Solution manual for college algebra 10th edition by larson
Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (1004.69 KB, 48 trang )
Solution Manual for College Algebra 10th Edition by Larson
CHAPTER P
Prerequisites
Section P.1
Review of Real Numbers and Their Properties..................................... 2
Section P.2
Exponents and Radicals ......................................................................... 5
Section P.3
Polynomials and Special Products ......................................................... 9
Section P.4
Factoring Polynomials..........................................................................16
Section P.5
Rational Expressions ............................................................................22
Section P.6
The Rectangular Coordinate System and Graphs ............................... 31
Review Exercises ..........................................................................................................37
Problem Solving ...........................................................................................................44
Practice Test ...............................................................................................................48
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
C H A P T E R
Prerequisites
P
Section P.1 Review of Real Numbers and Their Properties
11. (a)
1. irrational
2. origin
−2 −1
1
2
3
4
3
4
5
−5 −4 −3 −2 −1
0
1
(b)
−1
0
1
(c)
4. composite
5. terms
2
−
(d)
x
5
2
x
−5.2
6. Zero-Factor Property
7. −9, − 72 , 5, 23 ,
x
7
2
3. absolute value
x
−7 −6 −5 −4 −3 −2 −1
2, 0, 1, − 4, 2, −11
12. (a)
8.5
6
(a) Natural numbers: 5, 1, 2
8
−1
(d) Rational numbers: −9, − 72 , 5, 23 , 0, 1, − 4, 2, −11
9
10
11
12
2
3
4
5
x
4
3
(c) Integers: −9, 5, 0, 1, − 4, 2, −11
(e) Irrational numbers:
7
(b)
(b) Whole numbers: 0, 5, 1, 2
8.
0
0
(c)
1
x
−4.75
x
−7 −6 −5 −4 −3 −2 −1
2
(d)
5, − 7, − 73 , 0, 3.14, 54 , − 3, 12, 5
−8
3
−5 −4 −3 −2 −1
0
x
1
(a) Natural numbers: 12, 5
(b) Whole numbers: 0, 12, 5
13. −4 > −8
(c) Integers: −7, 0, − 3, 12, 5
−8
(d) Rational numbers: −7, − 73 , 0, 3.14, 54 , − 3, 12, 5
(e) Irrational numbers:
−7
14. 1 <
5
−6
−5
16
3
16
3
0
9. 2.01, 0.6, −13, 0.010110111 . . ., 1, − 6
15.
(a) Natural numbers: 1
5
6
1
>
2
3
0
(d) Rational numbers: 2.01, 0.6, −13, 1, − 6
9, 3.12, 12 π , 7, −11.1, 13
(a) Natural numbers: 25,
(b) Whole numbers: 25,
9, 7, 13
9, 7, 13
9, 7, 13
(d) Rational numbers:
25, −17,
x
6
x
1
16. − 87 < − 73
(e) Irrational numbers: 0.010110111 . . .
− 12
,
5
5
2 5
3 6
(c) Integers: −13, 1, − 6
(c) Integers: 25, −17,
4
2
3
(b) Whole numbers: 1
,
10. 25, −17, − 12
5
x
−4
− 87
−2
− 37
−1
x
0
17. (a) The inequality x ≤ 5 denotes the set of all real
numbers less than or equal to 5.
(b)
x
0
1
2
3
4
5
6
(c) The interval is unbounded.
9, 3.12, 7, −11.1, 13
(e) Irrational numbers: 12 π
2
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.1
18. (a) The inequality x < 0 denotes the set of all real
numbers less than zero.
(b)
x
−2
−1
0
1
2
19. (a) The inequality −2 < x < 2 denotes the set of all
real numbers greater than −2 and less than 2.
−1
0
1
2
20. (a) The inequality 0 < x ≤ 6 denotes the set of all real
numbers greater than zero and less than or equal to 6.
x
0
1
2
3
4
5
6
(c) The interval is bounded.
numbers greater than or equal to 4.
2
3
4
5
6
7
(c) The interval is unbounded.
22. (a)
than 2.
(b)
x
1
2
3
4
(c) The interval is unbounded.
23. (a) The interval [−5, 2) denotes the set of all real
numbers greater than or equal to − 5 and less than 2.
(b)
x
−5
−3
−1
1
3
24. (a) The interval ( −1, 2] denotes the set of all real
numbers greater than −1 and less than or equal to 2.
x
−2
−1
0
1
2
(c) The interval is bounded.
25. y ≥ 0; [0, ∞ )
26. y ≤ 25; ( −∞, 25]
x + 2
=
x + 2
−( x + 2 )
x + 2
= −1.
38. If x > 1, then x − 1 is positive.
x −1
=
x −1
x −1
= 1.
x −1
40. −5 = − 5 because −5 = −5.
41. − −6 < −6 because −6 = 6 and
− −6 = −(6) = −6.
42. − −2 = − 2 because −2 = −2.
43. d (126, 75) = 75 − 126 = 51
44. d ( − 20, 30) = 30 − ( − 20) = 50 = 50
(c) The interval is bounded.
(b)
37. If x < −2, then x + 2 is negative.
39. −4 = 4 because −4 = 4 and 4 = 4.
(−∞, 2) denotes the set of all real numbers less
0
36. − 4 − 4 = − 4( 4) = −16
So,
x
1
33. −1 − −2 = 1 − 2 = −1
So,
21. (a) The interval [4, ∞) denotes the set of all real
(b)
31. 3 − 8 = −5 = −( −5) = 5
35. 5 − 5 = 5(5) = 25
(c) The interval is bounded.
(b)
30. 0 = 0
34. −3 − −3 = −3 − (3) = −6
x
−2
3
32. 6 − 2 = 4 = 4
(c) The interval is unbounded.
(b)
Review of Real Numbers and Their Properties
(
)
( )
45. d − 52 , 0 = 0 − − 52
(
)
=
5
2
( )
= − 11
− − 14
46. d − 14 , − 11
4
4
= − 52 =
5
2
47. d ( x, 5) = x − 5 and d ( x, 5) ≤ 3, so x − 5 ≤ 3.
48. d ( x, −10) = x + 10 , and d ( x, −10) ≥ 6, so
x + 10 ≥ 6.
27. 10 ≤ t ≤ 22; [10, 22]
28. −3 ≤ k < 5; [−3, 5)
29. −10 = −( −10) = 10
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
4
Chapter P
Prerequisites
Expenditures, E
R − E
49. $2524.0 billion
$2982.5 billion
2524.0 − 2982.5 = $458.5 billion
50. $2162.7 billion
$3457.1 billion
2162.7 − 3457.1 = $1294.4 billion
51. $2450.0 billion
$3537.0 billion
2450.0 − 3537.0 = $1087.0 billion
52. $3021.5 billion
$3506.1 billion
3021.5 − 3506.1 = $484.6 billion
Receipts, R
53. 7 x + 4
60. 9 − 7x
Terms: 7 x, 4
(a) 9 − 7( −3) = 9 + 21 = 30
Coefficient: 7
(b) 9 − 7(3) = 9 − 21 = −12
54. 2 x − 3
61. x 2 − 3x + 2
Terms: 2 x, − 3
Coefficient: 2
3
55. 6 x − 5 x
(a)
(0)
(b)
(−1)
2
(b) −(1) + 5(1) − 4 = −1 + 5 − 4 = 0
2
56. 4 x 3 + 0.5 x − 5
Terms: 4 x 3 , 0.5 x, − 5
Coefficients: 4, 0.5
63.
x +1
x −1
(a)
57. 3 3 x 2 + 1
Terms: 3 3 x 2 , 1
(b)
Coefficient: 3 3
2x2 − 3
2
2x , − 3
Coefficient: 2
− 3( −1) + 2 = 1 + 3 + 2 = 6
(a) −( −1) + 5( −1) − 4 = −1 − 5 − 4 = −10
Coefficients: 6, − 5
Terms: 2
2
62. − x 2 + 5 x − 4
Terms: 6 x 3 , − 5 x
58. 2
− 3(0) + 2 = 2
2
64.
2
59. 4 x − 6
1+1
2
=
1−1
0
Division by zero is undefined.
−1 + 1
0
=
= 0
−1 − 1
−2
x − 2
x + 2
(a)
2 − 2
0
=
= 0
2 + 2
4
(b)
−2 − 2
−4
=
−2 + 2
0
(a) 4( −1) − 6 = −4 − 6 = −10
Division by zero is undefined.
(b) 4(0) − 6 = 0 − 6 = −6
65.
1
(h + 6) = 1, h ≠ −6
( h + 6)
Multiplicative Inverse Property
66.
(x
+ 3) − ( x + 3) = 0
Additive Inverse Property
67. x(3 y ) = ( x ⋅ 3) y
= (3x) y
Associative Property of Multiplication
Commutative Property of Multiplication
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.2
68.
