STUDENT’S
SOLUTIONS MANUAL
J
UDITH
A. P
ENNA
Indiana University Purdue University Indianapolis
C
OLLEGE
A
LGEBRA
:
G
RAPHS AND
M
ODELS
F
IFTH
E
DITION
Marvin L. Bittinger
Indiana University Purdue University Indianapolis
Judith A. Beecher
Indiana University Purdue University Indianapolis
David J. Ellenbogen
Community College of Vermont
Judith A. Penna
Indiana University Purdue University Indianapolis
Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson from electronic files supplied by the author. Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Publishing as Pearson, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-79125-2 ISBN-10: 0-321-79125-8 1 2 3 4 5 6 BRR 15 14 13 12 11
Contents
Chapter R . . . 1
Chapter 1 . . . 25
Chapter 2
. . . 63
Chapter 3
. . . 93
Chapter 4 . . . 129
Chapter 5
. . . 181
Chapter 6
. . . 215
Chapter 7
. . . 275
0 5 ⫺5
0
⫺3 ⫺1
0
⫺2
0 3.8
0 7
Chapter R
Basic Concepts of Algebra
Exercise Set R.1
1. Rational numbers: 2
3, 6,−2.45, 18.4, −11, 3 √
27, 51 6,−
8 7,
0,√16
3. Irrational numbers: √3, √6
26, 7.151551555. . .,−√35,√5 3
(Although there is a pattern in 7.151551555. . ., there is
no repeating block of digits.)
5. Whole numbers: 6,√3
27, 0,√16
7. Integers but not natural numbers: −11, 0
9. Rational numbers but not integers: 2
3, −2.45, 18.4, 5 1 6, −8
7
11. This is a closed interval, so we use brackets. Interval
no-tation is [−5,5].
13. This is a half-open interval. We use a parenthesis on
the left and a bracket on the right. Interval notation is (−3,−1].
15. This interval is of unlimited extent in the negative
direc-tion, and the endpoint−2 is included. Interval notation is
(−∞,−2].
17. This interval is of unlimited extent in the positive
direc-tion, and the endpoint 3.8 is not included. Interval
nota-tion is (3.8,∞).
19. {x|7< x}, or{x|x >7}.
This interval is of unlimited extent in the positive direction and the endpoint 7 is not included. Interval notation is (7,∞).
21. The endpoints 0 and 5 are not included in the interval, so
we use parentheses. Interval notation is (0,5).
23. The endpoint−9 is included in the interval, so we use a
bracket before the−9. The endpoint−4 is not included,
so we use a parenthesis after the−4. Interval notation is
[−9,−4).
25. Both endpoints are included in the interval, so we use
brackets. Interval notation is [x, x+h].
27. The endpointpis not included in the interval, so we use a
parenthesis before thep. The interval is of unlimited
ex-tent in the positive direction, so we use the infinity symbol ∞. Interval notation is (p,∞).
29. Since 6 is an element of the set of natural numbers, the
statement is true.
31. Since 3.2 is not an element of the set of integers, the state-ment is false.
33. Since−11
5 is an element of the set of rational numbers,
the statement is true.
35. Since√11 is an element of the set of real numbers, the
statement is false.
37. Since 24 is an element of the set of whole numbers, the
statement is false.
39. Since 1.089 is not an element of the set of irrational
num-bers, the statement is true.
41. Since every whole number is an integer, the statement is
true.
43. Since every rational number is a real number, the
state-ment is true.
45. Since there are real numbers that are not integers, the
statement is false.
47. The sentence 3 +y = y+ 3 illustrates the commutative
property of addition.
49. The sentence −3·1 = −3 illustrates the multiplicative
identity property.
51. The sentence 5·x=x·5 illustrates the commutative
prop-erty of multiplication.
53. The sentence 2(a+b) = (a+b)2 illustrates the commutative property of multiplication.
55. The sentence−6(m+n) =−6(n+m) illustrates the
com-mutative property of addition.
57. The sentence 8·1
8= 1 illustrates the multiplicative inverse
2 Chapter R: Basic Concepts of Algebra
59. The distance of−8.15 from 0 is 8.15, so| −8.15|= 8.15.
61. The distance 295 from 0 is 295, so|295|= 295.
63. The distance of−√97 from 0 is√97, so| −√97|=√97.
65. The distance of 0 from 0 is 0, so|0|= 0.
67. The distance of 5
4 from 0 is
5 4, so
5
4
= 5
4.
69. |14−(−8)|=|14 + 8|=|22|= 22, or | −8−14|=| −22|= 22
71. | −3−(−9)|=| −3 + 9|=|6|= 6, or | −9−(−3)|=| −9 + 3|=| −6|= 6
73. |12.1−6.7|=|5.4|= 5.4, or |6.7−12.1|=| −5.4|= 5.4
75. −3
4−
15 8
=−6
8−
15 8
=−21
8
= 21
8, or
15
8 −
−3
4
=15
8 +
3 4
=15
8 +
6 8
=21
8
= 21
8
77. | −7−0|=| −7|= 7, or |0−(−7)|=|0 + 7|=|7|= 7
79. Answers may vary. One such number is
0.124124412444. . ..
