



a 2b2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 1000 x 6  39.   3    27 y 

1/3

Solution  1000 x 6    3    27 y 

1/3

 10 x 2 3        3 y     

 49t 2  40.  4    100z 

1/3

10 x 2 3y

1/2

Solution

 49t 2   100z 4   

1/2

 7t 2       10z 2    

1/2

 

7t 10z 2 7t 10z 2

Simplify each expression. Write all answers without using negative exponents. 41. 43/2

Solution

  2 8

43/2  41/2

42. 82/3

Solution

  2 4

82/3  81/3

43. 163/2

Solution



163/2   161/2 44.  8

   4  64 3

2/3

Solution

 8 

2/3

   8  

1/3

  2 2  4   

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

45.  1000 2/3

Solution



10002/3   10001/3

    10 2

 100 46. 100 3/2

Solution



1003/2  1001/2

  10  1,000 3

47. 64 1/2

Solution 64 1/2 

1 1  1/2 8 64

48. 25 1/2

Solution 251/2 

1 1  251/2 5

49. 64 3/2

Solution 643/2 

1 64

3/2



64  1/2



1 3



1 512



1 343

50. 493/2

Solution 493/2 

1 493/2



 49  1/2



1 73

51. 93/2

Solution 93/2  

1 3/2



9  1/2



1 3



1 27

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

52.  27 

2/3

Solution

 27  4 53.   9

2/3



 27 



2/3

1  27 1/3     



 3 



1 9

5/2

Solution 5

5  4 1/2  2 32         243  9   3  

5/2

4   9

 25  54.    81 

3/2

Solution 3

3  25 1/2  5 125         81 9 729       

3/2

 25     81 

 27  55.     64 

2/3

Solution

 27     64   125  56.    8 

2/3

 64      27 

2/3

2  64 1/3   4 16          9  27    3  

4/3

Solution

 125     8 

4/3

4/3

 8     125 

4  8 1/3  2 16         5 625  125      

Simplify each expression. Assume that all variables represent positive numbers. Write all answers without using negative exponents.



57. 100s4



1/2

Solution

 100s  4

1/2

 

 1001/2 s4

1/2

 10s2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



58. 64u6v 3



1/3

Solution

64u v 

1/3





59. 32 y 10 z 5

  v  1/3

 641/3 u6

1/3

 4u2v

1/5

Solution

32 y z 

1/5



1/4

32 y z  10



60. 625a4 b8



1/5



  z 

321/5 y 10

1/5



1/5

1 2 y 2z

Solution

625a b  4

61.

x y  10

1/4



625a b  4

1/4

 625

1/4

a   b  1/4

1/4



1 5ab2

3/5

Solution

x y  10

3/5



62. 64a6 b12



 x 30/5 y 15/5  x 6 y 3

5/6

Solution

64a b  6



63. r 8 s 16



5/6



 645/6 a30/6 b60/6  641/6

 a b  2 a b  32a b 5

3/4

Solution

r s  8

3/4



 r 24/4 s 48/4  r 6 s 12 



2/3

 8 x y 

2/3

64. 8 x 9 y 12

1 r s 12 6

Solution 9

  8 

2/3

x 18/3 y 24/3 

 8 

2/3

x 6 y 8 

 2  x y 2



1 6

4x y 8

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 8a6  65.   9    125b 

2/3

Solution  8a6    9    125b 

 16 x 4  66.  8    625 y 

2/3

 8  

2/3

 2 a  4a  2

a 12/3

1252/3 b18/3

52 b6

25b6

3/4

Solution  16 x 4   625 y 8   

3/4

 27r 6  67.  12    1000s 



163/4 x 12/4 23 x 3 8x 3  3 6  3/4 24/4 625 y 5 y 125 y 6

2/3

Solution  27r 6   12    1000s 

2/3

 32m10  68.    15   243n 

 1000s 12     6  27r 

2/3



10002/3 s24/3 102 s8 100s8  2 4  272/3 r 12/3 3 r 9r 4

2/5

Solution  32m10    15    243n 

69.

2/5

 243n15    10    32m 

2/5

 243 

2/5

n30/5

322/5 m20/5

 3 n  9n  2

22 m4

4m4

a2/5a4/5 a1/5

Solution a 2/5a 4/5 a6/5  1/5  a5/5  a a 1/5 a

70.

x 6/7 x 3/7 x 2/7 x 5/7

Solution x 6/7 x 3/7 x 9/7   x 2/7 x 2/7 x 5/7 x 7/7

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Simplify each radical expression.

71.

Solution

49  72  7 81

72.

Solution

81  92  9 73.

125

Solution

74.

125  3 53  5

64

Solution 3

75.

64  3  4   4 3

125

Solution

76.

125  3  5  5

243

Solution 5

243  5  3   3 5

77. 5 

32 100, 000

Solution 5

78. 4

 2  32 2 1   5      100, 000 10 5  10 

256 625

Solution 4

256 4  4  4     625 5 5

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Simplify each expression, using absolute value symbols when necessary. Write answers without using negative exponents.

36x2

79.

Solution 36 x 2 

6 x   6 x  6 x 2

80.  25y 2

Solution

 5 y    5 y  5 y 2

 25 y 2  

9y 4

81.

Solution

3 y   3 y  3 y

9y4 

a4b8

82.

Solution a4 b8 

a b   a b  a b 2

83. 3 8 y 3

Solution 3

84.

8 y 3  3 2 y   2 y 3

27z9

Solution 3

85. 4



27 z 9  3 3z 3

  3 z 3

x4 y 8 z 12

Solution 4

x4 y 8 z 12

x y  xy 2  xy 2  4  3   3  z z3  z 

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

86. 5

a 10 b5 c 15

Solution 5

a10 b5 5  a2 b  a2 b      c3  c 15 c3  

Simplify each expression. Assume that all variables represent positive numbers so that no absolute value symbols are needed. 87.

8 2 Solution

8 2  4 2 2 2 2 2  2 88.

75  2 27 Solution 75  2 27  25 3  2 9 3  5 3  2  3  3  5 3  6 3   3

89.

200x2  98x2 Solution

200x2  98x2  100x2 2  49x2 2  10x 2  7 x 2  17 x 2 90.

128a3  a 162a Solution

128a3  a 162a  64a2 2a  a 81 2a  8a 2a  9a 2a  a 2a 91. 2 48 y 5  3 y 12 y 3

Solution



2 48 y 5  3 y 12 y 3  2 16 y 4 3 y  3 y 4 y 2 3 y  2 4 y 2

 3 y  3 y 2 y  3 y

 8y2 3y  6y2 3y  2y2 3y 92. y 112 y  4 175 y 3

Solution

y 112 y  4 175 y 3  y 16 7 y  4 25 y 2 7 y  y  4  7 y  4  5 y  7 y  4 y 7 y  20 y 7 y  24 y 7 y

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

93. 2 81  3 24 3

Solution 2 3 81  3 3 24  2 3 27 3 3  3 3 8 3 3  2  3  3 3  3  2  3 3  6 3 3  6 3 3  12 3 3

94. 3 32  2 162 4

Solution 3 4 32  2 4 162  3 4 16 4 2  2 4 81 4 2  3  2  4 2  2  3  4 2  6 4 2  6 4 2  0

95.

768z5  4 48z5

Solution 4

768z5  4 48z5  4 256z4 4 3z  4 16z4 4 3z  4z 4 3z  2z 4 3z  6z 4 3z

96. 2 5 64 y 2  3 5 486 y 2

Solution

2 5 64 y 2  3 5 486 y 2  2 5 32 5 2 y 2  3 5 243 5 2 y 2  2  2  5 2 y 2  3  3  5 2 y 2  4 5 2 y 2  9 5 2 y 2  5 5 2 y 2

8 x 2 y  x 2 y  50 x 2 y

97.

Solution 8 x 2 y  x 2 y  50 x 2 y  4 x 2 2 y  x 2 y  25 x 2 2 y  2x 2 y  x 2 y  5x 2 y  6x 2 y 3 3 98. 3x 18x  2 2x  72x

Solution 3 x 18 x  2 2 x 3  72 x 3  3 x 9 2 x  2 x 2 2 x  36 x 2 2 x  3 x  3 2 x  2 x 2 x  6 x 2 x  9x 2x  2x 2x  6x 2x  5x 2x

99. 3 16 xy 4  y 3 2 xy  3 54 xy 4

Solution 3

16 xy 4  y 3 2 xy  3 54 xy 4  3 8 y 3 3 2 xy  y 3 2 xy  3 27 y 3 3 2 xy  2 y 3 2 xy  y 3 2 xy  3 y 3 2 xy  0

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

100.

512x5  4 32x5  4 1250x5

Solution 4

512 x 5  4 32 x 5  4 1, 250 x 5  4 256 x 4 4 2 x  4 16 x 4 4 2 x  4 625 x 4 4 2 x  4 x 4 2x  2x 4 2x  5x 4 2x  7 x 4 2x

Rationalize each denominator and simplify. Assume that all variables represent positive numbers. 3 101. 3 Solution

3 3 102.