1
7
(7 ⋅ 12)
=
( 17 ⋅ 7)12
Exponents and Radicals
5
Associative Property of Multiplication
= 1 ⋅ 12
Multiplicative Inverse Property
= 12
Multiplicative Identity Property
69.
2x
x
8x
3x
5x
−
=
−
=
3
4
12
12
12
70.
3x
x
15 x
4x
19 x
+
=
+
=
4
5
20
20
20
71.
3x 5
x 1
x
⋅
=
⋅
=
10 6
2 2
4
72.
2x
6
2x 7
x 7
7x
÷
=
⋅
=
⋅
=
3
7
3 6
3 3
9
73. False. Because zero is nonnegative but not positive, not
every nonnegative number is positive.
76. (a) Because the price can only be a positive rational
number with at most two decimal places, the
description matches graph (ii).
(b) Because the distance is a positive real number, the
description matches graph (i).
A range of prices can only include zero and positive
numbers with at most two decimal places. So, a
range of prices can be represented by whole
numbers and some noninteger positive fractions.
A range of lengths can only include positive
numbers. So, a range of lengths can be represented
by positive real numbers.
77. (a)
74. False. Two numbers with different signs will always
have a product less than zero.
75. The product of two negative numbers is positive.
n
0.0001
0.01
1
100
10,000
5n
50,000
500
5
0.05
0.0005
(b) (i) As n approaches 0, the value of 5 n increases
without bound (approaches infinity).
(ii) As n increases without bound (approaches
infinity), the value of 5 n approaches 0.
Section P.2 Exponents and Radicals
1. exponent; base
11. (a)
(23 ⋅ 32 )
2
= 23 ⋅ 2 ⋅ 32 ⋅ 2
2. scientific notation
3. square root
4. index; radicand
= 26 ⋅ 34 = 64 ⋅ 81 = 5184
3
(b) −
5
3
5. like radicals
6. conjugates
12. (a)
7. rationalizing
8. power; index
9. (a) 5 ⋅ 53 = 54 = 625
2
(b)
10. (a)
5
1
1
= 5 −2 = 2 =
54
5
25
(33 )
0
=1
3
= 3 ⋅ 34 = 35 = 243
3− 4
(b) 48( −4)
13. (a)
(b)
−3
=
(−2)0
(−4)
= −
3
48
3
= −
64
4
=1
14. (a) 3−1 + 2−2 =
(b)
48
4 ⋅ 3−2
16
= 4 ⋅ 22 ⋅ 3−2 − (−1) = 4 ⋅ 4 ⋅ 3−1 =
2−2 ⋅ 3−1
3
2
(b) −3 = −9
2
3
52
33
5
3− 2
⋅ 52 − 3
= ( −1) 3 ⋅ 2 = −1 ⋅ 3
5
3
3
3
= −3 ⋅ 5−1 = −
5
(3−2 )
2
1
3
= 3−4 =
+
1
4
1
34
=
=
4
12
+
3
12
=
7
12
1
81
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
6
Chapter P
Prerequisites
15. When x = 2,
24. (a)
−3x3 = −3( 2) = −24.
3
(b)
16. When x = 4,
7 x −2 = 7 ( 4 )
−2
=
7
7
.
=
42
16
7 x2
7
= 7 x 2 − 3 = 7 x −1 =
x3
x
12( x + y )
3
3
4
4 3
43 34
64 ⋅ 81
5184
25. (a) = 3 ⋅ 4 =
=
3+ 4
y
y
y
y
y
y7
17. When x = 10,
2
b −2 b
a 2 b 2
(b) −2 = 2 2 = 1, a ≠ 0, b ≠ 0
a a
b a
6 x0 = 6(10) = 6(1) = 6.
0
18. When x = −3,
−1
26. (a) ( x 2 y −2 )
2 x3 = 2( −3) = 2( −27) = −54.
3
−1
−3 x = −3( −2)
= ( x −2 y 2 )
x2
y2
=
4
= −3(16) = −48.
(b)
(5 x 2 z 6 ) (5 x 2 z 6 )
3
−3
( 13 )
3
21. (a)
(5 z )3
3
( 271 ) =
= 12
= (1)(1)(1)
4.
9
=1
= 5 3 z 3 = 125z 3
27. (a)
( )
(b) 5x 4 x 2 = 5x 4 + 2 = 5x6
22. (a)
(b)
( 4 x3 )
2
0
28. (a)
= 40 ⋅ x 0 = 1, x ≠ 0
0
2
( − z ) (3 z 4 )
3
( x + 5)
(2 x2 )
0
−2
= 1, x ≠ −5
1
=
(2x )
2 2
2
( )
23. (a) 6 y 2 y
(b)
(b)
(− 2 x) = (− 2) x2 = 4x2
2
= (53 )( x 6 )( z18 )(5−3 )( x −6 )( z −18 )
= 50 ( x 0 )( z 0 )
20. When x = − 13 ,
12( − x) = 12
−1
= x 2 y −2
19. When x = −2,
4
4
4
3 −1
2
( x + y) = ( x + y)
3
3
=
9( x + y )
(b)
= 6 y ( 2 ⋅ 1) = 6 y ( 4) = 24 y
2
2
2
2
= ( −1) ( z 3 )3z 4
3
= −1 ⋅ 3 ⋅ z 3 + 4 = −3z 7
=
1
4 x4
(4 y −2 )(8 y4 ) = (4)(8)( y 4 − 2 ) = 32 y 2
(z
+ 2)
−3
x −3 y 4
29. (a)
5
(z
−3
+ 2)
−1
= ( z + 2)
−4
=
1
(z
+ 2)
4
3
5 x3
125 x 9
= 4 =
y12
y
3
a −2 b
b 2 b3
b5
(b) −2 = 2 3 = 5
a
b a
a a
30. (a)
(b)
3n ⋅ 3 2 n
3 n + 2n
3 3n
1
1
= 3n + 2 = 3n + 2 = 3 3n − (3n + 2) = 3 − 2 = 2 =
3n
2
3 ⋅3
3
3
3
9
x2 ⋅ xn
x2 + n
1
= 3 + n = x 2 + n − 3 − n = x −1 =
n
3
x ⋅x
x
x
31. 10,250.4 = 1.02504 × 104
35. 9.46 × 1012 = 9,460,000,000,000 kilometers
32. −0.000125 = −1.25 × 10−4
36. 9.0 × 10 − 6 = 0.000009 meter
33. 3.14 × 10−4 = 0.000314
37. (a)
34. − 2.058 × 106 = − 2,058,000
(b)
(2.0 × 109 )(3.4 × 10−4 ) = 6.8 × 105
(1.2 × 107 )(5.0 × 10−3 ) = 6.0 × 104
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.2
38. (a)
6.0 × 108
= 2.0 × 1011
3.0 × 10−3
(b)
3
3
27 = 3
(b)
(
)
(b)
36
( 2)
5
5
4
5/5
= 2
32 x5 =
= 2 = 2
(2 x)5 = 2 x
5
3 =
4
47. (a)
3
16 x5 =
20 =
(b)
44. (a)
3
(b)
128 =
64 ⋅ 2
=
3
3
75
=
4
3
64
3
33
52 ⋅ 3
2
2
3
=
=
18 z
z2
⋅2
2
3
⋅ 2 x2
75 x 2
y4
25 x 2 ⋅ 3
y4
(5 x ) 2
=
2 = 4 2
23 ⋅ 2
=
2
( 2 x)
=
5 = 2 5
4
z
b
75 x 2 y −4 =
(b)
34 x8
3
16
=
27
4a
2
4⋅5
=
3
b
3
= 3x 2
43. (a)
( 4a 2 )
z
= 2 x 3 2 x2
36 = 6
4
=
z ⋅ z
2
18
=
2
=
1
2
6x
182
32a 4
=
b2
(b)
= (6) = 216
(3 x 2 )
6 ⋅ 32 ⋅ x ⋅ y 2
182
=
z3
46. (a)
3
12 ⋅
42. (a)
(b)
5
( )
54 xy 4 =
(b)
27
3
=
2
8
3
40. (a)
3
36 x 2 ⋅ 2 x
= 3y2
27
=
8
7
= 6x 2x
9 = 3
3
41. (a)
72 x3 =
45. (a)
2.5 × 10−3
= 0.5 × 10−5 = 5.0 × 10−6
(b)
5.0 × 102
39. (a)
Exponents and Radicals
=
23 2
3
48. (a)
4
5 3
2
3x 4 y 2 =
4
( y2 )
5 x
y
(b)
5
4
3
2
x4 ⋅ 3 y2
= x ⋅
=
⋅3
2
4
3y2
3 x
y
160 x8 z 4 =
5
32 x 5 ⋅ 5 x 3 z 4
=
5
( 2 x)
5
⋅ 5 x3 z 4
= 2 x 5 5 x3 z 4
49. (a) 2
20 x 2 + 5 125 x 2 = 2
4x2 ⋅ 5 + 5
( 2 x)
= 2
= 4 x
= 29 x
(b) 8 147 x − 3
48 x = 8
2
⋅5 +5
5 + 25 x
25 x 2 ⋅ 5
(5 x)
2
⋅5
5
5
49 ⋅ 3 x − 3 16 ⋅ 3 x
= 8
7 2 ⋅ 3x − 3
= 56
3 x − 12
= 44
3x
42 ⋅ 3x
3x
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Full file8 at />Chapter P Prerequisites
50. (a) 3 3 54 x3 +
3
16 x 3 = 3 3 27 x3 ⋅ 2 +
3
8 x3 ⋅ 2
= 3 3 (3 x ) ⋅ 2 +
3
( 2 x)
3
= 9x
3
= 11x
(b)
3
3
64 x −
2 + 2x
3
27 x 4 =
3
64 ⋅ x −
=
3
(4)
3
3
⋅ x −
= ( 4 − 3 x)
1
=
3
52.
