C H A P T E R 3
Polynomial Functions
Section 3.1
Quadratic Functions and Models . . . . . . . . . . . . . 296
Section 3.2
Polynomial Functions of Higher Degree . . . . . . . . . 311
Section 3.3
Polynomial and Synthetic Division . . . . . . . . . . . . 328
Section 3.4
Zeros of Polynomial Functions . . . . . . . . . . . . . . 340
Section 3.5
Mathematical Modeling and Variation
Review Exercises
. . . . . . . . . .358
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 367
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381
Practice Test
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
C H A P T E R 3
Polynomial Functions
Section 3.1
Quadratic Functions and Models
You should know the following facts about parabolas.
■
f x ax2 bx c, a 0, is a quadratic function, and its graph is a parabola.
■
If a > 0, the parabola opens upward and the vertex is the point with the minimum y-value.
If a < 0, the parabola opens downward and the vertex is the point with the maximum y-value.
■
The vertex is b2a, f b2a.
■
To find the x-intercepts (if any), solve
ax2 bx c 0.
■
The standard form of the equation of a parabola is
f x ax h2 k
where a 0.
(a) The vertex is h, k.
(b) The axis is the vertical line x h.
Vocabulary Check
1. nonnegative integer; real
2. quadratic; parabola
4. positive; minimum
5. negative; maximum
3. axis or axis of symmetry
1. f x x 22 opens upward and has vertex 2, 0.
Matches graph (g).
2. f x x 42 opens upward and has vertex 4, 0.
Matches graph (c).
3. f x x2 2 opens upward and has vertex 0, 2.
Matches graph (b).
4. f x 3 x2 opens downward and has vertex 0, 3.
Matches graph (h).
5. f x 4 x 22 x 22 4 opens downward
and has vertex 2, 4. Matches graph (f).
6. f x x 12 2 opens upward and has vertex
1, 2. Matches graph (a).
7. f x x 32 2 opens downward and has
vertex 3, 2. Matches graph (e).
8. f x x 42 opens downward and has vertex 4, 0.
Matches graph (d).
296
Section 3.1
1
9. (a) y 2x2
Quadratic Functions and Models
1
(b) y 8 x2
y
y
5
6
4
4
3
2
2
−6
x
−4
−3
−2
−1
2
3
−6
Vertical shrink and reflection in the x-axis
Vertical shrink
(c) y 6
−4
x
1
−1
4
−2
1
3 2
2x
(d) y 3x2
y
y
5
6
4
4
3
2
2
−6
−4
x
−2
2
4
6
1
−3
−2
−1
x
1
−1
2
3
Vertical stretch and reflection in the x-axis
Vertical stretch
10. (a) y x 2 1
(b) y x2 1
y
y
5
4
4
3
3
2
2
1
−3
−3
−2
−1
1
−1
2
3
(c) y 2
3
Vertical translation one unit downward
(d) y x2 3
y
y
10
8
8
6
6
4
−6
−6
−4
−2
3
−2
Vertical translation one unit upward
x2
x
−2
x
x
–4
4
6
x
−2
2
4
6
Vertical translation three units upward
−4
Vertical translation three units downward
297
298
Chapter 3
Polynomial Functions
(b) y 3x2 1
11. (a) y x 12
y
−2
−1
y
5
5
4
4
3
3
x
1
−1
2
3
−3
4
Horizontal translation one unit to the right
(c) y 1 2
3x
−2
−1
x
2
1
3
−1
Horizontal shrink and a vertical translation one unit upward
(d) y x 32
3
y
y
8
10
6
8
4
2
−6
−2
x
2
−2
2
6
−8
−6
−4
−2
−4
Horizontal stretch and a vertical translation three units
downward
x
2
−2
4
Horizontal translation three units to the left
1
(b) y 2x 1 3
12. (a) y 12 x 22 1
2
y
y
8
10
6
8
4
6
4
− 6 −4 − 2
x
2
6
8 10
x
−8 − 6 − 4
2
6
8
−4
−6
Horizontal translation two units to right, vertical
shrink each y-value is multiplied by 12 , reflection in
the x-axis, and vertical translation one unit upward
1
(c) y 2x 22 1
Horizontal translation one unit to the right, horizontal
stretch (each x-value is multiplied by 2), and vertical
translation three units downward
(d) y 2x 12 4
y
y
6
7
4
2
x
−8 −6 −4
2
4
6
4
3
2
−4
−6
−8
Horizontal translation two units to left, vertical shrink
each y-value is multiplied by 12 , reflection in x-axis,
and vertical translation one unit downward
1
− 4 − 3 − 2 −1
−1
x
1
2
3
4
Horizontal translation one unit to left, horizontal shrink
each x-value is multiplied by 12 , and vertical translation
four units upward
Section 3.1
13. f x x2 5
Vertex: 0, 25
Axis of symmetry: x 0 or the y-axis
Axis of symmetry: x 0
y
Find x-intercepts:
− 4 −3
x
−1
x ± 5
1
3
x2 25
4
−2
x ±5
−3
x-intercepts: ± 5, 0
x-intercepts:
x
− 10
10
20
1
1
16. f x 16 4 x2 4 x2 16
Vertex: 0, 4
Vertex: 0, 16
Axis of symmetry: x 0 or the y-axis
Axis of symmetry: x 0
y
Find x-intercepts:
40
3
x2 8
1
x ± 8 ± 22
− 20
−6
15. f x 12 x2 4 12x 02 4
1 2
2x
30
25 x2 0
1
x2 5
y
Find x-intercepts:
2
x2 5 0
299
14. hx 25 x2
Vertex: 0, 5
5, 0 , 5, 0
Quadratic Functions and Models
x
−1
1
2
3
4
−3
22, 0, 22, 0
−5
9
6
3
x
−9 − 6 − 3
−3
18. f x x 62 3
y
20
16