3 3  3 3



6 5 5

6 5

Solution

6 5



6 5



103.

Solution 2 x







x x

104.

y Solution 8 y

105.







8 y y

2 3

Solution

2 3



2 3





23 4 3



23 4 3  4 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

106.

4d 3

Solution

4d 3

107.



4d 3



4d 3 3



4d 3 3 3



5a 3

25a

Solution

5a 3

108.

25a



5a 3

25a



5a2

6c2

27a2



5a 3 5a2 3

125a3



5a 3 5a2  3 5a2 5a



7 3 6c2 6c

7 3

36c

Solution

7 3

109.

36c



7 3

36c





7 3 6c2 3

216c3

2b 4

3a2

Solution

2b 4

3a2



2b 4

3a2





2b 4 27a2 4

81a4



2b 4 27a2 3a

x 2y

110.

Solution x  2y

111.

x 2y



x 2y



2y 2y



2 xy 2y

2u4 9v

Solution 3

3 3 3 3 2u4 2u4 u 2u 3 3v 2 u 3 6uv 2 u 3 6uv 2      3 3 3 3 9v 3v 9v 9v 3v 2 27v 3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

112.



3s 5 4r 2

Solution



113.

3 3 3s5 3s5 s3 3 s2 3 2r s 3 6rs2 s 3 6rs2       3 3 3 3 2r 4r 2 2r 4r 2 4r 2 8r 3

5 10

Solution

5 5 5 5 1     10 10 5 10 5 2 5 y 3

114.

Solution y y   3 3

y y



y 3 y

9 3

115.

Solution

9 3 9 3 3 3 27 3 1      3 3 3 3 33 3 33 3 3 3

116.

16b2 16

Solution 3

3 16b2 16b2 3 4b 3 64b3 4b b      3 3 3 3 16 16 4b 16 4b 16 4b 4 4b

16b3 64a

117.

Solution 5

5 5 16b3 16b3 5 2b2 32b5 2b b      5 5 5 64a 64a 2b2 64a 2b2 64a 2b2 32a 5 2b2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3x 57

118.

Solution

3x  57

3x 57







171x

9 19 x



3x 3 19 x



x 19 x

Rationalize each denominator and simplify. 1 1  3 27

119.

Solution 1 1   3 27

120. 3

1 3



1 27



1 3



3 3







3 3 3 3    3 3 9 81 3 3 3 2 3    9 9 9

1 3 1  2 16

Solution 3

3 3 3 3 3 1 3 1 1 1 1 34 1 4 34 4 4 34            2 16 3 2 3 16 3 2 3 4 3 16 3 4 3 8 3 64 2 4



x  8

121.

23 4 3 4 33 4   4 4 4

x x  2 32

Solution x  8

x  2

x  32

x 8



x 2



x 32

 

x 8 2x



x 2



2 2 2x



x 32



2 2

 16 4 64 2x 2x 2x    4 2 8 2 2x 4 2x 2x 2x     8 8 8 8

122. 3



y 3 y y  3 4 32 500

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 3 3 3 3 3 3 3 y y y y 32 y 32 y y 3 y y 2  3          3 3 3 3 3 3 3 3 3 4 32 500 4 32 500 4 2 32 2 500 2



2y 3





3 3

1, 000

2y 2y 2y   2 4 10 10 3 2 y 5 3 2 y 2 3 2 y 13 3 2 y     20 20 20 20 

Simplify each radical expression. 123.

Solution

124.

 

9  91/4  32

1/4

 32/4  31/2  3

Solution

125.

 

27  27 1/6  33

16x6

1/6

 33/6  31/2  3

Solution

126.









16 x 6  16 x 6

27x9

1/10









 24 x 6

1/10



 24/10 x 6/10  22/5 x 3/5  22 x 3



1/5

 5 4x3

Solution 6

27 x 9  27 x 9

1/6

 33 x 9

1/6



 33/6 x 9/6  31/2 x 3/2  3 x 3



1/2

 3x 3  x 3x

Fix It In Exercises 127 and 128, identify the step the first error is made and fix it. 127. Use the properties of exponents to simplify –10004/3.

Solution Step 2 was incorrect. Step 1: 

 1000  3

Step 2:  10 4 Step 3: 10,000

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

In Exercise 128, identify the step the first error is made and fix it. 3 128. Simplify the radical expression 5x 27 x  48x .

Solution Step 5 was incorrect.

 

  

Step 1: 5 x 9 3 x  16 3 x 2 x 2 Step 2: 5x 9 3x  16 x 3x

Step 3: 5 x  3  3 x  4 x 3 x Step 4: 15x 3x  4x 3x Step 5: 11x 3x

Applications 129. Hiking collage A square-shaped hiking collage of photos has an area of 120 square inches. What is the length of each of its sides?

Solution

s  120  4 30  2 30 inches 130. Volume The volume of a cube-shaped box is 2000 square inches. What is the length of each of its sides?

Solution

s  3 2000  3 1000 3 2  103 2 inches Discovery and Writing We often can multiply and divide radicals with different indices. For example, to multiply

3 by 3 5, we first write each radical as a sixth root 3 = 3 1/2 = 33/6 = 6 33 = 6 27 3

5 = 5 1/3 = 5 2/6 = 6 5 2 = 6 25

and then multiply the sixth roots. 3 3 5 = 6 27 6 25 = 6  27  25  = 6 625

Division is similar. Use this idea to write each of the following expressions as a single radical. 131.

23 2 Solution 6

2 3 2  21/2  21/3  23/6  22/6  23 22  6 8 6 4  6 32

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

33 5

132.

Solution

3 3 5  31/251/3  33/652/6  6 33 6 52  6 27 6 25  6 675 133.

3 2

Solution 4

3 2

134.



4 4 3 3 4 3 4 4 4 12 4 12       4 2 22/4 4 22 4 4 4 4 4 4 16

31/4 2

31/4

 1/2

2 5

Solution 3

2 5



21/3 5

 1/2

6 2 6 6 6 6 2 4 4 6 125 500 500       3/6 6 6 6 6 6 3 5 5 125 125 125 15, 625 5

22/6

135. Explain why a1/n is undefined if n is even and a represents a negative number.

Solution If a1/ n  x, then x n  a. However, if n is even, xn cannot be negative. 136. For what values of x does

x4 = x? Explain.

Solution 4

x 4  x . Since x  x if x  0, then 4 x 4  x if x  0.

137. Explain what is meant by rationalizing the denominator of a radical expression.

Solution To rationalize a denominator means to write an equivalent fraction with a denominator equal to a rational number. 138. If all of the radicals involved represent real numbers and y  0, explain why

n x x  n y y

Solution x x n   y y

1/ n



x 1/ n

 1/ n

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

139. If all of the radicals involved represent real numbers and there is no division by 0, explain why

x   y

 m/ n

n

ym xm

Solution

x   y

 m/ n



x  m/ n

x m/ n y m/ n

y m/ n

x m/ n y

x m/ n

  m/ n

  m/ n

 m/ n

y   x  m

1/ n

 ym    1/ n  x m  m  

1/ n

n

ym xm

140. The definition of x m / n requires that n x be a real number. Explain why this is important. (Hint: Consider what happens when n is even, m is odd, and x is negative.)

Solution Consider the case when n is even, m is odd and x is negative. Then

    x  . Thus,

x m/ n  x 1/ n

x must be a real number for the expression to be

defined.

Critical Thinking In Exercises 141–148, match each expression on the left with an equivalent expression on the right. Assume all variables represent positive numbers. 141.

 16 

1/4

142.  1024 

1/10

2

x 87 x

143. 0111/19

c. 2

144.  1

d. 0

12/19

x 86 x

145.

1

146.

x88

undefined

147.

148.

1 87

3 3

512

h. –1

Solution 141.

 16 

142.

 1024

1/4

is undefined. g

1/10

 

 210

1/10

 2. c

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

143. 0111/19  0. d 144.

 1

   1. f

12/19

 19 1

145.

1  1. h

146.

x88 

147.

148.

3 3



x87 87 x  x 87 x. b 

x 86



x 86 . e x

512  3 8  2. a

EXERCISES R.4 Getting Ready

Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Combine like terms and write in descending powers of x. 7  3 x  2 x 2  5 x 2  4 x  5

Solution





7  3 x  2 x 2  5 x 2  4 x  5  2 x 2  5 x 2   3 x  4 x    7  5   3 x 2  x  2



2. Remove parentheses and write in descending powers of  3 y 2  2 y 3  5  y

Solution









 3 y 2  2 y 3  5  y  3 y 2  2 y 3   5     y   3 y  2 y  5  y 2

 2y3  3y2  y  5



3. Use the Distributive Property to multiply. 3 x 4 x 2  7 x  2

Solution









3 x 4 x 2  7 x  2  3 x 4 x 2   3 x  7 x    3 x  2   12 x 3  21x 2  6 x



4. Use the Distributive Property to multiply. 5 yz 2 yz 2  7 yz  2

Solution









5 yz 2 yz 2  7 yz  2  5 yz 2 yz 2  5 yz 2  7 yz   5 yz 2  2   5 y z  35 y 2 z 3  10 yz 2 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

5. Use the Distributive Property to multiply.

Solution





5 1  5  5  1  5



5 1 5



 5   5  25  5  5

6. Identify the conjugate of 3  11.

Solution The conjugate of 3  11 is 3  11 , since a  b and a  b are conjugates.

