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CopyrightbyM
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Algebra 2
Worked-Out Solution Key
Prerequisite Skills (p. 328)
1. 23 and 1
2. 4
3. y5 ax21 bx1 c
4.
2
x
y
21
(1, 4)
x51
5.3
x
y
21
x521
2
(2 , 218 )1
2
3
4
6.
1
x
y
21
(22, 8)
x
5
2
2
7. x21 9x1 205 (x1 4)(x1 5)
8. 2x21 5x2 35 (2x2 1)(x1 3)
9. 9x22 645 (3x2 8)(3x1 8)
10. 2x21x1 65 0
x521 6 }}
(1)22 4(2)(6)}}
2(2) 5
21 6}
247}
4 5
21 6i}
47}
4
11. 10x21 13x5 3
10x21 13x2 35 0
(5x2 1)(2x1 3)5 0
5x2 15 0 or 2x1 35 0
5x5 1 or 2x5 23
x51
}
5
or x5 23}
2
12. x21 6x1 25 20
x21 6x2 185 0
x526 6 }}
(6)22 4(1)(218)}}
2(1)
526 6
}
108}
2
526 66
}
3}
2
5 23 63}
3
Lesson 5.1
5.1 Guided Practice (pp. 331333)
1. (42)35 46 Power of a power property
5 4096
2. (28)(28)35 (28)11 3Product of powers property
5 (28)4
5 4096
3. 12}9235
23
}
93 Power of a quotient property
58
}
729
4.6 p1024
}
9 p107 5
6
}
9p102427 Quotient of powers property
56
}
9
p10211
52
}
3 p1011
Negative exponent property
5.x26x5x35x261 51 3 Product of powers property
=x2
6. (7y2z5)(y24z21)5 7y21 (24)z51 (21) Product ofpowers property
5 7y22z4
57z4
}
y2
Negative exponent property
7. 1s3
}
t242
2
5
(s3)2
}
(t24)2
Power of a quotient property
5s6
}
t28
Power of a power property
5s6t8 Negative exponent property
8. 1x4y22
}
x3y6 2
3
5
(x4y22)3
}
(x3y6)3 Power of a quotient property
5x4 p3y22 p3
}
x3 p3y6 p3 Power of a product property
5x12y26
}
x9y18
5x1229y26218 Quotient of powers property
5x3
}
y24
Negative exponent property
5.1 Exercises (pp. 333335)
Skill Practice
1. a.Product of powers property
b.Negative exponent property
c.Power of a product property
2. No, 25.2 is not less then 10; the number should be
2.523 1022.
3. 33p325 3(31 2) Product of powers property
5 35
5 243 4. (422)35 426 Power of a power property
51}
46 Negative exponent property
51
}
4096
Chapter 5
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288
Algebra 2
Worked-Out Solution Key
5. (25)(25)45 (25)(11 4) Product of powers property
5 (25)5
5 23125
6. (24)25 28 Power of a power property
5 256
7.52
}
555 5(2 2 5) Quotient of powers property
5 523
51
}
53
Negative exponent property
51
}
125
8. 13}5245
34
}
54
Power of a quotient property
581
}
625
9. 12}7223
5223
}
723 Power of a quotient property
5
73
}
23
Negative exponent property
5343
}
8
10. 93p9215 9(31 (21)) Product of powers property
5 92
5 81
11.34
}
322
5 342 (22) Quotient of powers property
5 36
5 729
12. 12}3225
12}3245 12}32
251 4 Product of powers property
5 12}3221
53
}
2
Negative exponent property
13. 63p60p6255 631 02 5 Product of powers property
5 622
51
}
62
Negative exponent property
51
}
36
14. 111}2225
22511}22210
Power of a power property
5
1210
}
2210 Power of a quotient property
5 210 Negative exponent property
5 1024
15. (4.23 103)(1.53 106)5 (4.23 1.5)(1033 106)
5 6.33 109
16. (1.23 1023)(6.73 1027)5 (1.23 6.7)(10233 1027)
5 8.043 10210
17. (6.33 105)(8.93 10212)5 (6.33 8.9)(1053 10212)
5 56.073 1027
5 5.6073 1013 102 7
5 5.6073 1026
18. (7.23 109)(9.43 108)5 (7.23 9.4)(1093 108)
5 67.683 1017
5 6.7683 1013 1017
5 6.7683 1018
19. (2.13 1024)35 2.133 (1024)3
5 9.2613 10212
20. (4.03 103)45 4.043 (103)4
5 2563 1012
5 2.563 1023 1012
5 2.563 1014
21.8.13 1012
}
5.43 109 5
8.1
}
5.43
1012
}
109
5 1.53 103
22. 1.13 1023
}
5.53 1028
5 1.1
}5.5
3 1023
}1028
5 0.23 105
5 2.03 10213 105
5 2.03 104
23.(7.53 108)(4.53 1024)
}}
1.53 107 5
(7.53 4.5)(1083 1024)
}}
1.53 107
533.753 104
}
1.53 107
53.3753 1013 104
}}
1.53 107
53.3753 105
}
1.53
107
53.375
}
1.5
3105
}
107
5 2.253 1022
24.w22
}
w6 5 w22 2 6 Quotient of powers property
5 w28
51
}
w8
Negative exponent property
25.(22y3)55 (22)5(y3)5 Power of a product property
5 210y15 Power of a power property
5 1024y15
26.(p3q2)215p23q22 Power of a product property
51
}
p3q2 Negative exponent property
27.(w3x22)(w6x21)5 w31 6x221 (21) Product of powersproperty
5 w9x23
5w9
}
x3
Negative exponent
property
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
28.(5s22t4)235 523(s22)23(t4)23 Power of a productproperty
51
}
125
ps22(23)pt4(23) Power of a power
property
51
}
125ps6pt212
5s6
}
125t12
Negative exponent
property
29. 13a3b52235 (3)23(a3)23(b5)23 Power of a productproperty
51
}
27
a3(23)b5(23) Power of a power
property
51
}
27
a29b215
51
}
27a9b15 Negative exponent property
30.x21y2
}
x2y215x2122y22 (21) Quotient of powers property
5x23y3
5y3
}x3
Negative exponent property
31.3c3d
}
9cd215
3
}
9
c32 1d12 (21) Quotient of powers property
5c2d2
}
3
32.4r4s5
}
24r4s255
4
}
24
r42 4s52 (25) Quotient of powers property
51
}
6
r0s10
51
}
6
p1 ps10 Zero exponent property
5
s10
}
6
33.2a3b24
}
3a5b225
2
}
3
a32 5b242 (22) Quotient of powers property
52
}
3
a22b22
52
}
3a2b2
Negative exponent property
34.y11
}
4z3p
8z7
}
y7 5
8y11z7
}
4y7z3
5 2y112 7z72 3 Quotient of powers property
5 2y4z4
35.x2y
2
3
}3y2 p y
2
}x245
x2y2
1
}
3x24y2
Product of powers property
5x22 (24)y2122
}}
3
Quotient of powers property
5x6y23
}
3
5x6
}
3y3
Negative exponent property
36. B;
2x2y
}
6xy215
x22 1y12 (21)
}
3
5xy2
}
3
37. The error is that the exponents were divided and not
subtracted.
x10
}
x2 5x102 2
5x8
38. The error is that the exponents were multiplied instead
of added.
x5px35x51 3
5x8
39. The error is that the bases were multiplied.
(23)2p(23)45 (23)6
40.A5
}
3}
4
s2 s5x
}
3
5 }
3}4 1x}322
5
}
3}
4 1x
2
}32
25
}
3x2
}
4 p9
5
}
3x2
}
36
41. v5 r2h r = x h5x
}
2
5 (x)21x}22
5 x2px
}
2
5x3
}
2
42. v5 lpwph l5 2x, w55x
}
3
, h5x
5(2x)15x}32(x)
510x3
}
3
43.x15y12z85x4y7z11p?
x15y12z8
}
x4y7z11
5 ?
x152 4y122 7z82 115 ?
x11y5z235x11y5
}
z3
5 ?
44. 3x3y2512x2y5
}
?
?512x2y5
}
3x3y2
5 4x22 3y52 2
5 4x21y3
54y3
}
x
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
45. (a5b4)25 a14b21p?
a5 p2b4 p25 a14b21p?
a10b85 a14b21p?
a10b8
}
a14b215 ?
a102 14p
b82 (21)5 a24b95b9
}
a
45 ?
46. Sample answer:
x12x165 (x4y4)(x8y12)5 (x2y15)(x10y1)5 (x25y7)(x17y9)
47.1}
am5
a0
}
am5 a02 m5 a2m
48.am
}
an5 amp
1}
an
a2np
am5 a2n1 m5 am2 n
Problem Solving
49. Pacific: V5 (1.563 1014)(4.033 103)
5 (1.563 4.03)(10143 103)
5 6.28683 1017meters3
Atlantic: V5 (7.683 1013)(3.933 103)
5 (7.683 3.93)(10133 103)
5 30.1824 3 1016
5 3.018243 1013 1016
5 3.018243 1017meters3
Indian: V5 (6.863 1013)(3.963 103)
5 (6.863 3.96)(10133 103)
5 27.16563 1016
5 2.716563 1013 1016
5 2.716563 1017meters3
Arctic: V5 (1.41 3 1013)(1.213 103)
5 (1.413 1.21)(10133 103)
5 1.70613 1016meters3
50. (0.000022)(125,000,000)5 (2.23 1025)(1.253 108)
5 (2.23 1.25)(10253 108)
5 2.753 103
The continent has moved 2.753 103miles.
51. Bead: d5 6 mm
r5 3 mm
V54
}
3
r354
}
3
(3)35 36
Pearl : d5 9 mm
r59
}
2
mm
V54
}
3r35
4
}
319}22
35
243}
2
Volume of pearl
}}
Volume of bead 5
243}
2
}
365
243
}
725 3.375
The volume of the pearl is about 3.375 times greater
than the volume of the bead.
52. a. V5 314}3r325 4r3
b.V5 r2h
c.h5 3(2r)5 6r
d.Volume of cylinder5 r2(6r)5 6r3
Occupied volume54r3
}
6r35
2
}
3
The tennis balls take up 2
}
3
of the cans volume.
53. a.V5 r2h5 (9.53)2(1.55) 442 mm35
4.4231027m3
b. Answers will vary.
c.pennies5Volume of classroom
}}
Volume of 1 penny
The answer is an overestimate because of the shape of
the pennies. There will be gaps between them.
54. a. Volume of inner core54
}
3
r3
Earths total volume54
}
3(5r)3
54
}
3
(125r3)
5 125 p4
}
3
r3
5 125 pVolume of inner core
ratio5 125:1
b.Volume of outer core54
}
3111}6r1 r2
32
4
}
3r3
54913
}
21614}3r322
4
}
3
r3
54697
}
21614}3r32
ratio5
4697
}216
p4
}
3r3
}4
}
3
r3
54697
}
216
Mixed Review
55.
1
x
y
21
56.1
x
y
21
57.
1
x
y
21
58.1
x
y
21
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
59.
1
x
y
21
60.
2
x
y
21
61. x1y5 2
7x1 8y5 21
A X B
F1 17 8GpFx
yG5F221G
A2151}
82 7F 8 2127 1G5F 8 2127 1G
X5A21B5F 8 2127 1G F221G5F25
7G5 FxyG
The solution is (25, 7).
62. 2x2 2y5 3
2x1 8y5 1
A X B
F21 22 2 8GpFx
yG5 F31G
A21
51
}
282 (24)F 8 222 21G5F22 21}21
}
2
1
}
4
G X5A21B5
F22 2
1
}
2
1
}
2
1
}
4
GF31G5F
213}2
7
}
4 G
5FxyG
The solution is 1213}2,7
}
42.
63. 4x1 3y5 6
6x2 2y5 10
A X B
F4 36 22G Fx
yG5F610G
A21
51
}
282 18 F22 2326 4G5
F
1
}
13
3
}
26
3
}
13
2
}
13
G X5A21B5F1}13 3}263
}
13 2
2
}
13
GF610G5F 21}1322}13
G5FxyG The solution is 121}13, 2
2
}
132.
64. (81 3i)2 (71 4i)5 81 3i2 72 4i5 12 i
65. (52 2i)2 (291 6i)5 52 2i1 92 6i5 142 8i
66. i(31 i)5 3i1 i2 5 3i1 (21)5 211 3i
67. (121 5i)2 (72 8i)5 121 5i2 71 8i5 51 13i
68. (51 4i)(2 + 3i)5 101 15i1 8i1 12i2
5 101 23i1 12(21)
5 221 23i
69. (82 4i)(11 6i)5 81 48i2 4i2 24i2
5 81 44i2 24(21)
5 321 44i
Lesson 5.2
Investigating Algebra Activity 5.2 (p. 336)
1. Asxapproaches 2`,
f(x) approaches 2`.