81. Answers may vary. Since− 1
101 = 0.0099 and
− 1
100=−0.01, one such number is−0.00999.
83. Since 12+ 32= 10, the hypotenuse of a right triangle with
legs of lengths 1 unit and 3 units has a length of√10 units.
✏✏✏✏
✏✏✏✏
3
1
c c2= 12+ 32
c2= 10 c=√10
Exercise Set R.2
1. 3−7= 1 37
a−m= 1
am, a= 0
3. Observe that each exponent is negative. We move each
factor to the other side of the fraction bar and change the sign of each exponent.
x−5 y−4 =
y4 x5
5. Observe that each exponent is negative. We move each
factor to the other side of the fraction bar and change the sign of each exponent.
m−1n−12
t−6 =
t6 m1n12, or
t6 mn12
7. 230= 1 (For any nonzero real number,a0= 1.)
9. z0·z7=z0+7=z7, or
z0·z7= 1·z7=z7
11. 58·5−6= 58+(−6)= 52, or 25
13. m−5·m5=m−5+5=m0= 1
15. y3·y−7=y3+(−7)=y−4, or 1 y4
17. (x+ 3)4(x+ 3)−2= (x+ 3)4+(−2)= (x+ 3)2
19. 3−3·38·3 = 3−3+8+1= 36, or 729
21. 2x3·3x2= 2·3·x3+2= 6x5
23. (−3a−5)(5a−7) =−3·5·a−5+(−7)=−15a−12, or
−15
a12
25. (6x−3y5)(−7x2y−9) = 6(−7)x−3+2y5+(−9)= −42x−1y−4, or− 42
xy4
27. (2x)4(3x)3= 24·x4·33·x3= 16·27·x4+3= 432x7
29. (−2n)3(5n)2= (−2)3n3·52n2=−8·25·n3+2= −200n5
31. y
35 y31 =y
35−31=y4
33. b
−7 b12 =b
−7−12=b−19, or 1 b19
35. x
2y−2 x−1y =x
2−(−1)y−2−1=x3y−3, or x3 y3
37. 32x
−4y3 4x−5y8 =
32
4x
−4−(−5)y3−8= 8xy−5, or 8x y5
39. (2x2y)4= 24(x2)4y4= 16x2·4y4= 16x8y4 41. (−2x3)5= (−2)5(x3)5= (−2)5x3·5=−32x15
43. (−5c−1d−2)−2= (−5)−2c−1(−2)d−2(−2)=
c2d4 (−5)2 =
c2d4 25
45. (3m4)3(2m−5)4= 33m12·24m−20= 27·16m12+(−20)= 432m−8, or 432
m8
47.
2x−3y7 z−1
3 =(2x
−3y7)3 (z−1)3 =
23x−9y21
z−3 =
8x−9y21
z−3 , or
8y21z3 x9
49.
24a10b−8c7 12a6b−3c5
−5
= (2a4b−5c2)−5= 2−5a−20b25c−10,
or b
25 32a20c10
Exercise Set R.2 3
51. Convert 16,500,000 to scientific notation.
We want the decimal point to be positioned between the 1 and the 6, so we move it 7 places to the left. Since 16,500,000 is greater than 10, the exponent must be posi-tive.
16,500,000 = 1.65×107
53. Convert 0.000000437 to scientific notation.
We want the decimal point to be positioned between the 4 and the 3, so we move it 7 places to the right. Since
0.000000437 is a number between 0 and 1, the exponent
must be negative.
0.000000437 = 4.37×10−7
55. Convert 234,600,000,000 to scientific notation. We want
the decimal point to be positioned between the 2 and the 3, so we move it 11 places to the left. Since 234,600,000,000 is greater than 10, the exponent must be positive.
234,600,000,000 = 2.346×1011
57. Convert 0.00104 to scientific notation. We want the
deci-mal point to be positioned between the 1 and the last 0, so we move it 3 places to the right. Since 0.00104 is a number between 0 and 1, the exponent must be negative.
0.00104 = 1.04×10−3
59. Convert 0.00000000000000000000000000167 to scientific
notation.
We want the decimal point to be positioned between the 1 and the 6, so we move it 27 places to the right. Since
0.00000000000000000000000000167 is a number between 0
and 1, the exponent must be negative.
0.00000000000000000000000000167 = 1.67×10−27
61. Convert 7.6×105 to decimal notation.
The exponent is positive, so the number is greater than 10. We move the decimal point 5 places to the right.
7.6×105= 760,000
63. Convert 1.09×10−7 to decimal notation.
The exponent is negative, so the number is between 0 and 1. We move the decimal point 7 places to the left.
1.09×10−7= 0.000000109
65. Convert 3.496×1010to decimal notation.
The exponent is positive, so the number is greater than 10. We move the decimal point 10 places to the right.
3.496×1010= 34,960,000,000
67. Convert 5.41×10−8 to decimal notation.
The exponent is negative, so the number is between 0 and 1. We move the decimal point 8 places to the left.
5.41×10−8= 0.0000000541
69. Convert 2.319×108to decimal notation.
The exponent is positive, so the number is greater than 10. We move the decimal point 8 places to the right.