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A __________ is a real number or the product of a real number and one or more ________.

Solution monomial, variables 8. The _________ of a monomial is the sum of the exponents of its _________.

Solution degree, variables 9. A _________ is a polynomial with three terms.

Solution trinomial 10. A _________ is a polynomial with two terms.

Solution binomial 11. A monomial is a polynomial with _________ term.

Solution one 12. The constant 0 is called the _________ polynomial.

Solution zero 13 Terms with the same variables with the same exponents are called _________ terms.

Solution like

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

14. The ________ of a polynomial is the same as the degree of its term of highest degree.

Solution degree 15. To combine like terms, we add their _________ and keep the same ________ and the same exponents.

Solution coefficients, variables 16. The conjugate of 3 x  2 is _________.

Solution

3 x 2 Determine whether the given expression is a polynomial. If so, tell whether it is a monomial, a binomial, or a trinomial, and give its degree. 17. x 2  3 x  4

Solution yes, trinomial, 2nd degree 18. 5xy  x

Solution yes, binomial, 3rd degree 3 1/2 19. x  y

Solution no 20. x

3

 5 y 2

Solution no 2 3 21. 4x  5x

Solution yes, binomial, 3rd degree 2

22. x y

Solution yes, monomial, 5th degree 23.

15 Solution yes, monomial, 0th degree

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

24.

5 x  5 x 5

Solution no 25. 0

Solution yes, monomial, no degree 3 2 26. 3 y  4 y  2 y

Solution yes, none, 3rd degree Practice Perform the operations and simplify. 27.

 x  3x   5x  8x  3

Solution

 x  3x   5x  8x   x  3x  5x  8x  x  5x  3x  8x  6x  3x  8x 3



 

28. 2 x 4  5 x 3  7 x 3  x 4  2 x



Solution

 2 x  5x    7 x  x  2x   2 x  5 x  7 x  x  2x 4

 2 x 4  x 4  5x 3  7 x 3  2 x  x 4  2 x 3  2x 29.

 y  2 y  7   y  2 y  7 5

Solution

 y  2 y  7   y  2 y  7  y  2 y  7  y  2 y  7 5

 y 5  y 5  2 y 3  2 y 3  7  7  4 y 3  14



 

30. 3t 7  7t 3  3  7t 7  3t 3  7



Solution

3t  7t  3   7t  3t  7  3t  7t  3  7t  3t  7 7

 3t 7  7t 7  7t 3  3t 3  3  7  4t 7  4t 3  4

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



 



 



31. 2 x 2  3 x  1  3 x 2  2 x  4  4

Solution



 

2 x 2  3x  1  3 x 2  2x  4  4  2 x 2  2  3x   2  1  3 x 2  3  2x   3  4   4  2x  6 x  2  3x  6 x  12  4 2

 2x 2  3x 2  6 x  6 x  2  12  4   x 2  14



 



 



32. 5 x 3  8 x  3  2 3 x 2  5 x  7

Solution



 

5 x 3  8x  3  2 3x 2  5x  7  5 x 3  5  8x   5  3  2 3 x 2  2  5x   7  5x  40 x  15  6 x  10 x  7 3

 5x 3  6 x 2  40 x  10 x  15  7  5x 3  6 x 2  30 x  8







 





 



33. 8 t 2  2t  5  4 t 2  3t  2  6 2t 2  8

Solution





8 t 2  2t  5  4 t 2  3t  2  6 2t 2  8

   8  2t   8 5  4 t   4  3t   4  2  6 2t   6  8

8 t

 8t 2  16t  40  4t 2  12t  8  12t 2  48  8t 2  4t 2  12t 2  16t  12t  40  8  48  28t  96



 



 



34. 3 x 3  x  2 x 2  x  3 x 3  2 x

Solution



 

3 x 3  x  2 x 2  x  3 x 3  2 x  3 x 3  3   x   2 x 2  2  x   3 x 3  3  2 x   3 x  3 x  2 x  2x  3x  6 x 3

 3 x 3  3 x 3  2x 2  3x  2x  6 x  2 x 2  x





35. y y 2  1  y 2  y  2   y  2 y  2 

Solution





 

y y 2  1  y 2  y  2   y  2 y  2   y y 2  y  1  y 2  y   y 2  2   y  2 y   y  2   y  y  y3  2y2  2y2  2y 3

 y 3  y 3  2 y 2  2 y 2  y  2 y  4 y 2  y





36. 4a2  a  1  3a a2  4  a2  a  2 

Solution





 4a2  a  1  3a a2  4  a2  a  2 

 

 4a  a   4a2  1  3a a2  3a  4   a2  a   a2  2  2

 4a  4a  3a  12a  a3  2a2 3

 4a3  3a3  a3  4a2  2a2  12a  2a3  6a2  12a

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra









37. xy  x  4 y   y x 2  3 xy  xy  2 x  3 y 

Solution

xy  x  4 y   y x 2  3 xy  xy  2 x  3 y 

 

 xy  x   xy  4 y   y x 2  y  3 xy   xy  2 x   xy  3 y   x y  4 xy  x y  3 xy 2  2 x 2 y  3 xy 2 2

 x 2 y  x 2 y  2 x 2 y  4 xy 2  3 xy 2  3 xy 2  2 x 2 y  4 xy 2













38. 3mn m  2n  6m 3mn  1  2n 4mn  1

Solution

3mn  m  2n   6m  3mn  1  2n  4mn  1

 3mn  m   3mn  2n   6m  3mn   6m  1  2n  4mn   2n  1  3m2 n  6mn2  18m2 n  6m  8mn2  2n  3m2 n  18m2 n  6mn2  8mn2  6m  2n  15m2 n  2mn2  6m  2n



39. 2 x 2 y 3 4 xy 4

Solution





2 x 2 y 3 4 xy 4  2  4  x 2 xy 3 y 4  8 x 3 y 7



40. 15a3 b 2a2 b3

Solution



 

15a 3 b 2a 2 b3  15  2  a 3a 2 bb3  30a 5 b4

 mn  41. 3m2 n 2mn2     12 





Solution  mn   1  2 6 4 4 m 4 n4 2 3m2 n 2mn2   m n     3  2     m mmnn n  12 2  12   12 



42. 



3r 2 s 3  2r 2 s   15rs 2     5  3   2 

Solution 3r 2 s 3  2r 2 s   15rs 2   3   2   15  2 2 3 2 5 6             r r rs ss  3r s 5  3   2   5   3   2 

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



43. 4rs r 2  s2



Solution





 

4rs r 2  s 2  4rs r 2  4rs s 2  4r 3 s  4rs 3



44. 6u2v 2uv 2  y

Solution



 





6u2v 2uv 2  y  6u2v 2uv 2  6u2v   y   12u3v 3  6u2vy



45. 6ab2c 2ac  3bc 2  4ab2c

Solution



 







6ab2c 2ac  3bc 2  4ab2c  6ab2c  2ac   6ab2c 3bc 2  6ab2c 4ab2c  12a b c  18ab c  24a b c 2

46. 

mn2 4mn  6m2  8 2





Solution 



mn2 mn2 mn2 mn2 4mn  6m2  8   4mn   6m2    8 2 2 2 2  2m2 n3  3m3 n2  4mn2







47. a  2 a  2





Solution

a  2a  2  a  2a  2a  4 2



48. y  5

 y  5

 a2  4a  4

Solution  y  5  y  5   y 2  5 y  5 y  25  y 2  10 y  25

49.  a  6 

Solution

a  6  a  6a  6 2

 a2  6a  6a  36  a2  12a  36

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

50.  t  9 

Solution

t  9  t  9t  9 2

 t 2  9t  9t  81  t 2  18t  81 51.

 x  4 x  4  Solution

 x  4  x  4   x  4 x  4 x  16 2

 x 2  16





52. z  7 z  7



Solution

 z  7  z  7   z  7z  7 z  49 2

 z 2  49





53. x  3 x  5



Solution

 x  3 x  5  x  5 x  3x  15 2

 x 2  2 x  15





54. z  4 z  6



Solution

 z  4  z  6  z  6z  4z  24 2

 z 2  2z  24





55. u  2 3u  2



Solution

u  2 3u  2  3u  2u  6u  4 2

 3u2  4u  4





56. 4 x  1 2 x  3



Solution

 4 x  1 2x  3  8x  12 x  2 x  3 2

 8 x 2  10 x  3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra





57. 5 x  1 2 x  3



Solution

5x  1 2x  3  10 x  15x  2x  3 2

 10 x 2  13 x  3





58. 4 x  1 2x  7



Solution

 4 x  1 2x  7   8x  28x  2x  7 2

 8 x 2  30 x  7 59.  3a  2b 

Solution

 3a  2b    3a  2b  3a  2b   9a  6ab  6ab  4b  9a  12ab  4b 2





60. 4a  5b 4a  5b



Solution

 4a  5b 4a  5b  16a  20ab  20ab  25b  16a  25b 2

61.

