STUDENT'S SOLUTIONS MANUAL COLLEGE ALGEBRA:

333 Pages·2011·9.08 MB·English

by  Stephanie Lindsey

S ’
TUDENT S
S M
OLUTIONS ANUAL
J A. P
UDITH ENNA
Indiana University Purdue University Indianapolis
C A :
OLLEGE LGEBRA
GRAPHS AND MODELS
F E
IFTH 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

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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

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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
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 307
Copyright © 2013 Pearson Education, Inc.
Chapter R
Basic Concepts of Algebra
23. The endpoint −9 is included in the interval, so we use a
Exercise Set R.1 bracket before the −9. The endpoint −4 is not included,
so we use a parenthesis after the −4. Interval notation is
2 √ 1 8 [−9,−4).
1. Rational numbers: , 6,−2.45, 18. 4, −11, 327, 5 , − ,
√ 3 6 7
0, 16 25. Both endpoints are included in the interval, so we use
√ √ √ √ brackets. Intervalnotationis[x,x+h].
3. Irrationalnumbers: 3, 626,7.151551555...,− 35, 53
27. Theendpointpisnotincludedintheinterval,soweusea
(Although there is a pattern in 7.151551555..., there is
parenthesis before the p. The interval is of unlimited ex-
norepeatingblockofdigits.)
tentinthepositivedirection,soweusetheinfinitysymbol
√ √
5. Wholenumbers: 6, 327,0, 16 ∞. Intervalnotationis(p,∞).
7. Integersbutnotnaturalnumbers: −11,0 29. Since 6 is an element of the set of natural numbers, the
statementistrue.
2 1
9. Rational numbers but not integers: , −2.45, 18.4, 5 , 31. Since3.2isnotanelementofthesetofintegers,thestate-
3 6
8 mentisfalse.
−
7
11
33. Since − is an element of the set of rational numbers,
11. This is a closed interval, so we use brackets. Interval no- 5
tationis[−5,5]. thestatementistrue.
√
35. Since 11 is an element of the set of real numbers, the
(cid:1)5 0 5 statementisfalse.
13. This is a half-open interval. We use a parenthesis on 37. Since 24 is an element of the set of whole numbers, the
the left and a bracket on the right. Interval notation is statementisfalse.
(−3,−1].
39. Since1.089isnotanelementofthesetofirrationalnum-
bers,thestatementistrue.
(cid:1)3 (cid:1)1 0
41. Since every whole number is an integer, the statement is
15. This interval is of unlimited extent in the negative direc- true.
tion,andtheendpoint−2isincluded. Intervalnotationis
(−∞,−2]. 43. Since every rational number is a real number, the state-
mentistrue.
(cid:1)2 0 45. Since there are real numbers that are not integers, the
statementisfalse.
17. This interval is of unlimited extent in the positive direc-
tion, and the endpoint 3.8 is not included. Interval nota- 47. The sentence 3+y = y+3 illustrates the commutative
tionis(3.8,∞). propertyofaddition.
49. The sentence −3·1 =−3 illustrates the multiplicative
0 3.8 identityproperty.
19. {x|7<x},or{x|x>7}. 51. Thesentence5·x=x·5illustratesthecommutativeprop-
ertyofmultiplication.
Thisintervalisofunlimitedextentinthepositivedirection
and the endpoint 7 is not included. Interval notation is
53. Thesentence2(a+b)=(a+b)2illustratesthecommutative
(7,∞).
propertyofmultiplication.
55. Thesentence−6(m+n)=−6(n +m)illustratesthecom-
0 7
mutativepropertyofaddition.
21. Theendpoints0and5arenotincludedintheinterval,so
1
weuseparentheses. Intervalnotationis(0,5). 57. Thesentence8· =1illustratesthemultiplicativeinverse
8
property.
Copyright © 2013 Pearson Education, Inc.