2.319×108= 231,900,000
71. (4.2×107)(3.2×10−2) = (4.2×3.2)×(107×10−2)
= 13.44×105 This is not scientific notation.
= (1.344×10)×105
= 1.344×106 Writing scientific notation
73. (2.6×10−18)(8.5×107) = (2.6×8.5)×(10−18×107)
= 22.1×10−11 This is not scientific notation.
= (2.21×10)×10−11 = 2.21×10−10
75. 6.4×10−7
8.0×106 = 6.4 8.0×
10−7 106
= 0.8×10−13 This is not scientific
notation. = (8×10−1)×10−13
= 8×10−14 Writing scientific
notation
77. 1.8×10−3
7.2×10−9 = 1.8
7.2× 10−3 10−9
= 0.25×106 This is not scientific notation.
= (2.5×10−1)×106 = 2.5×105
79. The average number of pieces of trash per mile is the total
number of pieces of trash divided by the number of miles. 51.2 billion
76 million =
5.12×1010 7.6×107 ≈0.6737×103 ≈(6.737×10−1)×103 ≈6.737×102
On average, there are about 6.737×102pieces of trash on
each mile of roadway.
81. The number of people per square mile is the total number
of people divided by the number of square miles. 38,000
0.75 =
3.8×104 7.5×10−1 ≈0.50667×105 ≈(5.0667×10−1)×105 ≈5.0667×104
There are about 5.0667×104people per square mile.
83. We multiply the number of light years by the number of
miles in a light year.
4.22×5.88×1012= 24.8136×1012 = (2.48136×10)×1012 = 2.48136×1013 The distance from Earth to Alpha Centauri C is 2.48136×1013mi.
4 Chapter R: Basic Concepts of Algebra
85. First find the number of seconds in 1 hour:
1 hour = 1 hr✧×60 min
✦
1 hr✧ ×
60 sec
1 min
✦
= 3600 secThe number of disintegrations produced in 1 hour is the number of disintegrations per second times the number of seconds in 1 hour.
37 billion×3600
= 37,000,000,000×3600
= 3.7×1010×3.6×103 Writing scientific notation
= (3.7×3.6)×(1010×103)
= 13.32×1013 Multiplying
= (1.332×10)×1013 = 1.332×1014
One gram of radium produces 1.332×1014disintegrations
in 1 hour.
87. = 5·3 + 8·32+ 4(6−2)
= 5·3 + 8·32+ 4·4 Working inside parentheses
= 5·3 + 8·9 + 4·4 Evaluating 32
= 15 + 72 + 16 Multiplying
= 87 + 16 Adding in order
= 103 from left to right
89. 16÷4·4÷2·256
= 4·4÷2·256 Multiplying and dividing
in order from left to right
= 16÷2·256
= 8·256
= 2048
91. 4(8−6)2−4·3 + 2·8
31+ 190
= 4·2
2−4·3 + 2·8
3 + 1 Calculating in the
numerator and in the denominator = 4·4−4·3 + 2·8
4
= 16−12 + 16
4
= 4 + 16
4
= 20
4 = 5
93. Since interest is compounded semiannually,n= 2.
Substi-tute $3225 forP, 3.1% or 0.031 fori, 2 forn, and 4 fort
in the compound interest formula.
A=P
1 + i
n nt
= $3225
1 +0.031 2
2·4
Substituting = $3225(1 + 0.0155)2·4Dividing
= $3225(1.0155)2·4 Adding
= $3225(1.0155)8 Multiplying 2 and 4
≈$3225(1.130939628) Evaluating the
exponential expression
≈$3647.2803 Multiplying
≈$3647.28 Rounding to the nearest cent
95. Since interest is compounded quarterly,n= 4. Substitute
$4100 forP, 2.3% or 0.023 fori, 4 forn, and 6 fortin the compound interest formula.
A=P
1 + i
n nt
= $4100
1 +0.023 4
4·6
Substituting
= $4100(1 + 0.00575)4·6 Dividing
= $4100(1.00575)4·6 Adding
= $4100(1.00575)24 Multiplying 4 and 6
≈$4100(1.147521919) Evaluating the
exponential expression
≈$4704.839868 Multiplying
≈$4704.84 Rounding to the nearest cent
97. Substitute $250 forP, 0.05 forrand 27 fortand perform
the resulting computation.
S =P
1 + r
12
12·t
−1
r
12
= $250
1 +0.05 12
12·27
−1
0.05 12
≈$170,797.30
99. Substitute $120,000 forS, 0.03 forr, and 18 fortand solve
forP.
S=P
1 + r
12
12·t
−1
r
12
$120,000 =P
1 +0.03 12
12·18
−1
0.03 12
$120,000 =P
(1.0025)216−1 0.0025
$120,000≈P(285.94035) $419.67≈P
Exercise Set R.3 5
101. (xt·x3t)2= (x4t)2=x4t·2=x8t 103. (ta+x·tx−a)4= (t2x)4=t2x·4=t8x 105.