Asxapproaches 1`,
f(x) approaches 1`.
2. Asxapproaches 2`,
f(x) approaches1`.
Asxapproaches1`,
f(x) approaches 2`.
3. Asxapproaches 2`,
f(x) approaches1`.
Asxapproaches1`,
f(x) approaches1`.
4. Asxapproaches 2`,
f(x) approaches 2`.
Asxapproaches1`,
f(x) approaches 2`.
5. a.f (x)5xn, wherexis odd: Asxapproaches1`,
f (x) approaches1`. Asxapproaches 2`,
f (x) approaches 2`.
b.f (x)5 2xn, where nis odd: Asxapproaches2`,
f (x) approaches1`. Asxapproaches1 `,
f (x) approaches 2`.
c.f (x)5xn, where nis even : Asxapproaches2`,
f (x) approaches1`. Asxapproaches1 `,
f (x) approaches1`.
d.f (x)5 2xn, wherexis odd : Asxapproaches2`,
f (x) approaches2`. Asxapproaches1 `,
f (x) approaches2`.
6. Asxapproaches 2`,f (x) approaches1`. Asx
approaches1`,f (x) approaches 1`.f(x) is going to
resemble its highest power. Becausex6is the highest
power, the end behavior will be the same asf(x)5x6.
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
5.2 Guided Practice (pp. 338341)
1. The function is a polynomial function written as
f (x)5 22x 1 13 in standard form. It has degree 1 (linear)
and a leading coefficient of 22.
2. The function is not a polynomial function because the term
5x22has an exponent that is not a whole number.
3. The function is a polynomial function written as
h(x)5 6x22 3x1 in standard form. It has a degree
of 2 (quadratic) and a leading coefficient of 6.
4. f (x)5x41 2x31 3x227
f (22)5 (22)41 2(22)31 3(22)22 7
5 162 161 122 7
5 5
5. g(x)5x32 5x21 6x1 1
g(4)5(4)32 5(4)21 6(4)1 1
5 642 801 241 1
5 9
6. 2 5 3 21 7
10 26 50
5 13 25 57
f (2)5 57
7. 21 22 21 0 4 25
2 21 1 25
22 1 21 5 210
f (21)5 210
8. The degree is odd and the leading coefficient is negative.
9. x 23 22 21 0 1 2 3
y 132 37 4 23 4 37 132
x
y
21
1
10. x 23 22 21 0 1 2 3
y 32 9 0 21 0 23 216
x
y
22
1
11. x 23 22 21 0 1 2 3
y 58 20 6 4 2 212 250
x
y
21
1
12. s 0 5 10 15 20 25 30
E 0 3.2 51 258.2 816 1992.2 4131
The wind speed needed to generate the wave is about 25
miles per hour.
Wind speed
(miles per hour)
Energy
pe
rsquare
foot
(foot-pounds)
0 10 20 30 s
E
0
1000
2000
3000
5.2 Exercises (pp. 341344)
Skill Practice
1.f (x) is a degree 4 (quartic) polynomial function. The
leading coefficient is 25 while the constant term is 6.
2. The end behavior of a polynomial is the behavior the
function demonstrates as it approaches 6`.
3. The function is a polynomial function written asf (x)5 2x21 8 in standard form. It has a degree 2
(quadratic) and a leading coefficient of 21.
4. The function is a polynomial function written as
f (x)5 8x41 6x2 3 in standard form. It has a degree
4 (quartic) and a leading coefficient of 8.
5. The function is a polynomial that is already written in
standard form. It has degree 4 (quartic) and a leading
coefficient of .
6. The function is not a polynomial function because the term
5x22has an exponent that is not a whole number.
7. The function is a polynomial function already written
in standard form. It has degree 3 (cubic) and a leading
coefficient of 25}2
.
8. The function is not a polynomial function because the
term2
}
x has an exponent that is not a whole number.
9. f (x)5 5x32 2x21 10x2 15
f (21)5 5(21)32 2(21)21 10(21)2 15
5 252 22 102 15
5 232
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
10. f (x)5 5x42x32 3x21 8x
f (2)5 5(2)42 (2)32 3(2)21 8(2)
5 802 82 121 16
5 76
11. g(x)5 22x51 4x3
g(23)5 22(23)51 4(23)35 4862 1085 378
12. h(x)5 6x32 25x1 20
h(5)5 6(5)32 25(5)1 205 7502 1251 205 645
13. h(x)51
}
2
x423
}
4
x31x1 10
h(24)51
}
2(24)42
3
}
4(24)31 (24)1 10
5 1281 482 41 10
5 182
14. g(x)5 4x51 6x31x2210x1 5
g(22)5 4(22)51 6(22)31 (22)22 10(22)1 5
5 21282 481 41 201 5
5 2147
15. 3 5 22 28 16
15 39 93
5 13 31 109
f (3)5 109
16. 22 8 12 6 25 9
216 8 228 66
8 24 14 233 75
f (22)5 75
17. 26 1 8 27 35
26 212 114
1 2 219 149
f (26)5 149
18. 4 28 0 14 235
232 2128 2456
28 232 2114 2491
f (4)5 2491
19. 2 22 3 0 28 13
24 22 24 224
22 21 22 212 211
f(2)5 211
20. 23 6 0 10 0 0 227
218 54 2192 576 21728
6 218 64 2192 576 21755
f (23)5 21755
21. 32
7 11 4 0
221 230 278
27 210 226 278
f (3)5 278
22. 4 1 0 0 3 220
4 16 64 268
1 4 16 67 248
f (4)5 248
23. The error is that a zero was not placed as the coefficient ofthe missingx3term.
22 24 0 9 221 7
8 216 14 14
24 8 27 27 21
f (22)5 21 117
24. A; The ends approach opposite directions so the function
must be odd. It imitates an odd function whose coefficient
is positive.
25. The degree is even and the leading coefficient is positive.
26. The degree is odd and the leading coefficient is negative.27. The degree is even and the leading coefficient is negative.
28. f (x) 1`asx 2`
f (x) 1`asx 1`
29. f (x) 2`asx 2`
f (x) 2`asx 1`
30. f (x) 1`asx 2`
f (x) 2`asx 1`
31. f (x) 2`asx 2`
f (x) 1`asx 1`
32. f (x) 1`asx 2`
f (x) 1`asx 1`
33. f (x) 1`asx 2`
f (x) 2`asx 1`
34. f (x) 2`asx 2`
f (x) 1`asx 1`
35. f (x) 1`asx 2`
f (x) 1`asx 1`
36. f (x) 1`asx 2`
f (x) 2`asx 1`
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
37. Sample answer:f (x)5 2x52 4x21 3x2 1
x
y
211
38. x 23 22 21 0 1 2 3
y 227 28 21 0 1 8 27
2
x
y
22
39. x 23 22 21 0 1 2 3
y 281 216 21 0 21 216 281
2
x22
y
40. x 23 22 21 0 1 2 3
y 2240 229 2 3 4 35 246
1
x
y
22
41. x 23 22 21 0 1 2 3
y79 14
2
12
22
1 14 79
x
y
4
22
42. x 23 22 21 0 1 2 3
y 32 13 6 5 4 23 222
1
x
y
21
43. x 23 22 21 0 1 2 3
y 212 2 4 0 24 22 12
1
x
y
21
44. x 23 22 21 0 1 2 3
y 2105 232 29 0 7 0 257
1
x
y
21
45. x 23 22 21 0 1 2 3
y 2246 234 22 0 2 34 246
1
x
y
21
46. x 23 22 21 0 1 2 3
y 65 29 11 5 5 5 21
x
y
21
1
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
47. x 23 22 21 0 1 2 3
y 2238 232 24 24 22 32 248
x
y
21
1
48. x 23 22 21 0 1 2 3
y 42 2 2 6 2 2 42
x
y
21
1
49. x 23 22 21 0 1 2 3
y 2158 237 22 1 2 7 22
x
y
22
2
50. B;f(x) 5 21
}
3
x31 1
From the graph,f(x) 1` asx2`;and
f(x) 2` asx1`. So, the degree is odd
and the leading coefficient is negative. The
constant term 5f(0)5 1.
51. Wheng(x)5 2f(x):
g(x) 2` asx2` andg(x) 1` asx1`.
52.f(x)5 a3x31 a
2x21 a
1x1 a
0
When a35 2 and a
05 25:
f(x)5 2x31 a2x21 a
1x2 5
Whenf(1)5 0:
2(1)31 a2(1)21 a
1(1)2 55 0
a21 a
15 3
Whenf(2)5 3:
2(2)31 a2(2)21 a
1(2)2 55 3
4a21 2a
15 28
Solve the system of equations:
a21 a
15 3 324 24a
22 4a
15 212
4a21 2a
15 28 4a
21 2a
15 28
22a15 220
a15 10
Substitute a1into one of the original equations. Solve for a
2.
a21 105 3 a
25 27
The cubic function isf(x)5 2x32 7x21 10x2 5.
So,f(25)5 2(25)32 7(25)21 10(25) 2 55 2480.
53. a.x f(x) g(x)
f(x)}
g(x)
10 1000 840 1.19
20 8000 7280 1.10
50 125,000 120,200 1.04
100 1,000,000 980,400 1.02
200 8,000,000 7,920,800 1.01
b.Asx1`,f(x)
}
g(x)
1.
c.For the graph of a polynomial function, the end behavior
is determined by the functions degree and the sign of its
leading coefficient.
Problem Solving
54. w5 0.00071d32 0.090d21 0.48d
w5 0.0071(15)32 0.090(15)21 0.48(15)
5 23.96252 20.251 7.2 5 10.9125 carats
55. t 0 2 4 6 8
S 5.5 5.1 4.7 6.7 10.5
Skateboarding
participants(millions)
02 4 6 810 3 5 7 109
2
4
6
8
10
14
12
Years since 1992
t
S
There were 8 million skateboarders 6 years after
1992 (1998).
Chapter 5, continued
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56. a.The function is degree 3 (cubic).
b. t 0 2 4 6 8
M 21,600 21,396 22,800 25,284 28,320
t 10 12 14 16
M 31,380 33,936 35,460 35,424
c.
Indoormoviescreens
(thousands)
04 8 12 160
40
20
10
30
Years since 1987
t
M
57. t 0 2 4 6 8 10 12
s 1.2 1.46 1.68 1.59 1.44 1.97 4.41
Snowboarding
participants(millions)
02 4 6 810 3 5 7 109
2
1
Years since 1992
t
S
The number of snowboarders was more than 2 million
in 2002.
58. a. As t1`,P(x) 1`.
As t2`,P(x) 1`.
b.
06 12 18 240
4000
2000
1000
3000
t
P
Quarterlyperiodicals
Years since 1980
c. p(t)5 0.138t42 6.24t31 86.8t22 239t1 1450
p(30)5 0.138(30)42 6.24(30)31 86.8(30)2
2 239(30)1 1450
5 111,7802 168,4801 78,1202 71701 1450
5 15,700 periodicals
It is not appropriate to use this model to predict the
number of periodicals in 2010 beacuse the model was
made for 1980 to 2002.
59. a. S5 20.122t31 3.49t22 14.6t1 136
S(5)5 20.122(5)31 3.49(5)22 14.6(5)1 136
5 215.251 87.252 731 136
5 135 grams
H5 20.115t31 3.71t22 20.6t1 124
H(5)5 20.115(5)31 3.71(5)22 20.6(5)1 124
5 214.3751 92.752 1031 124
5 99.375 grams
Difference5 S(5)2H(5)
5 1352 99.375
5 35.625 grams
b.
Weight(grams)
0 2 4 6 80
60
120
240
180
Days after hatching
t
W
S
H
c. The chick is more likely to be a sarus chick than a
hooded chick. At 3 days the sarus chicks weight is about
120 grams, higher than the hooded chicks. Even though
neither chicks average weight at 3 days equals 130
grams, the sarus chick is closer.
60. a. y5 0.000304x3
2.20y5 0.000304(0.394x)3
y50.000304(0.394x)3
}}
2.2
(8.4523 1026)x3
b.
100 20 30 40 50
1
0
2
3y50.000304x3
Length (cm and in.)
Width(kgandlb)
x
y
y 5 (8.45231026)x3
The function from part (a) is a vertical shrink of the
original function.