 3m  4n 3m  4n Solution

 3m  4n 3m  4n  9m  12mn  12mn  16n  9m  16n 2

62.  4r  3s 

Solution

 4r  3s    4r  3s  4r  3s   16r  12rs  12rs  9s  16r  24rs  9s 2





63. 2 y  4 x 3 y  2 x



Solution

 2 y  4 x  3 y  2x   6 y  4 xy  12xy  8x  6 y  16xy  8x 2





64. 2 x  3 y 3 x  y



Solution

 2x  3 y  3x  y   6x  2xy  9xy  3 y  6x  7 xy  3 y 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



65.  9 x  y  x 2  3 y



Solution

9x  y   x  3 y   9x  27 xy  x y  3 y  9x  x y  27 xy  3 y 2





66. 8a2  b  a  2b 

Solution

8a  b  a  2b   8a  16a b  ab  2b 2



67.  5z  2t  z 2  t



Solution

5z  2t   z  t   5z  5tz  2tz  2t  5z  2tz  5tz  2t 2



68. y  2 x 2

 x  3 y  2

Solution

 y  2 x  x  3 y   x y  3 y  2x  6 x y  2 x  5x y  3 y 2

69.  3 x  1

Solution

 3x  1   3x  1 3x  1 3x  1   9 x  3 x  3 x  1  3 x  1   9 x  6 x  1  3 x  1  9 x  3 x   9 x  1  6 x  3 x   6 x  1  1  3 x   1  1 3

2 2

 27 x 3  9 x 2  18 x 2  6 x  3 x  1  27 x 3  27 x 2  9 x  1 70.  2 x  3 

Solution

 2x  3   2x  3 2x  3 2x  3   4 x  6 x  6 x  9  2 x  3   4 x  12 x  9  2 x  3   4 x  2 x   4 x  3   12 x  2 x   12 x  3   9  2 x   9  3  3

2 2

 8 x 3  12 x 2  24 x 2  36 x  18 x  27  8 x 3  36 x 2  54 x  27

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

71.

 3x  1  2x  4 x  3 2

Solution

 3 x  1  2 x  4 x  3   3 x  2 x   3 x  4 x   3 x   3   1  2 x   1  4 x   1   3  2

 6 x 3  12 x 2  9 x  2 x 2  4 x  3  6 x 3  14 x 2  5 x  3



72.  2 x  5 x 2  3 x  2



Solution

 2 x  5   x  3 x  2   2 x  x   2 x  3 x   2 x  2   5  x   5  3 x   5  2  2

 2 x 3  6 x 2  4 x  5 x 2  15 x  10  2 x 3  11x 2  19 x  10



73.  3 x  2 y  2 x 2  3 xy  4 y 2



Solution

 3x  2 y   2x  3xy  4 y  2











 3 x 2 x 2  3 x  3 xy   3 x 4 y 2  2 y 2 x 2  2 y  3 xy   2 y 4 y 2



 6 x  9 x y  12 xy  4 x y  6 xy  8 y  6 x  5 x y  6 xy  8 y 3 3



74.  4r  3s  2r 2  4rs  2s2



Solution

 4r  3s   2r  4rs  2s  2



 



 



 4r 2r 2  4r  4rs   4r 2s 2  3s 2r 2  3s  4rs   3s 2s 2



 8r  16r s  8rs  6r s  12rs  6s  8r  10r s  20rs  6s 3 3

Multiply the expressions as you would multiply polynomials.



75. 5  6 x

5  6 x 

Solution

5  6x 5  6x   5 5  5   6x   5  6x   6x 6x  25  5 6 x  5 6 x  36 x 2  25  6 x





76. 2 x  6 7 x  2



Solution

2x  6 7 x  2   2x(7 x )  2x 2  7 x 6  6 2  14 x 2  2 x 2  7 x 6  12  14 x 2  2 x 2  7 x 6  4  3  14 x 2  2 x 2  7 x 6  2 3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



77. 3  x



Solution

3  x   3  x 3  x   9  3 x  3 x  x  9  6 x  x 2



78. 5 y  2 x



Solution

5 y  2 x   5 y  2 x 5 y  2 x   25 y   5 y   2 x    5 y   2 x    2 x  2 x  2

 25 y 2  10 y x  10 y x  4 x 2  25 y 2  20 y x  4 x 79.

 5  3x 2  5x  Solution

 5  3x 2  5x   2 5  5x  6x  3 5x  3 5x  x  2 5 2

80.

 2  x 3  2x  Solution

 2  x 3  2x   3 2  2x  3x  2x  2x  2x  5x  3 2 2



81. 2 y n 3 y n  y  n

Solution



 





 

2 y n 3 y n  y  n  2 y n 3 y n  2 y n y  n  6 y n n  2 y n



82. 3a  n 2a n  3a n  1

   n

 6 y 2n  2 y 0  6 y 2n  2



Solution













3a  n 2a n  3a n  1  3a  n 2a n  3a  n 3a n  1  6a  n  n  9a  n  n  1  6a0  9a 1  6 



83. 5 x 2n y n 2 x 2 n y  n  3 x 2 n y n

Solution



 







 5 x 2 n y n 2 x 2 n y  n  3 x 2 n y n   5 x 2 n y n 2 x 2 n y  n  5 x 2 n y n 3 x 2 n y n  10 x

2n 2n

n    n

 15 x

 10 x y  15 x y 4n

2 n   2 n 

 10 x

9 a



nn

 15 y 2 n

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra









84. 2a3n b2n 5a 3n b  ab2 n

Solution







2a3n b2n 5a 3n b  ab2n  2a3n b2n 5a 3n b  2a3n b2n ab2n  10a

3 n   3 n

2n 1

 2a

3n 1

2 n   2 n



 10a0 b2n  1  2a3n  1b0  10b2n 1  2a3n  1





85. x n  3 x n  4



Solution

 x  3 x  4   x x  4 x  3x  12  x  x  12 n





86. a n  5 an  3



Solution

a  5a  3  a a  3a  5a  15  a  8a  15 n





87. 2r n  7 3r n  2



Solution

 2r  7  3r  2   2r 3r   2r  2   7  3r   14 n

 6r 2 n  4r n  21r n  14  6r 2 n  25r n  14







88. 4 z n  3 3z n  1

Solution

 4 z  3 3z  1  4 z 3z   4 z  1  3  3z   3 n

 12 z 2 n  4 z n  9 z n  3  12 z 2 n  13 z n  3



89. x 1/2 x 1/2 y  xy 1/2

Solution



 

x 1/2 x 1/2 y  xy 1/2  x 1/2 x 1/2 y  x 1/2 xy 1/2  x 2/2 y  x 3/2 y 1/2  xy  x 3/2 y 1/2



90. ab1/2 a 1/2 b1/2  b1/2

Solution



 

ab1/2 a 1/2 b1/2  b1/2  ab1/2a 1/2 b1/2  ab1/2 b1/2  a3/2 b2/2  ab2/2  a3/2 b  ab

91.

a

1/2

 b 1/2

Solution

a

1/2

 b 1/2

a

1/2

 b 1/2



a

1/2

 b 1/2  a 1/2a 1/2  a 1/2 b 1/2  a 1/2 b 1/2  b 1/2 b 1/2



 a 2/2  b2/2  a  b

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



92. x 3/2  y 1/2



Solution

x

3/2

 y 1/2

  x 2

3/2

 y 1/2

 x



 y 1/2  x 3/2 x 3/2  x 3/2 y 1/2  x 3/2 y 1/2  y 1/2 y 1/2

3/2

 x 6/2  2 x 3/2 y 1/2  y 2/2  x 3  2 x 3/2 y 1/2  y

Rationalize each denominator. 93.

2 31

Solution 2 31

94.



31





52

1 52



 5  2  5  2  5  2  5  2 5 2 5 2  5  2 5  4 1 1



5 2



7 2

3x 7 2



3x 7 2



7 2 7 2



 7  2  3 x  7  2  3 x  7  2  x 7  2   74 3 7  2  

14 y 2 3 Solution 14 y 2 3

97.

Solution

96.

Solution

95.

 3  1  2  3  1  2  3  1  3  1 2 31 31  3  1 3  1 2



14 y 2 3



23 23



 2  3  14 y  2  3  2 y 2  3   29  2  3

14 y

x x 3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution x x 3

98.





x 3

x 3 x 3





x x 3 x2 

 3

  x x  3

x2  3

y 2y  7

Solution

  y 2 y  7  2 y  7 2 y  7 2 y  7 2 y   7   4y  7 y

99.