(3xayb)3 (−3xayb)2
2 =
27x3ay3b
9x2ay2b
2 =3xayb2
= 9x2ay2b
Exercise Set R.3
1. 7x3−4x2+ 8x+ 5 = 7x3+ (−4x2) + 8x+ 5 Terms: 7x3,−4x2, 8x, 5
The degree of the term of highest degree, 7x3, is 3. Thus,
the degree of the polynomial is 3.
3. 3a4b−7a3b3+ 5ab−2 = 3a4b+ (−7a3b3) + 5ab+ (−2) Terms: 3a4b,−7a3b3, 5ab,−2
The degrees of the terms are 5, 6, 2, and, 0, respectively, so the degree of the polynomial is 6.
5. (3ab2−4a2b−2ab+ 6)+
(−ab2−5a2b+ 8ab+ 4)
= (3−1)ab2+ (−4−5)a2b+ (−2 + 8)ab+ (6 + 4) = 2ab2−9a2b+ 6ab+ 10
7. (2x+ 3y+z−7) + (4x−2y−z+ 8)+
(−3x+y−2z−4)
= (2 + 4−3)x+ (3−2 + 1)y+ (1−1−2)z+
(−7 + 8−4) = 3x+ 2y−2z−3
9. (3x2−2x−x3+ 2)−(5x2−8x−x3+ 4) = (3x2−2x−x3+ 2) + (−5x2+ 8x+x3−4) = (3−5)x2+ (−2 + 8)x+ (−1 + 1)x3+ (2−4) =−2x2+ 6x−2
11. (x4−3x2+ 4x)−(3x3+x2−5x+ 3) = (x4−3x2+ 4x) + (−3x3−x2+ 5x−3) =x4−3x3+ (−3−1)x2+ (4 + 5)x−3 =x4−3x3−4x2+ 9x−3
13. (3a2)(−7a4) = [3(−7)](a2·a4) =−21a6
15. (6xy3)(9x4y2) = (6·9)(x·x4)(y3·y2) = 54x5y5
17. (a−b)(2a3−ab+ 3b2)
= (a−b)(2a3) + (a−b)(−ab) + (a−b)(3b2) Using the distributive property = 2a4−2a3b−a2b+ab2+ 3ab2−3b3
Using the distributive property three more times
= 2a4−2a3b−a2b+ 4ab2−3b3 Collecting like terms
19. (y−3)(y+ 5)
=y2+ 5y−3y−15 Using FOIL
=y2+ 2y−15 Collecting like terms
21. (x+ 6)(x+ 3)
=x2+ 3x+ 6x+ 18 Using FOIL
=x2+ 9x+ 18 Collecting like terms
23. (2a+ 3)(a+ 5)
= 2a2+ 10a+ 3a+ 15 Using FOIL
= 2a2+ 13a+ 15 Collecting like terms
25. (2x+ 3y)(2x+y)
= 4x2+ 2xy+ 6xy+ 3y2 Using FOIL
= 4x2+ 8xy+ 3y2 27. (x+ 3)2
=x2+ 2·x·3 + 32
[(A+B)2=A2+ 2AB+B2] =x2+ 6x+ 9
29. (y−5)2
=y2−2·y·5 + 52
[(A−B)2=A2−2AB+B2]
=y2−10y+ 25
31. (5x−3)2
= (5x)2−2·5x·3 + 32
[(A−B)2=A2−2AB+B2] = 25x2−30x+ 9
33. (2x+ 3y)2
= (2x)2+ 2(2x)(3y) + (3y)2
[(A+B)2=A2+2AB+B2] = 4x2+ 12xy+ 9y2
35. (2x2−3y)2
= (2x2)2−2(2x2)(3y) + (3y)2
[(A−B)2=A2−2AB+B2] = 4x4−12x2y+ 9y2
37. (n+ 6)(n−6)
=n2−62 [(A+B)(A−B) =A2−B2]
=n2−36
39. (3y+ 4)(3y−4)
= (3y)2−42 [(A+B)(A−B) =A2−B2]
= 9y2−16
41. (3x−2y)(3x+ 2y)
= (3x)2−(2y)2 [(A−B)(A+B) =A2−B2] = 9x2−4y2
6 Chapter R: Basic Concepts of Algebra
43. (2x+ 3y+ 4)(2x+ 3y−4)
= [(2x+ 3y) + 4][(2x+ 3y)−4] = (2x+ 3y)2−42
= 4x2+ 12xy+ 9y2−16
45. (x+ 1)(x−1)(x2+ 1)
= (x2−1)(x2+ 1)
=x4−1
47. (an+bn)(an−bn) = (an)2−(bn)2 =a2n−b2n
49. (an+bn)2= (an)2+ 2·an·bn+ (bn)2 =a2n+ 2anbn+b2n
51. (x−1)(x2+x+ 1)(x3+ 1)
= [(x−1)x2+ (x−1)x+ (x−1)·1](x3+ 1) = (x3−x2+x2−x+x−1)(x3+ 1) = (x3−1)(x3+ 1)
= (x3)2−12
=x6−1
53. (xa−b)a+b
=x(a−b)(a+b) =xa2−b2
55. (a+b+c)2
= (a+b+c)(a+b+c)
= (a+b+c)(a) + (a+b+c)(b) + (a+b+c)(c)
=a2+ab+ac+ab+b2+bc+ac+bc+c2
=a2+b2+c2+ 2ab+ 2ac+ 2bc
Exercise Set R.4
1. 3x+ 18 = 3·x+ 3·6 = 3(x+ 6)
3. 2z3−8z2= 2z2·z−2z2·4 = 2z2(z−4)
5. 4a2−12a+ 16 = 4·a2−4·3a+ 4·4 = 4(a2−3a+ 4)
7. a(b−2) +c(b−2) = (b−2)(a+c)
9. 3x3−x2+ 18x−6 =x2(3x−1) + 6(3x−1) = (3x−1)(x2+ 6)
11. y3−y2+ 2y−2 =y2(y−1) + 2(y−1) = (y−1)(y2+ 2)
13. 24x3−36x2+ 72x−108 = 12(2x3−3x2+ 6x−9) = 12[x2(2x−3) + 3(2x−3)] = 12(2x−3)(x2+ 3)
15. x3−x2−5x+ 5 =x2(x−1)−5(x−1) = (x−1)(x2−5)
17. a3−3a2−2a+ 6 =a2(a−3)−2(a−3) = (a−3)(a2−2)
19. w2−7w+ 10
We look for two numbers with a product of 10 and a sum
of−7. By trial, we determine that they are−5 and−2.