Axis of symmetry: x 5
12
x-intercepts: ± 8, 0
17. f x x 52 6
Vertex: 5, 6
x2 64
x ±8
−2
x-intercepts:
18
16 14 x2 0
2
− 4 −3
y
Find x-intercepts:
12
3
6
9
y
Vertex: 6, 3
50
Axis of symmetry: x 6
40
30
Find x-intercepts:
x 52 6 0
Find x-intercepts:
− 20
x
− 12
4
x 52 6
8
x 62 3
−8
x 5 ± 6
20
x 62 3 0
10
x
− 20 − 10
10
20
30
1
2
Not possible for real x
x 5 ± 6
No x-intercepts
x-intercepts: 5 6, 0, 5 6, 0
19. h x x2 8x 16 x 42
20. gx x2 2x 1 x 12
Vertex: 4, 0
Vertex: 1, 0
y
Axis of symmetry: x 4
20
Axis of symmetry: x 1
x-intercept: 4, 0
16
x-intercept: 1, 0
y
6
5
4
12
3
8
2
4
−4
1
x
4
8
12
16
−4
−3
−2
−1
x
300
Chapter 3
Polynomial Functions
21. f x x 2 x 5
4
x2 x x
Vertex:
1
2
1
4
22. f x x 2 3x 1
1 5
4
4 4
x 2 3x 1
2
x
12, 1
Vertex:
Axis of symmetry: x 1
2
3
2
9
9 1
4
4 4
2
2
23, 2
y
4
Axis of symmetry: x y
3
2
3
2
1
5
Find x-intercepts:
x2 x Find x-intercepts:
4
5
0
4
x 2 3x 3
1 ± 1 5
x
2
−1
x
1
2
1
0
4
x
1
2
−2
−3
3 ± 9 1
x
2
1
−2
− 5 − 4 − 3 −2 − 1
3
Not a real number
No x-intercepts
x-intercepts:
3
± 2
2
23 ± 2, 0
24. f x x2 4x 1 x2 4x 1
23. f x x2 2x 5
x2 2x 1 1 5
x2 4x 4 4 1
x 12 6
x 22 5
Vertex: 1, 6
Vertex: 2, 5
Axis of symmetry: x 1
Axis of symmetry: x 2
Find x-intercepts:
Find x-intercepts: x2 4x 1 0
x2 4x 1 0
x2 2x 5 0
x2 2x 5 0
x
x
2 ± 4 20
2
1 ± 6
2 ± 5
x-intercepts: 2 ± 5, 0
x-intercepts: 1 6, 0, 1 6, 0
y
5
y
4
6
2
1
−6 −5
x
−4
2
−2
−4
6
4 ± 16 4
2
− 3 −2 − 1
−2
−3
x
1
2
Section 3.1
4 x
Vertex:
1
2
1
2 x2 x 1
2
1
1
4
21
4
4
20
2
12, 20
2 x
1
4
2
1
4
2
2 x
Axis of symmetry: x 301
26. f x 2x2 x 1
25. h x 4x2 4x 21
4 x2 x Quadratic Functions and Models
1
2
Vertex:
Find x-intercepts:
2
161 1
7
8
14, 78
Axis of symmetry: x 4x2 4x 21 0
1
4
Find x-intercepts: 2x2 x 1 0
4 ± 16 336
x
24
x
Not a real number ⇒ No x-intercepts
1 ± 1 8
22
y
Not a real number
y
6
No x-intercepts
5
4
3
20
1
10
−8
−4
−3
x
4
1
2
3
1
28. f x 3 x2 3x 6
14x2 8x 16 1416 12
13 x2 9x 6
14x 42 16
1 81
13 x2 9x 81
4 3 4 6
13 x 92 34
2
Vertex: 4, 16
Axis of symmetry: x 4
Vertex:
2x 12 0
x2 8x 48 0
4
4
8
16
x 4x 12 0
− 12
or x 12
− 16
x-intercepts: 4, 0, 12, 0
29. f x x2 2x 3 x 12 4
x
4
6
8
10
−2
x2 9x 18 0
−4
x 3x 6 0
−6
30. f x x2 x 30
x2 x 30
5
x2 x 14 14 30
Axis of symmetry: x 1
x-intercepts: 3, 0, 1, 0
−2
13 x2 3x 6 0
x-intercepts: 3, 0, 6, 0
− 20
Vertex: 1, 4
y
2
Find x-intercepts:
x
−8
92, 34 Axis of symmetry: x 92
y
Find x-intercepts:
x 4
x
−1
8
1
27. f x 4x 2 2x 12
1 2
4x
−2
−8
7
−5
x 12 121
4
2
35
1 121
Vertex: 2, 4 Axis of symmetry: x − 10
10
12
x-intercepts: 6, 0, 5, 0
− 80
302
Chapter 3
Polynomial Functions
31. gx x2 8x 11 x 42 5
32. f x x2 10x 14
Vertex: 4, 5
x2 10x 25 25 14
14
x 52 11
Axis of symmetry: x 4
x-intercepts: 4 ± 5, 0
−18
12
Vertex: 5, 11
5
−20
10
Axis of symmetry: x 5
−6
x-intercepts: 5 ± 11, 0
33. f x 2x2 16x 31
34. f x 4x2 24x 41
48
4x2 6x 41
2x 42 1
Vertex: 4, 1
−6
Axis of symmetry: x 4
x-intercepts: 4 ±
1
2 2,
− 15
12
4x2 6x 9 36 41
4x 32 5
−12
0
0
0
6
Vertex: 3, 5
Axis of symmetry: x 3
No x-intercepts
35. gx 12x2 4x 2 12x 22 3
Vertex: 2, 3
36. f x 35 x2 6x 5
35 x2 6x 9 27
5 3
4
35 x 32 42
5
Axis of symmetry: x 2
−8
x-intercepts: 2 ± 6, 0
−20
4
−4
42
Vertex: 3, 5 6
− 14
x-intercepts: 3 ± 14, 0
37. 1, 0 is the vertex.
38. 0, 1 is the vertex.
y ax 1 0 a x 1
f x ax 02 1 ax2 1
Since the graph passes through the point 0, 1, we have:
Since the graph passes through 1, 0,
2
2
1 a0 12
0 a12 1
1a
1 a.
y 1x 12 x 12
So, y x2 1.
39. 1, 4 is the vertex.
40. 2, 1 is the vertex.
y ax 1 4
f x a x 22 1
Since the graph passes through the point 1, 0, we have:
Since the graph passes through 0, 3,
2
0 a1 12 4
3 a 0 22 1
4 4a
3 4a 1
1 a
4 4a
y 1x 12 4 x 12 4
10
Axis of symmetry: x 3
1 a.
So, y x 22 1.
− 10
Section 3.1
41. 2, 2 is the vertex.
Quadratic Functions and Models
42. 2, 0 is the vertex.
y ax 22 2
f x a x 22 0 a x 22
Since the graph passes through the point 1, 0, we have:
Since the graph passes through 3, 2,
2 a 3 22
0 a1 2 2
2
2 a.
2 a
So, y 2x 22.
y 2x 22 2
43. 2, 5 is the vertex.
44. 4, 1 is the vertex.
f x ax 22 5
f x a x 42 1
Since the graph passes through the point 0, 9, we have:
Since the graph passes through 2, 3,
9 a0 2 5
3 a 2 42 1
4 4a
3 4a 1
1a
4 4a
2
1 a.
f x 1x 22 5 x 22 5
So, f x x 42 1.
45. 3, 4 is the vertex.
46. 2, 3 is the vertex.
f x ax 3 4
f x a x 22 3
Since the graph passes through the point 1, 2, we have:
Since the graph passes through 0, 2,
2
2 a 0 22 3
2 a1 32 4
2 4a 3
2 4a
12
1 4a
a
f x 12x
14 a.