2y  7





y 2y  7

y 2



   



   

2  3 1 3    1 3 1 3 1 3

2 6 3

y 2 y 2



y 2 y 2 y 2 y 2 y2  2y 2  2    2 y2  2 y 2 y 2 y2  2

x 3 x 3 Solution x 3 x 3

101.



y 2

Solution

100.



x 3 x 3 x 3 x 3 x2  2x 3  3    2 x2  3 x 3 x 3 x2  3

2 3 1 3 Solution 2 3

12 

 3

 3  2  6  3  3 2

13

  

2  6  3 3 2  2  6  3 3





2 3 3 2  6 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

102.

3 2 1 2

Solution

3 2 1 2

3 6 2

3  2 1 2   1 2 1 2



12 

 2

 2  3  6  2  2 2

12

3 6 2 2 1   3 6 22 





 6  2  3 2 103.

x

x

Solution x x

104.

y y



x x



x x

y y

x 2  xy  xy 



 x  y 2



x  2 xy  y xy

2x  y 2x  y Solution 2x  y 2x  y



2x  y 2x  y



2x  y 2x  y

4 x 2  y 2x  y 2x  y 2



 2x   y 2



2x  2 y 2x  y 2 2x  y 2

Rationalize each numerator. 105.

21 2 Solution

    2

21  2

106.

2  12 21 21 21     2 21 2 21 2 21 2



  2  1

x 3 3 Solution

x 3 x 3 x 3  x 3 x 9     3 3 x  3 3  x  3 3  x  3 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

107.

y 3 y 3 Solution

y 3 y 3 108.



y2 

  3

y 3 y 3 y2  3    y  3 y  3 y2  y 3  y 3  9 y2  2y 3  3

a b a b Solution

 a   b 2

a b a b 109.

x3 3

a b



a b



a b a b



a  ab  ab  b 2

ab



a  2 ab  b

Solution

 x  3   x   x3 x 3 x  3  x  2

x3 x  3

x3 x  3

x 3 x



 x 3  x 3



110.

x3 x



 x 3  x

1 x3 x

2h  2 h Solution 2h  2  h



     2

2h  2 2h  2 2h  2   h 2h  2 h 2h  2  

2h2

 2  h  2 h

 2  h  2



1 2h  2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Perform each division and write all answers without using negative exponents. 111.

36a2 b3 18ab6

Solution 36a2 b3 2a  2a2  1b3 6  2a 1b 3  3 6 18ab b 112.

45r 2 s5t 3 27r 6 s 2t 8

Solution 45r 2 s5t 3 5 2 6 5 2 3 8 5 4 3 5 5s 3   r s t   r s t   3 3 27r 6 s 2t 8 3r 4 t 5 113.

16 x 6 y 4 z 9 24 x 9 y 6 z 0

Solution 16 x 6 y 4 z 9 24 x y z 9

114.



2 6 9 4 6 90 2 2z 9   x 3 y 2 z 9   3 2 x y z 3 3 3x y

32m6 n4 p2 26m6 n7 p2 Solution 32m6 n4 p2 26m6 n7 p2

115.



16 6 6 4 7 2 2 16 0 3 0 16 m n p  mn p  13 13 13n3

5 x 3 y 2  15 x 3 y 4 10 x 2 y 3 Solution

5x 3 y 2  15x 3 y 4 5x 3 y 2 15x 3 y 4   10 x 2 y 3 10 x 2 y 3 10 x 2 y 3 x 3xy   2y 2 116.

9m4 n9  6m3 n4 12m3 n3

Solution

9m4 n9  6m3 n4 9m4 n9 6m3 n4   12m3 n3 12m3 n3 12m3 n3 3mn6 n   4 2 117.

24 x 5 y 7  36 x 2 y 5  12 xy 60 x 5 y 4

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 24 x 5 y 7  36 x 2 y 5  12 xy 60 x 5 y 4 118.



24 x 5 y 7

36 x 2 y 5

12 xy

60 x 5 y

 4

 4

 4

2y3 3y 1  3  4 3 5 5x 5x y

9a 3 b4  27a 2 b4  18a 2 b3 18a 2 b7

Solution 9a 3 b4  27a2 b4  18a2 b3 9a3 b4 27a 2 b4 18a 2 b3 a 3 1       4 2 7 2 7 2 7 2 7 3 3 18a b 18a b 18a b 18a b 2b 2b b Perform each division. If there is a nonzero remainder, write the answer in quotient +

remainder divisor

form. 119. x  3 3x 2  11x  6

Solution 3x 

x  3 3 x 2  11x  6 3x 2  9x 2x  6 2x  6 0 120. 3 x  2 3 x 2  11x  6

Solution x  3 3 x  2 3 x 2  11x  6 3x 2  2x 9x  6 9x  6 0 121. 2x  5 2x 2  19x  37

Solution x 

7  2 x25

2 x  5 2 x 2  19 x  37 2x 2  5x  14 x  37  14 x  35 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

122. x  7 2 x 2  19 x  35

Solution 2x 

x  7 2 x  19 x  35 2

2 x 2  14 x  5 x  35  5 x  35 0

123.

2x 3  1 x1 Solution 2 x 2  2 x  2  X3 1 x  1 2x 3  0x 2  0x  2x  2x 3

2x 2  0 x 

2x  2x 2

2x 

2x 

2 3

124.

2 x 3  9x 2  13x  20 2x  7 Solution x 2  x  3  2 x1 7 2 x  7 2 x 3  9 x 2  13 x  20 2x 3  7 x 2  2 x 2  13 x  20  2x 2  7 x 6 x  20 6 x  21 1

125. x 2  x  1 x 3  2 x 2  4 x  3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution x

x  x  1 x  2x 2  4 x  3 2

x3  x2  x  3x 2  3x  3  3x 2  3x  3 0 126. x 2  3 x 3  2x 2  4 x  5

Solution x  2  x2x31 x 2  3 x 3  2x 2  4 x  5  3x

 2x  x  5 2

 2x 2

6  x 1

127.

x 5  2x 3  3x 2  9 x3  2

Solution

x2 

 x2 5 x3 2

2 

x 3  2 x5  0x 4  2x 3  3x 2  0x  9  2x 2

 2x 3  x 2  0x  9  2x 3

4  x

128.

5

x 5  2x 3  3x 2  9 x3  3

Solution

x2 

2

3 x3 3

x 3  3 x 5  0 x 4  2x 3  3x 2  0x  9  3x 2

x5  2x 3

 0x  9

 2x

6

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

129.

x 5  32 x 2 Solution x 4  2 x 3  4 x 2  8 x  16 x  2 x 5  0 x 4  0 x 3  0 x 2  0 x  32 x 5  2x 4 2x 4  0x 3 2x 4  4 x 3 4x 3  0x 2 4 x 3  8x 2 8x 2  0 x 8 x 2  16 x 16 x  32 16 x  32 0

130.

x4  1 x1 Solution

x3  x2  x 

x  1 x 4  0x 3  0x 2  0x  1 x4  x3  x 3  0x 2  x3  x2 x 2  0x x2  x  x1  x1 0 131. 11x  10  6 x 2 36 x 4  121x 2  120  72x 3  142x

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 6 x 2  x  12 6 x 2  11x  10 36 x 4  72 x 3  121x 2  142 x  120 36 x 4  66 x 3  60 x 2 6 x 3  61x 2  142 x 6 x 3  11x 2  10 x  72 x 2  132 x  120  72 x 2  132 x  120 0 132. x  6 x 2  12 121x 2  72 x 3  142 x  120  36 x 4

Solution 6 x 2  11x  10 6 x 2  x  12 36 x 4  72 x 3 36 x 4  6 x 3

 121x 2  142 x  120  72 x 2

66 x 3

 49 x 2  142 x

66 x 3

 11x 2  132 x  60 x 2  10 x  120  60 x 2  10 x  120 0

Fix It In Exercises 133 and 134, identify the step the first error is made and fix it. 133. Use the Special Product Formula  x  y   x 2  2 xy  y 2 to square the given binomial 2

difference.  4 x  7 y 

Solution Step 2 was incorrect. Step 1: Square 4x and get 16 x 2 .

  

Step 2: Multiply 2 4 x 7 y and get 56xy Step 3: Square 7y and get 49y

2 2 Step 4: Combine the results of Steps 1, 2, and 3, and get 16x  56xy  49 y

134. Rationalize the denominator.

4 2 3

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution Step 4 was incorrect. Step 1:

Step 2:

5 3    5  3  5  3  4

20  4 3 25  5 3  5 3  3

Step 3:

20  4 3 22

Step 4:

10  4 3 11

Applications 135. Geometry Find an expression that represents the area of the brick wall.

Solution









Area  length  width   x  5  x  2  ft 2  x 2  2 x  5 x  10 ft 2  x 2  3 x  10 ft 2

136. Geometry The area of the triangle shown in the illustration is represented as

 x  3 x  40  square feet. Find an expression that represents its height. 2

Solution

2 x  10 1  base  height 2 x  8 2 x 2  6 x  80 1 x 2  3 x  40   x  8   height 2 x 2  16 x 2  10 x  80 2 x 2  3 x  40   x  8   height  10 x  80 2 x 2  6 x  80   x  8   height 0 2 x 2  6 x  80  height The height is  2 x  10 ft. x 8 Area =





Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

137. Gift boxes The corners of a 12 in.-by-12 in. piece of cardboard are folded inward and glued to make a box. Write a polynomial that represents the volume of the resulting box. Crease here and fold inward

Solution Volume  l  w  h

  12  2 x  12  2 x  x in.3

    144 x  48 x  4 x  in.   4 x  48 x  144 x  in.