Mixed Review
61. 2b1 115 152 6b
2b5 42 6b
8b5 4
b51
}
2
62. 2.7n1 4.35 12.94 63. 27< 6y2 1 < 5
2.7n5 8.64 26 < 6y< 6
n5 3.2 21
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64. x22 14x1 485 0
(x2 6)(x2 8)5 0
x2 65 0 or x2 85 0
x5 6 or x5 8
65. 224q22 90q5 21
224q22 90q2 215 0
8q21 30q1 75 0
(4q1 1)(2q1 7)5 0
4q1 15 0 or 2q1 75 0
4q5 21 or 2q5 27
q5 21
}
4
or q5 27
}
2
66. z21 5z< 36
z21 5z2 36 < 0
z21 5z2 365 0
(z1 9)(z2 4)5 0
z1 95 0 (z2 4)5 0
z5 29 z5 4
Part ofz-axis Chosen pt.,x z21 5z 2 36
z< 29 210 (210)21 5(210)2 365 14
29 4 5 (5)21 5(5)2 365 14
Is z215z236
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Graphing Calculator Activity 5.2 (p. 345)
1. 210x 10, 210y 50
2. 220x 50, 22000y 8000
3. 210x 10, 210x 10
4. 25x 5, 210x 30
5. 25x 10, 210x 50
6. 25x 10, 240x 40
7. 25x 10, 220x 50
8. 25x 5, 220x 20
9. Sample answer:The window forg(x) should be the same
x-interval, but they-interval will be shifted by cunits.
10. Sample answer: 210x 20, 0y 3000
Lesson 5.3
5.3 Guided Practice (pp. 346348)
1. (t22 6t1 2)1 (5t22 t2 8)
5 t21 5t22 6t2 t1 22 8
5 6t22 7t2 6 2. (8d2 31 9d3)2 (d32 13d22 4)
5 8d2 31 9d32 d31 13d21 4
5 9d32 d31 13d21 8d2 31 4
5 8d31 13d21 8d1 1
3. (x1 2)(3x22x2 5)
5 (x1 2)3x22 (x1 2)(x)2 (x1 2)(5)
5 3x31 6x22x22 2x2 5x2 10
5 3x31 5x22 7x2 10
4. (a2 5)(a1 2)(a1 6)5 (a22 3a2 10)(a1 6)
5 (a22 3a2 10)a1 (a22 3a2 10)6
5 a32 3a22 10a1 6a22 18a2 60
5 a31 3a22 28a2 60
5. (xy2 4)35 (xy)32 3(xy)2(4)1 3(xy)(4)22 (4)3
5x3y32 12x2y21 48xy2 64
6. T5 CpD
5 0.542t2 2 7.16t 1 79.4
3 109t 1 4010
2173.42t22 28711.6t 1 318,394
5.9078t32 780.44t21 8654.6t
59.078t31 1392.98t22 20,057t1 318,394
T5 59.078t31 1392.98t22 20,057t1 318,394
5.3 Exercises (pp. 349352)
Skill Practice
1. When you add or subtract polynomials, you add or subtract
the coefficients of like terms.
2. Subtracting two polynomials is simply adding the opposite
of the subtracted polynomial to the first polynomial.
3. (3x22 5)1 (7x22 3)5 3x21 7x22 52 3
5 10x22 8
4. (x22 3x1 5)2 (24x21 8x1 9)
5x22 3x1 51 4x22 8x2 9
5x21 4x22 3x2 8x1 52 9
5 5x22 11x2 4
5. (4y21 9y2 5)2 (4y22 5y1 3) 5 4y21 9y2 52 4y21 5y2 3
5 4y22 4y21 9y1 5y2 52 3
5 14y2 8
6. (z21 5z2 7)1 (5z22 11z2 6)
5z21 5z21 5z2 11z2 72 6
5 6z22 6z2 13
7. (3s31s)1 (4s32 2s21 7s1 10)
5 3s31 4s32 2s21s1 7s1 10
5 7s32 2s21 8s1 10
8. (2a22 8)2 (a31 4a22 12a1 4)
5 2a22 82 a32 4a21 12a2 4
5 2a31 2a22 4a21 12a2 82 4
5 2a32 2a21 12a2 12
9. (5c21 7c1 1)1 (2c32 6c1 8)
5 2c31 5c21 7c2 6c1 11 8
5 2c31 5c21 c1 9
10. (4t32 11t21 4t)2 (2 7t22 5t1 8)
5 4t32 11t21 4t1 7t21 5t2 8
5 4t32 11t21 7t21 4t1 5t2 8
5 4t32 4t21 9t2 8
11. (5b2 6b31 2b4)2 (9b31 4b42 7)
5 5b2 6b31 2b42 9b32 4b41 7
5 2b42 4b42 6b32 9b31 5b1 7
5 22b42 15b31 5b1 7
12. (3y22 6y41 52 6y)1 (5y42 6y31 4y)
5 26y41 5y42 6y31 3y22 6y1 4y1 5
5 2y42 6y31 3y22 2y1 5
13. (x42x31x22x1 1)1 (x1x42 12x2)
5x41x42x31x22x22x1x1 12 1
5 2x42x3
14. (8v42 2v21 v2 4)2 (3v32 12v21 8v)
5 8v42 2v21 v2 42 3v31 12v22 8v
5 8v42 3v32 2v21 12v21 v2 8v2 4
5 8v42 3v31 10v22 7v2 4
15. B;
(8x42 4x32x1 2)2 (2x42 8x22x1 10)
5 8x42 4x32x1 22 2x41 8x21x2 10
5 8x42 2x42 4x31 8x22x1x1 22 10
5 6x42 4x31 8x22 8
Chapter 5, continued
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Chapter 5, continued
16. x(2x22 5x1 7)5 2x2(x)2 5x(x)1 7(x)
5 2x32 5x21 7x
17. 5x2(6x1 2)5 6x(5x2)1 2(5x2)
5 30x31 10x2
18. (y2 7)(y1 6)5 (y2 7)(y)1 (y2 7)(6)
5y22 7y1 6y2 42
5y22y2 42
19. (3z1 1)(z2 3)5 (3z1 1)(z)2 (3z1 1)(3)
5 3z21z2 9z2 3
5 3z22 8z2 3
20. (w1 4)(w21 6w2 11)
5 (w1 4)(w2)1 (w1 4)(6w)2 (w1 4)(11)
5 w31 4w21 6w21 24w2 11w2 44
5 w31 10w21 13w2 44
21. (2a2 3)(a22 10a2 2)
5 (2a2 3)(a2)2 (2a2 3)(10a)2 (2a2 3)(2)
5 2a32 3a22 20a21 30a2 4a1 6
5 2a32 23a21 26a1 6
22. (5c22 4)(2c21 c2 3)
5 (5c22 4)(2c2)1 (5c22 4)(c)2 (5c22 4)(3)
5 10c42 8c21 5c32 4c2 15c21 12
5 10c41 5c32 23c22 4c1 12
23. (2x21 4x1 1)(x22 8x1 3)
5 (2x21 4x1 1)(x2)2 (2x21 4x1 1)(8x)
1 (2x21 4x1 1)(3)
5 2x41 4x31x21 8x32 32x22 8x2 3x2
1 12x1 3
5 2x41 12x32 34x21 4x1 3
24. (2d21 4d1 3)(3d22 7d1 6)
5 (2d21 4d1 3)(3d2)2 (2d21 4d1 3)(7d)
1 (2d21 4d1 3)(6)
5 23d41 12d31 9d21 7d32 28d22 21d
2 6d21 24d1 18
5 23d41 19d32 25d21 3d1 18
25. (3y21 6y2 1)(4y22 11y2 5)
5 (3y21 6y2 1)(4y2)2 (3y21 6y2 1)(11y)
2 (3y21 6y2 1)(5)
5 12y41 24y32 4y22 33y32 66y21 11y
2 15y22 30y1 5
5 12y42 9y32 85y22 19y1 5
26. The error is that the negative sign was not carried through
the subtracted polynomial (x31 7x2 2).
(x22 3x1 4)2 (x31 7x2 2)
5x22 3x1 42x32 7x1 2
5 2x31x22 10x1 6
27. (2x2 7)3is not the difference of the cubes of (2x) and (7).
It is the quantity (2x2 7) multiplied together three times.
(2x2 7)35 (2x)32 3(2x)2(7)1 3(2x)(7)22 (7)3
5 8x32 21(4x2)1 6x(49)2 343
5 8x32 84x21 294x2 343
28. (x1 4)(x2 6)(x2 5)5 (x22 2x2 24)(x2 5)
5 (x2
2 2x2 24)(x)2 (x2
2 2x2 24)(5) 5x32 2x22 24x2 5x21 10x1 120
5x32 7x22 14x1 120
29. (x1 1)(x2 7)(x1 3)5 (x22 6x2 7)(x1 3)
5 (x22 6x2 7)(x)2 (x22 6x2 7)(3)
5x32 6x22 7x1 3x22 18x2 21
5x32 3x22 25x2 21
30. (z2 4)(2z1 2)(z1 8)5 (2z21 6z2 8)(z1 8)
5 (2z21 6z2 8)(z)1 (2z21 6z2 8)(8)
5 2z31 6z22 8z2 8z21 48z2 64
5 2z32 2z21 40z2 64
31. (a2 6)(2a1 5)(a1 1)5 (2a22 7a2 30)(a1 1) 5 (2a22 7a2 30)(a)1 (2a22 7a2 30)(1)
5 2a32 7a22 30a1 2a22 7a2 30
5 2a32 5a22 37a2 30
32. (3p1 1)(p1 3)(p1 1)5 (3p21 10p1 3)(p1 1)
5 (3p21 10p1 3)(p)1 (3p21 10p1 3)(1)
5 3p31 10p21 3p1 3p21 10p1 3
5 3p31 13p21 13p1 3
33. (b2 2)(2b2 1)(2b1 1)5 (2b22 5b1 2)(2b1 1)
5 2(2b22 5b1 2)(b)1 (2b22 5b1 2)(1)
5 22b31 5b22 2b1 2b22 5b1 2
5 22b31 7b22 7b1 2
34. (2s1 1)(3s2 2)(4s2 3)5 (6s22s2 2)(4s2 3)
5 (6s22s2 2)(4s)2 (6s22s2 2)(3)
5 24s32 4s22 8s2 18s21 3s1 6
5 24s32 22s22 5s1 6
35. (w2 6)(4w2 1)(23w1 5)
5 (4w22 25w1 6)(23w1 5)
5 2(4w22 25w1 6)(3w)1 (4w22 25w1 6)(5)
5 212w31 75w22 18w1 20w22 125w1 30
5 212w31 95w22 143w1 30
36. (4x2 1)(22x2 7)(25x2 4)
5 (28x22 26x1 7)(25x2 4) 5 2(28x22 26x1 7)(5x)2 (28x22 26x1 7)(4)
5 40x31 130x22 35x1 32x21 104x2 28
5 40x31 162x21 69x2 28
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37. (3q2 8)(29q1 2)(q2 2)
5 (227q21 78q2 16)(q2 2)
5 (227q21 78q2 16)(q)2 (227q21 78q2 16)(2)
5 227q31 78q22 16q1 54q22 156q1 32
5 227q31 132q22 172q1 32
38. (x1 5)(x2 5)5 (x)22 (5)25x22 25
39. (w2 9)25 (w)22 2(wp9)1 (9)2
5 w22 18w1 81
40. (y1 4)35 (y)31 3(y)2(4)1 3(y)(4)21 (4)3
5y31 12y21 48y1 64
41. (2c1 5)25 (2c)21 2(2cp5)1 (5)2
5 4c21 20c1 25
42. (3t2 4)35 (3t)32 3(3t)2(4)1 3(3t)(4)22 (4)3
5 27t32 108t21 144t2 64
43. (5p2 3)(5p1 3)5 (5p)22 (3)2
5 25p22 9
44. (7x2y)35 (7x)32 3(7x)2(y)1 3(7x)(y)22 (y)3
5 343x32 147x2y1 21xy22y3
45. (2a1 9b)(2a2 9b)5 (2a)22 (9b)2
5 4a22 81b2
46. (3z1 7y)35 (3z)31 3(3z)2(7y)1 3(3z)(7y)21 (7y)3
5 27z31 189z2y1 441zy21 343y3
47. D;
(3x2 2y)25 (3x)22 2(3xp2y)1 (2y)2
5 9x22 12xy1 4y2
48. V5 lwh l5 3x1 1; w5x; h5x1 3
5 (3x1 1)(x)(x1 3)
5 (3x21x)(x1 3)
5 (3x21x)(x)1 (3x21x)(3)
5 3x31x21 9x21 3x
5 3x31 10x21 3x