32 4
1
So, f x 4 x 22 3.
47. 5, 12 is the vertex.
48. 2, 2 is the vertex.
f x ax 5 12
f x a x 22 2
Since the graph passes through the point 7, 15, we have:
Since the graph passes through 1, 0,
2
0 a 1 22 2
15 a7 52 12
3 4a ⇒ a 34
f x 3
4 x
0a2
2 a.
5 12
2
So, f x 2x 22 2.
1 3
49. 4, 2 is the vertex.
50.
f x ax 14 32
2
Since the graph passes through the point 2, 0,
we have:
0 a2 32
f x 1 2
4
49
16 a
⇒ a
24
49
x 1 2
4
3
2
24
49
3
2
52, 34 is the vertex.
2
f x a x 52 34
Since the graph passes through 2, 4,
4 a2 52 34
2
4
81
4a
19
4
81
4a
19
81
a.
34
19
5
3
So, f x 81 x 2 4.
2
303
304
Chapter 3
Polynomial Functions
51. 52, 0 is the vertex.
f x ax 52. 6, 6 is the vertex.
5 2
2
f x ax 62 6
Since the graph passes through 10, 2 ,
Since the graph passes through the point 2, 3 ,
we have:
7
16
3
a
16
3
a
72
f x 16
3
61 3
16
5 2
2
3
2
a 61
10 6 6
3
2
1
100
a6
2
9
1
2 100 a
x 5 2
2
450 a.
So, f x 450x 62 6.
53. y x2 16
54. y x2 6x 9
0 x2 16
x-intercepts: ± 4, 0
x-intercept: 3, 0
x2 16
0 x2 6x 9
x ±4
0 x 32
x30 ⇒ x3
55. y x2 4x 5
x-intercepts: 5, 0, 1, 0
0 x2 4x 5
56. y 2x2 5x 3
0 x 5x 1
x5
or
x-intercepts:
x 1
12, 0, 3, 0
0 2x2 5x 3
0 2x 1x 3
2 x 1 0 ⇒ x 12
x 3 0 ⇒ x 3
57. f x x2 4x
58. f x 2x2 10x
4
x-intercepts: 0, 0, (4,0
0
x2
x-intercepts: 0, 0, 5, 0
−4
4x
0 xx 4)
x0
or
14
8
0 2x2 10x
−1
0 2xx 5
−4
6
−6
2x 0 ⇒ x 0
x4
The x-intercepts and the solutions of f x 0 are the same.
x50 ⇒ x5
The x-intercepts and the solutions of f x 0 are the same.
59. f x x2 9x 18
60. f x x2 8x 20
12
x-intercepts: 3, 0, 6, 0
0 x2 9x 18
0 x 3)x 6
x3
or
x-intercepts: 2, 0, 10, 0
−8
16
−4
x6
The x-intercepts and the solutions of f x 0 are the same.
10
−4
12
0 x2 8x 20
0 x 2x 10
−40
x 2 0 ⇒ x 2
x 10 0 ⇒ x 10
The x-intercepts and the solutions of f x 0 are the same.
Section 3.1
61. f x 2x2 7x 30
x-intercepts:
52,
0, 6, 0
62. f x 4x2 25x 21
10
−5
10
0 2x 5)x 6
x
10
−9
0 x 74x 3
−40
2
− 70
x 7 0 ⇒ x 7
x6
or
x-intercepts: 7, 0, 0
3
4,
305
0 4x2 25x 21
0 2x2 7x 30
52
Quadratic Functions and Models
The x-intercepts and the solutions of f x 0 are the same.
4x 3 0 ⇒ x 43
The x-intercepts and the solutions of f x 0 are the same.
63. f x 12x2 6x 7
7
64. f x 10 x2 12x 45
10
x-intercepts: 15, 0, 3, 0
x-intercepts: 1, 0, 7, 0
0 12x2 6x 7
0 x2 6x 7
−10
14
−6
or
7 2
0 10
x 12x 45
0 x 15x 3
The x-intercepts and the solutions of f x 0 are the same.
opens upward
The x-intercepts and the solutions of f x 0 are the same.
66. f x x 5x 5
x 1x 3
x 5x 5
x2 2x 3
x2 25, opens upward
gx x 1x 3
− 60
x30 ⇒ x3
x7
65. f x x 1x 3
4
x 15 0 ⇒ x 15
0 x 1x 7
x 1
10
−18
opens downward
gx f x, opens downward
x 1x 3
gx x2 25
x2 2x 3
Note: f x a x2 25 has x-intercepts 5, 0 and
5, 0 for all real numbers a 0.
x2 2x 3
Note: f x ax 1x 3 has x-intercepts 1, 0
and 3, 0 for all real numbers a 0.
67. f x x 0x 10
opens upward
x2 12x 32, opens upward
x2 10x
gx x 0x10
x2
opens downward
gx f x, opens downward
gx x2 12x 32
10x
Note: f x ax 0x 10 axx 10 has
x-intercepts 0, 0 and 10, 0 for all real numbers
a 0.
1
69. f x x 3x 2 2
68. f x x 4x 8
opens upward
Note: f x a x 4x 8 has x-intercepts 4, 0 and
8, 0 for all real numbers a 0.
5
70. f x 2x 2 x 2
x 3x 12 2
2x 52 x 2
x 32x 1
2x2 12 x 5
2x2 7x 3
2x2 x 10, opens upward
gx 2x2 7x 3
opens downward
2x2 7x 3
Note: f x ax 32x 1 has x-intercepts
3, 0 and 12, 0 for all real numbers a 0.
gx f x, opens downward
gx 2x2 x 10
5
5
Note: f x ax 2 x 2 has x-intercepts 2, 0
and 2, 0 for all real numbers a 0.
306
Chapter 3
Polynomial Functions
71. Let x the first number and y the second number.
Then the sum is
72. Let x first number and y second number.
Then, x y S, y S x. The product is
x y 110 ⇒ y 110 x.
Px xy xS x.
The product is Px xy x110 x 110x x2.
Px Sx x2
Px x 110x
2
x2 Sx
x2 110x 3025 3025
x2 Sx x 552 3025
x 552 3025
x
The maximum value of the product occurs at the vertex of
Px and is 3025. This happens when x y 55.
73. Let x the first number and y the second number.
Then the sum is
1
1
x 212 441 x 212 147
3
3
The maximum value of the product occurs at the vertex
of Px and is 147. This happens when x 21 and
42 21
y
7. Thus, the numbers are 21 and 7.
3
(b)
y
4
8
8x50 x
50 x x50 x 3
3
3
2000
0
60
0
This area is maximum when x 25 feet and
1
y 100
3 33 3 feet.
—CONTINUED—
This area is maximum when
1
x 25 feet and y 100
3 33 3 feet.
x
A
5
600
10
10663
15
1400
20
1600
25
16663
30
1600
x
1
4
4x 3y 200 ⇒ y 200 4x 50 x
3
3
42 3 x.