 144  48 x  4 x 2 x in.3 2

3 3

138. Travel Complete the following table, which shows the rate (mph), time traveled (hr), and distance traveled (mi) by a family on vacation.

r 3x + 4

∙

d 3x2 + 19x + 20

Solution t

d 3 x 2  19 x  20  3x  4 r x  5

3 x  4 3 x 2  19 x  20 3x 2  4 x 15 x  20 15 x  20 0 t  x 5

Discovery and Writing 139. Show that a trinomial can be squared by using the formula

a  b  c   a  b  c  2ab  2bc  2ac. 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

a  b  c   a  b  c a  b  c   a a  b  c   b a  b  c   c a  b  c  2

 a2  ab  ac  ab  b2  bc  ac  bc  c2  a2  b2  c2  2ab  2bc  2ac 140. Show that  a  b  c  d   a2  b2  c 2  d 2  2ab  2ac  2ad  2bc  2bd  2cd . 2

Solution

a  b  c  d   a  b  c  d a  b  c  d   a a  b  c  d   b a  b  c  d   c a  b  c  d   d a  b  c  d  2

 a2  ab  ac  ad  ab  b2  bc  bd  ac  bc  c2  cd  ad  bd  cd  d 2  a2  b2  c2  d 2  2ab  2ac  2ad  2bc  2bd  2cd 141. Explain the FOIL method.

Solution Answers may vary. 142. Explain how to rationalize the numerator of

X 2 . X

Solution Multiply the numerator and denominator by the conjugate of the numerator

 x  2 .

143. Explain why  a  b   a 2  b2 . 2

Solution Check the formula with a  1 and b  2 . 144. Explain why

a2  b2  a2  b2 .

Solution Check the formula with a  3 and b  4 . Critical Thinking In Exercises 145–150, determine if the statement is true or false. If the statement is false, then correct it and make it true. 145. All polynomials are trinomials.

Solution False. Some polynomials are trinomials. 146. All binomials are polynomials.

Solution True.

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

147.  12 x  5 y   144 x 2  120 xy  25 y 2 2

Solution True.  12 x  5 y    12 x  5 y  12 x  5 y   144 x 2  60 xy  60 xy  25 y 2 2

 144 x 2  120 xy  25 y 2

148.  6 x  y   36 x 2  y 2 2

Solution False.  6 x  y    6 x  y  6 x  y   36 x 2  6 xy  6 xy  y 2  36 x 2  12 xy  y 2 2







149. x 1/3  6 4 x 1/3  7  4 x 1/9  17 x 1/3  42

Solution







False. x 1/3  6 4 x 1/3  7  4 x 2/3  7 x 1/3  24 x 1/3  42  4 x 2/3  17 x 1/3  42







150. x 3  5 x 3  5  x 9  25

Solution







False. x 3  5 x 3  5  x 6  5 x 3  5 x 3  25  x 6  25  x16  25

0 0 2 +

2 x

In Exercises 151 and 152, the revenue associated with selling x units of a product is dollars, and the cost associated with producing x units of the product is –200x + 500 dollars.

151. Determine the polynomial that represents the profit in dollars of making x units of the product.

Solution





Profit  Revenue  Cost  x 2  200 x   200 x  500   x 2  200 x  200 x  500  x 2  400 x  500

152. If 100 units are produced and sold, would the profit exceed $50,000?

Solution Use the answer to #139. x 2  400 x  500   100   400  100   500  49, 500  False. 2

EXERCISES R.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

The prime factorization of three terms are shown. Find their greatest common factor.

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

12x3 y2 = 2  2  3 x  x  x  y  y 18xy 4  2  3  3  x  y  y  y  y

30x2 y 3  2  3  5  x  x  y  y  y Solution All terms have 2, 3, x, and y 2 in common. The greatest common factor of all 3 terms 2 2 is 2  3  x  y  6xy

2. Find the greatest common factor of 10 x 5  y  2  , 20 x 4  y  2  , and 15x 4  y  2  . 2

Solution The greatest common factor of 10, 20, and 15 is 5. The greatest common factor of

x5 , x4 , and x4 is x 4 . The greatest common factor of  y  2  ,  y  2  and  y  2  is 2

 y  2  . The greatest common factor of all 3 terms is 5 x  y  2 . 2

3. Fill in the boxes to complete each factorization.



8 x 3  6 x 2  10 x  2 x 4 x 2    5

8 x  11  1  8 x 11

Solution





8 x 3  6 x 2  10 x  2 x 4 x 2  3 x  5

8 x  11  1  8 x  11



4. Fill in the boxes to complete each factorization. a.

x 2  10 x  9   x  9  x   

12 x 2  5 x  2   3 x  2   1

Solution a.

x 2  10 x  9   x  9 x  1

12 x 2  5 x  2   3 x  2 4 x  1

5. Fill in each set of parentheses. a.

49 x 2  100z 4      

125 x 3  8z 3      

27 y 6  64 z 3      

Solution a.



49 x 2  100z 4   7 x   10z 2 2



Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

125 x 3  8z 3   5 x    2z 

27 y 6  64z 3  3 y 2



6. Fill in the box. x  7

   4z  3

  x  7  x  7 2/3

Solution

 x  7

2/3

 x  7

5/3

 x7

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. When polynomials are multiplied together, each polynomial is a __________ of the product.

Solution factor 8. If a polynomial cannot be factored using __________ coefficients, it is called a _________ polynomial.

Solution integer, prime Complete each factoring formula. 9.

ax  bx  ________

Solution

ax  bx  x  a  b 

2 2 10. x  y  ________

Solution

x 2  y 2   x  y  x  y 

2 2 11. x  2xy  y  ________

Solution x 2  2 xy  y 2   x  y  x  y    x  y 

2 2 12. x  2xy  y  _________

Solution x 2  2 xy  y 2   x  y  x  y    x  y 

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3 3 13. x  y  ________

Solution



x 3  y 3   x  y  x 2  xy  y 2



3 3 14. x  y  ________.

Solution



x 3  y 3   x  y  x 2  xy  y 2



Practice In each expression, factor out the greatest common monomial. 15. 3x  6 _________.

Solution

3x  6  3  x  2

16. 5 y  15 _________.

Solution

5 y  15  5  y  3

17. 8 x 2  4 x 3 _________.

Solution

8x 2  4 x 3  4 x 2  2  x 

3 2 18. 9 y  6 y _________.

Solution

9 y 3  6 y 2  3 y 2  3 y  2

2 2 3 2 19. 7 x y  14x y ________.

Solution

7 x 2 y 2  14 x 3 y 2  7 x 2 y 2  1  2 x 

20. 25 y z  15 yz ________. 2

Solution

25 y 2 z  15 yz 2  5 yz  5 y  3z 

In each expression, factor by grouping.







21. a x  y  b x  y



Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

a  x  y   b  x  y    x  y  a  b 







22. b x  y  a x  y



Solution

b  x  y   a  x  y    x  y  b  a 

23. 4a  b  12a 2  3ab

Solution

4a  b  12a2  3ab  4a  b  3a  4a  b   1  4a  b   3a  4a  b    4a  b  1  3a 

2 24. x  4x  xy  4 y

Solution

x 2  4 x  xy  4 y  x  x  4   y  x  4    x  4  x  y 

In each expression, factor the difference of two squares. 25. 4 x 2  9

Solution 4 x 2  9   2 x   32   2 x  3  2 x  3  2

26. 36 z 2  49

Solution 36 z 2  49   6z   7 2   6 z  7  6 z  7  2

27. 4  9r 2

Solution 4  9r 2  22   3r    2  3r  2  3r  2

28. 16  49x 2

Solution 16  49 x 2  42   7 x    4  7 x  4  7 x  2

29. 81x 4  1

Solution



81x 4  1  9 x 2

  1  9 x  19 x  1  9 x  1 3 x  1 3 x  1 2

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

30. 81  x 4

Solution

   9  x 9  x   9  x   3  x  3  x 

81  x 4  92  x 2 31.