49. V5 r2h r5x2 4 ; h5 2x1 3
5 (x2 4)2(2x1 3)
5 (x22 8x1 16)(2x1 3)
5 ((x22 8x1 16)(2x)1 (x22 8x1 16)(3))
5 (2x32 16x21 32x1 3x22 24x1 48)
5 (2x32 13x21 8x1 48)
5 2x32 13x21 8x1 48
50. V5s3 s5x2 5
5 (x2 5)3
5(x)32 3(x)2(5)1 3(x)(5)22 (5)3
5x32 15x21 75x2 125
51. V51
}
3
Bh B5 (2x2 3)2; h5 (3x1 4)
51
}
3(2x2 3)2(3x1 4)
51
}
3
(4x22 12x1 9)(3x1 4)
51
}
3
((4x22 12x1 9)(3x)1 (4x22 12x1 9)(4))
51
}
3
(12x32 36x21 27x1 16x22 48x1 36)
51
}
3
(12x32 20x22 21x1 36)
5 4x3220
}
3
x22 7x1 12
52. (a1 b)(a2 b)0 a22 b2
(a1 b)(a)2 (a1 b)(b)0 a22 b2
a21 ab2 ab2 b20 a22 b2
a22 b25 a22 b2
53. (a1 b)20 a21 2ab1 b2
(a1 b)(a1 b)0 a21 2ab1 b2
(a1 b)(a)1 (a1 b)(b)0 a21 2ab1 b2
a21 ab1 ab1 b20 a21 2ab1 b2
a21 2ab1 b25 a21 2ab1 b2
54. (a1 b)30 a31 3a2b1 3ab21 b3
(a1 b)(a1 b)(a1 b)0 a31 3a2b1 3ab21 b3
(a21 2ab1 b2)(a1 b)0 a31 3a2b1 3ab21 b3
(a21 2ab1 b2)(a)1 (a21 2ab1 b2)(b)
0 a31 3a2b1 3ab21 b3
a31 2a2b1 ab21 a2b1 2ab21 b3
0 a31 3a2b1 3ab21 b3
a31 3a2b1 3ab21 b35 a31 3a2b1 3ab21 b3
55. (a2 b)30 a32 3a2b1 3ab22 b3
(a2 b)(a2 b)(a2 b)0 a32 3a2b1 3ab22 b3
(a22 2ab1 b2)(a2 b)0 a32 3a2b1 3ab22 b3
(a22 2ab1 b2)(a)2 (a22 2ab1 b2)(b)
0 a32 3a2b1 3ab22 b3
a32 2a2b1 ab22 a2b1 2ab21 b3
0 a32 3a2b1 3ab22 b3
a32 3a2b1 3ab21 b35 a32 3a2b1 3ab22 b3
56. a.p(x)1 q(x)5 (x42 7x1 14)1 (x22 5)
5x41x22 7x1 9
degree 4 b.p(x)2 q(x)5 (x42 7x1 14)2 (x22 5)
5x42x22 7x1 19
degree 4
c.p(x) pq(x)5 (x42 7x1 14)(x22 5)
5 (x42 7x1 14)(x2)2 (x42 7x1 14)(5)
5x62 7x31 14x22 5x41 35x2 70
5x62 5x42 7x31 14x21 35x2 70
degree 6
Chapter 5, continued
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Worked-Out Solution Key
d.p(x)1 q(x) : degree m
p(x)2 q(x):degree m
p(x) pq(x) : degree (m1 n)
57. a.x52 15 (x2 1)(x41x31x21x1 1)
Check: (x41x31x21x1 1)(x)
2 (x41x31x21x1 1)
5x51x41x31x21x2x42x3
2x22x2 1
5x52 1
x62 15 (x2 1)(x51x41x31x21x1 1)
Check: (x51x41x31x21x1 1)(x)
2 (x51x41x31x21x1 1)
5x61x51x41x31x21x
2x52x42x32x22x2 1
5x62 1
b.xn2 15 (x2 1)(xn2 11xn2 21xn2 31 . . .1xn2 n)
5 (x2 1)(xn2 11xn2 21xn2 31 . . .1 1)
Check: (xn2 11xn2 21xn2 31 . . .1 1)(x) 2 (xn2 11xn2 21xn2 31 . . .1 1)
5xn2 11 11xn 2 21 11xn2 31 11 . . .1x
2xn2 12xn2 22xn2 32 . . .2 1
5xn1xn2 11xn2 21 . . .xn2 (n2 1)
2xn2 12xn2 22xn2 3. . .21
5xn2 1
58.f(x)5 (x1 a)(x1 b)(x1 c)(x1 d)
5 (x21 ax1 bx1 ab)(x1 c)(x1 d)
5 ((x21 ax1 bx1 ab)(x)
1 (x21 ax1 bx1 ab)(c))(x1 d)
5 (x31 ax21 bx21 abx1 cx2
1 acx1 bcx1 abc)(x1 d)
5 (x31 ax21 bx21 abx1 cx21 acx
1 bcx1 abc)(x)1 (x31 ax21 bx21 abx 1 cx21 acx1 bcx1 abc)(d)
5x41 ax31 bx31 abx21 cx31 acx2
1 bcx21 abcx1 dx31 adx21 bdx21 abdx
1 cdx21 acdx1 bcdx1 abcd
5x41 (a1 b1 c1 d )x3
1 (ab1 ac1 bc1 ad1 bd1 cd )x2
1 (abc1 abd1 acd1 bcd )x1 abcd
coefficient ofx35 (a1 b1 c1 d )
constant term5 abcd
Problem Solving
59. T5M1F
5 (0.091t32 4.8t21 110t1 5000)1 (0.19t32 12t2
1 350t1 3600)
5 0.281t32 16.8t21 460t1 8600
60. R5PpD
5 (6.82t22 61.7t1 265)(4.11t1 4.44)
5 (6.82t22 61.7t1 265)(4.11t)1 (6.82t22 61.7t
1 265)(4.44)
5 28.0302t32 253.587t21 1089.15t1 30.2808t2
2 273.948t1 1176.6
5 28.0302t32 223.3062t21 815.202t1 1176.6
R(3)5 28.0302(3)32 223.3062(3)21 815.202(3)
1 1176.6
5 756.81542 2009.75581 2445.6061 1176.6
5 2369.2656
The total revenue in 2002 from DVD sales was 2369.2656
million dollars, or $2,369,265,600.
61. F5 0.0116s21 0.789
P5 0.00267sF
5 0.00267s(0.0116s21 0.789)
5 0.00003s31 0.00211s
The model isP5 0.00003s31 0.00211s.
P(10)5 0.00003(10)31 0.00211(10)
5 0.031 0.0211
5 0.0511
So, you need 0.0511 horsepower to keep the bicycle
moving 10 miles per hour.
62.
r11
r
0.5 cm
1 cm
M(r)5 volume of cookie 1 volume of marshmallow
5 (r1 1)2h11
}
214}3r32
5 (r1 1)211}221}
2
p4
}
3
r3
5}
2
(r21 2r1 1)1
}
2
p4
}
3
r3
5}
214}3r31 r21 2r1 12
52}
3r31
}2r21 r1
}
2
D(r)51
}
214}3(r1 1)321 (r1 1)21
1
}
22
5
}21
4
}3
(r1 1)321}2(r1 1)2
5}
2 p
4
}
3
(r31 3r21 3r1 1)1}2(r21 2r 1 1)
5}
214}3r31 4r21 4r1
4
}
31 r21 2r 1 12
52}
3r31
5}2r21 3r1
7}
6
Chapter 5, continued
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Company.
302
Algebra 2
Worked-Out Solution Key
C(r)5D(r)2M(r)
52}
3r31
5}
2r21 3r1
7}
6
2 12}3r31}2r21 r1
}22
52}
3r31
5}
2r21 3r1
7}
62
2}
3r3
2
}2r22 r2
}2
5 2r21 2r12}
3
63.N5 total mens players1 total womens players
5 (Lm
pSm
)1 (LwpS
w)
Lm
pSm
5 (20.127t31 0.822t22 1.02t1 31.5)p
(5.57t1 182)
5 (20.127t31 0.822t22 1.02t1 31.5)(5.57t)
1 (20.127t31 0.822t22 1.02t1 31.5)(182)
5 20.70739t42 18.53546t31 143.9226t2
2 10.185t1 5733
LwpS
w5 (20.0662t31 0.437t22 0.725t1 22.3)p
(12.2t1 185)
5 (20.0662t31 0.437t22 0.725t1 22.3)p
(12.2t)1 (20.0662t31 0.437t2
2 0.725t1 22.3)(185)
5 20.80764t42 6.9156t31 72t21 137.935t
1 4125.5
N5 20.70739t42 18.53546t31 143.9226t2
2 10.185t1 57332 0.80764t4
2 6.9156t31 72t21 137.935t1 4125.5
5 21.51503t42 25.45106t31 215.9226t2
1 127.751 9858.5
The model for the total number of people is written by
adding the number of men players to the number of
women players. The number of men and women players
is written by multiplying the number of teams by the
average team size for each group.
64.s5 t2 5
t5s1 5
C5 20.00105t31 0.0281t21 0.465t1 48.8
C5 20.00105(s1 5)31 0.0281(s1 5)21
0.465(s1 5)1 48.8
5 20.00105(s31 3s2(5)1 3s(25)1 125)
1 0.0281(s21 10s1 25)1 0.465s1 2.3251 48.8
5 20.00105s32 0.01575s22 0.07875s2 0.13125
1 0.0281s21 0.281s1 0.70251 0.465s1 2.325
1 48.8
5 20.00105s31 0.01235s21 0.66725s1 51.69625
Mixed Review
65. 2x2 75 11 66. 102 3x5 25
2x5 18 23x5 15
x5 9 x5 25
67. 4t2 75 2t
275 22t
7
}
25 t
68. y22 2y2 485 0
(y1 6)(y2 8)5 0
y1 65 0 or y2 85 0
y5 26 or y5 8
69. w22 15w1 545 0
(w2 6)(w2 9)5 0
w2 65 0 or w2 95 0
w5 6 or w5 9
70. x21 9x1 145 0
(x1 2)(x1 7)5 0 x1 25 0 or x1 75 0
x5 22 or x5 27
71. 4z21 21z2 185 0
(4z2 3)(z1 6)5 0
4z2 35 0 or z1 65 0
4z5 3 or z5 26
z53
}
4
72. 9a22 30a1 255 0
(3a2 5)25 0
3a2
55
0
3a5 5
a55
}
3
73. x1y2 2z5 24 y5 2x1 2z2 4
3x2y1z5 22
2x1 2y1 3z5 29
3x 2 (2x1 2z2 4)1z5 22
3x1x2 2z1 41z5 22
4x2z5 18
2x1 2(2x1 2z2 4)1 3z5 29
2x2 2x1 4z2 81 3z5 29 23x1 7z5 21
4x2z5 18 33 12x2 3z5 54
23x1 7z5 21 34 212x1 28z5 24
25z5 50
z5 2
23x1 7(2)5 21 x5 5
y5 251 2(2)2 45 25
The solution is (5, 25, 2).
Chapter 5, continued
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CopyrightbyM
cDougalLittell,adivisionofHoughtonMifflin
Company.
Algebra 2
Worked-Out Solution Key
74. x2 2y1z5 213 x5 2y2z2 13
2x1 4y1z5 35
3x1 2y1 4z5 28
2(2y 2z 2 13)1 4y1z5 35
22y1z1 131 4y1z5 35
2y1 2z5 22
y1z5 11
3(2y2z2 13)1 2y1 4z5 28
6y2 3z2 391 2y1 4z5 28
8y1z5 67
y1z5 11 321 2y2z5 211
8y1z5 67 8y1z5 67
7y 5 56
y 5 8
81z5 11 z5 3
x5 2(8)2 (3) 2 135 0
The solution is (0, 8, 3).
75. 3x2y2 2z5 20 y5 3x2 2z2 20
2x1 3y2z5 216
22x2y1 3z5 25
2x1 3(3x2 2z2 20)2z5 216
2x1 9x2 6z2 602z5 216
8x2 7z5 44
22x2(3x2 2z2 20)1 3z5 25
22x2 3x1 2z1 201 3z5 25
25x1 5z5 225
2x1z5 25
8x2 7z5 44 8x2 7z5 44
2x1z5 25 38 28x1 8z5 240
z5 4
2x1 45 25 x5 9
y5 3(9)2 2(4)2 205 21
The solution is (9, 21, 4).
