1112
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
Solutions to Odd.Numbered Exercises
1. Y2=4+~
4
x--1
4
X+2
3o Y2=X--2+~
(x- 2)(x + 2) + 4
4(x- 1)+4
x-1
4x-4+4
x-1
x+2
xZ-4+4
x+2
x+2
= Yl
= Yl
-21
-17
5.
y2 "- x3 -- 4x + 2.2 + 1
(x3 - 4x)(~ + I) + 4x
x2+ 1
4
x~ + .P - 4.P - 4x + 4x
x~+l
x+3)
2x+ 4
2
2x + lOx + 12
(2~+ 6x)
4x +12
- (4~ + 1~)
0
2x2 + lOx + 12
=2x+4
x+3
4x+5)
xz- 3x+l
3~
4x 7x -llx+5
(4.P + 5x~)
- 12x2 - 1 lx
-(- 12x:~ - 15x)
4x+5
- (4x + 5)
0
4x3 - 7x2 - 1 lx + 5 =x~-3x+ 1
4x + 5
103
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
7
7x+ 3
11o x+2)
13.
3x+5
3+
2+
6x
l Ox x +8
2x2+Ox+ 1)
- (7x + 14)
-11
- (6x3 + Ox2 + 3x)
lOx2 -- 2X +8
- (lOx~ + Ox +5)
7x + 3
11
-7-~
x+2
x+2
-2x +3
6x3+ l Oxz+x+8
2x-3
=3x+5
2
2x + 1
2x~ + 1
15.
x2 + 2x + 4
x2-2x+3) x4+Ox3+3x2 +Ox + 1
+ 3X2 + 1
2X -- 11
=X2+2X+4+
X~-- 2X+ 3
.~-- 2X+ 3
X4
-- (x4 -- 2x3 + 3x2)
2x3 + Oxa +Ox
-- (2x3 -- 4x2 + 6x)
4x2 --6x + 1
-(4x2 -8x + 12)
2x - 11
2x
17. x2-2x+ 1) 2x3-4x2- 15x +5
-(2x3 -- 4X2 + 2X)
--17x +5
2x3-4x2- 15x+5
=2x
(x- 1)2
21. 3
67
-1
75
18
6 25
17x-5
(x- 1)2
26
222
1
0 512
-8
64 -512
-8
64
0
x3 + 512
-x~- 8x+ 64
x+8
32
-32
0 - 16
0
9x3 - 18x2 - 16x+ 32 2
=9x - 16
x -- 2
4
4
16
-23
-15
-2
-7
15
14
-30
0
4x3+ 16x2-23x- 15
= 4x2 + 14x - 30
x+½
29. f(x)=x3 - xa- 14x + 11, k = 4
1
-16
0
9
0
1 -1
9 -18
18
74 248
1
--10 12 --22
12 8 80
3 2 20 58
3X3- IOx2+ 12X--22
58
= 3x2 + 2x + 20 + ~
x-4
x-4
23. 2
6x3 + 7x2 - x + 26
= 6x2 + 25x + 74 248
+~
x-3
x-3
25. -8
19. 4 3
-14
11
4
12
-8
3
-2
3
f(x) = (x- 4)(x2+3x-2) +3
f(4) = (0)(26) + 3 = 3
31. f(x)=x3+3x2-2x- 14, k=
1
3
-2
~
2+3-v/’~
1
3+ v/~
+ (3 + +
f(-v~) = (0)(4 + 6-V~)- 8 = -8
- 14
6
-8
-8
104
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
-6
-12 -4
4 - 4ff~ 10 - 2~/~ 4
4 -2 - 4~/~ -2 - 2~/~ 0
f(x) = (x- 1 + .,/~)[4x2 -(2 + 4~/~)x- (2 + 2x/~)]
f(X- ~/~) = 0
35. f(x) = 4x3 - 13x + 10
10
0 -13
(a) 1 4
4
4 -9
4
4 -9
1 =f(1)
(c) ½
40
-13
10
1
-6
2
42
- 12
(b) -2
0
-13
10
-8
16
-6
4
-8
3
4
0
-13
10
32
256
1944
32
243
1954 = f(8)
4
(d) 8
4
_~_4= f(½)
4 = f(-2)
37. h(x)=3x3+5x2- lOx+ 1
(a) 3 [ 3
5
9
I