 x  z   25 2

Solution

 x  z   25   x  z   5   x  z  5 x  z  5 2

32.  x  y   9 2

Solution

x  y  9  x  y  3   x  y  3 x  y  3 2

In each expression, factor the trinomial. 33. x 2  8 x  16

Solution x 2  8 x  16   x  4  x  4    x  4 

34. a 2  12a  36

Solution a2  12a  36   a  6  a  6    a  6 

35. b2  10b  25

Solution b2  10b  25   b  5  b  5    b  5 

2 36. y  14 y  49

Solution y 2  14 y  49   y  7  y  7    y  7 

37. m2  4mn  4n2

Solution

m2  4mn  4n2   m  2n m  2n

  m  2n

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

38. r 2  8rs  16 s 2

Solution

r 2  8rs  16s2   r  4s  r  4s    r  4s 

2 2 39. 12x  xy  6 y

Solution

12 x 2  xy  6 y 2   4 x  3 y  3x  2 y 

2 2 40. 8x  10xy  3 y

Solution

8 x 2  10 xy  3 y 2   4 x  y  2 x  3 y 

In each expression, factor the trinomial by grouping. 41. x 2  10 x  21 Solution x 2  10 x  21 : a  1, b  10, c  21

key number  ac  1  21  21

x 2  10 x  21  x 2  7 x  3 x  21

 x  x  7  3  x  7   x  7  x  3 

42. x 2  7 x  10

Solution x 2  7 x  10 : a  1, b  7, c  10

key number  ac  1  10   10

x 2  7 x  10  x 2  5 x  2 x  10

 x  x  5  2  x  5   x  5  x  2 

43. x 2  4 x  12

Solution x 2  4 x  12 : a  1, b  4, c  12

key number  ac  1  12   12 x 2  4 x  12  x 2  6 x  2 x  12

 x  x  6  2  x  6   x  6  x  2 

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

44. x 2  2 x  63

Solution x 2  2 x  63 : a  1, b  2, c  63

key number  ac  1  63   63 x 2  2 x  63  x 2  9 x  7 x  63

 x  x  9  7  x  9   x  9  x  7 

45. 6p  7 p  3 2

Solution 6 p2  7 p  3 : a  6, b  7, c  3

key number  ac  6  3   18 6 p2  7 p  3  6 p2  9 p  2 p  3

 3p  2p  3    2p  3    2 p  3  3 p  1

46. 4q  19q  12 2

Solution 4q2  19q  12 : a  4, b  19, c  12

key number  ac  4  12   48

4q2  19q  12  4q2  3q  16q  12

 q  4q  3   4  4q  3    4q  3  q  4 

In each expression, factor the sum of two cubes. 47. t 3  343 Solution



t 3  343  t 3  73   t  7  [t 2   t  7   72 ]   t  7  t 2  7t  49



48. r 3  8s 3

Solution



3 2 r 3  8s 3  r 3   2s    r  2s  r 2   r  2s    2s     r  2s  r 2  2rs  4s2  



Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3 3 49. 125 y  216z

Solution 3 3 2 2 125 y 3  216z 3   5 y    6z    5 y  6z   5 y    5 y  6z    6z    



  5 y  6z  25 y 2  30 yz  36z 2



3 3 50. 27 y  1000z

Solution 3 3 2 2 27 y 3  1000z 3   3 y    10z    3 y  10z   3 y    3 y  10z    10z    



  3 y  10z  9 y 2  30 yz  100z 2



In each expression, factor the difference of two cubes. 51. 8 z 3  27

Solution



3 2 8z 3  27   2z   33   2z  3   2z    2z  3   32    2z  3  4 z 2  6z  9  



52. 125a 3  64

Solution



3 2 125a3  64   5a   43  (5a  4)  5a    5a  4   42    5a  4  25a2  20a  16  



3 3 53. 343y  z

Solution



343 y 3  z 3   7 y   z 3   7 y  z  (7 y )2   7 y  z   z 2    7 y  z  49 y 2  7 yz  z 2 3



3 3 54. 27 y  512z

Solution 3 3 2 2 27 y 3  512z 3   3 y    8z    3 y  8z   3 y    3 y  8z    8z     2 2   3 y  8z  9 y  24 yz  64z





Factor each expression completely. If an expression is prime, so indicate. 55. 3a 2 bc  6ab 2c  9abc 2

Solution

3a2 bc  6ab2c  9abc2  3abc  a  2b  3c 

3 3 3 2 2 2 56. 5x y z  25x y z  125xyz

Solution



5 x 3 y 3 z 3  25 x 2 y 2 z 2  125 xyz  5 xyz x 2 y 2 z 2  5 xyz  25



100

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

57. 3 x 3  3 x 2  x  1

Solution





3 x 3  3 x 2  x  1  3 x 2  x  1  1  x  1   x  1 3 x 2  1

58. 4 x  6 xy  9 y  6

Solution

4 x  6 xy  9 y  6  2 x  2  3 y   3  3 y  2   3 y  2 2 x  3

59. 2txy  2ctx  3ty  3ct

Solution

2txy  2ctx  3ty  3ct  t  2 xy  2cx  3 y  3c   t 2 x  y  c   3  y  c    t  y  c  2 x  3 

60. 2ax  4ay  bx  2by

Solution

2ax  4ay  bx  2by  2a  x  2 y   b  x  2 y    x  2 y  2a  b 

61. ax  bx  ay  by  az  bz

Solution

ax  bx  ay  by  az  bz  x  a  b   y  a  b   z  a  b    a  b  x  y  z 

2 3 2 2 62. 6x y  18xy  3x y  9x

Solution







 



6 x 2 y 3  18 xy  3 x 2 y 2  9 x  3x 2 xy 3  6 y  xy 2  3  3x 2 y xy 2  3  1 xy 2  3    2  3 x xy  3  2 y  1 63. x 2   y  z 





Solution

x 2   y  z    x   y  z    x   y  z     x  y  z  x  y  z  2

64. z 2   y  3 

Solution

z 2   y  3   z   y  3   z   y  3    z  y  3 z  y  3 2

101

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

65.  x  y    x  y  2

Solution

 x  y    x  y    x  y    x  y   x  y    x  y    x  y  x  y  x  y  x  y    2 x  2 y   4 xy 2

66.  2a  3    2a  3  2

Solution

 2a  3   2a  3   2a  3   2a  3  2a  3   2a  3   2a  3  2a  3 2a  3  2a  3   4a  6   24a 2

4 4 67. x  y

Solution

    y    x  y  x  y    x  y   x  y  x  y  2

x4  y 4  x2

68. z 4  81

Solution

   9   z  9 z  9   z  9 z  3    z  9  z  3 z  3

z 4  81  z 2

69. 3 x 2  12

Solution





3 x 2  12  3 x 2  4  3  x  2  x  2 

70. 3x y  3xy 3

Solution





3 x 3 y  3 xy  3 xy x 2  1

 3 xy  x  1 x  1 2 71. 18xy  8x

Solution



18 xy 2  8 x  2 x 9 y 2  4



 2 x  3 y  2  3 y  2 

102

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

72. 27 x 2  12

Solution



27 x 2  12  3 9 x 2  4



 3  3 x  2  3 x  2 

73. x 2  2 x  15

Solution

x2  2x  15  prime 74. x 2  x  2

Solution

x2  x  2  prime 75. 15  2a  24a 2

Solution

15  2a  24a2  24a2  2a  15

  6a  5 4a  3

76.  32  68 x  9 x 2

Solution

32  68 x  9 x 2  9 x 2  68 x  32   9 x  4  x  8

2 2 77. 6x  29xy  35 y

Solution

6 x 2  29 xy  35 y 2   3 x  7 y  2 x  5 y 

2 2 78. 10x  17 xy  6 y

Solution

10 x 2  17 xy  6 y 2   5 x  6 y  2 x  y 

79. 12p  58pq  70q 2

Solution





12 p2  58pq  70q2  2 6 p2  29pq  35q2  2  6 p  35q  p  q 

2 2 80. 3x  6xy  9 y

Solution





3 x 2  6 xy  9 y 2  3 x 2  2 xy  3 y 2  3  x  3 y  x  y 

103

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

81. 6m2  47 mn  35n2

Solution





6m2  47mn  35n2   6m2  47mn  35n2    6m  5n  m  7n 

82.  14 r 2  11rs  15 s 2

Solution





14r 2  11rs  15s2   14r 2  11rs  15s2    7r  5s  2r  3s 

83. 6 x 3  23 x 2  35 x

Solution





6 x 3  23 x 2  35 x   x 6 x 2  23 x  35   x  6 x  7  x  5 

3 2 84.  y  y  90 y

Solution





 y 3  y 2  90 y   y y 2  y  90   y  y  10  y  9 

85. 6 x 4  11x 3  35 x 2

Solution





6 x 4  11x 3  35 x 2  x 2 6 x 2  11x  35  x 2  2 x  7  3 x  5 

86. 12 x  17 x 2  7 x 3

Solution





12 x  17 x 2  7 x 3  7 x 3  17 x 2  12 x   x 7 x 2  17 x  12   x  x  3  7 x  4 

87. x 4  2 x 2  15

Solution





x 4  2 x 2  15  x 2  5 x 2  3



88. x 4  x 2  6

Solution





x4  x2  6  x2  3 x2  2



89. a 2 n  2a n  3

Solution







a2n  2an  3  an  3 an  1

104

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

90. a 2 n  6a n  8

Solution





a2n  6an  8  an  4 an  2



91. 6 x 2 n  7 x n  2

Solution













6 x 2n  7 x n  2  3 x n  2 2 x n  1

92. 9 x 2 n  9 x n  2

Solution

9 x 2n  9 x n  2  3 x n  2 3 x n  1

93. 4x

 9 y 2n

Solution

   3 y    2 x  3 y  2 x  3 y 

4 x 2n  9 y 2n  2 x n

94. 8 x 2 n  2 x n  3

Solution







8 x 2n  2 x n  3  4 x n  3 2 x n  1

95. 10 y

 11 y n  6

Solution





10 y 2n  11 y n  6  5 y n  2 2 y n  3

96. 16 y



 25 y 2n

Solution



16 y 4 n  25 y 2n  y 2n 16 y 2 n  25





 y 2n  4 y n  52    n 2n  y 4y  5 4yn  5







97. 2 x 3  2000

Solution











2 x 3  2000  2 x 3  1000  2 x 3  103  2  x  10  x 2  10 x  100



105

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3 98. 3 y  648

Solution











3 y 3  648  3 y 3  216  3 y 3  63  3  y  6  y 2  6 y  36



99.  x  y   64 3

Solution

 x  y   64   x  y   4   x  y   4  x  y   4  x  y   4    x  y  4   x  2xy  y  4 x  4 y  16 3