2 Chapter R: Basic Concepts of Algebra
59. Thedistanceof−8.15from0is8.15,so|−8.15| =8.15. 9. z0·z7=z0+7=z7,or
61. Thedistance295from0is295,so |295|=295. z0·z7=1·z7=z7
63. Thedistanceof−√97from0is√97,so|−√97|=√97. 11. 58·5−6=58+(−6)=52,or25
65. Thedistanceof0from0is0,so |0|=0. 13. m−5·m5=m−5+5=m0=1
(cid:1) (cid:1) 1
5 5 (cid:1)5(cid:1) 5 15. y3·y−7=y3+(−7)=y−4,or
67. Thedistanceof from0is ,so(cid:1) (cid:1)= . y4
4 4 4 4
69. |14−(−8)|=|14+8|=|22|=22,or 17. (x+3)4(x+3)−2=(x+3)4+(−2)=(x+3)2
|−8 −14|=|−22| =22 19. 3−3·38·3=3−3+8+1=36,or729
71. |−3 −(−9)|=|−3+9|=|6|=6,or 21. 2x3·3x2=2·3·x3+2=6x5
|−9 −(−3)|=|−9+3|=|−6| =6
23. (−3a−5)(5a−7)=−3 ·5·a−5+(−7)=−15a−12,or
73. |12.1−6.7|=|5.4|=5.4,or 15
−
|6.7−12.1|=|−5.4| =5.4 a12
(cid:1)(cid:1) (cid:1)(cid:1) (cid:1)(cid:1) (cid:1)(cid:1) (cid:1)(cid:1) (cid:1)(cid:1) 25. (6x−3y5)(−7x2y−9)=6(−7)x−3+2y5+(−9)=
75. (cid:1)(cid:1)− 3 − 15(cid:1)(cid:1)=(cid:1)(cid:1)− 6 − 15(cid:1)(cid:1)=(cid:1)(cid:1)− 21(cid:1)(cid:1)= 21, or 42
(cid:1) 4 (cid:2) 8 (cid:3)(cid:1) (cid:1)8 8 (cid:1) (cid:1) 8 (cid:1) 8(cid:1) (cid:1) −42x−1y−4,or−
(cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) xy4
(cid:1)(cid:1)15 − − 3 (cid:1)(cid:1)=(cid:1)(cid:1)15 + 3(cid:1)(cid:1)=(cid:1)(cid:1)15 + 6(cid:1)(cid:1)=(cid:1)(cid:1)21(cid:1)(cid:1)= 21
8 4 8 4 8 8 8 8 27. (2x)4(3x)3=24·x4·33·x3=16·27·x4+3=432x7
77. |−7 −0|=|−7| =7, or 29. (−2n)3(5n)2=(−2)3n3·52n2=−8·25·n3+2=
|0−(−7)|=|0+7|=|7|=7 −200n5
79. A0.n12sw41e2rs44m12ay44v4a.r.y...Onesuchnumberis 31. y35 =y35−31=y4
y31
1
81. Answersmayvary. Since −101 =0.0099and 33. b−7 =b−7−12=b−19,or 1
− 1 =−0.01,onesuchnumberis−0.00999. b12 b19
100
35. x2y−2 =x2−(−1)y−2−1=x3y−3,or x3
83. Since12+32=10,thehypotenuseofarighttri√anglewith x−1y y3
legsoflengths1unitand3unitshasalengthof 10units.
✏✏✏✏c✏✏✏✏1 cc22 ==√1120+32 3379.. (3422xxx−2−y54)yy483==243(42xx2)−44y−4(−=5)1y63x−28·4=y48=xy1−65x,8oyr48yx5
3 c= 10
41. (−2x3)5=(−2)5(x3)5=(−2)5x3·5=−32x15
Exercise Set R.2 43. (−5c−1d−2)−2=(−5)−2c−1(−2)d−2(−2)=
(cid:2) (cid:3) c2d4 c2d4
=
1. 3−7= 1 a−m= 1 , a(cid:9)=0 (−5)2 25
37 am
45. (3m4)3(2m−5)4=33m12·24m−20=
3. Observe that each exponent is negative. We move each 432
factortotheothersideofthefractionbarandchangethe 27·16m12+(−20)=432m−8,or m8
signofeachexponent. (cid:2) (cid:3)
x−5 y4 2x−3y7 3 (2x−3y7)3 23x−9y21
y−4 = x5 47. z−1 = (z−1)3 = z−3 =
8x−9y21 8y21z3
5. Observe that each exponent is negative. We move each , or
z−3 x9
factortotheothersideofthefractionbarandchangethe
(cid:2) (cid:3)
signmof−e1anc−h1e2xponent6t. t6 49. 2142aa160bb−−38cc57 −5=(2a4b−5c2)−5=2−5a−20b25c−10,
= ,or
t−6 m1n12 mn12
b25
or
7. 230=1 (Foranynonzerorealnumber,a0=1.) 32a20c10
Copyright © 2013 Pearson Education, Inc.