1
x2 42x 441 441
3
The maximum value of the product occurs at the vertex
of Px and is 72. This happens when x 12 and
y 24 122 6. Thus, the numbers are 12 and 6.
75. (a)
42 x
.
3
1
Px x2 42x
3
1
1
x 122 144 x 122 72
2
2
(c)
S2
4
The product is Px xy x
1
x2 24x 144 144
2
Then the sum is x 3y 42 ⇒ y 1
Px x2 24x
2
A 2xy 2x
2
74. Let x the first number and y the second number.
24 x
The product is Px xy x
.
2
x
The maximum value of the product occurs at the vertex
of Px and is S 24. This happens when x y S2.
24 x
x 2y 24 ⇒ y .
2
S
2
S2 S2
4
4
2
2
Section 3.1
Quadratic Functions and Models
75. — CONTINUED —
8
(d) A x50 x
3
(e) They are all identical.
x 25 feet and y 3313 feet
8
x2 50x
3
8
x2 50x 625 625
3
8
x 252 625
3
8
5000
x 252 3
3
The maximum area occurs at the vertex and is 50003
square feet. This happens when x 25 feet and
y 200 4253 1003 feet. The dimensions
are 2x 50 feet by 3313 feet.
1
76. (a) Radius of semicircular ends of track: r 2 y
Distance around two semicircular parts of track:
(c) Area of rectangular region:
A xy x
1
d 2 r 2 y y
2
(b) Distance traveled around track in one lap:
d y 2x 200
1
200x 2x2
2
x2 100x
y 200 2x
y
200 2x
2
x2 100x 2500 2500
200 2x
2
5000
x 502 The area is maximum when x 50 and
y
200 250 100
.
4
24
77. y x2 x 12
9
9
The vertex occurs at 78. y b
4
24
249
3. The maximum height is y3 32 3 12 16 feet.
2a 249
9
9
16 2 9
x x 1.5
2025
5
(a) The ball height when it is punted is the y-intercept.
y
16
9
02 0 1.5 1.5 feet
2025
5
(b) The vertex occurs at x The maximum height is f
b
95
3645
.
2a
2162025
32
16 3645
3645
32 2025 32 —CONTINUED—
2
9 3645
1.5
5 32
6561 13,122 96 6657
6561 6561
1.5 feet 104.02 feet.
64
32
64
64
64
64
307
308
Chapter 3
Polynomial Functions
78. —CONTINUED—
(c) The length of the punt is the positive x-intercept.
0
x
16 2 9
x x 1.5
2025
5
95 ± 952 41.5162025 1.8 ± 1.81312
322025
0.01580247
x 0.83031 or x 228.64
The punt is approximately 228.64 ft.
79. C 800 10x 0.25x2 0.25x2 10x 800
The vertex occurs at x 10
b
20.
2a
20.25
The cost is minimum when x 20 fixtures.
81. P 0.0002x2 140x 250,000
The vertex occurs at x 140
b
350,000.
2a
20.0002
The profit is maximum when x 350,000 units.
83. R p 25p 2 1200p
80. C 100,000 110x 0.045x2
The vertex occurs at x 110
1222.
20.045
The cost is minimum when x 1222 units.
82. P 230 20x 0.5x2
The vertex occurs at x b
20
20.
2a
20.5
Because x is in hundreds of dollars, 20 100 2000
dollars is the amount spent on advertising that gives
maximum profit.
84. R p 12p2 150p
(a) R20 $14,000 thousand
(a) R$4 12$42 150$4 $408
R25 $14,375 thousand
R$6 12$62 150$6 $468
R30 $13,500 thousand
R$8 12$82 150$8 $432
(b) The revenue is a maximum at the vertex
b
1200
24
2a 225
(b) The vertex occurs at
p
b
150
$6.25
2a
212
R24 14,400
Revenue is maximum when price $6.25 per pet.
The unit price that will yield a maximum revenue of
$14,400 thousand is $24.
The maximum revenue is
f $6.25 12$6.252 150$6.25 $468.75.
85. C 4299 1.8t 1.36t 2, 0 ≤ t ≤ 43
(a)
(b) Vertex 0, 4299
5000
0
43
0
(c) C 40 2051
Annually:
Daily:
209,128,0942051
8879 cigarettes
48,308,590
8879
24 cigarettes
366
The vertex occurs when y 4299 which is the
maximum average annual consumption. The warnings
may not have had an immediate effect, but over time
they and other findings about the health risks and the
increased cost of cigarettes have had an effect.
Section 3.1
86. (a) and (c)
87. (a)
Quadratic Functions and Models
309
25
950
0
100
−5
4
650
12
(b) 0.002s2 0.005s 0.029 10
(b) y 4.303x 49.948x 886.28
2
2s2 5s 29 10,000
(d) 1996
2s2 5s 10,029 0
(e) Vertex occurs at
a 2, b 5, c 10,029
b
49.948
x 5.8
2a 24.303
Minimum occurs at year 1996.
(f) x 18
s
5 ± 52 4210,029
22
s
5 ± 80,257
4
y 4.303182 49.94818 886.28 1381.388
s 72.1, 69.6
There will be approximately 1,381,000 hairdressers
and cosmetologists in 2008.
The maximum speed if power is not to exceed
10 horsepower is 69.6 miles per hour.
(b) y 0.0082x2 0.746x 13.47
88. (a) and (c)
(d) The maximum of the graph is at x 45.5, or about 45.5 mi/h.
Algebraically, the maximum occurs at
31
x
10
0.746
b
45.5 mi/h.
2a 20.0082
80
20
90. True. The vertex of f x is 4,
5
71
is 4, 4 .
5 53
4
89. True. The equation 12x2 1 0 has no real solution,
so the graph has no x-intercepts.
91. f x ax2 bx c
b
a x2 x c
a
b
b2
b2
a x2 x 2 2 c
a
4a
4a
a x
b
2a
a x f 2
b
2a
b2
c
4a
2
4ac b2
4a
b2
b
b
a
b c
2a
4a2
2a
b2
b2
c
4a 2a
b2 2b2 4ac 4ac b2
4a
4a
So, the vertex occurs at b 4ac b2
b
b
,
, f 2a
4a
2a
2a
.
and the vertex of gx
310
Chapter 3
Polynomial Functions
92. Conditions (a) and (d) are preferable because profits
would be increasing.
93. Yes. A graph of a quadratic equation whose vertex is
0, 0 has only one x-intercept.
94. If f x ax2 bx c has two real zeros, then by the Quadratic Formula they are
x
b ± b2 4ac
.
2a
The average of the zeros of f is
2b
b b2 4ac b b2 4ac
2a
2a
2a
b
.
2
2
2a
This is the x-coordinate of the vertex of the graph.