76. detF3 243 1 G5 3(1)2 3(24)5 31 125 15
77. det
F
5 7
24 9G
5 5(9)2 (24)(7)5 451 285 73
78. detF21 8 0 3 4 2325 2 1G5 (241 1201 0)2 (01 61 24) 5 1162 305 86
79. detF 2 3 2426 1 523 21 22
G5 (242 452 24)2 (122 101 36) 5 2732 385 2111
Quiz 5.15.3 (p. 352)
1. 35p3215 351 (21)5 345 81
2. (24)25 24 p25 285 256
3. 1 2}322225
22
}
(322)25
22
}
3245 22p345 4 p815 324
4. 13}5222
5 15}3225
52
}32
525
}
9
5. (x4y22)(x23y8)5x41 (23)y(22)1 85xy6
6. (a2b25)235 (a2)23(b25)235 a26pb155b15
}
a6
7.x3y7
}
x24y05x32(24)y72 05x7y7
8.
c3d22
}
c5d215
c
32 5
pd
222 (21)5
c
22
d
215
1
}
c2d
9.x 23 22 21 0 1 2 3
y 244 29 2 1 0 11 46
4
x
y
22
10.
x 23 22 21 0 1 2 3
y 95 26 7 2 21 10 71
1
x
y
21
11.x 23 22 21 0 1 2 3
y 58 15 22 25 26 217 250
1
x
y
22
Chapter 5, continued
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CopyrightbyM
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Company.
304
Algebra 2
Worked-Out Solution Key
12. (x31x22 6)2 (2x21 4x2 8)
5x31x22 62 2x22 4x1 8
5x32x22 4x1 2
13. (23x21 4x2 10)1 (x22 9x1 15)
5 23x21x21 4x2 9x2 101 15
5 22x22 5x1 5
14. (x2 5)(x22 5x1 7)
5 (x22 5x1 7)(x)2 (x22 5x1 7)(5)
5x32 5x21 7x2 5x21 25x2 35
5x32 10x21 32x2 35
15. (x1 3)(x2 6)(3x2 1)5 (x22 3x2 18)(3x2 1)
5 (x22 3x2 18)(3x)2 (x22 3x2 18)
5 3x32 9x22 54x2x21 3x1 18
5 3x32 10x22 51x1 18
16. (7,282,000,000,000)4 (294,000,000)5
7.282 3 1012
}
2.94 3 108 5
7.282
}
2.94 3
102
}
108
2.48 3104
5 $24,800
Lesson 5.4
5.4 Guided Practice (pp. 354-356)
1. x32 7x21 10x5x(x22 7x1 10)5x(x2 5)(x2 2)
2. 3y52 75y35 3y3(y22 25)5 3y3(y2 5)(y1 5)
3. 16b51 686b25 2b2(8b31 343)
5 2b2(2b1 7)(4b22 14b1 49)
4. w32 275 (w2 3)(w21 3w1 9)
5. x31 7x22 9x2 635x2(x1 7) 2 9(x1 7)
5 (x1 7)(x22 9) 5 (x1 7)(x2 3)(x1 3)
6. 16g42 6255 (4g2)22 (25)2
5 (4g22 25)(4g21 25)
5 (2g2 5)(2g1 5)(4g21 25)
7. 4t62 20t41 24t25 4t2(t42 5t21 6)
5 4t2(t22 2)(t22 3)
8. 4x52 40x31 36x5 0
4x(x42 10x21 9)5 0
4x(x22 9)(x22 1)5 0
4x(x2 3)(x1 3)(x2 1)(x1 1)5 0
x5 0,x5 3,x5 23,x5 1, orx5 21
9. 2x51 24x5 14x3
2x52 14x31 24x5 0
2x(x42 7x21 12)5 0
2x(x22 3)(x22 4)5 0
2x(x22 3)(x2 2)(x1 2)5 0
x5 0,x5 2}
3 ,x5 }
3 ,x5 2, orx5 22
10. 227x31 15x25 26x4
6x42 27x31 15x25 0
3x2(2x22 9x1 5)5 0
3x25 0 or 2x22 9x1 55 0
x5 0 orx59 6 }}
922 4 p2 p5}}
2 p2 5
9 6 }
41}
4
11. Volume
(ft3)5
Interiorlength
(ft)
p
Interiorwidth
(ft)
p
Interiorheight
(ft)
40 5 (6x2 2) p (3x2 2) p (x2 1)
405 18x32 36x21 22x2 4
05 18x32 36x21 22x2 44
05 18x2(x2 2)1 22(x2 2)
05 (18x21 22)(x2 2)
The only real solutions isx5 2. The basin is 12 ft long,
6 ft wide, and 2 ft high.
5.4 Exercises (pp. 356359)
Skill Practice
1. The expression 8x61 10x32 3 is in quadraticform
because it can be written as 2u21 5u2 3 where u5 2x3.
2. The factorization of a polynomial is factored completely if
it is written as a product of unfactorable polynomials with
integer coefficients.
3. 14x22 21x5 7x(2x2 3)
4. 30b32 54b25 6b2(5b2 9)
5. c31 9c21 18c5 c(c21 9c1 18)5 c(c1 3)(c1 6)
6.z32 6z22 72z5z (z22 6z2 72)5z(z2 12)(z1 6)
7. 3y52 48y35 3y3(y22 16)5 3y3(y2 4)(y1 4)
8. 54m51 18m41 9m35 9m3(6m21 2m1 1) 9. A;
(2x72 32x3)5 2x3(x42 16)
5 2x3(x22 4)(x21 4)
5 2x3(x1 2)(x2 2)(x21 4)
10.x31 85x31 235 (x1 2)(x22 2x1 4)
11.y32 645y32 435 (y2 4)(y21 4y1 16)
12. 27m31 15 (3m)31 135 (3m1 1)(9m22 3m1 1)
13. 125n31 2165 (5n)31 635 (5n1 6)(25n22 30n1 36)
14. 27a32 10005 (3a)32 103
5 (3a2 10)(9a21 30a1 100)
15. 8c31 3435 (2c)31 735 (2c1 7)(4c22 14c1 49)16. 192w32 35 3(64w32 1)
5 3((4w)32 13)
5 3(4w2 1)(16w21 4w1 1)
17. 25z31 3205 25(z32 64)
5 25(z32 43)
5 25(z2 4)(z21 4z1 16)
Chapter 5, continued
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CopyrightbyM
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Company.
Algebra 2
Worked-Out Solution Key
18. x31x21x1 15x2(x1 1)1 (x1 1)5 (x1 1)(x21 1)
19.y32 7y21 4y2 285y2(y2 7)1 4(y2 7)
5 (y2 7)(y21 4)
20. n31 5n22 9n2 455 n2(n1 5)2 9(n1 5)
5 (n1 5)(n22 9)
5 (n1 5)(n2 3)(n1 3)
21. 3m32 m21 9m2 35 3m31 9m2 m22 3
5 3m(m21 3)2 (m21 3)
5 (m21 3)(3m2 1)
22. 25s32 100s22s1 45 25s2(s2 4)2 (s2 4)
5 (s2 4)(25s22 1)
5 (s2 4)(5s2 1)(5s1 1)
23. 4c31 8c22 9c2 185 4c2(c1 2)2 9(c1 2)
5 (c1 2)(4c22 9)
5 (c1 2)(2c2 3)(2c1 3)
24.x42 255 (x2)22 525 (x22 5)(x21 5)
25. a41 7a21 65 (a21 1)(a21 6)
26. 3s42s22 245 (3s21 8)(s22 3)
27. 32z52 2z5 2z (16z42 1)
5 2z (4z21 1)(4z22 1)
5 2z (4z21 1)(2z1 1)(2z2 1)
28. 36m61 12m41 m25 m2(36m41 12m21 1)
5 m2(6m21 1)(6m21 1)
5 m2(6m21 1)2
29. 15x52 72x32 108x5 3x(5x42 24x22 36)
5 3x(5x21 6)(x22 6)
30. The error is that when the binomial was factored, it was
not factored properly.
8x32 275 0
(2x2 3)(4x21 6x1 9)5 0
2x2 35 0
x53
}
2
23
}
2
31. The error is thatx5 0 is not included in the solutions,
andx22 16 was not factored completely before finding
the zeros.
3x(x22 16)5 0
3x(x2 4)(x1 4)5 0
x5 0,x5 4, orx5 24
32. y32 5y25 0
y2(y2 5)5 0
y5 0 ory5 5
33. 18s35 50s
18s32 50s5 0
2s(9s22 25)5 0
2s(3s2 5)(3s1 5)5 0
s5 0,s55
}
3, ors5 2
5
}
3
34. g31 3g22g2 35 0
g2(g1 3)2 (g1 3)5 0
(g1 3)(g22 1)5 0
(g1 3)(g1 1)(g2 1)5 0
g5 23,g5 21, org5 1
35. m31 6m22 4m2 245 0
m2(m1 6)2 4(m1 6)5 0
(m1 6)(m22 4)5 0
(m1 6)(m1 2)(m2 2)5 0
m5 26, m5 22, or m5 2
36. 4w41 40w22 445 0
4(w41 10w22 11)5 0
4(w21 11)(w22 1)5 0
4(w21 11)(w1 1)(w2 1)5 0
w5 21 or w5 1
37. 4z55 84z3
4z52 84z35 0
4z3(z22 21)5 0 z5 0,z5
}
21 , orz5 2}
21
38. 5b31 15b21 12b5 236
5b31 15b21 12b1 365 0
5b2(b1 3)1 12(b1 3)5 0
(b1 3)(5b21 12)5 0
b5 23
39. x62 4x42 9x21 365 0
x4(x22 4)2 9(x22 4)5 0
(x22 4)(x42 9)5 0
(x2 2)(x1 2)(x22 3)(x21 3)5 0
x5 2,x5 22,x5 }
3 , orx5 2}
3
40. 48p55 27p3
48p52 27p35 0
3p3(16p22 9)5 0
(3p3)(4p1 3)(4p2 3)5 0
p5 0,p5 23
}
4
, orp53
}
4
41. C;
3x42 27x21 9x5x3
3x42x32 27x21 9x5 0
x(3x32 27x2x21 9)5 0
x(3x(x22 9)2 (x22 9))5 0 x(x22 9)(3x2 1)5 0
x(x2 3)(x1 3)(3x2 1)5 0
x5 0,x5 3,x5 23, orx51
}
3
42. 16x32 44x22 42x5 2x(8x22 22x2 21)
5 2x(4x1 3)(2x2 7)
43. n42 4n22 605 (n21 6)(n22 10)
Chapter 5, continued
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CopyrightbyM
cDougalLittell,adivisionofHoughtonMifflin
Company.
306
Algebra 2
Worked-Out Solution Key
44. 24b42 500b5 24b(b31 125)
5 24b(b1 5)(b22 5b1 25)
45. 36a32 15a21 84a2 355 36a31 84a2 15a22 35
5 12a(3a21 7)2 5(3a21 7)
5 (3a21 7)(12a2 5)
46. 18c41 57c32 10c25 c2(18c21 57c2 10)
5 c2(6c2 1)(3c1 10)
47. 2d42 13d 22 455 (2d 21 5)(d 22 9)
5 (2d 21 5)(d1 3)(d2 3)
48. 32x52 108x25 4x2(8x32 27)
5 4x2(2x2 3)(4x21 6x1 9)
49. 8y62 38y42 10y25 2y2(4y42 19y22 5)
5 2y2(4y21 1)(y22 5)
50.z52 3z42 16z1 485z4(z2 3)2 16(z2 3)
5 (z2 3)(z42 16)
5 (z2 3)(z21 4)(z22 4)
5 (z2 3)(z21 4)(z2 2)(z1 2)
51.A5 lpw l5 3x1 2, w5x1 4
485 (3x1 2)(x1 4)
485 3x21 14x1 8
05 3x21 14x2 40
05 (3x1 20)(x2 2)
x5 220
}
3 orx5 2, but a side length cannot be negative
sox5 2.
52. V5 lpwph l5 2x, w5x2 1, h5x2 4
405 (2x)(x2 1)(x2 4)
405 (2x22 2x)(x2 4)
405 2x32 10x21 8x
05 2x32 10x21 8x2 40
05 2x2(x2 5)1 8(x2 5)
05 (x2 5)(2x21 8)
The only real solution isx5 5.