Exercise Set R.2 3
51. Convert16,500,000toscientificnotation. 71. (4.2×107)(3.2×10−2)
We want the decimal point to be positioned between the =(4.2×3.2)×(107×10−2)
1 and the 6, so we move it 7 places to the left. Since =13.44×105 Thisisnotscientificnotation.
16,500,000 is greater than 10, the exponent must be posi-
=(1.344×10)×105
tive.
16,500,000=1.65×107 =1.344×106 Writingscientificnotation
73. (2.6×10−18)(8.5×107)
53. Convert0.000000437toscientificnotation.
=(2.6×8.5)×(10−18×107)
We want the decimal point to be positioned between the
4 and the 3, so we move it 7 places to the right. Since =22.1×10−11 Thisisnotscientificnotation.
0.000000437 is a number between 0 and 1, the exponent =(2.21×10)×10−11
mustbenegative. =2.21×10−10
0.000000437=4.37×10−7
6.4×10−7 6.4 10−7
55. Convert 234,600,000,000 to scientific notation. We want 75. 8.0×106 = 8.0 × 106
the decimal point to be positioned between the 2 and the =0.8×10−13 Thisisnotscientific
3,sowemoveit11placestotheleft. Since234,600,000,000
notation.
isgreaterthan10,theexponentmustbepositive. =(8×10−1)×10−13
234,600,000,000=2.346×1011 =8×10−14 Writingscientific
notation
57. Convert 0.00104 to scientific notation. We want the deci-
malpointtobepositionedbetweenthe1andthelast0,so 1.8×10−3
wemoveit3placestotheright. Since0.00104isanumber 77.
7.2×10−9
between0and1,theexponentmustbenegative.
1.8 10−3
0.00104=1.04×10−3 = ×
7.2 10−9
59. Convert 0.00000000000000000000000000167 to scientific =0.25×106 Thisisnotscientificnotation.
notation. =(2.5×10−1)×106
We want the decimal point to be positioned between the =2.5×105
1 and the 6, so we move it 27 places to the right. Since
0.00000000000000000000000000167isanumberbetween0
79. Theaveragenumberofpiecesoftrashpermileisthetotal
and1,theexponentmustbenegative.
numberofpiecesoftrashdividedbythenumberofmiles.
0.00000000000000000000000000167=1.67×10−27 51.2billion 5.12×1010
=
61. Convert7.6×105 todecimalnotation. 76million 7.6×107
≈0.6737×103
The exponent is positive, so the number is greater than
10. Wemovethedecimalpoint5placestotheright. ≈(6.737×10−1)×103
7.6×105=760,000 ≈6.737×102
Onaverage,thereareabout6.737×102 piecesoftrashon
63. Convert1.09×10−7 todecimalnotation.
eachmileofroadway.
Theexponentisnegative,sothenumberisbetween0and
1. Wemovethedecimalpoint7placestotheleft. 81. Thenumberofpeoplepersquaremileisthetotalnumber
1.09×10−7=0.000000109 ofpeopledividedbythenumberofsquaremiles.
38,000 3.8×104
65. Convert3.496×1010 todecimalnotation. 0.75 = 7.5×10−1
The exponent is positive, so the number is greater than ≈0.50667×105
10. Wemovethedecimalpoint10placestotheright. ≈(5.0667×10−1)×105
3.496×1010=34,960,000,000 ≈5.0667×104
67. Convert5.41×10−8 todecimalnotation. Thereareabout5.0667×104 peoplepersquaremile.
Theexponentisnegative,sothenumberisbetween0and
83. We multiply the number of light years by the number of
1. Wemovethedecimalpoint8placestotheleft.
milesinalightyear.