95. 4, 3 and 2, 1
m
96.
1
13
2
2 4
6
3
72, 2, m 23
y2
1
y 1 x 2
3
3
7
x
2
2
3
21
y2 x
2
4
1
2
y1 x
3
3
3
13
y x
2
4
1
5
y x
3
3
97. 4x 5y 10 ⇒ y 45x 2 and m 45
98. y 3x 2
The slope of the perpendicular line through 0, 3 is
m 54 and the y-intercept is b 3.
m 3
For a parallel line, m 3. So, for 8, 4, the line is
y 54x 3
y 4 3x 8
y 4 3x 24
y 3x 20.
For Exercises 99–104, let f x 14x 3, and gx 8x2.
99. f g3 f 3 g3
100. g f 2 822 142 3 32 28 3 7
143 3 832 27
74 f 74g 74
101. fg 102.
3 24
4
g 1.5 1481.5
1.5
18
3
f
2
74 3
8 74 14 11
2
1408
128
49 49
103. f g1 f g1 f 8 148 3 109
104. g f 0 g f 0 g140 3 g3
832 72
Section 3.2
Section 3.2
Polynomial Functions of Higher Degree
311
Polynomial Functions of Higher Degree
You should know the following basic principles about polynomials.
■ f x a xn a xn1 . . . a x2 a x a , a 0, is a polynomial function of degree n.
n
■
n1
2
1
(a) an > 0, then
■
n
(b) an < 0, then
1. f x → as x → .
1. f x → as x → .
2. f x → as x → .
2. f x → as x → .
If f is of even degree and
(a) an > 0, then
■
0
If f is of odd degree and
(b) an < 0, then
1. f x → as x → .
1. f x → as x → .
2. f x → as x → .
2. f x → as x → .
The following are equivalent for a polynomial function.
(a) x a is a zero of a function.
(b) x a is a solution of the polynomial equation f x 0.
(c) x a is a factor of the polynomial.
(d) a, 0 is an x-intercept of the graph of f.
■
A polynomial of degree n has at most n distinct zeros and at most n 1 turning points.
■
A factor x ak, k > 1, yields a repeated zero of x a of multiplicity k.
(a) If k is odd, the graph crosses the x-axis at x a.
(b) If k is even, the graph just touches the x-axis at x a.
■
If f is a polynomial function such that a < b and f a f b, then f takes on every value between f a and f b in the
interval a, b.
■
If you can find a value where a polynomial is positive and another value where it is negative, then there is at least one real
zero between the values.
Vocabulary Check
1. continuous
2. Leading Coefficient Test
3. n; n 1
4. solution; x a; x-intercept
5. touches; crosses
6. standard
7. Intermediate Value
1. f x 2x 3 is a line with y-intercept 0, 3. Matches
graph (c).
2. f x x2 4x is a parabola with intercepts 0, 0 and
4, 0 and opens upward. Matches graph (g).
3. f x 2x2 5x is a parabola with x-intercepts
0, 0 and 52, 0 and opens downward. Matches
graph (h).
4. f x 2x3 3x 1 has intercepts 0, 1, 1, 0,
12 123, 0 and 12 123, 0. Matches graph (f).
1
5. f x 4x4 3x2 has intercepts 0, 0 and ± 23, 0.
Matches graph (a).
6. f x 13 x 3 x 2 43 has y-intercept 0, 43 .
Matches graph (e).
7. f x x4 2x3 has intercepts 0, 0 and 2, 0.
Matches graph (d).
1
9
8. f x 5 x 5 2x 3 5 x has intercepts 0, 0, 1, 0,
1, 0, 3, 0, 3, 0. Matches graph (b).
312
Chapter 3
Polynomial Functions
9. y x3
(a) f x x 23
(b) f x x3 2
y
y
4
3
3
2
2
1
1
x
−3 −2
2
3
4
x
−4 −3 −2
2
3
4
5
−2
−3
−4
−4
−5
Horizontal shift two units to the right
(c) f x Vertical shift two units downward
(d) f x x 23 2
12x3
y
y
4
3
3
2
2
1
1
x
−4 −3 −2
2
3
4
x
−3 −2
1
2
4
5
−2
−2
−3
−3
−4
−4
−5
Reflection in the x-axis and a vertical shrink
Horizontal shift two units to the right and
a vertical shift two units downward
10. y x5
(a) f x x 15
(b) f x x5 1
y
y
4
4
3
3
2
2
1
x
− 4 −3
1
2
3
4
x
− 4 − 3 −2
1
−3
−3
−4
−4
Horizontal shift one unit to the left
(c) f x 1 2
3
4
Vertical shift one unit upward
1
(d) f x 2 x 15
1 5
2x
y
y
4
4
3
3
2
2
1
x
− 4 − 3 −2
2
3
4
x
−5 −4 −3 −2
1
−3
−3
−4
−4
Reflection in the x-axis, vertical shrink each y-value
1
is multiplied by 2 , and vertical shift one unit upward
2
3
Reflection in the x-axis, vertical shrink each y-value is
1
multiplied by 2 , and horizontal shift one unit to the left
Section 3.2
Polynomial Functions of Higher Degree
313
11. y x4
(a) f x x 34
(b) f x x4 3
y
y
6
4
5
3
4
2
1
3
2
1
2
3
4
x
−5 −4 −3 −2 −1
1
2
3
−2
−4
Horizontal shift three units to the left
(c) f x 4 x
− 4 − 3 −2
Vertical shift three units downward
(d) f x 12x 14
x4
y
y
6
5
3
2
1
− 4 − 3 −2
x
1
−1
2
3
x
−4 −3 −2 −1
4
1
2
3
4
−2
−2
Reflection in the x-axis and then a vertical
shift four units upward
Horizontal shift one unit to the right and a vertical shrink
each y-value is multiplied by 12 (f) f x 12x 2
4
(e) f x 2x4 1
y
y
6
6
5
5
4
3
2
1
x
− 4 −3 − 2 − 1
−1
1
2
3
−4 −3
4
x
−1
−1
1
3
4
−2
Vertical shift one unit upward and a horizontal shrink
each y-value is multiplied by 12 Vertical shift two units downward and a horizontal stretch
each y-value is multiplied by 12 12. y x 6
1
(a) f x 8 x6
(b) f x x 2 6 4
y
y
4
3
2
1
−4 −3 −2
x
−1
2
3
−5 −4
4
x
−2
1
2
3
−2
−3
−4
−4
Vertical shrink each y-value is multiplied by 8 and
reflection in the x-axis
1
—CONTINUED—
Horizontal shift two units to the left and vertical shift four
units downward
314
Chapter 3
Polynomial Functions
12. —CONTINUED—
(c) f x x 6 4
1
(d) f x 4 x 6 1
y
y
4
4
3
3
2
2
1
x
−4 −3 −2
2
3
4
−4 −3 −2
x
2
−1
3
4
−2
−3
−4
Vertical shift four units downward
(e) f x 1 6
4x
2
Reflection in the x-axis, vertical shrink each y-value is
1
multiplied by 4 , and vertical shift one unit upward
(f) f x 2x6 1
y
y
−8 − 6
−2
x
2
6
8
− 4 −3 − 2 − 1
−1
−4
Horizontal stretch (each x-value is multiplied by 4),
and vertical shift two units downward
2
3
4
−2
Horizontal shrink each x-value is multiplied by 2 , and
vertical shift one unit downward
1
14. f x 2x2 3x 1
1
13. f x 3x3 5x
Degree: 2
Degree: 3
Leading coefficient:
x
1
1
3
The degree is odd and the leading coefficient is positive.