w2−7w+ 10 = (w−5)(w−2)
21. x2+ 6x+ 5
We look for two numbers with a product of 5 and a sum of 6. By trial, we determine that they are 1 and 5.
x2+ 6x+ 5 = (x+ 1)(x+ 5)
23. t2+ 8t+ 15
We look for two numbers with a product of 15 and a sum of 8. By trial, we determine that they are 3 and 5.
t2+ 8t+ 15 = (t+ 3)(t+ 5)
25. x2−6xy−27y2
We look for two numbers with a product of−27 and a sum
of−6. By trial, we determine that they are 3 and−9.
x2−6xy−27y2= (x+ 3y)(x−9y)
27. 2n2−20n−48 = 2(n2−10n−24)
Now factorn2−10n−24. We look for two numbers with a
product of−24 and a sum of−10. By trial, we determine
that they are 2 and−12. Thenn2−10n−24 =
(n+ 2)(n−12). We must include the common factor, 2,
to have a factorization of the original trinomial. 2n2−20n−48 = 2(n+ 2)(n−12)
29. y2−4y−21
We look for two numbers with a product of−21 and a sum
of−4. By trial, we determine that they are 3 and−7.
y2−4y−21 = (y+ 3)(y−7)
31. y4−9y3+ 14y2=y2(y2−9y+ 14)
Now factory2−9y+ 14. Look for two numbers with a
product of 14 and a sum of−9. The numbers are−2 and
−7. Theny2−9y+ 14 = (y−2)(y−7). We must include
the common factor,y2, in order to have a factorization of
the original trinomial.
y4−9y3+ 14y2=y2(y−2)(y−7)
33. 2x3−2x2y−24xy2= 2x(x2−xy−12y2)
Now factorx2−xy−12y2. Look for two numbers with a
product of−12 and a sum of−1. The numbers are−4 and
3. Thenx2−xy−12y2= (x−4y)(x+3y). We must include
the common factor, 2x, in order to have a factorization of
the original trinomial.
Exercise Set R.4 7
35. 2n2+ 9n−56
We use the FOIL method.
1. There is no common factor other than 1 or−1.
2. The factorization must be of the form
(2n+ )(n+ ).
3. Factor the constant term, −56. The possibilities
are−1·56, 1(−56),−2·28, 2(−28),−4·16, 4(−16),
−7·8, and 7(−8). The factors can be written in
the opposite order as well: 56(−1),−56·1, 28(−2), −28·2, 16(−4),−16·4, 8(−7), and−8·7. 4. Find a pair of factors for which the sum of the
out-side and the inout-side products is the middle term,
9n. By trial, we determine that the factorization
is (2n−7)(n+ 8).
37. 12x2+ 11x+ 2
We use the grouping method.
1. There is no common factor other than 1 or−1.
2. Multiply the leading coefficient and the constant: 12·2 = 24.
3. Try to factor 24 so that the sum of the factors is the coefficient of the middle term, 11. The factors we want are 3 and 8.
4. Split the middle term using the numbers found in step (3):
11x= 3x+ 8x
5. Factor by grouping.
12x2+ 11x+ 2 = 12x2+ 3x+ 8x+ 2 = 3x(4x+ 1) + 2(4x+ 1)
= (4x+ 1)(3x+ 2)
39. 4x2+ 15x+ 9
We use the FOIL method.
1. There is no common factor other than 1 or−1.
2. The factorization must be of the form
(4x+ )(x+ ) or (2x+ )(2x+ ).
3. Factor the constant term, 9. The possibilities are 1·9,−1(−9), 3·3, and−3(−3). The first two pairs of factors can be written in the opposite order as well: 9·1,−9(−1).
4. Find a pair of factors for which the sum of the out-side and the inout-side products is the middle term,
15x. By trial, we determine that the factorization
is (4x+ 3)(x+ 3).