3
1
⋅
3
8
=
2
3
8
⋅
2
3
22
3
2
5
=
14 − 2
53.
54.
3
5 +
55.
5 +
3
56.
7 −3
=
4
57.
3
=
3
3
58. x 2
2
3
27 x3 ⋅ x
3
(3 x )
3
⋅ x
x
x
3
3
=
83 4
83 4
=
= 43 4
3
2
8
14 + 2
=
14 + 2
5
⋅
14 − 2
=
3
5 +
=
5 +
3
6
3
2
6
3
7 −3
⋅
4
(
14 + 2
5
(
14
)
2
(
⋅
5 −
5 −
3
6
=
6
⋅
5 −
5 −
3
=
3
3
=
5 −
5
)
16
(b)
81
x2 3
3
)
−2
14 + 2
=
=
14 − 4
(
5 −
(
5 −
3
3
)
7 +3
)
2
3
(
2
14 + 2
x
2
,x ≠ 0
1
32
35
=
81
=
16
1
(
34
5
32
)
3
=
1
(2)
3
=
1
)
(
= −3
14 + 2
2
=
)
5 −
(
6 = 3
1
5
)
)
7 +3
(
−1 2
100
)
−3
= 10 − 3 =
12
4
=
9
4
32 = 32 4 = 31 2 =
(b)
6
( x + 1)4
64. (a)
6
x3 = x3 6 = x1 2 =
(b)
4
(3 x 2 )
4
1
1000
4
2
=
3
9
=
63. (a)
= ( x + 1)
3
46
= ( x + 1)
23
=
3
( x + 1)2
x
= 3x 2
32 = (321 2 )
12
65. (a)
= 321 4 =
3
(b)
6 −
)
8
3
81
27
3
= 4
= =
16
2
8
)
10
6
−1
= −
(
5
=
9
(b)
4
a2
−3 4
(
)
(
5
62. (a) 100 − 3 2 =
3
61. (a) 32 −3 5 =
3
7 +3
7 −9
=
=
7 +3
4 7 +3
4
60. a 0.4 = a 2 5
=
6
5−3
52
3
=
5−6
(
(
2
5 −
x = x 2 ⋅ x1 2
59. 3 x − 2 3 =
)
− ( 2)
64 = 641 3
= x
⋅2
2
= 4 3 x − 3x
51.
3
3
4
2x =
((2 x) )
4
14 12
32 =
4
= ( 2 x)
16 ⋅ 2 = 2 4 2
18
=
8
2x
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.3
(
12
243( x + 1) = 243( x + 1)
66. (a)
)
Polynomials and Special Products
52
69. t = 0.03125 2 − (12 − h) , 0 ≤ h ≤ 12
12
= ( 243( x + 1))
14
h (in centimeters)
t (in seconds)
=
4
243( x + 1)
0
0
=
4
3 ⋅ 81( x + 1)
1
2.93
2
5.48
3
7.67
4
9.53
5
11.08
6
12.32
7
13.29
8
14.00
9
14.50
= ( x − 1) −1
10
14.80
1
x −1
11
14.93
12
14.96
= 3 4 3( x + 1)
(b)
3
10a 7b =
((10a b) )
13 12
7
9
= (10a 7b)
16
=
6
= a
6
10a ⋅ a ⋅ b
6
10ab
67. (a) ( x − 1)1/ 3 ( x − 1) 2 / 3 = ( x − 1)3 / 3 = x − 1
(b) ( x − 1)1/ 3 ( x − 1) −4 / 3 = ( x − 1) −3 / 3
=
68. (a) (4 x + 3)
5/ 2
(4 x + 3)
−5/3
15/6
= (4 x + 3)
(4 x + 3)
= (4 x + 3)5/6 , x ≠ −
−10/6
3
4
70. Package A: x 3 = 500
x =
(b) (4 x + 3) −5/ 2 (4 x + 3) 2/3 = (4 x + 3) −15/ 6 (4 x + 3) 4/ 6
3
500 ≈ 7.9 in.
Package B: x 3 = 250
= (4 x + 3) −11/6
x =
1
=
(4 x + 3)11 6
3
250 ≈ 6.3 in.
So, 2 x = 2(6.3) = 12.6 in.
Because 7.9 < 12.6, the length of a side of package A is
less than twice the length of a side of package B.
71. False. When x = 0, the expressions are not equal.
( )
72. False. When a power is raised to a power, you multiply the exponents: a n
k
= a nk .
73. False. When a sum is raised to a power, you multiply the sum by itself using the Distributive Property.
(a + b) 2 = a 2 + 2ab + b 2 ≠ a 2 + b 2
74. False. Raising the numerator and denominator to the second power changes the expression. To rationalize the denominator,
b
= 1.
multiply the fraction by
b
Section P.3 Polynomials and Special Products
1. n; an ; a0
5. (a) Standard form: 7x
2. monomial; binomial; trinomial
(b) Degree: 1
Leading coefficient: 7
3. like terms
(c) Monomial
4. First terms; Outer terms; Inner terms; Last terms
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
10
Chapter P
Prerequisites
6. (a) Standard form: 3
13.
(b) Degree: 0
Leading coefficient: 3
(c) Monomial
14.
7. (a) Standard form: − 12 x5 + 14 x
3x + 4
4
= 3+
= 3 + 4 x −1 is not a polynomial
x
x
because it includes a term with a negative exponent.
x2 + 2 x − 3
is a polynomial.
2
Standard form:
(b) Degree: 5
Leading coefficient: − 12
1 2
3
x + x −
2
2
15. y 2 − y 4 + y 3 is a polynomial.
(c) Binomial
Standard form: − y 4 + y 3 + y 2
8. (a) Standard form: 2 x + 3
(b) Degree: 1
Leading coefficient: 2
16. y 4 −
y is not a polynomial because it includes a term
with a square root.
(c) Binomial
17. (6 x + 5) − (8 x + 15) = 6 x + 5 − 8 x − 15
9. (a) Standard form: −4 x 5 + 6 x 4 + 1
= (6 x − 8 x) + (5 − 15)
(b) Degree: 5
Leading coefficient: −4
= −2 x − 10
18. ( 2 x 2 + 1) − ( x 2 − 2 x + 1) = 2 x 2 + 1 − x 2 + 2 x − 1
(c) Trinomial
= ( 2 x 2 − x 2 ) + 2 x + (1 − 1)
10. (a) Standard form: 25 y 2 − y + 1
(b) Degree: 2
Leading coefficient: 25
= x2 + 2x
19. (t 3 − 1) + (6t 3 − 5t ) = t 3 − 1 + 6t 3 − 5t
(c) Trinomial
= (t 3 + 6t 3 ) − 5t − 1
11. 2 x − 3x 3 + 8 is a polynomial.
= 7t 3 − 5t − 1
Standard form: −3x3 + 2 x + 8
12. 5 x 4 − 2 x 2 + x −2 is not a polynomial because it includes
a term with a negative exponent.
20. ( 4 y 2 − 3) + ( − 7 y 2 + 9) = 4 y 2 − 3 − 7 y 2 + 9
= ( 4 y 2 − 7 y 2 ) + ( − 3 + 9)
= − 3y2 + 6
21. (15 x 2 − 6) + ( − 8.3 x 3 − 14.7 x 2 − 17) = 15 x 2 − 6 − 8.3 x 3 − 14.7 x 2 − 17
= − 8.3 x 3 + (15 x 2 − 14.7 x 2 ) + ( − 6 − 17)
= − 8.3 x 3 + 0.3 x 2 − 23
22. (15.6 w4 − 14 w − 17.4) + (16.9 w4 − 9.2 w + 13) = 15.6 w4 − 14 w − 17.4 + 16.9w4 − 9.2 w + 13
= (15.6 w4 + 16.9 w4 ) + ( −14 w − 9.2 w) + ( −17.4 + 13)
= 32.5w4 − 23.2w − 4.4
23. 5 z − 3z − (10 z + 8) = 5 z − (3z − 10 z − 8)
= 5 z − 3z + 10 z + 8
= (5 z − 3z + 10 z ) + 8
= 12 z + 8
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.3
24.