The graph falls to the left and rises to the right.
15. gx 5 72x 3x2
Leading coefficient: 2
The degree is even and the leading coefficient is positive.
The graph rises to the left and rises to the right.
16. hx 1 x6
Degree: 2
Degree: 6
Leading coefficient: 3
Leading coefficient: 1
The degree is even and the leading coefficient is negative.
The graph falls to the left and falls to the right.
The degree is even and the leading coefficient is negative.
The graph falls to the left and falls to the right.
17. f x 2.1x5 4x3 2
18. f x 2x5 5x 7.5
Degree: 5
Degree: 5
Leading coefficient: 2.1
Leading coefficient: 2
The degree is odd and the leading coefficient is negative.
The graph rises to the left and falls to the right.
The degree is odd and the leading coefficient is positive.
The graph falls to the left and rises to the right.
19. f x 6 2x 4x2 5x3
20. f x 3x4 2x 5
4
Degree: 3
Degree: 4
Leading coefficient: 5
Leading coefficient:
The degree is odd and the leading coefficient is negative.
The graph rises to the left and falls to the right.
The degree is even and the leading coefficient is positive.
The graph rises to the left and rises to the right.
3
4
Section 3.2
Polynomial Functions of Higher Degree
7
22. f s 8 s3 5s2 7s 1
2
21. ht 3t2 5t 3
Degree: 3
Degree: 2
Leading coefficient:
7
Leading coefficient: 8
23
The degree is even and the leading coefficient is
negative. The graph falls to the left and falls to the right.
23. f x 3x3 9x 1; gx 3x3
The degree is odd and the leading coefficient is negative.
The graph rises to the left and falls to the right.
24. f x 13 x3 3x 2, gx 13 x3
6
8
g
f
g
f
−4
−9
4
9
−8
−6
25. f x x4 4x3 16x; gx x4
26. f x 3x 4 6x 2, gx 3x 4
5
12
−8
f
g
8
−6
g
6
f
−3
−20
27. f x x2 25
28. (a) f x 49 x2
(a) 0 x 2 25 x 5x 5
0 7 x7 x
Zeros: x ± 5
x ± 7, both with multiplicity 1
(b) Each zero has a multiplicity of 1. (odd multiplicity)
(b) Multiplicity of x 7is 1.
Multiplicity of x 7 is 1.
Turning point: 1 (the vertex of the parabola)
(c)
There is one turning point.
10
−30
30
(c)
55
−30
−30
30
−5
29. ht t 2 6t 9
30. (a) f x x2 10x 25
(a) 0 t2 6t 9 t 32
0 x 5 2
Zero: t 3
x 5, with multiplicity 2
(b) t 3 has a multiplicity of 2 (even multiplicity).
(b) The multiplicity of x 5 is 2.
Turning point: 1 (the vertex of the parabola)
(c)
315
(c)
4
−18
There is one turning point.
25
18
−25
−20
15
−5
316
Chapter 3
Polynomial Functions
31. f x 13x2 13x 23
(a) 0 13 x 2 13 x 23
(c)
13 x 2 x 2
4
−6
3 x 2x 1
1
6
Zeros: x 2, x 1
−4
(b) Each zero has a multiplicity of 1 (odd multiplicity).
Turning point: 1 (the vertex of the parabola)
1
5
3
32. (a) f x x 2 x 2
2
2
(b) The multiplicity of
3
1
5
a ,b ,c
2
2
2
x
5
2 ±
522 4 12 32 1
5
±
2
The multiplicity of
5 37
is 1.
2
5 37
is 1.
2
There is one turning point.
(c)
37
4
3
−8
4
5 ± 37
, both with multiplicity 1
2
−5
33. f x 3x 3 12x 2 3x
(a) 0 3x3 12 x 2 3x 3xx 2 4x 1
(c)
8
−6
Zeros: x 0, x 2 ± 3 (by the Quadratic
Formula)
6
(b) Each zero has a multiplicity of 1 (odd multiplicity).
−24
Turning points: 2
34. (a) gx 5x x 2 2x 1
(b) The multiplicity of x 0 is 1.
0 5x x 2 2x 1
The multiplicity of x 1 2 is 1.
0 x
The multiplicity of x 1 2 is 1.
x2
2x 1
For x2 2x 1, a 1, b 2, c 1.
x
2 ± 2 2 411
21
There are two turning points.
(c)
12
−1
2 ± 8
2
3
−16
1 ± 2
The zeros are 0, 1 2, and 1 2, all with
multiplicity 1.
35. f t t3 4t2 4t
(a) 0 t 3 4t 2 4t tt 2 4t 4 t t 22
(c)
5
Zeros: t 0, t 2
(b) t 0 has a multiplicity of 1 (odd multiplicity)
t 2 has a multiplicity of 2 (even multiplicity)
Turning points: 2
−7
8
−5
Section 3.2
36. (a) f x x 4 x 3 20x 2
0 x 2 x 2 x 20
Polynomial Functions of Higher Degree
37. gt t5 6t3 9t
(a) 0 t 5 6t 3 9t t t 4 6t 2 9 t t 2 32
t t 3 t 3 2
0 x 2 x 4x 5
x 0, 4, 5
(b) t 0 has a multiplicity of 1 (odd multiplicity).
(b) The multiplicity of x 0 is 2.
t ± 3 each have a multiplicity of 2
(even multiplicity)
The multiplicity of x 5 is 1.
Turning points: 4
The multiplicity of x 4 is 1.
There are three turning points.
(c)
25
−6
2
Zeros: t 0, t ± 3
0 with multiplicity 2, 4 and 5 with multiplicity 1.
(c)
317
6
−9
9
6
−6
−150
39. f x 5x4 15x2 10
38. (a) f x x 5 x 3 6x
0 x x 4 x 2 6
(a) 0 5x 4 15x 2 10
5x 4 3x2 2
0 x x 2 3x 2 2
5x 2 1x 2 2
x 0, ± 2, all with multiplicity 1
(b) The multiplicity of x 0 is 1.
No real zeros
The multiplicity of x 2 is 1.