41. 2y2+y−6
We use the grouping method.
1. There is no common factor other than 1 or−1.
2. Multiply the leading coefficient and the constant: 2(−6) =−12.
3. Try to factor−12 so that the sum of the factors is
the coefficient of the middle term, 1. The factors we
want are 4 and−3.
4. Split the middle term using the numbers found in step (3):
y= 4y−3y
5. Factor by grouping.
2y2+y−6 = 2y2+ 4y−3y−6 = 2y(y+ 2)−3(y+ 2) = (y+ 2)(2y−3)
43. 6a2−29ab+ 28b2
We use the FOIL method.
1. There is no common factor other than 1 or−1.
2. The factorization must be of the form
(6x+ )(x+ ) or (3x+ )(2x+ ).
3. Factor the coefficient of the last term, 28. The pos-sibilities are 1·28,−1(−28), 2·14,−2(−14), 4·7,
and−4(−7). The factors can be written in the
op-posite order as well: 28·1,−28(−1), 14·2,−14(−2), 7·4, and−7(−4).
4. Find a pair of factors for which the sum of the out-side and the inout-side products is the middle term,
−29. Observe that the second term of each
bino-mial factor will contain a factor ofb. By trial, we
determine that the factorization is (3a−4b)(2a−7b).
45. 12a2−4a−16
We will use the grouping method. 1. Factor out the common factor, 4.
12a2−4a−16 = 4(3a2−a−4)
2. Now consider 3a2−a−4. Multiply the leading
coefficient and the constant: 3(−4) =−12.
3. Try to factor−12 so that the sum of the factors is
the coefficient of the middle term, −1. The factors
we want are−4 and 3.
4. Split the middle term using the numbers found in step (3):
−a=−4a+ 3a
5. Factor by grouping.
3a2−a−4 = 3a2−4a+ 3a−4 =a(3a−4) + (3a−4) = (3a−4)(a+ 1)
We must include the common factor to get a factor-ization of the original trinomial.
12a2−4a−16 = 4(3a−4)(a+ 1)
47. z2−81 =z2−92= (z+ 9)(z−9)
49. 16x2−9 = (4x)2−32= (4x+ 3)(4x−3)
51. 6x2−6y2= 6(x2−y2) = 6(x+y)(x−y)
53. 4xy4−4xz2= 4x(y4−z2) = 4x[(y2)2−z2] = 4x(y2+z)(y2−z)
8 Chapter R: Basic Concepts of Algebra
55. 7pq4−7py4= 7p(q4−y4) = 7p[(q2)2−(y2)2] = 7p(q2+y2)(q2−y2) = 7p(q2+y2)(q+y)(q−y)
57. x2+ 12x+ 36 =x2+ 2·x·6 + 62 = (x+ 6)2
59. 9z2−12z+ 4 = (3z)2−2·3z·2 + 22= (3z−2)2
61. 1−8x+ 16x2= 12−2·1·4x+ (4x)2 = (1−4x)2
63. a3+ 24a2+ 144a
=a(a2+ 24a+ 144) =a(a2+ 2·a·12 + 122) =a(a+ 12)2
65. 4p2−8pq+ 4q2 = 4(p2−2pq+q2) = 4(p−q)2
67. x3+ 64 =x3+ 43
= (x+ 4)(x2−4x+ 16)
69. m3−216 =m3−63
= (m−6)(m2+ 6m+ 36)
71. 8t3+ 8 = 8(t3+ 1) = 8(t3+ 13)
= 8(t+ 1)(t2−t+ 1)
73. 3a5−24a2 = 3a2(a3−8) = 3a2(a3−23)
= 3a2(a−2)(a2+ 2a+ 4)
75. t6+ 1 = (t2)3+ 13
= (t2+ 1)(t4−t2+ 1)
77. 18a2b−15ab2= 3ab·6a−3ab·5b = 3ab(6a−5b)
79. x3−4x2+ 5x−20 =x2(x−4) + 5(x−4) = (x−4)(x2+ 5)
81. 8x2−32 = 8(x2−4)
= 8(x+ 2)(x−2)
83. 4y2−5
There are no common factors. We might try to factor this polynomial as a difference of squares, but there is no integer which yields 5 when squared. Thus, the polynomial is prime.
85. m2−9n2=m2−(3n)2 = (m+ 3n)(m−3n)
87. x2+ 9x+ 20
We look for two numbers with a product of 20 and a sum of 9. They are 4 and 5.
x2+ 9x+ 20 = (x+ 4)(x+ 5)
89. y2−6y+ 5
We look for two numbers with a product of 5 and a sum
of−6. They are−5 and−1.
y2−6y+ 5 = (y−5)(y−1)
91. 2a2+ 9a+ 4
We use the FOIL method.
1. There is no common factor other than 1 or−1.
2. The factorization must be of the form
(2a+ )(a+ ).
3. Factor the constant term, 4. The possibilities are 1·4,−1(−4), and 2·2. The first two pairs of factors
can be written in the opposite order as well: 4·1,
−4(−1).