( y3
Polynomials and Special Products
11
+ 1) − ( y 2 + 1) + (3 y − 7) = y 3 + 1 − ( y 2 + 1) − (3 y − 7)
= y3 + 1 − y 2 − 1 − 3 y + 7
= y 3 − y 2 − 3 y + (1 − 1 + 7)
= y3 − y 2 − 3 y + 7
25. 3 x( x 2 − 2 x + 1) = 3 x( x 2 ) + 3 x( −2 x) + 3 x(1)
3
(
)(
2
= 3x − 6 x + 3x
(
)
x3 − x 2 + 2 x
x2 −
4
(
2
= −15 x − 6 x
2 x 4 − x3 + 4 x 2
6 x 3 − 3 x 2 + 12 x
3
4x2 − 2x + 8
= 4 y3 − 7 y 4
2 x 4 + 5 x 3 + 5 x 2 + 10 x + 8
3
39.
31. −2 x(0.1x + 17) = ( −2 x)(0.1x) + ( −2 x)(17)
= −0.2 x 2 − 34 x
) = (6 y)(5) − (6 y)( 83 y)
= 30 y −
9 2
y
4
= − 94 y 2 + 30 y
33.
x + 4
× x + 3x + 2
( ) = 2(2 y ) + (−3.5 y)(2 y )
3
y
8
( x + 7)( x + 5) = x 2 + 5 x + 7 x + 35
( x + 10)( x − 10)
(x
− 8)( x + 4) = x 2 + 4 x − 8 x − 32
2
41.
( x + 2 y)( x − 2 y)
= x2 − ( 2 y) = x2 − 4 y 2
2
42. ( 4a + 5b)( 4a − 5b) = ( 4a) − (5b) = 16a 2 − 25b2
2
2
43. ( 2 x + 3) = ( 2 x ) + 2( 2 x )(3) + 32
2
2
= 4 x 2 + 12 x + 9
44. (5 − 8 x ) = 52 + ( 2)(5)( −8 x ) + ( −8 x )
2
2
= 25 − 80 x + 64 x 2
= x 2 − 4 x − 32
35. (3 x − 5)( 2 x + 1) = 6 x + 3x − 10 x − 5
= x2 − 102 = x2 − 100
40. ( 2 x + 3)( 2 x − 3) = ( 2 x) − 32 = 4 x 2 − 9
= x 2 + 12 x + 35
34.
)
2
= −4.5t 3 − 15t
(
x + 2 = x4 + 2 x2 + x + 2
)(
2x2 −
29. (1.5t 2 + 5)( −3t ) = (1.5t 2 )( −3t ) + (5)( −3t )
32. 6 y 5 −
3
38. 2 x 2 − x + 4 x 2 + 3x + 2
28. −3 x(5 x + 2) = −3 x(5 x) + ( −3 x)( 2)
3
x + 2
x + 0x + 2x2 +
= −15 z 2 + 5 z
= −7 y + 4 y
x + 1
x 4 − x3 + 2 x 2
27. −5 z (3 z − 1) = −5 z (3 z ) + ( −5 z )( −1)
4
x + 2
2
× x +
( )
= 4 y 4 + 2 y3 − 3 y 2
30. ( 2 − 3.5 y ) 2 y
2
x −
26. y 2 4 y 2 + 2 y − 3 = y 2 4 y 2 + y 2 ( 2 y ) + y 2 ( −3)
3
)
37. x 2 − x + 2 x 2 + x + 1
= 64 x 2 − 80 x + 25
2
= 6x2 − 7 x − 5
36. (7 x − 2)( 4 x − 3) = 28 x − 21x − 8 x + 6
(
)
45. 4 x3 − 3
= ( 4 x3 ) − 2( 4 x3 )(3) + (3)
2
2
= 16 x 6 − 24 x3 + 9
2
= 28 x 2 − 29 x + 6
2
46. (8 x + 3) = (8 x) + 2(8 x)(3) + 32
2
2
= 64 x 2 + 48 x + 9
47.
(x
+ 3) = x 3 + 3( x ) (3) + 3( x)(3) + 33
3
2
2
= x 3 + 9 x 2 + 27 x + 27
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
12
48.
Chapter P
(x
Prerequisites
− 2) = x 3 − 3 x 2 ( 2) + 3 x ( 2) − 23
3
56. ( 2.4 x + 3) = ( 2.4 x) + 2( 2.4 x )(3) + (3)
2
2
= x 3 − 6 x 2 + 12 x − 8
3
2
2
57. ( x − 3) + y = ( x − 3) + 2 y ( x − 3) + y 2
2
= 8 x3 − 12 x 2 y + 6 xy 2 − y 3
= x 2 − 6 x + 9 + 2 xy − 6 y + y 2
50. (3x + 2 y ) = (3x) + 3(3x) ( 2 y ) + 3(3x)( 2 y ) + ( 2 y )
3
3
2
2
(
1
x
5
)(
−3
1
x
5
) ( )
1
x
5
+3 =
2
1 2
x
25
=
− (3)
= x 2 + 2 xy + y 2 − 6 x − 6 y + 9
3
= 27 x3 + 54 x 2 y + 36 xy 2 + 8 y 3
51.
2
58. ( x + 1) − y = ( x + 1) + 2( x + 1)( − y ) + ( − y )
2
= x 2 − 2 xy + y 2 + 2 x − 2 y + 1
−9
59. ( m − 3) + n
( m − 3) − n = ( m − 3) − ( n)
2
2
2
= 2.25 x − 16
= m 2 − n 2 − 6m + 9
53. ( − 6 x + 3 y )( − 6 x − 3 y ) = ( − 6 x) − (3 y )
2
2
60. ( x − 3 y ) + z
( x − 3 y ) − z = ( x − 3 y ) − ( z )
2
= 36 x 2 − 9 y 2
)(
) ( )
2
( 14 x − 5)
2
=
( 14 x)
=
1 2
x
16
2
− ( 4b 2 )
( )
− 2 14 x (5) + ( −5)
−
5
x
2
2
= x 2 − 6 xy + 9 y 2 − z 2
2
61. (u + 2)(u − 2)(u 2 + 4) = (u 2 − 4)(u 2 + 4)
= 9a 6 − 16b 4
55.
2
= m 2 − 6m + 9 − n 2
2
54. 3a3 − 4b 2 3a 3 + 4b 2 = 3a3
2
= x 2 + 2 x + 1 − 2 xy − 2 y + y 2
2
52. (1.5 x − 4)(1.5 x + 4) = (1.5 x ) − ( 4)
(
2
= 5.76 x 2 + 14.4 x + 9
49. ( 2 x − y ) = ( 2 x) − 3( 2 x) y + 3( 2 x) y 2 − y 3
3
2
2
= u 4 − 16
62.
(x
+ y )( x − y )( x 2 + y 2 ) = ( x 2 − y 2 )( x 2 + y 2 )
= ( x2 ) − ( y 2 ) = x4 − y 4
2
+ 25
2
63. ( − 3 x 3 + x 2 + 9) − ( 4 x 2 − 5) = − 3 x 3 + x 2 + 9 − 4 x 2 + 5
= − 3 x 3 + ( x 2 − 4 x 2 ) + (9 + 5)
= − 3 x 3 − 3 x 2 + 14
64. ( 2t 4 − 10t 3 − 4t ) − ( − 7t 4 + 5t 2 − 1) = 2t 4 − 10t 3 − 4t + 7t 4 − 5t 2 + 1
= ( 2t 4 + 7t 4 ) − 10t 3 − 5t 2 − 4t + 1
= 9t 4 − 10t 3 − 5t 2 − 4t + 1
65.
( y2 + 3 y − 5)( y2 − 6 y + 4)
(
y2 + 3y − 5
× x2 − x + 3
× y − 6y + 4
y 4 + 3 y3 − 5 y 2
x 4 + 4 x3 − x 2
2
− x3
6 y − 18 y + 30 y
− 4 x2 + x
+ 4 y 2 + 12 y − 20
4
3
)
x2 + 4x − 1
2
3
)(
66. x 2 + 4 x − 1 x2 − x + 3
+ 3x 2 + 12 x − 3
2
x + 3 x − 2 x 2 + 13x − 3
y − 3 y − 19 y + 42 y − 20
67.
(
4
3
x +
y
)(
x −
) ( x) − ( y)
y =
2
2
= x − y
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.3
(
)(
68. 5 +
)
( x)
x = (5) −
x 5−
2
(
69. x −
5
)
( 5) + ( 5)
= x − 2( x)
2
P = 135 x − (93x + 35,000)
= 135 x − 93 x − 35,000
2
= 42 x − 35,000
(b) For x = 5000:
= x2 − 2 5x + 5
70.