100.  x  y   27 3

Solution

 x  y   27   x  y   3   x  y   3  x  y   3  x  y   3    x  y  3  x  2 xy  y  3x  3 y  9 3

6 6 101. 64a  y

Solution

    y   8a  y 8a  y    2a  y   4a  2ay  y   2a  y   4a  2ay  y    2a  y  2a  y   4a  2ay  y  4a  2ay  y 

64a6  y 6  8a3

102. a 6  b6

Solution

    b   a  b (a )  a b  (b )   a  b a  a b  b 

a6  b6  a2

2 2

103. a 3  b3  a  b

Solution









a3  b3  a  b   a  b  a2  ab  b2   a  b  1   a  b  a2  ab  b2  1





104. a2  y 2  5  a  y 

Solution

a  y   5 a  y   a  y a  y   5 a  y   a  y a  y  5 2

106

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

6 6 105. 64x  y

Solution



64 x 6  y 6  4 x 2

   y    4 x  y (4 x )  4 x y  ( y )    4 x  y  16 x  4 x y  y  3

2 2

2 2 106. z  6z  9  225 y

Solution

z 2  6z  9  225 y 2   z  3 z  3  225 y 2   z  3   15 y  2

  z  3  15 y  z  3  15 y 

2 2 107. x  6x  9  144 y

Solution

x 2  6 x  9  144 y 2   x  3 x  3  144 y 2   x  3   12 y  2

  x  3  12 y  x  3  12 y 

2 2 108. x  2x  9 y  1

Solution

x 2  2 x  9 y 2  1  x 2  2 x  1  9 y 2   x  1 x  1  9 y 2   x  1   3 y    x  1  3 y  x  1  3 y  2

109.  a  b   3  a  b   10 2

Solution

a  b  3 a  b  10  a  b  5 a  b  2  a  b  5a  b  2 2

110. 2  a  b   5  a  b   3 2

Solution

2  a  b   5  a  b   3  2  a  b   1  a  b   3   2a  2b  1 a  b  3 2

111.

x6  7x3  8

Solution















x 6  7 x 3  8  x 3  8 x 3  1   x  2  x 2  2 x  4  x  1 x 2  x  1

112. x 6  13 x 4  36 x 2

107

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution











x 6  13 x 4  36 x 2  x 2 x 4  13 x 2  36  x 2 x 2  9 x 2  4  x 2  x  3  x  3  x  2  x  2 

113. x 4  x 2  1

Solution

x 4  x 2  1  x 4  2x 2  1  x 2

     x  1  x   x  1  x  x  1  x    x  x  1 x  x  1  x2  1 x2  1  x2 2

114. x 4  3 x 2  4

Solution

x 4  3x 2  4  x 4  4 x 2  4  x 2

     x  2  x   x  2  x  x  2  x    x  x  2  x  x  2   x2  2 x2  2  x2 2

115. x 4  7 x 2  16

Solution

x 4  7 x 2  16  x 4  8 x 2  16  x 2

     x  4  x   x  4  x  x  4  x    x  x  4  x  x  4   x2  4 x2  4  x2 2

4 2 116. y  2 y  9

Solution

y4  2y2  9  y4  6y2  9  4y2

  y  3  4 y   y  3   2 y    y  3  2 y  y  3  2 y    y  2 y  3  y  2 y  3   y2  3 2

108

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

117. 4a 4  1  3a 2

Solution









 a   2a  1  a  2a  1  a    2a  a  1 2a  a  1

4a4  1  3a2  4a4  4a2  1  a2  2a2  1 2a2  1  a2  2a2  1

118. x 4  25  6 x 2

Solution









  2x    x  5  2 x  x  5  2 x    x  2 x  5  x  2 x  2 

x 4  25  6 x 2  x 4  10 x 2  25  4 x 2  x 2  5 x 2  5  4 x 2  x 2  5

Factor each expression by grouping three terms and two terms. 2 119. x  x  6  xy  2 y

Solution

x 2  x  6  xy  2 y   x  3 x  2  y  x  2   x  2 x  3  y 

2 120. 2x  5x  2  xy  2 y

Solution

2x 2  5 x  2  xy  2 y   2x  1 x  2  y  x  2   x  2 2x  1  y 

121. a 4  2a 3  a 2  a  1

Solution





a4  2a3  a2  a  1  a2 a2  2a  1  a  1  a2  a  1 a  1  1  a  1   a  1 a2  a  1  1





 (a  1) a3  a2  1 122. a 4  a 3  2a 2  a  1

Solution





a4  a3  2a2  a  1  a2 a2  a  2  a  1  a2  a  2 a  1  1  a  1

  a  1 [a (a  2)  1] 2





 (a  1) a3  2a2  1

Factor the indicated monomial from the given expression. 123. 3 x  2; 2

109

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

3x  2  2  32x  22   2  32 x  1

124. 5 x  3; 5

Solution

5 x  3  5  55x  53   5  x  53 

125. x 2  2x  4;2

Solution

   2  x  x  2 2

x 2  2 x  4  2 x2  22x  42 1 2

126. 3 x 2  2 x  5;3

Solution

   3x  x   2

3 x 2  2 x  5  3 33x  23x  53 2

2 3

5 3

127. a  b; a

Solution

a  b  a  aa  ab   a  3  ab 

128. a  b; b

Solution

a  b  b  ab  bb   b  ab  1

129. x  x 1/2 ; x 1/2

Solution

  x x

x  x 1/2  x 1/2 x 1 1/2  x 1/2 1/2 1/2

1/2



1

130. x 3/2  x 1/2 ; x 1/2

Solution



x 3/2  x 1/2  x 1/2 x 3/2  1/2  x 1/2  1/2 x

1/2

 x  1



110

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

131. 2x  2 y ; 2

Solution

 2x 2y    2x  2 y  2   2  2    2 132.

 2x  y 

3a  3b; 3 Solution

 3a 3b    3a  3b  3   3 3  



 3 a  3b



133. ab3/2  a3/2b; ab

Solution  ab3/2 a3/2 b  ab3/2  a3/2 b  ab    ab   ab  ab b1/2  a 1/2





134. ab2  b; b1

Solution  ab2 b  ab2  b  b 1  1  1  b   b 1 3  b ab  b2





Factor completely and simplify each algebraic expression. Write answers using positive exponents. 135. 2 x 4 /5  8 x 2/5

Solution









2 x 4/5  8 x 2/5  2 x 2/5 x 2/5  4

136. 9 x 3/7  18 x 1/7

Solution

9 x 3/7  18 x 1/7  9 x 1/7 x 2/7  2

137. x 6  3 x 2

Solution x 6  3 x 2  x 6 

3 x8 3 x8  3  2  2  2 x x x x2

111

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

138. 5 x 3  10 x 7

Solution 5 x  10 x 3

139.  x  1

1/2

7

 5x  3

  x  1



5 x 10 x



10 x



5 x 10  10 x





5 x 10  2 x



3/2

Solution

 x  1

1/2

140.  x  6 

  x  1

2/3

  x  1

3/2

1/2

[1   x  1]   x  x  1

  x  6

 1   x  6    x  6  

1/2

  x  6

5/3

Solution

 x  6

2/3

141. 2  x  1

  x  6

5/3

3/5

 4 x  x  1

2/3

8/5

Solution

2  x  1



142. x 2  5



3/5

1/5

 4 x  x  1









 x2  5

 x  7

 2  x  1

8/5

2   x  1

 x  1  2 x    

 x  1

8/5



2  x  1

 x  1

8/5

4/5

Solution

 x  5 2

1/5

 x2  5

4/5



 x2  5



4/5





 x 2  5  1   

x2  6

 x  5 2

4/5

Fix It In Exercises 143 and 144, identify the step the first error is made and fix it. 143. Factor completely: 2x2 − 2xy − 8x + 8y

Solution Step 2 was incorrect.



Step 1: 2 x 2  xy  4 x  4 y



Step 2: 2  x  x  y   4  x  y  



Step 3: 2 x  y

 x  4

112