53. V51
}
3r2h r5 2x2 5, h5 3x
12551
}
3(2x2 5)2(3x)
3755 (4x22 20x1 25)(3x)
3755 12x32 60x21 75x
05
12x32
60x21
75x2
375 05 4x32 20x21 25x2 125
05 4x2(x2 5)1 25(x2 5)
05 (x2 5)(4x21 25)
The only real solution isx5 5.
54.x3y62 275 (xy2)32 33
5 (xy22 3)(x2y41 3xy21 9)
55. 7ac21 bc22 7ad22 bd25 c2(7a1 b)2d2(7a1 b)
5 (7a1 b)(c22 d2)
5 (7a1 b)(c2 d )(c1 d)
56.x2n2 2xn1 15 (xn2 1)(xn2 1)
5 (xn2 1)2
57. a5b22 a2b41 2a4b2 2ab31 a32 b2
5 a2b2(a32 b2)1 2ab(a32 b2)1 (a32 b2) 5 (a32 b2)(a2b21 2ab1 1)
5 (a32 b2)(ab1 1)(ab1 1)
5 (a32 b2)(ab1 1)2
Problem Solving
58. V5 lpwph l5 12x2 15, w5 12x2 21, h5x
9455 (12x2 15)(12x2 21)(x)
9455 (144x22 432x1 315)x
05 144x32 432x21 315x2 945
05 16x32 48x21 35x2 105
05
16x2
(x2
3)1
35(x2
3) 05 (x2 3)(16x21 35)
The only real solution isx5 3. The block is
3 meters high.
59. V5 lpwph l5x2 2, w5 3x2 2, h5 6x2 2
1125 (x2 2)(3x2 2)(6x2 2)
1125 (3x22 8x1 4)(6x2 2)
1125 18x32 48x21 24x2 6x21 16x2 8
1125 18x32 54x21 40x2 8
05 18x32 54x21 40x2 120
05 9x32 27x21 20x2 60
05 9x2(x2 3)1 20(x2 3)
05 (9x21 20)(x2 3)
x5 3 cm
3x5 9 cm
6x5 18 cm
The outer dimensions of the mold should be
3 cm3 9 cm3 18 cm.
60. a. V5 lpwph
Volume of platform 15 (8x)(6x)(x) 5 48x3
Volume of platform 25 (6x)(4x)(x) 5 24x3
Volume of platform 35 (4x)(2x)(x)5 8x3
b. 12505 48x31 24x31 8x3
12505 80x3
c. 12505 80x3
15.6255x3
2.5 ft5x
Platform 1: 20 ft3 15 ft3 2.5 ft
Platform 2: 15 ft3 10 ft3 2.5 ft
Platform 3: 10 ft3 5 ft3 2.5 ft
Chapter 5, continued
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CopyrightbyM
cDougalLittell,adivisionofHoughtonMifflin
Company.
Algebra 2
Worked-Out Solution Key
61. l5x
x
x25
x25
h5x2 5
w5x2 5
V5 lpwph
2505 (x)(x2 5)(x2 5)
2505x(x22 10x1 25)
2505x32 10x21 25x
05x32 10x21 25x2 250
05x2(x2 10)1 25(x2 10)
05 (x21 25)(x2 10)
x5 10
x2 55 5
The dimensions of the prism should be
10 in.3 5 in.3 5 in.
62. Volume of steel 5 volume of outside 2 volume of inside
6.425 (x)(x2 2)(x1 8)2 89.58
6.425x(x21 6x2 16)2 89.58
6.425x31 6x22 16x2 89.58
05x31 6x22 16x2 96
05x2(x1 6)2 16(x1 6)
05 (x1 6)(x22 16)
05 (x1 6)(x2 4)(x1 4)
x5 4 feet
The outer dimensions of the tank should be
4 ft3 2 ft3 12 ft.
63. V5 lpwph l5x2 2, w5 32 2x, h5 3x1 4
7
}
35 (x2 2)(32 2x)(3x1 4)
7
}3
5 (x2 2)(9x1 122 6x22 8x)
7
}
35 9x21 12x2 6x32 8x22 18x2 241 12x21 16x
7
}
35 26x31 13x21 10x2 24
75 218x31 39x21 30x2 72
05 218x31 39x21 30x2 79
From the graph, the only real real solution is 21.37. This
result would yield a negative dimension of the platform.
So, the volume cannot be7
}
3
cubic feet.
x
y
10
1(2
1.37, 0)
64. a.y y31y2
1 11 15 2
2 81 45 12
3 271 95 36
4 641 165 80
5 1251 255 150
6 2161 365 252
7 3431 495 392
8 5121 645 576
9 7291 815 810
10 10001 1005 1100
b.x31 2x25 96
x3
}8
1x2
}4
5 12 1x}2231 1x}22
25 12
y5x
}
2 25
x
}
2
x5 4
c.3x31 2x25 512
27x3
}
8 1
9x2
}
4 5 576 13x}22
31 13x}22
25 576
y53x
}
2 85
3x
}
2
x516
}
3
d. ax41 bx35 c Original equation
a4x4
}
b4 1
a3x3
}
b3 5
a3c
}
b4 Multiply each side by
a3
}
b4.
1ax}b241 1ax}b2
35
a3c
}
b4 Rewrite equation.
Then, create a table with values fory41y3.
65. a. The three solids form a cube with a smaller cube
taken out of the corner. If the solids were connected,
V5 a32 b3because the side of the large cube5
aand the side of the cube taken out5 b.
b. I: V5 apap(a2 b)5 a2(a2 b)
II: V5 ap(b)(a2 b) 5 ab(a2 b)
III: V5 bpb(a2 b)5 b2(a2 b)
c. V5 I1 II1 II
5 a2(a2 b)1 ab(a2 b)1 b2(a2 b)
5 (a2 b)(a21 ab1 b2)
5 a32 b3
Mixed Review
66.
1
x
y
21
67.
1
x
y
21
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
68.
2
x
y
21
69.1
x
y
21
70.
1
x
y
21
71.1
x
y
21
72.
1
x
y
21
73.
1
x
y
21
74.1
x
y
21
75.
1
x
y
21
76. 2 5 23 4 21 10
10 14 36 70
5 7 18 35 80
f (2)5 80
77. 23 23 0 1 26 2 4
9 227 78 2216 642
23 9 226 72 2214 646
f (23)5 646
78. 22 5 0 24 12 0 20
210 20 232 40 280
5 210 16 220 40 260
f (22)5 260
79. 4 26 0 0 9 215
224 296 2384 21500
26 224 296 2375 21515
f (4)5 21515
Problem Solving Workshop 5.4 (p. 361)
1.y5x31 4x22 8x
x y
1 23
2 8
3 39
4 96
x5 4
2.y5x32 9x22 14x1 7
y5 233
Intersections : (22.56, 233)
(1.56, 233)
(10,233)
x5 22.56, 1.56, or 10
3.y5 2x32 11x21 3x1 5
x y
1 21
2 217
3 231
4 231
5 25
6 59
x5 6
4.y5x41x32 15x22 8x1 6
y5 245 Intersections: (23.43, 245)
(22.53, 245)
(1.96, 245)
(3, 245)
x5 23.43, 22.53, 1.96, or 3
5.y5 2x41 2x3 1 6x21 17x2 4
y5 32
Intersections: (1.45, 32)
(4, 32)
x5 1.45 or 4
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
6. y5 23x41 4x31 8x21 4x2 11
y5 13
Intersections: (1.64, 13)
(2, 13)
x5 1.64 or 2
7. y5 4x42 16x31 29x22 95x
x y
0 0
0.5 242
1 278
1.5 2111
2 2138
2.52
150
x5 2.5
8. y5 (2x2 2)(x2 2)(x2 1)
x y
2 0
3 8
4 36
5 96
6 200
The volume is 200 cubic
feet whenx5 6.
Length5 2x5 12 ft
Width5x5 6 ft
Height5x5 6 ft
Intersection:x5 6,y5 200
Length5 2x5 12 ft
Width5x5 6 ft
Height5x5 6 ft
9. x5 height
x2 45 width
x1 65 length
V5 lpwph
y5 (x16)(x2 4)(x)
x y
6 144
7 273
8 448
9 675
10 960
11 1309
12 1728
Intersection at:x5 12,
y5 1728
x5 12
height5x5 12 ft
width5x2 45 8 ft
length5x1 65 18 ft
10. y5 0.0000984x42 0.00712x31 0.162x22 1.11x1 12.3
x y
1 11.35
2 10.67
3 10.24
4 10.02
5 9.97
x5 5 Intersection:x5 5,y5 9.97
In 1975, pineapple consumption was about 9.97 pounds
per person.
Lesson 5.5
5.5 Guided Practice (pp. 362365)
1. 2x22 3x1 8
x21 2x2 1qww2x41x31 0x21x2 1
2x41
4x32
2x2
23x31 2x21x
23x32 6x21 3x
8x22 2x2 1
8x21 16x2 8
218x1 7
2x41x31x2 1
}}
x21 2x2 1
5 2x22 3x1 81218x1 7
}
x21 2x2 1
2. x22 3x1 10
x1 2qwwx32x21 4x2 10
x31 2x2
2
3x
21
4x 23x22 6x
10x2 10
10x1 20
230
x32x21 4x2 10
}}
x1 2
5x22 3x1 10230
}
x1 2
3. 23 1 4 21 21
23 23 12
1 1 24 11
x31 4x22x2 1
}}
x1 3
5x2
1x2 4111
}
x1 3
4. 1 4 1 23 7
4 5 2
4 5 2 9
4x31x22 3x1 7
}}
x2 1 5 4x21 5x1 21
9
}
x2 1
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
5. 4 1 26 5 12
4 28 212
1 22 23 0
f (x)5x32 6x21 5x1 12
5 (x2 4)(x22 2x2 3)
5 (x2 4)(x2 3)(x1 1)
6. 4 1 21 222 40
4 12 240
1 3 210 0
f (x)5x32x22 22x1 40
5 (x2 4)(x21 3x2 10)
5 (x2 4)(x1 5)(x2 2)
7. 22 1 2 29 218
22 0 18
1 0 29 0
f (x)5x31 2x22 9x2 18
5 (x1 2)(x22 9)
5 (x 1 2)(x2 3)(x1 3)
The other zeros are 3 and 23.
8. 22 1 8 5 214
22 212 14
1 6 27 0
f (x)5x3
1 8x2
1 5x2 14 5 (x1 2)(x21 6x2 7)
5 (x1 2)(x1 7)(x2 1)
The other zeros are 27 and 1.
9. 255 215x31 40x
05 215x31 40x2 25
1 215 0 40 225
215 215 25
215 215 25 0
(x2 1)(215x22 15x1 25)5 0
x 0.9 is the other positive solution. The companycould still make the same profit producing about
900,000 shoes.
5.5 Exercises (pp. 366368)
Skill Practice
1. If a polynomialf (x) is divided byx2 k,then the
remainder is r5f (k).
2. The red numbers represent the coefficients of the
quotient and the blue number represents the remainder.