4. Find a pair of factors for which the sum of the out-side and the inout-side products is the middle term,
9a. By trial, we determine that the factorization
is (2a+ 1)(a+ 4).
93. 6x2+ 7x−3
We use the grouping method.
1. There is no common factor other than 1 or−1.
2. Multiply the leading coefficient and the constant: 6(−3) =−18.
3. Try to factor−18 so that the sum of the factors is
the coefficient of the middle term, 7. The factors we
want are 9 and−2.
4. Split the middle term using the numbers found in step (3):
7x= 9x−2x
5. Factor by grouping.
6x2+ 7x−3 = 6x2+ 9x−2x−3 = 3x(2x+ 3)−(2x+ 3)
= (2x+ 3)(3x−1)
95. y2−18y+ 81 =y2−2·y·9 + 92 = (y−9)2
97. 9z2−24z+ 16 = (3z)2−2·3z·4 + 42 = (3z−4)2
99. x2y2−14xy+ 49 = (xy)2−2·xy·7 + 72 = (xy−7)2
101. 4ax2+ 20ax−56a= 4a(x2+ 5x−14) = 4a(x+ 7)(x−2)
103. 3z3−24 = 3(z3−8)
= 3(z3−23)
Exercise Set R.5 9
105. 16a7b+ 54ab7 = 2ab(8a6+ 27b6) = 2ab[(2a2)3+ (3b2)3]
= 2ab(2a2+ 3b2)(4a4−6a2b2+ 9b4)
107. y3−3y2−4y+ 12 =y2(y−3)−4(y−3) = (y−3)(y2−4) = (y−3)(y+ 2)(y−2)
109. x3−x2+x−1 =x2(x−1) + (x−1) = (x−1)(x2+ 1)
111. 5m4−20 = 5(m4−4)
= 5(m2+ 2)(m2−2)
113. 2x3+ 6x2−8x−24 = 2(x3+ 3x2−4x−12) = 2[x2(x+ 3)−4(x+ 3)] = 2(x+ 3)(x2−4) = 2(x+ 3)(x+ 2)(x−2)
115. 4c2−4cd−d2= (2c)2−2·2c·d−d2 = (2c−d)2
117. m6+ 8m3−20 = (m3)2+ 8m3−20
We look for two numbers with a product of−20 and a sum
of 8. They are 10 and−2.
m6+ 8m3−20 = (m3+ 10)(m3−2)
119. p−64p4=p(1−64p3) =p[13−(4p)3]
=p(1−4p)(1 + 4p+ 16p2)
121. y4−84 + 5y2
=y4+ 5y2−84
=u2+ 5u−84 Substitutingufory2
= (u+ 12)(u−7)
= (y2+ 12)(y2−7) Substitutingy2foru
123. y2− 8
49+
2
7y=y
2+2 7y−
8 49 =
y+4
7
y−2
7
125. x2+ 3x+9
4 =x
2+ 2·x·3
2+
3 2
2
=
x+ 3
2
2
127. x2−x+1
4 =x
2−2·x·1
2+
1 2
2
=
x−1
2
2
129. (x+h)3−x3
= [(x+h)−x][(x+h)2+x(x+h) +x2] = (x+h−x)(x2+ 2xh+h2+x2+xh+x2) =h(3x2+ 3xh+h2)
131. (y−4)2+ 5(y−4)−24
=u2+ 5u−24 Substitutingufory−4
= (u+ 8)(u−3)
= (y−4 + 8)(y−4−3) Substitutingy−4
foru
= (y+ 4)(y−7)
133. x2n+ 5xn−24 = (xn)2+ 5xn−24
= (xn+ 8)(xn−3)
135. x2+ax+bx+ab=x(x+a) +b(x+a) = (x+a)(x+b)
137. 25y2m−(x2n−2xn+ 1)
= (5ym)2−(xn−1)2
= [5ym+ (xn−1)][5ym−(xn−1)]
= (5ym+xn−1)(5ym−xn+ 1)
139. (y−1)4−(y−1)2 = (y−1)2[(y−1)2−1] = (y−1)2[y2−2y+ 1−1] = (y−1)2(y2−2y) =y(y−1)2(y−2)
Exercise Set R.5
1. x−5 = 7
x= 12 Adding 5
The solution is 12.
3. 3x+ 4 =−8
3x=−12 Subtracting 4
x=−4 Dividing by 3
The solution is−4.
5. 5y−12 = 3
5y= 15 Adding 12
y= 3 Dividing by 5
The solution is 3.
7. 6x−15 = 45
6x= 60 Adding 15
x= 10 Dividing by 6
The solution is 10.
9. 5x−10 = 45
5x= 55 Adding 10
x= 11 Dividing by 5
10 Chapter R: Basic Concepts of Algebra
11. 9t+ 4 =−5
9t=−9 Subtracting 4
t=−1 Dividing by 9
The solution is−1.
13. 8x+ 48 = 3x−12
5x+ 48 =−12 Subtracting 3x
5x=−60 Subtracting 48
x=−12 Dividing by 5
The solution is−12.
15. 7y−1 = 23−5y
12y−1 = 23 Adding 5y
12y= 24 Adding 1
y= 2 Dividing by 12
The solution is 2.