(x +
3
)
2
2
= x + 2x 3 +
( 3)
13
71. (a) Profit = Revenue − Cost
2
= 25 − x
2
Polynomials and Special Products
P = 42(5000) − 35,000
2
= 210,000 − 35,000
= 175,000
2
= x + 2x 3 + 3
So, the profit is $175,000.
72. (a) 500(1 + r ) = 500( r + 1)
2
2
= 500( r 2 + 2r + 1)
= 500r 2 + 1000r + 500
(b)
r
500(1 + r )
2
2 12 %
3%
4%
4 12 %
5%
$525.31
$530.45
$540.80
$546.01
$551.25
(c) As r increases, the amount increases.
73. (a) The possible gene combinations of an offspring with albino coloring is 14 , or 25%.
(b)
(0.5 N
+ 0.5a ) = (0.5 N ) + 2(0.5 N )(0.5a ) + (0.5a )
2
2
2
= 0.25 N 2 + 0.5 Na + 0.25a 2
(c) The probability of an offspring with albino coloring is represented by the coefficient of the a 2 term,
which is 0.25, or 25%.
74. (a) (100 − x)(100 + x) = (100) − ( x)
2
2
= 10,000 − x 2
(b) As x increases, the area of the foundation decreases.
(c) When x = 21 feet, the area of the new foundation is
A = 10,000 − ( 21)
2
= 9559 square feet.
75. Area of shaded region = Area of outer rectangle − Area of inner rectangle
A = 2 x( 2 x + 6) − x( x + 4)
= 4 x 2 + 12 x − x 2 − 4 x
= 3x 2 + 8 x
76. Area of shaded region = Area of outer triangle − Area of inner triangle
A =
1
2
(9 x)(12 x) − 12 (6 x)(8x)
= 54 x 2 − 24 x 2
= 30 x 2
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
14
Chapter P
Prerequisites
77. Area of shaded region = Area of larger square − Area of smaller square
A = ( 4 x + 2) − ( x − 1)
2
2
= 16 x 2 + 16 x + 4 − ( x 2 − 2 x + 1)
= 15 x 2 + 18 x + 3
78. Area of shaded region = Area of larger triangle − Area of smaller triangle
A =
1
2
(2 x
+ 8) −
2
=
1 2
4x
2
=
1
2
=
3 2
x
2
(3 x 2
1
2
(x
+ 4)
2
+ 32 x + 64 − ( x 2 + 8 x + 16)
+ 24 x + 48)
+ 12 x + 24
79. (a) V = l ⋅ w ⋅ h
= ( 26 − 2 x )(18 − 2 x)( x)
= 2(13 − x )( 2)(9 − x)( x)
= 4 x( −1)( x − 13)( −1)( x − 9)
= 4 x( x − 13)( x − 9)
= 4 x 3 − 88 x 2 + 468 x
(b) When x = 1: V = 4(1) − 88(1) + 468(1) = 384 cm3
3
2
x (cm )
When x = 2: V = 4( 2) − 88( 2) + 468( 2) = 616 cm
3
2
V (cm
3
3
)
1
2
3
384
616
720
When x = 3: V = 4(3) − 88(3) + 468(3) = 720 cm3
3
3
80. (a) V = l ⋅ w ⋅ h
=
1
2
( 45 − 3x)(15 −
=
1
2
( 45 − 3x)(15 x
=
1
2
=
2 x) x
− 2x2 )
(675x − 90 x 2 − 45x 2 + 6 x3 )
1 6 x3 − 135 x 2 + 675 x
)
2(
(b) When x = 3:
V =
1
6
2
(3)
3
3
− 135(3) + 675(3) =
=
1
2
1
2
(6 ⋅ 27
[162 − 1215 +
= 486 cm
When x = 5:
V =
1
6
2
(5)
3
V =
1
6
2
( 7)
3
2025] =
1
2
(972)
3
2
− 135(5) + 675(5) =
1
2
[6 ⋅ 125 − 135 ⋅ 25 + 3375]
=
1
2
[750 − 3375 + 3375]
= 375 cm
When x = 7:
− 135 ⋅ 9 + 2025)
=
1
2
(750)
3
2
− 135(7) + 675(7) =
1
2
[6 ⋅ 343 − 135 ⋅ 49 +
=
1
2
[2058 − 6615 +
4725]
4725] =
1
2
(168)
= 84 cm3
x (cm )
Volume (cm
3
)
3
5
7
486
375
84
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.3
Polynomials and Special Products
15
81. (a) Approximations will vary. Actual safe loads for x = 12:
2
2
S6 = 0.06(12) − 2.42(12) + 38.71 = 335.2561 (using a calculator)
2
2
S8 = 0.08(12) − 3.30(12) + 51.93 = 568.8225
(using a calculator)
Difference in safe loads = 568.8225 − 335.2561 = 233.5664 pounds
(b) The difference in safe loads decreases in magnitude as the span increases.
82. (a) T = R + B = 1.1x + (0.0475 x 2 − 0.001x + 0.23)
= 0.0475 x 2 + 1.099 x + 0.23
(b)
x ( mi hr )
30
40
55
T (ft )
75.95
120.19
204.36
(c) Stopping distance increases at an accelerating rate as speed increases
(
)
83. False. 4 x 2 + 1 (3x + 1) = 12 x3 + 4 x 2 + 3x + 1
(
) (
2
2
87. Because x m x n = x m + n , the degree of the product is
m + n.
)
84. False. x + 1 + − x + 3 = 4, which is not a
88. If the degree of one polynomial is m and the degree of
the second polynomial is n (and n > m ), the degree of
the sum of the polynomials is n.
second-degree polynomial.
85. False. ( 4 x + 3) + ( −4 x + 6) = 4 x + 3 − 4 x + 6
89. The middle term was omitted when squaring the
binomial.
= 3+ 6
= 9
(x
86. True. The leading coefficient of
(an x
n
− 3) = ( x) − 2( x)(3) + (3)
2
2
2
= x2 − 6x + 9
+ + a 0 )(bm x + + b0 ) is anbm .
m
≠ x2 + 9
90. (a) The cardboard was 52 inches long and 24 inches wide. The box was constructed by cutting a square of x inches by x inches
from each corner of the rectangular piece of cardboard and folding the side pieces up.
(b) The polynomial is a 3rd degree polynomial because the length, width, and height of the box are expressions in the
variable x.
(c) Volume = length × width × height
= (52 − 2 x)( 42 − 2 x) x
= 4 x 3 − 188 x 2 + 2184 x
The value of x (to the nearest tenth of an inch) that yields the maximum possible volume of the box may be found by
substituting values of x into the volume equation.
x
5
6
7
8
9
10
V
6720
7200
7448
7488
7344
7040
The maximum volume occurs for a value of x between 7 and 8.
x
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
V
7461.0
7471.9
7480.7
7487.6
7492.5
7495.4
7496.4
7495.5
7492.7
The maximum volume occurs when x ≈ 7.7 inches.
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Chapter P
16
Prerequisites
91. The unknown polynomial may be found by adding − x 3 + 3 x 2 + 2 x − 1 and 5 x 2 + 8:
(− x3
+ 3 x 2 + 2 x − 1) + (5 x 2 + 8) = − x 3 + (3 x 2 + 5 x 2 ) + 2 x + ( −1 + 8)
= − x3 + 8 x 2 + 2 x + 7
92.
(x + y)
2
≠ x2 + y 2
Let x = 3 and y = 4.
(3 + 4)2
= (7) = 49
2
32 + 42 = 9 + 16 = 25
>
Not Equal
If either x or y is zero, then (x + y ) would equal x 2 + y 2 .
2
Section P.4 Factoring Polynomials
1. factoring
15.
(x
− 1) − 4 = ( x − 1) − ( 2)
2
2
2
2. completely factored
= ( x − 1) + 2
( x − 1) − 2
3. perfect square binomial
= ( x + 1)( x − 3)
4. grouping
16. 25 − ( z + 5) = 52 − ( z + 5)
2
(
= 5 − ( z + 5)
5 + ( z + 5)
)
5. 2 x3 − 6 x = 2 x x 2 − 3
3
(
2
2
)
6. 3z − 6 z + 9 z = 3z z − 2 z + 3
7. 3 x( x − 5) + 8( x − 5) = ( x − 5)(3 x + 8)
8.