3. x1 5
x2 4qwwx21x2 17
x22 4x
5x2 17
5x2 20 3
x21x2 17
}
x2 4 5x1 51
3
}
x2 4
4. 3x1 4
x2 5qww3x22 11x2 26
3x22 15x
4x2 26
4x2 20
26
3x22 11x2 26
}}
x2 5 5 3x1 42
6
}
x2 5
5. x21 4x1 7
x2 1qww
x31 3x21 3x1 2
x32x2
4x21 3x
4x22 4x
7x1 2
7x2 7
9
x31 3x21 3x1 2
}}
x2 1 5x21 4x1 71
9
}
x2 1
6. 2x1 9
4x2 1qww
8x21 34x2 1
8x22 2x
36x2 1
36x2 9
8
8x21 34x2 1
}}
4x2 1 5 2x1 91
8
}
4x2 1
7. 3x1 8
x21xqww3x31 11x21 4x1 1
3x31 3x2
8x21 4x
8x21 8x
24x1 1
3x31 11x21 4x1 1
}}
x21x 5 3x1 81
24x1 1
}
x21x
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
8. 7x111
x21 1qww7x31 11x21 7x1 5
7x3 1 7x
11x21 0x1 5
11x21 0x1 11
26
7x31 11x21 7x1 5
}}
x21 1 5 7x1 112 6
}
x21 1
9. 5x22 12x1 37
x21 2x2 4qwww5x42 2x32 7x21 0x2 39
5x41 10x32 20x2
212x31 13x21 0x
212x22 24x22 48x
37x21 48x2 39
37x21 74x2 148
226x1 109
5x42 2x32 7x22 39
}}
x
21
2x2
4
5 5x22 12x1 371226x1 109
}
x21 2x2 4
10. 4x21 12x1 44
x22 3x2 2qwww
4x41 0x31 0x21 5x2 4
4x42 12x32 8x2
12x31 8x21 5x
12x32 36x22 24x
44x21 29x2 4
44x22 132x2 88
161x1 84
4x41 5x2 4
}
x22 3x2 2 5 4x21 12x1 441
161x1 84
}
x22 3x2 2
11. 5 2 27 10
10 15
2 3 25
2x22 7x1 10
}}
x2 5 5 2x1 31
25
}
x2 5
12. 2 4 213 25
8 210
4 25 215
4x21 13x2 5
}}
x2 2 5 4x2 52
15
}
x2 2
13. 24 1 8 1
24 216
1 4 215
x21 8x1 1
}
x1 4 5x1 42
15
}
x1 4
14. 3 1 0 9
3 9
1 3 18
x21 9
}
x2 3 5x1 31
18
}
x2 3
15. 4 1 25 0 22
4 24 216
1 21 24 218
x32 5x22 2
}
x2 4
5x22x2 4218
}
x2 4
16. 23 1 0 24 6
23 9 215
1 23 5 29
x32 4x1 6
}
x1 3
5x22 3x1 52 9
}
x1 3
17. 6 1 25 28 13 212
6 6 212 6
1 1 22 1 26
x42 5x32 8x21 13x2 12
}}}
x2 6 5x31x22 2x1 12
6
}
x2 6
18. 25 1 4 0 16 235
25 5 225 45
1 21 5 29 10
x41 4x31 16x2 35
}}
x1 5 5x32x21 5x2 91
10
}
x1 5
19. The division was done correctly, however, there is an
error in how the quotient is written.
2 1 0 25 3
2 4 22
1 2 21 1
x32 5x1 3
}
x2 2 5x21 2x2 11
1
}
x2 2
20. The error is that a zero was not used as a place holder for
the missingx2term.
2 1 0 25 3
2 4 22
1 2 21 1
x32 5x1 3
}
x2 2 5x21 2x2 11
1
}
x2 2
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
21. 6 1 210 19 30
6 224 230
1 24 25 0
f (x)5x32 10x22 19x1 30
5 (x2 6)(x22 4x2 5)
5 (x2 6)(x2 5)(x1 1)
22. 24 1 6 5 212
24 28 12
1 2 23 0
f (x)5x31 6x21 5x2 12
5 (x1 4)(x21 2x2 3)
5 (x1 4)(x1 3)(x2 1)
23. 8 1 22 240 264
8 48 64
1 6 8 0
f(x)5x32 2x22 40x2 64
5 (x2 8)(x21 6x1 8)
5 (x2 8)(x1 4)(x1 2)
24. 210 1 18 95 150
210 280 2150
1 8 15 0
f (x)5x31 18x21 95x1 150
5 (x1 10)(x21 8x1 15)
5 (x1 10)(x1 3)(x1 5)
25. 29 1 2 251 108
29 63 2108
1 27 12 0
f (x)5x31 2x22 51x1 108
5 (x1 9)(x22 7x1 12)
5 (x1 9)(x2 4)(x2 3)
26. 22 1 29 8 60
22 22 260
1 211 30 0 f (x)5x32 9x21 8x1 60
5 (x1 2)(x22 11x1 30)
5 (x1 2)(x2 5)(x2 6)
27. 1 2 215 34 221
2 213 21
2 213 21 0
f (x)5 2x32 15x21 34x2 21
5 (x2 1)(2x22 13x1 21)
5 (x2 1)(2x2 7)(x2 3)
28. 5 3 22 261 220
15 65 20
3 13 4 0
f(x)5 3x32 2x22 61x2 20
5 (x2 5)(3x21 13x1 4)
5 (x2 5)(3x1 1)(x1 4)
29. 23 1 22 221 218
23 15 18
1 25 26 0
f (x)5x32 2x22 21x2 18
5 (x1 3)(x22 5x2 6)
5 (x1 3)(x2 6)(x1 1)
The other zeros are 6 and 21.
30. 10 4 225 2154 40
40 150 240
4 15 24 0
f (x)5 4x32 25x22 154x1 40
5 (x2 10)(4x21 15x2 4)
5 (x2 10)(4x2 1)(x1 4)
The other zeros are1
}
4
and 24.
31. 7 10 281 71 42
70 277 242
10 211 26 0
f (x)5 10x32 81x21 71x1 42
5 (x2 7)(10x22 11x2 6)
5 (x2 7)(5x1 2)(2x2 3)
The other zeros are2
2
}
5
and
3
}
2
.
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
32. 24 3 34 72 264
212 288 64
3 22 216 0
f (x)5 3x31 34x21 72x2 64
5 (x1 4)(3x21 22x2 16)
5 (x1 4)(3x2 2)(x1 8)
The other zeros are2
}
3
and 28.
33. 9 2 210 271 29
18 72 9
2 8 1 0
f (x)5 2x32 10x22 71x2 9
5 (x2 9)(2x21 8x1 1)
x528 6 }
642 4(2)}}
2(2) 5
28x 6 }
56}
4
524 6
}
14}2
The other zeros are about 20.13 and 23.87.
34. 22 5 21 218 8
210 22 28
5 211 4 0
f (x)5 5x32x22 18x1 8
5 (x1 2)(5x22 11x1 4)
x5
11 6 }}
1212 4(5)(4)
}}
10
5
116 }
41
}
10
The other zeros are about 0.46 and 1.74.
35. D;
26 4 15 263 254
224 54 54
4 29 29 0
f (x)5 4x31 15x22 63x2 54
5 (x1 6)(4x22 9x2 9)
5 (x1 6)(4x1 3)(x2 3)
The zeros are 26, 23
}4
, and 3.
36. (x1 4)(x1 2)5x21 6x1 8
l52x31 17x21 46x1 40
}}
x21 6x1 8
2x1 5
x21 6x1 8qww2x31 17x21 46x1 40
2x31 12x21 16x
5x21 30x1 40
5x21 30x1 40
0
The missing dimension is 2x1 5.
37. (x2 1)(x1 6)5x21 5x2 6
w5x31 13x21 34x2 48
}}
x21 5x2 6
x1 8
x21 5x2 6qwwx31 13x21 34x2 48
x31 5x22 6x
8x21 40x2 48
8x21 40x2 48 0
The missing dimension isx1 8.
38. a. 2 1 25 212 36
2 26 236
1 23 218 0
f (x)5x32 5x22 12x1 36
5 (x2 2)(x22 3x2 18)
5 (x2 2)(x2 6)(x1 3)
The other zeros are 6 and 23.
b. The factors are (x2 2), (x2 6), and (x1 3).
c. The solutions arex5 2,x5 6,x5 23.
39. A;
5 1 21 k 230
5 20 30
1 4 k1 20 0
Forx2 5 to be a factor, the remainder must be zero. In
the last column 30 must be added to 230 to get zero.
k1 20must equal 6 in order to have a product of 30 to
add to the 230. k1 205 6 so, k5 214.
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
40. a.1}
2
b.1
}
2
30 7 239 14
15 11 214
30 22 228 0
f(x)5 1x2 1}22(30x21 22x2 28)
c.f(x)51x2 1}22(2)(15x21 11x2 14) 5 (2x2 1)(3x2 2)(5x1 7)
Problem Solving
41. P5 2x31 4x21x
45 2x31 4x21x
05 2x31 4x21x2 4
4 21 4 1 24
24 0 4
21 0 1 0
P5 (x2 4)(2x21 1)
5 (x2 4)(12x)(11x)
x5 1 is the only other positive solution. So, the company
could still make the same profit by producing only
1 million T-shirts.
42. P5 24x31 12x21 16x
485 24x31 12x21 16x
05 24x31 12x21 16x2 48
3 24 12 16 248
212 0 48
24 0 16 0
P5 (x2 3)(24x21 16)5 (x2 3)(42 2x)(41 2x)
x5 2 is the only other positive solution. The company
could produce 2 million MP3 players and still make the
same profit.
43. Let a5 average attendance per team.
a5A
}
T5
21.95x31 70.1x22 188x1 2150
}}}
14.8x1 725
20.132x21 11.2x2 561.35
14.8x1 725qwww21.95x31 70.1x22 188x1 2150
21.95x32 95.7x2
165.8x22 188x
165.8x21 8120x
28308x1 2150
28308x2 406,978.75
409,128.75
A function is
a5 20.132x21 11.2x2 5611409,129
}
14.8x1 725
44. a. Total revenue
5 (number of radios sold)(price per radio)
5 (x)(402 4x2)5 40x2 4x3
An expression is 40x2 4x3.
b.P5 (40x2 4x3)2 15x5 25x2 4x3
A function isP5 25x2 4x3.
c. WhenP5 24:
245 25x2 4x3
05 4x32 25x1 24
You know that 1.5 is a solution, sox2 1.5 is a factor
of 4x32 25x1 24.
1.5 4 0 225 24
6 9 224
4 6 216 0
So, (x2 1.5)(4x21 6x2 16)5 0. Use the quadraticformula to findx1.39, the other positive solution.
So, about 1.39 million radios also make a profit of$24,000,000.
d.No. The solutionx22.88 does not make sense
because you cannot produce a negative number of
radios.
45. LetP5 the precent of visits to the national park that
were overnight stays.
P5S
}
V5
20.00722x41 0.176x32 1.40x21 3.39x1 17.6
}}}}
3.10x1 256
20.
00233x
310.
249x
222
1x11
73
5.
287
3.
10x12
56qw
wwww
20.0
0
722x
410.
176x
321.
40x
213.3
9x
11
7.6
20.0
0722x
420.
596x
3
0.
772x
321.
40x
2
0.
772x
316
3.
7x
2
265.
1x
213.
39x
265.
1x
225
376x
5379.
39x11
7.
6
5379.
39x14
4,
233.
472
2
44215.
872
So,
P5 20.00233x31 0.249x22 21x1 17352444,216
}
3.10x1 256
.
The function for the percent of visits that were overnight
stays is the overnight stays divided by the total visits.
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
46. P5 26x31 72x
965 26x31 72x
05 26x31 72x2 96
2 26 0 72 296
212 224 96
26 212 48 0
P5 (26x22 12x1 48)(x2 2)
5 (23x1 6)(2x1 8)(x2 2)
5 23(x2 2)(2)(x1 4)(x2 2)
5 26(x2 2)(x1 4)(x2 2)
The zeros are at 2 and 24. Production levels cannot be
negative sox5 2 is the only solution.
Mixed Review
47. x2 4y< 5 x2 4y< 5
12 4(4) < 5 4 2 4(21) < 5
12
16 < 5 41
4 < 5 215 < 5 yes 8 < 5 no
(1, 4) is a solution.
48. 3x1 2y 1 3x1 2y 1
3(22)1 2(4) 1 3(1)1 2(23) 1
261 8 1 32 6 1
2 1 yes 23 1 no
(22, 4) is a solution.
49. 5x2 2y> 10 5x2 2y> 10
5(4)2 2(6) > 10 5(8)2 2(10) > 10
202 12 > 10 402 20 > 10
8 > 10 no 20 > 10 yes (8, 10) is a solution.
50. 6x1 5y 15 6x1 5y 15
6(25)1 5(10) 15 6(21)1 5(4) 15
2301 50 15 261 20 15
20 15 no 14 15 yes
(21, 4) is a solution.
51. x21 3x2 405 0
(x2 5)(x1 8)5 0
x2 55 0 or x1 85 0
x5 5 or x5 28
52. 5x2
1 13x1 65 0 (5x1 3)(x1 2)5 0
5x1 35 0 or x1 25 0
5x5 23
x5 23
}
5
or x5 22
53.x21 7x1 25 0
x527 6 }}
(7)22 4(1)(2)}}
2(1)
527 6
}
41}
2
20.30 and 26.70
54. 4x21 15x1 105 0
x5215 6 }}
(15)22 4(4)(10)}}
2(4) 5
215 6 }
65}
8
22.88 and 20.87
55. 2x21 15x1 315 0
x5215 6 }}
(15)22 4(2)(31)}}
2(2) 5
215 6 }
223}}
4 5
215 6 i}
23}
4
The solutions are215 1 i
}
23}
4 and
215 2 i}
23}
4 .
56.x21 2x1 105 0
x522 6
}}
(2)22 4(1)(10)
}}
2(1)
522 6
}
236
}
2
5 21 63i
The solutions are 21 1 3iand 21 2 3i.