5.41×10−8=0.0000000541 4.22×5.88×1012 =24.8136×1012
69. Convert2.319×108 todecimalnotation. =(2.48136×10)×1012
The exponent is positive, so the number is greater than =2.48136×1013
10. Wemovethedecimalpoint8placestotheright.
ThedistancefromEarthtoAlphaCentauriCis
2.319×108=231,900,000 2.48136×1013 mi.
Copyright © 2013 Pearson Education, Inc.
4 Chapter R: Basic Concepts of Algebra
85. Firstfindthenumberofsecondsin1hour: 93. Sinceinterestiscompoundedsemiannually,n=2. Substi-
✦
60min 60sec tute $3225 for P, 3.1% or 0.031 for i, 2 for n, and 4 for t
1hour=1✧hr× × ✦ =3600sec
1✧hr 1min inthecompoundinterestformula.
(cid:4) (cid:5)
The number of disintegrations produced in 1 hour is the i nt
A=P 1+
numberofdisintegrationspersecondtimesthenumberof n
(cid:4) (cid:5)
secondsin1hour. 0.031 2·4
37billion×3600 =$3225 1+ 2 Substituting
=37,000,000,000×3600 =$3225(1+0.0155)2·4Dividing
=3.7×1010×3.6×103 Writingscientific =$3225(1.0155)2·4 Adding
notation =$3225(1.0155)8 Multiplying2and4
=(3.7×3.6)×(1010×103)
≈$3225(1.130939628) Evaluatingthe
=13.32×1013 Multiplying exponentialexpression
=(1.332×10)×1013 ≈$3647.2803 Multiplying
=1.332×1014 ≈$3647.28 Roundingtothenearestcent
Onegramofradiumproduces1.332×1014 disintegrations
95. Sinceinterestiscompoundedquarterly,n=4. Substitute
in1hour.
$4100forP,2.3%or0.023fori,4forn,and6fortinthe
compoundinterestformula.
87. =5·3+8· 32+4(6−2) (cid:4) (cid:5)
i nt
=5·3+8· 32+4· 4 Workinginsideparentheses A=P 1+
n
=5·3+8· 9+4· 4 Evaluating32 (cid:4) 0.023(cid:5)4·6
=$4100 1+ Substituting
=15+72+16 Multiplying 4
=$4100(1+0.00575)4·6 Dividing
=87+16 Addinginorder
=$4100(1.00575)4·6 Adding
=103 fromlefttoright
=$4100(1.00575)24 Multiplying4and6
89. 16÷4·4÷2·256
≈$4100(1.147521919) Evaluatingthe
=4·4÷2·256 Multiplyinganddividing exponentialexpression
inorderfromlefttoright ≈$4704.839868 Multiplying
=16÷2·256
≈$4704.84 Roundingtothenearestcent
=8·256
=2048 97. Substitute$250forP,0.05forr and27fortandperform
theresultingcomputation.
91. 4(8−6)2−4·3+2· 8 (cid:4) r (cid:5)12·t 
31+190 1+ −1
4·22−4·3+2· 8 S =P 12r 
= Calculatinginthe
3+1 12
numeratorandin (cid:4) (cid:5) 
0.05 12·27
thedenominator 1+ −1
4·4−4·3+2· 8 =$250 12 
= 0.05
4
16−12+16 12
= 4 ≈$170,797.30
4+16
= 99. Substitute$120,000forS,0.03forr,and18fortandsolve
4
forP.
20 (cid:4) (cid:5) 
= r 12·t
4 1+ −1
=5 S =P 12r 
12
(cid:4) (cid:5) 
0.03 12·18
1+ −1
$120,000=P 12 
0.03
12
(cid:10) (cid:11)
(1.0025)216−1
$120,000=P
0.0025
$120,000≈P(285.94035)
$419.67≈P
Copyright © 2013 Pearson Education, Inc.