(x
2
+ 3) − 4( x + 3) = ( x + 3) ( x + 3) − 4
2
= ( x + 3)( x − 1)
9. x 2 − 81 = x 2 − 92
= ( x + 9)( x − 9)
10. x 2 − 64 = x 2 − 82
= ( x + 8)( x − 8)
= (5 − z − 5)(5 + z + 5)
= − z ( z + 10)
(
= (9u 2
)(
)
+ 1)(3u + 1)(3u − 1)
17. 81u 4 − 1 = 9u 2 + 1 9u 2 − 1
(
= ( x2
)(
)
+ 4 y 2 )( x + 2 y )( x − 2 y )
18. x 4 − 16 y 4 = x 2 + 4 y 2 x 2 − 4 y 2
19. x 2 − 4 x + 4 = x 2 − 2( 2) x + 22 = ( x − 2)
2
20. 4t 2 + 4t + 1 = ( 2t ) + 2( 2t )(1) + 12 = ( 2t + 1)
2
2
11. 25 y 2 − 4 = (5 y) − 22 = (5 y + 2)(5 y − 2)
21. 25z 2 − 30 z + 9 = (5z) − 2(5z )(3) + 32 = (5z − 3)
12. 4 y 2 − 49 = ( 2 y ) − 7 2 = ( 2 y + 7)( 2 y − 7)
22. 36 y 2 + 84 y + 49 = (6 y ) + 2(6 y )(7) + 7 2
2
2
64 − 9 z = 8 − (3z ) = (8 + 3z )(8 − 3z )
2
13.
2
2
2
2
= ( 6 y + 7)
2
23. 4 y 2 − 12 y + 9 = ( 2 y ) − 2( 2 y )(3) + (3)
2
14. 81 − 36 z 2 = 9(9 − 4 z 2 )
2
= 932 − ( 2 z )
= 9(3 + 2 z )(3 − 2 z )
= ( 2 y − 3)
2
2
24. 9u 2 + 24uv + 16v 2 = (3u ) + 2(3u )( 4v) + ( 4v)
2
= (3u + 4v)
(
2
2
)
25. x3 − 8 = x3 − 23 = ( x − 2) x 2 + 2 x + 4
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2
Solution Manual for College Algebra 10th Edition by Larson
Section P.4
(
)
Factoring Polynomials
17
26. x3 + 125 = x3 + 53 = ( x + 5) x 2 − 5 x + 25
29. 27 x 3 + 8 = (3 x) + 23 = (3 x + 2)(9 x 2 − 6 x + 4)
27. 8t 3 − 1 = ( 2t ) − 13 = ( 2t − 1)( 4t 2 + 2t + 1)
30. 64 y 3 − 125 = ( 4 y ) − 53
3
28. 27 z + 1 = (3 z ) + 1 = (3 z + 1)(9 z − 3 z + 1)
3
3
3
2
3
3
= ( 4 y − 5)(16 y 2 + 20 y + 25)
(
31. u 3 + 27v3 = u 3 + (3v) = (u + 3v) u 2 − 3uv + 9v 2
3
32.
(x
+ 2) − y 3 = ( x + 2) − ( y )
3
3
)
3
2
= ( x + 2 − y ) ( x + 2) + y ( x + 2) + y 2
= ( x − y + 2)( x 2 + xy + 4 x + y 2 + 2 y + 4)
(
33. x 2 + x − 2 = ( x + 2)( x − 1)
)
(
)
47. 3x5 + 6 x3 − 2 x 2 − 4 = 3x3 x 2 + 2 − 2 x 2 + 2
= (3x3 − 2)( x 2 + 2)
34. x2 + 5x + 6 = ( x + 2)( x + 3)
(
)
(
)
48. 8 x5 − 6 x 2 + 12 x3 − 9 = 2 x 2 4 x3 − 3 + 3 4 x3 − 3
35. s 2 − 5s + 6 = ( s − 3)( s − 2)
= ( 4 x − 3)( 2 x + 3)
3
36. t − t − 6 = (t + 2)(t − 3)
2
2
49. a ⋅ c = ( 2)(9) = 18. Rewrite the middle term,
37. 3x 2 + 10 x − 8 = (3x − 2)( x + 4)
9 x = 6 x + 3 x, because (6)(3) = 18 and 6 + 3 = 9.
38. 2 x 2 − 3x − 27 = ( 2 x − 9)( x + 3)
2 x 2 + 9 x + 9 = 2 x 2 + 6 x + 3x + 9
= 2 x( x + 3) + 3( x + 3)
39. 5x + 31x + 6 = (5 x + 1)( x + 6)
= ( x + 3)( 2 x + 3)
2
40. 8x2 + 51x + 18 = (8x + 3)( x + 6)
41. −5 y 2 − 8 y + 4 = −(5 y 2 + 8 y − 4)
= −(5 y − 2)( y + 2)
42. −6 z 2 + 17 z + 3 = −(6 z 2 − 17 z − 3)
= −(6 z + 1)( z − 3)
43. x3 − x 2 + 2 x − 2 = x 2 ( x − 1) + 2( x − 1)
= ( x − 1)( x 2 + 2)
44. x3 + 5 x 2 − 5 x − 25 = x 2 ( x + 5) − 5( x + 5)
= ( x + 5)( x − 5)
2
45. 2 x3 − x 2 − 6 x + 3 = x 2 ( 2 x − 1) − 3( 2 x − 1)
50. a ⋅ c = (6)( −2) = −12. Rewrite the middle term,
x = 4 x − 3 x, because 4( −3) = −12 and
4 + ( −3) = 1.
6 x 2 + x − 2 = 6 x 2 + 4 x − 3x − 2
= 2 x(3x + 2) − 1(3x + 2)
= ( 2 x − 1)(3x + 2)
51. a ⋅ c = (6)( −15) = −90. Rewrite the middle term,
− x = −10 x + 9 x, because ( −10)(9) = −90 and
−10 + 9 = −1.
6 x 2 − x − 15 = 6 x 2 − 10 x + 9 x − 15
= 2 x(3x − 5) + 3(3x − 5)
= ( 2 x + 3)(3x − 5)
= ( 2 x − 1)( x 2 − 3)
46. 3x3 + x 2 − 15 x − 5 = x 2 (3 x + 1) − 5(3x + 1)
= ( x 2 − 5)(3 x + 1)
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
18
Chapter P
Prerequisites
52. a ⋅ c = (12)(1) = 12. Rewrite the middle term,
60. 9 x 2 + 12 x − 3 x3 = −3x3 + 9 x 2 + 12 x
−13 x = −12 x − x, because ( −12)( −1) = 12 and
= −3x( x 2 − 3 x − 4)
−12 − 1 = −13.
= −3x( x − 4)( x + 1)
2
2
12 x − 13x + 1 = 12 x − 12 x − x + 1
61.
= 12 x( x − 1) − 1( x − 1)
( x2
2
+ 3) − 16 x 2 = ( x 2 + 3) + 4 x
( x + 3) − 4 x
2
= ( x 2 + 4 x + 3)( x 2 − 4 x + 3)
= ( x − 1)(12 x − 1)
(
= ( x + 3)( x + 1)( x − 3)( x − 1)
)
53. 6 x 2 − 54 = 6 x2 − 9 = 6( x + 3)( x − 3)
(
62.
)
54. 12 x 2 − 48 = 12 x 2 − 4 = 12( x + 2)( x − 2)
( x2
+ 8) − 36 x 2 = ( x 2 + 8) − (6 x)
2
2
2
= ( x 2 + 8) − 6 x
( x + 8) + 6 x
= ( x 2 − 6 x + 8)( x 2 + 6 x + 8)
55. x3 − x 2 = x2 ( x − 1)
(
2
= ( x − 4)( x − 2)( x + 4)( x + 2)
)
56. x3 − 16 x = x x 2 − 16 = x( x + 4)( x − 4)
57. 1 − 4 x + 4 x 2 = (1 − 2 x)
63. 2 x3 + x 2 − 8 x − 4 = x 2 ( 2 x + 1) − 4( 2 x + 1)
2
= ( 2 x + 1)( x 2 − 4)
= ( 2 x + 1)( x + 2)( x − 2)
58. −9 x 2 + 6 x − 1 = −(9 x 2 − 6 x + 1)
= −(9 x 2 − 3x − 3 x + 1)
64. 3 x 3 + x 2 − 27 x − 9 = x 2 (3 x + 1) − 9(3 x + 1)
= − 3 x(3 x − 1) − (3 x − 1)
= (3 x + 1)( x 2 − 9)
= −(3 x − 1)
= (3 x + 1)( x + 3)( x − 3)
2
59. 2 x 2 + 4 x − 2 x3 = −2 x( − x − 2 + x 2 )
65. 2 x(3 x + 1) + (3 x + 1) = (3 x + 1) 2 x + (3 x + 1)
2
= −2 x( x 2 − x − 2)
= (3 x + 1)(5 x + 1)
= −2 x( x + 1)( x − 2)
66. 4 x( 2 x − 1) + ( 2 x − 1) = ( 2 x − 1) 4 x + ( 2 x − 1)
2
= ( 2 x − 1)(6 x − 1)
67. 2( x − 2)( x + 1) − 3( x − 2) ( x + 1) = ( x − 2)( x + 1) 2( x + 1) − 3( x − 2)
2
2
= ( x − 2)( x + 1)[2 x + 2 − 3 x + 6]
= ( x − 2)( x + 1)( − x + 8)
= −( x − 2)( x + 1)( x − 8)
68. 2( x + 1)( x − 3) − 3( x + 1) ( x − 3) = ( x + 1)( x − 3) 2( x − 3) − 3( x + 1)
2
2
= ( x + 1)( x − 3)[2 x − 6 − 3x − 3]
= ( x + 1)( x − 3)( − x − 9)
= −( x + 1)( x − 3)( x + 9)
69. 5( 2 x + 1) ( x + 1) + ( 2 x + 1)( x + 1) = ( 2 x + 1)( x + 1) 5( 2 x + 1) + ( x + 1)
2
2
3
2
= ( 2 x + 1)( x + 1) (10 x + 5 + x + 1)
2
= ( 2 x + 1)( x + 1) (11x + 6)
2
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.4
Factoring Polynomials
19
70. 7(3 x + 2) (1 − x) + (3 x + 2)(1 − x) = (3 x + 2)(1 − x) 7(3 x + 2) + (1 − x)
2
2
3
2
= (3 x + 2)(1 − x) ( 21x + 14 + 1 − x)
2
= (3 x + 2)(1 − x) ( 20 x + 15)
2
= 5(3 x + 2)(1 − x) ( 4 x + 3)
2
72.