3 14
- 10
1
42 96
32
5
-6
(c) -2I 3
I
39. 2
1
-8
97 =f(3)
-10
1
2 16
3 -1
- 10 1
2 _8 3
5
1
(b) ½ 3
-8 17 =f(-2)
0
-7
6
2
4
-6
1
2
-3
0
x3-7x+6=(x-2)(x~+2x-3)
= (x- 2)(x + 3)(x- 1)
(d) -5 I 3
I3 - 10
2 -15
27
1
- 199 = f(-5)
-10
-7
20
2 -14
2x3- 15x~+27x- 10
= (x - ½)(2x~ - 14x + 20)
Zeros: ½, 2, 5
43.-2I 1 2 -2 -4
-2 0 4
0
0 -2
x3 + 2xz - 2x- 4 = (x + 2)(xz - 2)
=
Zeros: - 2,
40
= (~x- 1)(x- 2)(x- 5)
Zeros: 2, -3, 1
1
1
-200
5 -10
- 15 50
10
0
105
45. (a) - 2I 2
I
2
112 -3
2 -1
PART L" Solutions to Odd-Numbered Exercises and Practice Tests
1
-4
-5
6
2
-2
-3
1
0
2
47. (a) 5 I 1
[
1
1
-1
-411
0
!
-15
5
58
-50
-40
40
1
-10
8
0
1
-4
-10
12
8
-8
-3
2
0
(b) x2 - 3x + 2 = (x-.2)(x- 1),
(b) Remaining factor: (2x - 1)
Remaining factors: (x - 2), (x - 1)
(c) f(x) - (x + 2)(x- 1)(2x- 1)
. (c) f(x)’= (x- 5)(x + 4)(x- 2)(x- 1)
(d) Real zeros: - 2, 1, ½
(e)
-4
5
(d) Real zeros: 5, -4, 2, 1
7
(e)
49° (a) -½ I’ 6
I
41
-3
6 38
-9
- 19
-14
14
-28
0
2
g 6 38 -28
4 28
(b) 6x + 42 Remaining factor (or 6(x + 7))
(c) f(x) = (2x + l)(3x- 2)(x + 7)
Note: Use ~(6x + 42) = x + 7
12
(d) Real zeros: -g,
g, -7
(e)
a20
6 42 0
51. f(x) = x3 + 3x~- x- 3
Possible rational zeros: + 1, +3
Zeros shown on graph: -3, - 1, 1
53. f(x) = 2x4 - 17x3 + 35x2 + 9x - 45
Possible rational zeros: +1,_+3, _+5, _+9, _+15, _+45,
4-1_ +3 +5 +9 4.15 _+.~
--2’ --2’ --2, --2, --~’,
Zeros shown of graph: 1, ~, 3, 5
55. f(x) = x~ + x2 - 4x- 4
(a) Possible rational zeros: _+1, +2, +4
(b)
57. f(x) = -4x3 + 15x2 - 8x - 3
(a) Possible rational zeros: _+¼, +!_2, +-~, +1,_:~,+3_ _+3
1
-6~ 6
-7
(c) -2, -1, 2 on graph
(c) -¼, 1, 3 on graph
106
PART h Solutions to Odd-Numbered Exercises and Practice Tests
59. f(x) = -2x4 + 13x3 - 21x~ + 2x + 8
(a) Possible rational zeros:
(a) Possible rational zeros:
+~, +½, -2,
+~_±~,
2 _+1, +~, +2,+_~3, +4, +8
+½, +1, +2, +4, +8
(b)
61. f(x) = 6x3 - x~- 13x+8
16
-12
12
(c) Real zeros: 1,
12
(c) -½, 1, 2, 4 on graph
If(x) = (x - 1)(6x2 + 5x - 8); Use Quadratic
Formula]
63. Z4 -- Z3 -- 2z - 4 = 0
Possible rational zeros: +1, +2, +4
-1I
1
-1
0
-2
1
-1
-2
2
2
-2
-4
-2
2
-4
2
0
0
2
4
0
.