Exercise Set R.3 5
101. (xt·x3t)2=(x4t)2=x4t·2=x8t 19. (y−3)(y+5)
103. (ta+x·tx−a)4=(t2x)4=t2x·4=t8x =y2+5y−3y−15 UsingFOIL
(cid:12) (cid:13) (cid:12) (cid:13) =y2+2y−15 Collectingliketerms
(3xayb)3 2 27x3ay3b 2
105. =
(−3xayb)2 9x2ay2b 21. (x+6)(x+3)
(cid:14) (cid:15)
= 3xayb 2 =x2+3x+6x+18 UsingFOIL
=9x2ay2b =x2+9x+18 Collectingliketerms
23. (2a+3)(a+5)
Exercise Set R.3
=2a2+10a+3a+15 UsingFOIL
1. 7x3−4x2+8x+5=7x3+(−4x2)+8x+5 =2a2+13a+15 Collectingliketerms
Terms: 7x3,−4x2,8x,5
25. (2x+3y)(2x+y)
Thedegreeofthetermofhighestdegree,7x3,is3. Thus,
=4x2+2xy+6xy+3y2 UsingFOIL
thedegreeofthepolynomialis3.
=4x2+8xy+3y2
3. 3a4b−7a3b3+5ab−2=3a4b+(−7a3b3)+5ab+(−2)
Terms: 3a4b,−7a3b3,5ab,−2 27. (x+3)2
=x2+2· x·3+32
The degrees of the terms are 5, 6, 2, and, 0, respectively,
sothedegreeofthepolynomialis6. [(A+B)2=A2+2AB+B2]
=x2+6x+9
5. (3ab2−4a2b−2ab+6)+
(−ab2−5a2b+8ab+4) 29. (y−5)2
=(3−1)ab2+(−4−5)a2b+(−2+8)ab+(6+4) =y2−2·y·5+52
=2ab2−9a2b+6ab+10 [(A−B)2=A2−2AB+B2]
7. (2x+3y+z−7)+(4x−2y−z+8)+ =y2−10y+25
(−3x+y−2z−4) 31. (5x−3)2
=(2+4−3)x+(3− 2+1)y+(1− 1−2)z+ =(5x)2−2·5x·3+32
(−7+8− 4) [(A−B)2=A2−2AB+B2]
=3x+2y−2z−3 =25x2−30x+9
9. (3x2−2x−x3+2)− (5x2−8x−x3+4) 33. (2x+3y)2
=(3x2−2x−x3+2)+(−5x2+8x+x3−4) =(2x)2+2(2x)(3y)+(3y)2
=(3−5)x2+(−2+8)x+(−1+1)x3+(2− 4) [(A+B)2=A2+2AB+B2]
=−2x2+6x−2 =4x2+12xy+9y2
11. (x4−3x2+4x)−(3x3+x2−5x+3) 35. (2x2−3y)2
=(x4−3x2+4x)+(−3x3−x2+5x−3)
=(2x2)2−2(2x2)(3y)+(3y)2
=x4−3x3+(−3−1)x2+(4+5)x−3
[(A−B)2=A2−2AB+B2]
=x4−3x3−4x2+9x−3 =4x4−12x2y+9y2
13. (3a2)(−7a4)=[3(−7)](a2·a4) 37. (n+6)(n−6)
=−21a6 =n2−62 [(A+B)(A−B)=A2−B2]
15. (6xy3)(9x4y2)=(6·9)(x·x4)(y3·y2) =n2−36
=54x5y5
39. (3y+4)(3y−4)
17. (a−b)(2a3−ab+3b2) =(3y)2−42 [(A+B)(A−B)=A2−B2]
=(a−b)(2a3)+(a−b)(−ab)+(a−b)(3b2) =9y2−16
Usingthedistributiveproperty
41. (3x−2y)(3x+2y)
=2a4−2a3b−a2b+ab2+3ab2−3b3
=(3x)2−(2y)2 [(A−B)(A+B)=A2−B2]
Usingthedistributiveproperty
threemoretimes =9x2−4y2
=2a4−2a3b−a2b+4ab2−3b3 Collectinglike
terms
Copyright © 2013 Pearson Education, Inc.