57. (x22 4x1 15)1 (23x21 6x2 12)
5x22 4x1 152 3x21 6x2 12
5 22x21 2x1 3
58. (2x22 5x1 8)2 (5x22 7x2 7)
5 2x22 5x1 82 5x21 7x1 7
5 23x21 2x1 15
59. (3x2 4)(3x31 2x22 8)
5 3x(3x31 2x22 8)2 4(3x31 2x22 8)
5 9x41 6x
32 24x2 12x
32 8x
21 32
5 9x42 6x32 8x22 24x1 32
60. (3x2 5)35 (3x)32 3(3x)2(5)1 3(3x)(5)22 (5)3
5 27x32 135x21 225x2 125
Mixed Review of Problem Solving (p. 369)
1. a. 164,000,000,0005 1.643 1011yards
b.1.643 1011
}
1.203 102
51.64
}
1.20
31011
}
102
5 1.3673 109football fields
2. a. T(x)5 lpwph
5 (4x)(x)(2x)
5 8x3
b. C(x)5 (4x2 2)(x2 2)(2x2 4)
5 (4x22 10x1 4)(2x2 4)
5 8x32 36x21 48x2 16
c.I(x)5 T(x)2 C(x)
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
d.I(x)5 8x32 (8x32 36x21 48x2 16)
5 36x22 48x1 16
I(8)5 36(8)22 48(8)1 16
5 23042 3841 16
5 1936
The volume of the insulation is 1936 cubic inches.
3. Cube: V5x3A}V
56x2
}
x3 5
6
}x
A5 6x2
Sphere: V54
}
3
1x}223
A}V
5x2
}}
6x3
56
}x
5}
6
x3
A5 41x}222
5 x2
The cells exchange at the same rate.
4. f(x)5 2x41 2x21 1
x
y
211
5. a.Cis a degree 4 (quartic).
b.t, year 0 1 2 3 4
C, cost 51 46.14 42.33 40.50 40.97
t, year 5 6 7 8
C, cost 43.38 46.73 49.38 49.05
c.
1 2 3 4 5 6 7 8 90
20
10
0
40
60
50
30
Years since 1995
Cost(dollars)
t
C
No, the model will not accurately predict cell phone
bills beyond 2003. The model was not intended to be
used past 2003. From the graph you can see that it drops
sharply past 2003.
6. a.R5x(1002 10x)5 100x2 10x3
A function for the total revenue isR(x)5 100x2 10x3.
b.P5 (100x2 10x3)2 30x5 70x2 10x3
A function for the profit isP(x)5 70x2 10x3.
c. WhenP5 60:
605 70x2 10x3
05 10x32 70x1 60
You know that 2 is a solution, sox2 2 is a factor of
10x32 70x1 60.
2 10 0 270 60
20 40 260
10 20 230 0
So, (x2 2)(10x21 20x2 30)5 0. Use factoring tofind thatx5 1 andx5 23 are the other solutions.
d.No. The solutionx5 23 does not make sense because
you cannot produce a negative number of cameras.
7. V51
}
3
h(s2)
4851
}
3(x)(3x2 6)2
485x
}
3
(9x22 36x1 36)
485 3x32 12x21 12x
05 3x32 12x21 12x2 48
05 3x2(x2 4)1 12(x2 4)
05 (3x21 12)(x2 4)
x5 4
The height is 4 feet.
Lesson 5.6
5.6 Guided Practice (pp. 371373)
1. Factors of the constant term: 61, 63, 65, 615
Factors of the leading coefficient: 61
Possible rational zeros: 61
}
1, 6
3
}
1, 6
5
}
1, 6
15
}
1
Simplified list: 61, 63, 65, 615
2. Factors of the constant term: 61, 62, 63, 66
Factors of the leading coefficient: 61, 62
Possible rational zeros: 61
}
1
, 62
}
1
, 63
}
1
, 66
}
1
, 61
}
2
,
62
}
2
, 63
}
2
, 66
}
2
Simplified list: 61, 62, 63, 66, 61
}
2
, 63
}
2
3. Possible rational zeros: 61
}
1
, 62
}
1
, 63
}
1
, 66
}
1
, 69
}
1
, 618
}
1
Testx5 1:
1 1 24 215 18
1 23 218
1 23 218 0
Because 1 is a zero off, f(x) can be writen as:
f(x)5 (x2 1)(x22 3x2 18)5 (x2 1)(x2 6)(x1 3)
The zeros are 23, 1 and 6.
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
4. Possible rational zeros: 61, 62, 67, 614
Testx5 1:
1 1 28 5 14
1 27 22
1 27 22 212
1 is not a zero. Testx5 21:
21 1 28 5 14
21 9 214
1 29 14 0
Because 21 is a zero off,f(x) can be written as:
f(x)5 (x1 1)(x22 9x1 14)5 (x1 1)(x2 7)(x2 2)
The zeros offare 21, 2 and 7.
5. Possible rational zeros: 61, 63, 61
}
2, 6
3
}
2, 6
1
}
3, 6
1
}
4,
63
}4
, 61
}6
, 61
}8
, 63
}8
, 61
}12
, 61
}16
, 63
}16
, 61
}24
, 61
}48
Reasonable values:x 5 23
}
4,x5
3
}
16,x5
1
}
2
Check:x5 23
}
4:
23
}
4 48 4 220 3
236 24 23
48 232 4 0
23
}
4
is a zero.
f(x)5 1x1 3}42(48x22 32x1 4)
5 1x1 3}42(4)(12x22 8x1 1) 5 (4x1 3)(6x2 1)(2x2 1)
The real zeros offare 23
}
4
,1
}
6
, and1
}
2
.
6. Possible rational zeros: 61, 62, 63, 66, 67, 614,
621, 642, 61
}
2
, 63
}
2
, 67
}
2
, 621
}
2
Reasonable zeros:x5 27
}
2,x5 22,x5
3
}
2,x5 2
27
}
2
2 5 218 219 42
27 777
}
2
2273
}
4
2 22 21139
}
2
2105
}
4
22 2 5 218 219 42
24 22 40 242
2 1 220 21 0
f(x)5 (x1 2)(2x31x22 20x1 21)5 (x1 2) pg(x)
Possible rational zeros ofg(x): 61, 63, 67, 621, 61
}
2
,
63
}
2
, 67
}
2
, 621
}
2
Reasonable zeros:x53
}
2
,x5 2
3
}
2
2 1 220 21
3 6 221
2 4 214 0
f(x)5 (x1 2) pg (x)
5 (x1 2)1x2 3}22(2x21 4x2 14) 2x21 4x2 145 0
x524 6 }}
(4)22 4(2)(214)}}
2(2) 5
24 6 }
128}
4
5 21 6 2}
2
The real zeros offare: 22,3
}
2
, 211 2
}
2 , and 212 2
}
2 .
7. V 51
}
3
(x2)(x2 1)
651
}
3
x2(x2 1)
185x32x2
05x32x2218
Possible rational zeros: 61, 62, 63, 66, 69, 618
2 1 21 0 218
2 2 4
1 1 2 214
3 1 21 0 218
3 6 18
1 2 6 0
The other solutions, which satisfyx21 2x1 6, are
x5 21 6 i}
5 and can be discarded because they are
imaginary. The only real solution isx5 3. The base is
3 33 feet. The height isx2 1 or 2 feet.
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
5.6 Exercises (pp. 374377)
Skill Practice
1. If a polynomial function has integer coefficients, then
every rational zero of the function has the formp
}qwhere
pis a factor of the constant termand qis a factor of the
leading coefficient.
2. To shorten the list, draw the graph to see approximatelywhere the zeros are.
3. f(x)5x32 3x1 28
Factors of constant term: 61, 62, 64, 67, 614, 628
Factors of leading coefficient: 61
Possible rational zeros: 61
}
1, 6
2
}
1, 6
4
}
1, 6
7
}
1, 6
14
}
1, 6
28
}
1
Simplified list: 61, 62, 64, 67, 614, 628
4.g (x)5x32 4x21x2 10
Factors of constant term: 61, 62, 65, 610
Factors of leading coefficient: 61
Possible rational zeros: 61
}
1
, 62
}
1
, 65
}
1
, 610
}1
Simplified list: 61, 62, 65, 610
5.f(x)5 2x41 6x32 7x1 9
Factors of constant term: 61, 63, 69
Factors of leading coefficient: 61, 62
Possible rational zeros: 61
}
1
, 63
}
1
, 69
}
1
, 61
}
2
, 63
}
2
, 69
}
2
Simplified list: 61, 63, 69, 61
}
2
, 63
}
2
, 69
}
2
6. h(x)5 2x31x22x2 18
Factors of constant term: 61, 62, 63, 66, 69, 618
Factors of leading coefficient: 61, 62
Possible rational zeros: 61
}
1
, 62
}
1
, 63
}
1
, 66
}
1
, 69
}
1
, 618
}
1
, 61
}
2
,
62
}
2
, 63
}
2
, 66
}
2
, 69
}
2
, 618
}
2
Simplified list: 61, 62, 63, 66, 69, 618, 61
}
2
, 63
}
2
, 69
}
2
7.g (x)5 4x51 3x32 2x2 14
Factors of the constant term: 61, 62, 67, 614
Factors of the leading coefficient: 61, 62, 64
Possible rational zeros: 61
}
1, 6
2
}
1, 6
7
}
1, 6
14
}
1, 6
1
}
2, 6
2
}
2, 6
7
}
2,
614
}
2, 6
1
}
4, 6
2
}
4, 6
7
}
4, 6
14
}
4
Simplified list: 61, 62, 67, 614, 61
}2
, 67
}2
, 61
}4
, 67
}4
8.f(x)5 3x41 5x32 3x1 42
Factors of the constant term: 61, 62, 63, 66, 67,
614, 621, 642
Factors of the leading coefficient: 61, 63
Possible rational zeros: 61
}
1
, 62
}
1
, 63
}
1
, 66
}
1
, 67
}
1
,
614
}1, 621
}1, 642
}1, 61}3, 62}3, 63}3, 66}3, 67}3, 614
}3
,
621
}
3
, 642
}
3
Simplified list: 61, 62, 63, 66, 67, 614,
621, 642, 61
}
3
, 62
}
3
, 67
}
3
, 614
}
3
9. h(x)5 8x41 4x32 10x1 15
Factors of the constant term: 61, 63, 65, 615
Factors of the leading coefficient: 61, 62, 64, 68
Possible rational zeros: 61
}
1, 6
3
}
1, 6
5
}
1, 6
15
}
1, 6
1
}
2,
6
3
}
2
,6
5
}
2
,6
15
}
2
,6
1
}
4
,6
3
}
4
,6
5
}
4
,6
15
}
4
,6
1
}
8
,6
3
}
8
,
65
}
8
, 615
}
8
Simplified list: 61, 63, 65, 615, 61
}
2
, 63
}
2
,
65
}
2
, 615
}
2
, 61
}
4
, 63
}
4
, 65
}
4
, 615
}
7
, 61
}
8
, 63
}
8
, 65
}
8
,
615
}
8
10. h(x)5 6x32 3x21 12
Factors of the constant term: 61, 62, 63, 64, 66, 612
Factors of the leading coefficient: 61, 62, 63, 66
Possible rational zeros: 61
}
1
, 62
}
1
, 63
}
1
, 64
}
1
, 66
}
1
,
612
}
1, 6
1
}
2, 6
2
}
2, 6
3
}
2, 6
4
}
2, 6
6
}
2, 6
12
}
2,
61
}
3
, 62
}
3
, 63
}
3
, 64
}
3
, 66
}
3
, 612
}
3
, 61
}
6
, 62
}
6
, 63
}
6
,
64
}
6, 6
6
}
6, 6
12
}
6
Simplified list: 61, 62, 63, 64, 66, 612,
61
}
2
, 63
}
2
, 61
}
3
, 62
}
3
, 64
}
3
, 61
}
6
11. Possible rational zeros: 61, 62, 63, 64, 66, 68, 612,
624
Testx5 1:
1 1 212 35 224
1 211 24
1 211 24 0
f(x)5 (x2 1)(x22 11x1 24)
5 (x2 1)(x2 3)(x2 8)
Real zeros are 1, 3, and 8.
Chapter 5, continued
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Algebra 2
Worked-Out Solution Key
12.Possible rational zeros: 61, 62, 64, 67, 68, 614, 628,
656
Testx5 1:
1 1 25 222 56
1 24 226
1 24 226 30 Testx5 21:
21 1 25 222