17. 3x−4 = 5 + 12x
−9x−4 = 5 Subtracting 12x
−9x= 9 Adding 4
x=−1 Dividing by−9
The solution is−1.
19. 5−4a=a−13
5−5a=−13 Subtractinga
−5a=−18 Subtracting 5
a= 18
5 Dividing by−5
The solution is18
5.
21. 3m−7 =−13 +m
2m−7 =−13 Subtractingm
2m=−6 Adding 7
m=−3 Dividing by 2
The solution is−3.
23. 11−3x= 5x+ 3
11−8x= 3 Subtracting 5x
−8x=−8 Subtracting 11
x= 1
The solution is 1.
25. 2(x+ 7) = 5x+ 14
2x+ 14 = 5x+ 14
−3x+ 14 = 14 Subtracting 5x
−3x= 0 Subtracting 14
x= 0
The solution is 0.
27. 24 = 5(2t+ 5)
24 = 10t+ 25
−1 = 10t Subtracting 25
− 1
10 =t Dividing by 10
The solution is−1
10.
29. 5y−4(2y−10) = 25
5y−8y+ 40 = 25
−3y+ 40 = 25 Collecting like terms
−3y=−15 Subtracting 40
y= 5 Dividing by−3
The solution is 5.
31. 7(3x+ 6) = 11−(x+ 2)
21x+ 42 = 11−x−2
21x+ 42 = 9−x Collecting like terms
22x+ 42 = 9 Addingx
22x=−33 Subtracting 42
x=−3
2 Dividing by 22
The solution is−3
2.
33. 4(3y−1)−6 = 5(y+ 2)
12y−4−6 = 5y+ 10
12y−10 = 5y+ 10 Collecting like terms
7y−10 = 10 Subtracting 5y
7y= 20 Adding 10
y= 20
7 Dividing by 7
The solution is20
7.
35. x2+ 3x−28 = 0
(x+ 7)(x−4) = 0 Factoring
x+ 7 = 0 or x−4 = 0 Principle of zero products
x=−7or x= 4
The solutions are−7 and 4.
37. x2+ 5x= 0
x(x+ 5) = 0 Factoring
x= 0or x+ 5 = 0 Principle of zero products
x= 0or x=−5
The solutions are 0 and−5.
39. y2+ 6y+ 9 = 0
(y+ 3)(y+ 3) = 0
y+ 3 = 0 or y+ 3 = 0
y=−3or y=−3
Exercise Set R.5 11
41. x2+ 100 = 20x
x2−20x+ 100 = 0 Subtracting 20x
(x−10)(x−10) = 0
x−10 = 0 or x−10 = 0
x= 10or x= 10
The solution is 10.
43. x2−4x−32 = 0
(x−8)(x+ 4) = 0
x−8 = 0or x+ 4 = 0
x= 8or x=−4
The solutions are 8 and−4.
45. 3y2+ 8y+ 4 = 0
(3y+ 2)(y+ 2) = 0
3y+ 2 = 0 or y+ 2 = 0
3y=−2 or y=−2
y=−2
3 or y=−2
The solutions are−2
3 and−2.
47. 12z2+z= 6
12z2+z−6 = 0
(4z+ 3)(3z−2) = 0
4z+ 3 = 0 or3z−2 = 0
4z =−3 or 3z= 2
z =−3
4 or z=
2 3
The solutions are−3
4 and
2 3.
49. 12a2−28 = 5a
12a2−5a−28 = 0
(3a+ 4)(4a−7) = 0
3a+ 4 = 0 or4a−7 = 0
3a=−4 or 4a= 7
a=−4
3 or a=
7 4
The solutions are−4
3 and
7 4.
51. 14 =x(x−5) 14 =x2−5x
0 =x2−5x−14
0 = (x−7)(x+ 2)
x−7 = 0or x+ 2 = 0
x= 7or x=−2
The solutions are 7 and−2.
53. x2−36 = 0
(x+ 6)(x−6) = 0
x+ 6 = 0 or x−6 = 0
x=−6or x= 6
The solutions are−6 and 6.
55. z2= 144
z2−144 = 0
(z+ 12)(z−12) = 0
z+ 12 = 0 or z−12 = 0
z =−12or z = 12
The solutions are−12 and 12.
57. 2x2−20 = 0
2x2= 20 x2= 10
x=√10 or x=−√10 Principle of square roots
The solutions are√10 and−√10, or±√10.
59. 6z2−18 = 0
6z2= 18 z2= 3
z=√3 or z=−√3
The solutions are√3 and−√3, or±√3.
61. A= 1
2bh
2A=bh Multiplying by 2 on both sides
2A
h =b Dividing byhon both sides
63. P = 2l+ 2w
P−2l= 2w Subtracting 2lon both sides
P−2l
2 =w Dividing by 2 on both sides
65. A= 1
2h(b1+b2) 2A
h =b1+b2 Multiplying by
2
hon both sides
2A
h −b1=b2, or
2A−b1h
h =b2
67. V = 4
3πr 3
3V = 4πr3 Multiplying by 3 on both sides
3V
4r3 =π Dividing by 4r