6 Chapter R: Basic Concepts of Algebra
43. (2x+3y+4)(2x+3y−4) 15. x3−x2−5x+5
=[(2x+3y)+4][(2x+3y)−4] =x2(x−1)−5(x−1)
=(2x+3y)2−42 =(x−1)(x2−5)
=4x2+12xy+9y2−16 17. a3−3a2−2a+6
45. (x+1)(x−1)(x2+1) =a2(a−3)−2(a−3)
=(x2−1)(x2+1) =(a−3)(a2−2)
=x4−1
19. w2−7w+10
47. (an+bn)(an−bn)=(an)2−(bn)2 Welookfortwonumberswithaproductof10andasum
=a2n−b2n of−7. Bytrial,wedeterminethattheyare−5and−2.
w2−7w+10=(w−5)(w−2)
49. (an+bn)2 =(an)2+2· an·bn+(bn)2
=a2n+2anbn+b2n 21. x2+6x+5
We look for two numbers with a product of 5 and a sum
51. (x−1)(x2+x+1)(x3+1)
of6. Bytrial,wedeterminethattheyare1and5.
=[(x−1)x2+(x−1)x+(x−1)·1](x3+1)
x2+6x+5=(x+1)(x+5)
=(x3−x2+x2−x+x−1)(x3+1)
=(x3−1)(x3+1) 23. t2+8t+15
=(x3)2−12 Welookfortwonumberswithaproductof15andasum
of8. Bytrial,wedeterminethattheyare3and5.
=x6−1
t2+8t+15=(t+3)(t+5)
53. (xa−b)a+b
25. x2−6xy−27y2
=x(a−b)(a+b)
Welookfortwonumberswithaproductof−27andasum
=xa2−b2 of−6. Bytrial,wedeterminethattheyare3and−9.
55. (a+b+c)2 x2−6xy−27y2=(x+3y)(x−9y)
=(a+b+c)(a+b+c) 27. 2n2−20n−48=2(n2−10n−24)
=(a+b+c)(a)+(a+b+c)(b)+(a+b+c)(c) Nowfactorn2−10n−24. Welookfortwonumberswitha
=a2+ab+ac+ab+b2+bc+ac+bc+c2 productof−24andasumof−10. Bytrial,wedetermine
=a2+b2+c2+2ab+2ac+2bc thattheyare2and−12. Thenn2−10n−24=
(n+2)(n−12). We must include the common factor, 2,
tohaveafactorizationoftheoriginaltrinomial.
Exercise Set R.4 2n2−20n−48=2(n+2)(n−12)
1. 3x+18=3· x+3· 6=3(x+6) 29. y2−4y−21
Welookfortwonumberswithaproductof−21andasum
3. 2z3−8z2=2z2·z−2z2·4=2z2(z−4) of−4. Bytrial,wedeterminethattheyare3and−7.
5. 4a2−12a+16=4· a2−4·3a+4· 4=4(a2−3a+4) y2−4y−21=(y+3)(y−7)
7. a(b−2)+c(b−2)=(b−2)(a+c) 31. y4−9y3+14y2=y2(y2−9y+14)
Now factor y2 −9y+14. Look for two numbers with a
9. 3x3−x2+18x−6 productof14andasumof−9. Thenumbersare−2and
=x2(3x−1)+6(3x−1) −7. Theny2−9y+14=(y−2)(y−7). Wemustinclude
=(3x−1)(x2+6) the common factor, y2, in order to have a factorization of
theoriginaltrinomial.
11. y3−y2+2y−2 y4−9y3+14y2=y2(y−2)(y−7)
=y2(y−1)+2(y−1)
33. 2x3−2x2y−24xy2=2x(x2−xy−12y2)
=(y−1)(y2+2)
Now factor x2−xy−12y2. Look for two numbers with a
13. 24x3−36x2+72x−108 productof−12andasumof−1. Thenumbersare−4and
=12(2x3−3x2+6x−9) 3. Thenx2−xy−12y2=(x−4y)(x+3y). Wemustinclude
thecommonfactor,2x,inordertohaveafactorizationof
=12[x2(2x−3)+3(2x−3)]
theoriginaltrinomial.
=12(2x−3)(x2+3) 2x3−2x2y−24xy2=2x(x−4y)(x+3y)
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Exercise Set R.4 7
35. 2n2+9n−56 4. Split the middle term using the numbers found in
step(3):
WeusetheFOILmethod.