I
1
-4
4
0
65. x4- 13xz- 12x=O
x(xa- 13x- 12)=0
x = 0 is a solution.
Possible rational zeros ofx3 - 13x - 12 = 0
are + 1, +2, +3, :!:4, :1:6 and :!: 12. Using a
graphing utility or synthetic division, you find
that the zeros are 0, - 1, -3, 4.
1
z4 - z3 -2z - 4 = (z + 1)(z - 2)(z2 + 2) = 0
The on.ly real zeros are -1 and 2. You can verify this
by graphing the function f(z) = z4 - z3 - 2z - 4.
67. 2x~ - l lx3 - 6x2 + 64x + 32 = 0
Using a graphing utility, you can see that there are
three zeros. Using synthetic division, you can verify
that these zeros are -2, -½, 4.
69. f(x) = x3 - 2x~- 5x + 10
(a) Zeros: 2, 2.236, -2.236
(b) 2 I 1 -2 -5 10
I
2
10
y(x) = (x-
= (x71. h(t) = t3- 2t~- 7t + 2
(a) zeros: -2, 3.732, 0.268
(b) -2I 1 -2 -7 2
-2
8 -2 t=-2isazero
1 -4
1
0
~
h(t) = (t + 2)(t - 4t + 1)
= (, + 2)[,- (~/~ + 2)][, + (V~- 2)]
0
-5
-10 x=2isazero
0
5)
+-
107
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
73. h(x) = x5 - 7x~ + 10xa + 14x2 - 24x
(a) h(x) = x(x4 - 7xa + 10x~ + 14x - 24)
From the calculator we have x = 0, 3, 4
and x ~- _+ 1.414.
10
(b) 3 1 -7
3 - 12
-2
1 -4
-4
4
0
14 -24
-6
24
0
8
-2
0
-2
75. f(x) = x4 - 4x3 q- 15
(a) 4
1 -4
4
10
0
0
0
4 is an upper bound.
0
1 -4
-1
8
-8
0
0 15
00
0 15
1 -5
5
5
0
-5
-5
15
5
20
- 1 is a lower bound.
f(x) = x(x- 3)(x- 4)(x~ - 2)
= x(x- 3)(x- 4)(x - ~/~)(x +
The exact roots are x = 0, 3, 4, + ~4~.
79. P(x) = x4 - ~x~ + 9
= ¼(4x4 - 25x~ -t- 36)
77. f(x) = x4- 4x3 + 16x- 16
(a) 5
1 -4
1
5
1,
0
16 -16
5
5
25 205
41 189
= ¼(4x~ - 9)(x~ - 4)
= ¼(2x + 3)(2x - 3)(x + 2)(x -, 2)
The zeros are __.~ and __.2.
5 is anupperbo~nd.
(b) -3 1 -4
0
16
-3
21
21
-63
-47
1
-7
-16
141
125
-3 is a lower bound.
=¼(4xa-x=-4x+ 1)
= ¼[x~(4x-i’)- 1(4x- 1)I
= ¼(4x- 1)(x~ - 1)
= ¼(4x - l)(x + 1)(x - I)
The zeros are ¼ and ¯ 1.
f(x) = x3 - x = x(x + 1)(x- 1)
Rational zeros: 3 (x = 0, + 1)
Irrational zeros: 0
Matches (b).