(b) Turning point: 1
The multiplicity of x 2 is 1.
(c)
40
There are two turning points.
(c)
6
−4
4
−5
−9
9
−6
40. (a) f x 2x4 2x2 40
0 2x 4 2x 2 40
41. gx x3 3x2 4x 12
(a) 0 x 3 3x 2 4x 12 x 2 x 3 4x 3
0 2x2 4x 5 x 5 x 2 4x 3 x 2x 2x 3
x ± 5, both with multiplicity 1
(b) The multiplicity of x 5 is 1.
Zeros: x ± 2, x 3
(b) Each zero has a multiplicity of 1 (odd multiplicity).
The multiplicity of x 5 is 1.
There is one turning point.
(c)
(c)
4
−8
20
−6
Turning points: 2
7
6
−16
−60
318
Chapter 3
Polynomial Functions
42. (a) f x x 3 4x 2 25x 100
0 x 2x 4 25x 4
43. y 4x3 20x2 25x
(a)
12
0 x 2 25x 4
0 x 5x 5x 4
−2
x ± 5, 4, all with multiplicity 1
(b) The multiplicity of x 5 is 1.
The multiplicity of x 5 is 1.
6
−4
5
(b) x-intercepts: 0, 0, 2, 0
(c) 0 4x3 20x2 25x
The multiplicity of x 4 is 1.
0 x2x 52
There are two turning points.
x 0 or x 2
(c)
140
−9
5
(d) The solutions are the same as the x-coordinates
of the x-intercepts.
9
−20
45. y x5 5x3 4x
44. y 4x 3 4x 2 8x 8
(a)
(a)
2
−3
4
3
−6
6
−4
−11
(b) (1, 0, 1.414214, 0, 1.414214, 0
(b) x-intercepts: 0, 0, ± 1, 0, ± 2, 0
(c) 0 4x 3 4x 2 8x 8
(c) 0 x5 5x3 4x
0 4x2x 1 8x 1
0 xx2 1x2 4
0 4x 2 8 x 1
0 xx 1x 1x 2x 2
0 4
x 0, ± 1, ± 2
x2
2x 1
x ± 2, 1
(d) The intercepts match part (b).
(d) The solutions are the same as the x-coordinates
of the x-intercepts.
46. y 14 x 3 x 2 9
(a)
(c) 0 14 x 3x 2 9
12
x 0, ± 3
−18
18
−12
x-intercepts: 0, 0, ± 3, 0
(d) The intercepts match part (b).
(b) 0, 0, 3, 0, 3, 0
47. f x x 0x 10
48. f x x 0x 3
f x x2 10x
xx 3
Note: f x ax 0x 10 axx 10
has zeros 0 and 10 for all real numbers a 0.
x 2 3x
Note: f x axx 3 has zeros 0 and 3 for all real
numbers a.
Section 3.2
49. f x x 2x 6
Polynomial Functions of Higher Degree
50. f x x 4x 5
f x x 2x 6
x 4x 5
f x x2 4x 12
x 2 x 20
Note: f x ax 2x 6 has zeros 2 and 6
for all real numbers a 0.
51. f x x 0x 2x 3
Note: f x a x 4x 5 has zeros 4 and 5 for all
real numbers a.
52. f x x 0x 2x 5
xx 2x 3
xx 2x 5
x x 2 7x 10
x3
6x
5x2
319
Note: f x axx 2x 3 has zeros 0, 2, 3 for
all real numbers a 0.
53. f x x 4x 3x 3x 0
x 3 7x 2 10x
Note: f x a xx 2x 5 has zeros 0, 2, 5 for all
real numbers a.
54. f x x 2x 1x 0x 1x 2
x 4x2 9x
xx 2x 1x 1x 2
x4 4x3 9x2 36x
xx 2 4x 2 1
Note: f x ax4 4x3 9x2 36x has these zeros
for all real numbers a 0.
xx 4 5x 2 4
x 5 5x 3 4x
Note: f x a xx 2x 1x 1x 2 has
zeros 2, 1, 0, 1, 2 for all real numbers a.
55. f x x 1 3 x 1 3 x 1 3 x 1 3
56. f x x 2 x 4 5 x 4 5 x 2 x 4 5x 4 5
x 1 3 x 2 x 4 2 5
x2 2x 1 3
xx 4 2 5x 2x 4 2 10
x2 2x 2
x 3 8x 2 16x 5x 2x 2 16x 32 10
2
2
Note: f x a
numbers a 0.
x2
2x 2 has these zeros for all real
x 3 10x 2 27x 22
Note: f x ax 3 10x 2 27x 22 has these zeros
for all real numbers a.
57. f x x 2x 2
58. f x x 8x 4
x 22 x2 4x 4
x 8x 4 x2 12x 32
Note: f x ax2 4x 4, a 0, has degree 2 and
zero x 2.
59. f x x 3x 0x 1
60. f x x 2x 4x 7
xx 3x 1 x 2x 3x
3
Note: f x ax2 12x 32, a 0, has degree 2 and
zeros x 8 and 4.
2
Note: f x ax3 2x2 3x, a 0, has degree 3 and
zeros x 3, 0, 1.
61. f x x 0x 3 x 3 xx 3x 3 x3 3x
Note: f x ax3 3x, a 0, has degree 3 and zeros
x 0, 3, 3.
x 2x2 11x 28 x 3 9x2 6x 56
Note: f x ax 3 9x2 6x 56, a 0, has degree
3 and zeros x 2, 4, and 7.
62. f x x 93 x 3 27x2 243x 729
Note: f x ax 3 27x2 243x 729, a 0, has
degree 3 and zero x 9.
320
Chapter 3
Polynomial Functions
63. f x x 52x 1x 2 x 4 7x3 3x2 55x 50
or f x x 5x 12x 2 x 4 x3 15x2 23x 10
or f x x 5x 1x 22 x4 17x2 36x 20
Note: Any nonzero scalar multiple of these functions would also have degree 4 and zeros x 5, 1, 2.
64. f x x 4x 1x 3x 6 x 4 4x 3 23x2 54x 72
Note: f x ax 4 4x 3 23x2 54x 72, a 0, has degree 4 and zeros x 4, 1, 3, and 6.
65. f x x4x 4 x5 4x 4
or f x x3x 42 x5 8x 4 16x3
or f x x2x 43 x5 12x4 48x3 64x2
or f x xx 44 x5 16x 4 96x3 256x2 256x
Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros x 0 and 4.
66. f x x 32x 1x 5x 6 x5 6x4 22x3 108x2 189x 270
or f x x 3x 12x 5x 6 x 5 10x 4 14x 3 88x2 183x 90
or f x x 3x 1x 52x 6 x 5 14x 4 50x 3 68x2 555x 450
or f x x 3x 1x 5x 62 x 5 15x 4 59x 3 63x2 648x 540
Note: Any nonzero multiple of these functions would also have degree 5 and zeros x 3, 1, 5, and 6.