4 y2
25
2
2
x
2
1
4
74. 9 y 2 −
3
y
2
+
2
()
= ( y + 23 )( y
= x3 −
1
4
)
−
2y
3
4
9
+
)
3
2
+
3
y
4
9
16
+
)
1
x
1
1
1
x
2
x
x
x
x
x
1
1
1
x
1
x
1
1
x
1
x
1
x
x
1
x
x
x
x
1
1
1
2
= π ( r + 2) − r 2
= π ( 4r + 4)
= 4π ( r + 1)
1
x
x
1
x
1
x
1
1
x
x
x
= π r 2 + 4r + 4 − r 2
x
1
x
x
1
2
x
1
x
1
81. A = π ( r + 2) − π r 2
77. x + 3x + 2 = ( x + 2)( x + 1)
x
1
1
2
1
1
x
2
1
2
3
( 43 )
= ( x − 34 )( x
27
64
( 14 ) + ( 14 )
2
2
3
x
1
80. 3x 2 + 7 x + 2 = (3x + 1)( x + 2)
= (3 y ) − 2(3 y )
= y3 +
1
)
(
8
27
1
1
2
1 2
2
= 3y −
75. y 3 +
1
x
x
( 12 ) + ( 12 )
1
16
x
1
= z 2 + 2( z )
= (z +
x
x
( 52 y) − (8)
= ( 52 y + 8)( 52 y − 8)
− 64 =
73. z 2 + z +
76. x 3 −
79. 2 x2 + 7 x + 3 = ( 2 x + 1)( x + 3)
( 13 )
= ( 4 x + 13 )( 4 x − 13 )
= ( 4 x) −
1
9
71. 16 x 2 −
82. Area =
1
2
=
5
8
=
( x2
5 2
x
8(
5 2
x
8(
=
5
8
=
1
78. x + 4 x + 3 = ( x + 3)( x + 1)
2
(x
(x
( 54 )( x + 3) − 12 (5)(4)
+ 3)
+ 6 x + 9) −
5
8
(16)
+ 6 x + 9 − 16)
+ 6 x − 7)
+ 7)( x − 1)
x
x
1
x
1
1
x
1
1
1
x
1
1
x
1
x
1
1
x
1
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1
Solution Manual for College Algebra 10th Edition by Larson
20
Chapter P
Prerequisites
83. (a) V = π R 2 h − π r 2 h
= π h( R 2 − r 2 )
= π h( R + r )( R − r )
(b) Let w = thickness of the shell and let p = average radius of the shell.
p
w
h
R
r
So, R = p +
1
1
w and r = p − w
2
2
V = π h( R + r )( R − r )
1
1
1
1
= π h p + w + p − w p + w − p − w
2
2
2
2
= π h(2 p)( w)
= 2π pwh
= 2π (average radius)(thickness of shell) h
84. kQx − kx 2 = kx(Q − x)
85. For x 2 + bx − 15 to be factorable, b must equal m + n where mn = −15.
Factors of −15
Sum of factors
(15)(−1)
15 + ( −1) = 14
(−15)(1)
−15 + 1 = −14
(3)( −5)
3 + ( −5) = −2
(−3)(5)
−3 + 5 = 2
The possible b-values are 14, −14, − 2, and 2.
86. For x 2 + bx + 24 to be factorable, b must equal m + n where mn = 24.
Factors of 24
Sum of factors
(1)(24)
1 + 24 = 25
( −1)( −24)
−1 + ( −24) = −25
(2)(12)
2 + 12 = 14
( −2)(−12)
−2 + ( −12) = −14
(3)(8)
3 + 8 = 11
( −3)(−8)
−3 + ( −8) = −11
(4)(6)
4 + 6 = 10
(−4)(−6)
−4 + ( −6) = −10
The possible b-values are 25, − 25, 14, −14, 11, −11, 10, −10.
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.4
Factoring Polynomials
21
87. For 2 x 2 + 5 x + c to be factorable, the factors of 2c must add up to 5.
Possible c-values
2c
Factors of 2c that add up to 5
2
4
3
6
−3
−6
−7
−14
−12
−24
(1)(4) = 4 and 1 + 4 = 5
(2)(3) = 6 and 2 + 3 = 5
(6)(−1) = −6 and 6 + (−1) = 5
(7)(−2) = −14 and 7 + (−2) = 5
(8)(−3) = −24 and 8 + (−3) = 5
These are a few possible c-values. There are many correct answers.
If c = 2:
2 x 2 + 5 x + 2 = ( 2 x + 1)( x + 2)
If c = 3:
2 x 2 + 5 x + 3 = ( 2 x + 3)( x + 1)
If c = −3:
2 x 2 + 5 x − 3 = ( 2 x − 1)( x + 3)
If c = −7:
2 x 2 + 5 x − 7 = ( 2 x + 7)( x − 1)
If c = −12:
2 x 2 + 5 x − 12 = ( 2 x − 3)( x + 4)
88. For 3x 2 − x + c to be factorable, the factors of 3c must add up to −1.
Possible c-values
3c
Factors of 3c must add up to −1
−2
−6
(2)(−3)
= −6 and 2 + ( −3) = −1
−4
−12
(3)(−4)
= −12 and 3 + ( −4) = −1
−10
−30
(5)(−6)
= −30 and 5 + ( −6) = −1
These are a few possible c-values. There are many correct answers.
If c = −2:
3x 2 − x − 2 = (3x + 2)( x − 1)
If c = −4:
3x 2 − x − 4 = (3x − 4)( x + 1)
If c = −10: 3x 2 − x − 10 = (3x + 5)( x − 2)
89. True. a 2 − b2 = (a + b)(a − b)
90. True. u 2 + 2uv + v 2 = (u + v) and u 2 − 2uv + v 2 = (u − v)
2
2
91. 3 should be factored out of both binomials to yield (3 x + 6)(3x − 9) = 3( x + 2)(3)( x − 3) = 9( x + 2)( x − 3) .
92. No, (3 x − 6)( x + 1) is not completely factored because (3x − 6) = 3( x − 2).
So, the completely factored form is 3( x − 2)( x + 1).
93. x 2 n − y 2 n = ( x n ) − ( y n ) = ( x n + y n )( x n − y n )
2
2
This is not completely factored unless n = 1.
(
)(
) (
)
For n = 3: ( x3 + y3 )( x3 − y3 ) = ( x + y)( x2 − xy + y 2 )( x − y)( x2 + xy + y 2 )
For n = 4: ( x 4 + y 4 )( x4 − y 4 ) = ( x4 + y 4 )( x2 + y 2 )( x + y)( x − y)
For n = 2: x 2 + y 2 x2 − y 2 = x2 + y 2 ( x + y)( x − y)
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
22
Chapter P
Prerequisites
94. x 3n + y 3n = ( x n ) + ( y n ) = ( x n + y n )( x 2 n − x n y n + y 2 n )
3
3
Depending on the value of n, this may factor further.
95. Answers will vary. Sample answer: x 2 − 3
a
a
96. (a) a 2 − b 2 =
−
a− b
=
b
b
a
= ( a + b) ( a − b)
=
a
a
b
b
(b)
(a − b)
2
b
a
a− b
=
a
=
a− b
+
a
−
b
= a 2 − 2ab + b 2
a
b
a
b
−
b
a
b
b
(c)
(a
b
+ b) =
2
a
a
b
a
=
+
a
b
+
a
b
b
+
b
a
= a 2 + ab + ab + b 2
= a 2 + 2ab + b 2
97. u 6 − v 6 = (u 3 ) − (v 3 )
2
2
= (u 3 + v3 )(u 3 − v 3 )
2
2
= (u + v)(u 2 − uv + v 2 )
(u − v)(u + uv + v )
= (u + v )(u − v)(u 2 + uv + v 2 )(u 2 − uv + v 2 )
x6 − 1 = ( x + 1)( x − 1)( x2 + x + 1)( x2 − x + 1)
x6 − 64 = x6 − 26 = ( x + 2)( x − 2)( x2 + 2 x + 4)( x2 − 2 x + 4)
Section P.5 Rational Expressions
1. domain
2. rational expression
3. complex
4. equivalent
5. The domain of 3x 2 − 4 x + 7 is the set of all real
numbers.