$3. f(x)=x3- 1 =(x- 1)(x~+x+ 1)
Rational zeros: 1 (x = 1)
Irrational zeros: 0
Matches (d).
108 PART I: Solution.s to Odd-Numbered Exercises a,nd Practice Tests
87. (a)
35
-2 ~7
10
(b) R = 0.01326t3 - 0.06765t2 + 1.2306t + 16.6770
(c)
t
-2
R
13.86
Model
13.84
0
1
2
3
15.21
16.78
18.10
19.08
19.39 21.62
15.37
16.68
17.85
18.97
20.12
(d) 12 1.01326
-0.06765
0.15912
1.2306
1.09764
16.6770
27.9389
.01326
0.09147
2.3282
44.6159
I
6
7
23.07
24.41
26.48
22.80
24.49
26.52
4
21.37
R(12) ~ 44.62. No. The model will turn sharply upward.
(b) V=l, w ¯ h=(15-2x)(9-2x)x
= x(9 - 2x)(15 - 2x)
Since length, width, and height cannot be negative,
we have 0 < x < ~ for the domain.
89. (a)
(C) 125
(d) 56 = x(9 - 2x)(15 - 2x)
56 = 135x - 48x~ + 4x3
0 =4x3- 48x~+ 135x - 56
¢
o
The volume is maximum when x ~ 1.82.
The dimensions are: length = 15 -2(1.82)= 11.36
width = 9 - 2(1.82) = 5.36
height = x = 1.82
1.82 cm x 5.36 cm x 11.36 cm
The zeros of this polynomial are ½, ~, and 8.
x cannot equal 8 since it is not in the domain of V.
[The length cannot equal - 1 and the width cannot
equal -7. The product of (8)(- 1)(-7) = 56 so it
showed up as an extraneous solution.]
109
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
91. y = -5.05x3 + 3857x - 38,411.25, 13 <x< 18
(a) 2,0°
(b) The second ah’-fuel ratio of 16.89 can be obtained
by finding the second point where the curves y and
Yl = 2400 intersect.
1~
o
(c) Solve -5.05x3 + 3857x- 38,411.25 = 2400 or-5.05x3 + 3857x- 40,811.25 = 0. By synthetic division,
15
-5.05
0
3857
-40,811.25
-75.75 -1136.25 40,811.25
-5.05 -75.75
2720.75
0
(d) The positive zero of the quadratic - 5.05xz - 75.75x + 2720.75 can be found by the Quadratic Formula.
x~
75.75 - ,/(75.75)2 - 4(-5.05)(2720.75)
~ 16.89
2(-5.05)
93. False, -~ is a root off.
95. ½
6
1
3
64
-92
2
45
-45
184
4
0 92
-48
48
-90
0
184 96
0
True.
97.
x2n -Xn+3
x" - 2 )’x3" - 3X2n + 5x~ - 6
x3n -- 2x2n
-x~’~ + 5x"
- x:z" + 2xn
3xn - 6
3xn - 6
0
3"
:zn
n
x - 3x + 5x - 6 zn n
Hence,
= x - x + 3.
xn-2
101. (a) (fo g)(x) = f(x2 + 1)
= 2(x2 + 1) - 5
99. You can check polynomial division by multiplying
the quotient by the divisor. This should yield the
original dividend if the multiplication was performed correctly.
103. (a) (f o g)(x) = f(4x + x~)
-.-.
= 2x~- 3
(b) (g of)(x) = g(2x - 5)
= (2x- 5)2 + 1
1
4x +xz
(b) (g °f)(x) = g@) = 4(¼) + (~)z
4x+l
= 4x~ - 20x + 26
105. f(x) = (x- 1)(x + 3)(x- 8)
=x~-6x~- 19x+24
x~
[answer not unique]
= (x- 2)2 - 3
=xZ-4x+4-3
=x2 - 4x+ 1
[answer not unique]