1. Thereisnocommonfactorotherthan1or−1. y=4y−3y
5. Factorbygrouping.
2. Thefactorizationmustbeoftheform
(2n+ )(n+ ). 2y2+y−6=2y2+4y−3y−6
3. Factor the constant term, −56. The possibilities =2y(y+2)− 3(y+2)
are−1·56,1(−56),−2·28,2(−28),−4·16,4(−16), =(y+2)(2y−3)
−7·8, and 7(−8). The factors can be written in
theoppositeorderaswell: 56(−1),−56·1,28(−2), 43. 6a2−29ab+28b2
−28·2,16(−4),−16·4,8(−7),and−8·7.
WeusetheFOILmethod.
4. Findapairoffactorsforwhichthesumoftheout- 1. Thereisnocommonfactorotherthan1or −1.
side and the inside products is the middle term,
9n. By trial, we determine that the factorization 2. Thefactorizationmustbeoftheform
is(2n−7)(n+8). (6x+ )(x+ )or(3x+ )(2x+ ).
3. Factorthecoefficientofthelastterm,28. Thepos-
37. 12x2+11x+2 sibilities are 1·28, −1(−28), 2·14, −2(−14), 4·7,
Weusethegroupingmethod. and −4(−7). The factors can be written in the op-
1. Thereisnocommonfactorotherthan1or−1. positeorderaswell: 28·1,−28(−1),14·2,−14(−2),
7·4,and−7(−4).
2. Multiply the leading coefficient and the constant:
12·2=24. 4. Findapairoffactorsforwhichthesumoftheout-
side and the inside products is the middle term,
3. Trytofactor24sothatthesumofthefactorsisthe −29. Observe that the second term of each bino-
coefficient of the middle term, 11. The factors we mial factor will contain a factor of b. By trial, we
wantare3and8. determinethatthefactorizationis(3a−4b)(2a−7b).
4. Split the middle term using the numbers found in
step(3): 45. 12a2−4a−16
11x=3x+8x Wewillusethegroupingmethod.
5. Factorbygrouping. 1. Factoroutthecommonfactor,4.
12x2+11x+2=12x2+3x+8x+2 12a2−4a−16=4(3a2−a−4)
=3x(4x+1)+2(4x+1) 2. Now consider 3a2 −a−4. Multiply the leading
coefficientandtheconstant: 3(−4)=−12.
=(4x+1)(3x+2)
3. Try to factor −12 so that the sum of the factors is
39. 4x2+15x+9 the coefficient of the middle term, −1. The factors
wewantare−4and3.
WeusetheFOILmethod.
1. Thereisnocommonfactorotherthan1or−1. 4. Split the middle term using the numbers found in
step(3):
2. Thefactorizationmustbeoftheform
−a=−4a+3a
(4x+ )(x+ )or(2x+ )(2x+ ).
5. Factorbygrouping.
3. Factor the constant term, 9. The possibilities are
1·9,−1(−9),3·3,and−3(−3). Thefirsttwopairs 3a2−a−4=3a2−4a+3a−4
of factors can be written in the opposite order as =a(3a−4)+(3a−4)
well: 9·1,−9(−1). =(3a−4)(a+1)
4. Findapairoffactorsforwhichthesumoftheout-
Wemustincludethecommonfactortogetafactor-
side and the inside products is the middle term,
izationoftheoriginaltrinomial.
15x. By trial, we determine that the factorization
12a2−4a−16=4(3a−4)(a+1)
is(4x+3)(x+3).
41. 2y2+y−6 47. z2−81=z2−92=(z+9)(z−9)
Weusethegroupingmethod. 49. 16x2−9=(4x)2−32=(4x+3)(4x−3)
1. Thereisnocommonfactorotherthan1or−1.
51. 6x2−6y2=6(x2−y2)=6(x+y)(x−y)
2. Multiply the leading coefficient and the constant:
2(−6)=−12. 53. 4xy4−4xz2 =4x(y4−z2)
3. Try to factor −12 so that the sum of the factors is =4x[(y2)2−z2]
thecoefficientofthemiddleterm,1. Thefactorswe =4x(y2+z)(y2−z)
wantare4and−3.
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