6. The domain of 6 x 2 − 9, x > 0 is the set of all positive
real numbers because the polynomial is restricted to that
set.
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.5
Rational Expressions
7. The domain of
20.
4(3 − x)
12 − 4 x
=
= − 4, x ≠ 3
x −3
x −3
1
8. The domain of
is the set of all real numbers x
x + 5
such that x ≠ −5 .
21.
y 2 − 16
( y + 4)( y − 4)
=
y + 4
y + 4
1
is the set of all real numbers x
3− x
such that x ≠ 3.
= y − 4, y ≠ −4
x 2 − 25
( x + 5)( x − 5)
=
5− x
−1( x − 5)
x + 6
is the set of all real numbers x
3x + 2
2
such that x ≠ − .
3
22.
x − 4
is the set of all real numbers x
1 − 2x
23.
3 y ( 3 y + 2)
6 y + 9 y2
3y
2
=
=
,y ≠ −
12 y + 8
4(3 y + 2)
4
3
24.
−4 y ( 2 y − 1)
4 y − 8 y2
=
10 y − 5
5( 2 y − 1)
9. The domain of
10. The domain of
1
2
such that x ≠
11. The domain of
= −( x + 5), x ≠ 5
.
( x − 2)( x − 3) is the
x2 − 5x + 6
=
x2 + 6 x + 8
( x + 4)( x + 2)
= −
set of all real numbers x such that x ≠ − 4, − 2.
( x − 1)( x + 1) is the
x2 − 1
12. The domain of 2
=
x + 3 x − 10
( x + 5)( x − 2)
set of all real numbers x such that x ≠ − 5, 2.
13. The domain of x − 7 is the set of all real numbers x
such that x ≥ 7.
25.
( x + 5)( x − 1) = x − 1 , x ≠ − 5
x2 + 4 x − 5
=
x 2 + 8 x + 15
( x + 5)( x + 3) x + 3
27.
x2 − x − 2
x2 − x − 2
=
2
10 − 3x − x
− ( x 2 + 3x − 10)
5
.
2
such that x ≥
15. The domain of
=
1
is the set of all real numbers x
x −3
such that x > 3.
16. The domain of
4y
1
,y ≠
5
2
2
( x + 10)( x − 2) = x − 2 , x ≠ −10
26. x2 + 8 x − 20 =
x + 11x + 10
( x + 10)( x + 1) x + 1
2 x − 5 is the set of all real numbers x
14. The domain of
28.
1
is the set of all real numbers x
x + 2
such that x > −2.
17.
5 x (3 x )
15 x 2
3x
,x ≠ 0
=
=
10 x
5 x ( 2)
2
18.
6 y (3)
18 y 2
3
=
=
5
60 y
10 y 3
6 y 2 (10 y 3 )
29.
( x + 1)( x − 2)
− ( x + 5)( x − 2)
1
= − ,x ≠ 5
2
x +1
,x ≠ 2
x +5
x 2 − 16
x 2 − 16
= 2
2
x + x − 16 x − 16
x ( x + 1) − 16( x + 1)
3
=
=
30.
= −
− ( x 2 − 3 x − 4)
4 + 3x − x 2
=
2 x2 − 7 x − 4
2 x2 − 7 x − 4
− ( x + 1)( x − 4)
x +1
=
= −
,x ≠ 4
2x + 1
(2 x + 1)( x − 4)
2
x −5
x −5
=
19.
10 − 2 x
− 2( x − 5)
23
(x
x 2 − 16
+ 1)( x 2 − 16)
1
, x ≠ ±4
x +1
x2 − 1
x2 − 1
=
x3 + x 2 + 9 x + 9
x 2 ( x + 1) + 9( x + 1)
=
( x − 1)( x + 1)
( x + 1)( x 2 + 9)
=
x −1
, x ≠ −1
x2 + 9
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
24
31.
Chapter P
Prerequisites
5 x3
5 x3
=
3
2x + 4
2( x3 + 2)
When simplifying fractions, only common factors can be divided out, not terms.
32.
x
x −3
x2 − x − 6
1
x + 2
0
1
2
3
1
2
1
2
1
3
1
3
1
4
1
4
Undef.
1
5
4
5
6
1
6
1
6
1
7
1
7
1
8
1
8
The expressions are not equivalent at x = 3.
33.
5
1
x −1
,x ≠1
⋅
=
5( x − 2)
x − 1 25( x − 2)
34.
r ( r + 1)( r − 1)
r
r2
r
r2 − 1
r +1
÷ 2
=
⋅
=
=
,r ≠ 1
r −1 r −1
r −1
r2
r 2 ( r − 1)
r
35.
x2 − 4
2− x
x2 − 4 2x + 4
÷
=
⋅
12
2x + 4
12
2− x
x + 2)( x − 2) 2( x + 2)
(
=
⋅
12
− ( x − 2)
= −
36.
(x
+ 2)
6
2
, x ≠ ±2
t2 − t − 6
t + 3
t + 3
(t − 3)(t + 2)
(t − 3)(t + 2)(t + 3)
⋅
=
⋅
=
t 2 + 6t + 9 t 2 − 4
(t + 3)(t + 3) (t − 2)(t + 2)
(t + 3)(t + 3)(t − 2)(t + 2)
=
t −3
, t ≠ −2
(t + 3)(t − 2)
37.
( x + 2 y)( x − y ) ⋅
x 2 + xy − 2 y 2
x
x
x − y
⋅ 2
=
=
, x ≠ −2 y
x3 + x 2 y
x + 3 xy + 2 y 2
x2 ( x + y)
( x + 2 y )( x + y ) x( x + y )2
38.
x 2 − 14 x + 49 3 x − 21 ( x − 7)( x − 7)
x+7
÷
=
⋅
x 2 − 49
x+7
( x + 7)( x − 7) 3( x − 7)
=
39.
40.
1
, x ≠ ±7
3
x − 1 − ( x − 4)
x −1
x − 4
−
=
x + 2
x + 2
x + 2
x −1− x + 4
=
x + 2
3
=
x + 2
2x − 1 1 − x
2x − 1 + 1 − x
x
+
=
=
x + 3
x + 3
x + 3
x + 3
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solution Manual for College Algebra 10th Edition by Larson
Section P.5
41.
42.
43.
44.
25
(1)( x + 1) + x(3x + 2)
x
1
+
=
x +1
3x + 2
(3x + 2)( x + 1) (3x + 2)( x + 1)
=
x + 1 + 3x 2 + 2 x
(3x + 2)( x + 1)
=
3x 2 + 3x + 1
(3x + 2)( x + 1)
x( x − 1)
6( x + 4)
x
6
−
=
−
x + 4
x −1
( x + 4)( x − 1) ( x + 4)( x − 1)
=
x 2 − x − 6 x − 24
( x + 4)( x − 1)
=
x 2 − 7 x − 24
( x + 4)( x − 1)
3
3
x
x
−
=
−
2x + 4
2( x + 2)
x + 2
x + 2
=
3
2x
−
2( x + 2)
2( x + 2)
=
3 − 2x
2( x + 2)
2
4
2
4
+
=
+
x −9
x +3
( x + 3)( x − 3) x + 3
2
=
=
=
=
45. −
(x
4( x − 3)
2
+
+ 3)( x − 3)
( x + 3)( x − 3)
2 + 4 x − 12
+ 3)( x − 3)
(x
(x
(x
4 x − 10
+ 3)( x − 3)
2( 2 x − 5)
+ 3)( x − 3)
− ( x 2 + 1)
1
2
1
2x
1
+ 2
+ 3
=
+
+
2
2
2
x
x +1 x + x
x( x + 1)
x( x + 1)
x( x + 1)
=
− x( x − 2)
− x2 − 1 + 2x + 1
− x2 + 2x
=
=
2
2
x( x + 1)
x( x + 1)
x( x 2 + 1)
= −
46.
Rational Expressions
x −2
2− x
= 2
,x ≠ 0
x2 + 1
x +1
2
2
1
2
2
1
+
+ 2
=
+
+
x +1
x −1
x −1
x +1
x − 1 ( x + 1)( x − 1)
=
=
(x
2( x − 1)
+ 1)( x − 1)
+
(x
2( x + 1)
+ 1)( x − 1)
+
(x
1
+ 1)( x − 1)
2x − 2 + 2x + 2 + 1
4x + 1
=
( x + 1)( x − 1)
( x + 1)( x − 1)
© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