68. gx x 4 4x2 x2x 2x 2
67. f x x3 9x xx2 9 xx 3x 3
(a) Falls to the left; rises to the right
(a) Rises to the left; rises to the right
(b) Zeros: 0, 3, 3
(b) Zeros: 2, 0, 2
(c)
x
3
2
1
0
1
2
3
f x
0
10
8
0
8
10
0
(d)
y
(c)
x
0.5
1
1.5
2.5
gx
0.94
3
3.94
14.1
y
(d)
4
12
3
(0, 0)
(−3, 0)
−12 − 8
4
2
(3, 0)
x
−4
4
8
1
(−2, 0)
−4 −3
12
−4
(0, 0)
(2, 0)
1
3
x
−1
4
−8
−4
1
1
7
69. f t 4t2 2t 15 4t 12 2
(d) The graph is a parabola with vertex 1, 72 .
(a) Rises to the left; rises to the right
y
(b) No real zero (no x-intercepts)
(c)
8
t
1
0
1
2
3
f t
4.5
3.75
3.5
3.75
4.5
6
2
−4
−2
t
2
4
Section 3.2
Polynomial Functions of Higher Degree
71. f x x3 3x2 x2x 3
70. gx x2 10x 16 x 2x 8
(a) Falls to the left; falls to the right
(a) Falls to the left; rises to the right
(b) Zeros: 2, 8
(b) Zeros: 0 and 3
(c)
(c)
x
1
3
5
7
9
gx
7
5
9
5
7
y
(d)
321
x
1
0
1
2
3
f x
4
0
2
4
0
y
(d)
1
10
(0, 0)
8
−1
(3, 0)
1
2
x
4
6
4
−3
2
(2, 0)
(8, 0)
−4
x
4
6
10
73. f x 3x3 15x2 18x 3xx 2x 3
72. f x 1 x 3
(a) Rises to the left; falls to the right
(a) Falls to the left; rises to the right
(b) Zero: 1
(b) Zeros: 0, 2, 3
(c)
(c)
x
2
1
0
1
2
f x
9
2
1
0
7
0
1
2
2.5
3
3.5
f x
0
6
0
1.875
0
7.875
y
(d)
y
(d)
x
7
3
6
5
2
4
3
2
(1, 0)
−2
−1
(0, 0) 1 (2, 0)
x
2
−1
1
4
5
6
−2
75. f x 5x2 x3 x25 x
74. f x 4x 3 4x2 15x
x4x2 4x 15
(a) Rises to the left; falls to the right
x2x 52x 3
(b) Zeros: 0, 5
(c)
(a) Rises to the left; falls to the right
(b) Zeros:
(c)
(3, 0)
x
−3 − 2 − 1
−1
32,
0,
5
2
x
3
2
1
0
1
2
3
f x
99
18
7
0
15
14
27
x
5
4
3
2
1
0
1
f x
0
16
18
12
4
0
6
y
(d)
5
(−5, 0)
−15
y
(d)
(0, 0)
− 10
5
20
16
12
8
(− 32, 0(
−4 −3 −2
( 52, 0(
4
(0, 0)
1
2
3
4
− 20
x
x
10
322
Chapter 3
Polynomial Functions
76. f x 48x2 3x 4
3x2x2 16
(a) Rises to the left; rises to the right
(d)
y
(− 4, 0)
(b) Zeros: 0, ± 4
100
(0, 0)
(c)
5 4 3
x
f x 675 0
2
1
0 1
2
3
−6
4 5
−2
(4, 0)
2
x
6
189 144 45 0 45 144 189 0 675
− 200
− 300
1
78. hx 3 x 3x 42
77. f x x2x 4
(a) Falls to the left; rises to the right
(a) Falls to the left; rises to the right
(b) Zeros: 0, 4
(b) Zeros: 0 and 4
(c)
(c)
x
1
0
1
2
3
4
5
f x
5
0
3
8
9
0
25
y
(d)
x
1
0
1
2
3
4
5
hx
3
25
0
3
32
3
9
0
125
3
y
(d)
14
2
(0, 0)
−4
−2
(4, 0)
2
6
12
x
10
8
8
6
4
(0, 0)
−4 −2
79. g t 14t 22t 22
(4, 0)
2
4
6
x
8 10 12
1
80. gx 10 x 12x 33
(a) Falls to the left; falls to the right
(a) Falls to the left; rises to the right
(b) Zeros: 2 and 2
(b) Zeros: 1, 3
(c)
t
3
gt
25
4
2
1
0
94
y
(d)
−3
−1
1
4
94
2
3
0
25
4
(c)
x
2
1
0
1
2
4
g x
12.5
0
2.7
3.2
0.9
2.5
y
(d)
(2, 0)
(−2, 0)
0
t
−1
−2
1
2
6
3
4
(−1, 0)
−6 −4 −2
−5
−6
2
(3, 0)
4
6
x
8
Section 3.2
81. f x x3 4x xx 2x 2
Polynomial Functions of Higher Degree
1
82. f x 4x 4 2x 2
6
−9
6
−9
9
9
−6
−6
Zeros: 0, 2, 2 all of multiplicity 1
Zeros: 2.828 and 2.828 with multiplicity 1; 0, with
multiplicity 2
84. h x 5x 2 2 3x 5 2
1
1
83. gx 5x 12x 32x 9
21
14
−12
323
18
− 12
12
−3
−6
Zeros: 2, 53, both with multiplicity 2
Zeros: 1 of multiplicity 2, 3 of multiplicity 1, and
9
2 of multiplicity 1
85. f x x3 3x2 3
86. f x 0.11x 3 2.07x 2 9.81x 6.88
x
y1
3
51
2
17
1
1
0
3
The function has three zeros.
They are in the intervals
1, 0, 1, 2 and 2, 3.
1
1
2
1
They are x 0.879, 1.347, 2.532 .
3
3
4
19
10
−5
5
−10
87. gx 3x4 4x3 3
10
−5
5
−10
x
y1
4
509
3
132
2
13
The function has three zeros. They are in the intervals
0, 1, 6, 7, and 11, 12. They are approximately 0.845,
6.385, and 11.588.
x
y
x
y
10
−4
16
0
6.88
7
1.91
1
0.97
8
4.56
2
5.34
9
6.07
3
6.89
10
5.78
4
6.28
11
3.03
5
4.17
12
2.84
6
1.12
−10
88. h x x 4 10x 2 3
The function has four zeros. They are
in the intervals 4, 3, 1, 0, 0, 1,
and 3, 4. They are approximately
± 3.113 and ± 0.556.
x
y
4
99
3
6
2
21
1
6
0
3
1
6
10
1
4
The function has two zeros.
They are in the intervals
2, 1 and 0, 1.
0
3
1
4
They are x 1.585, 0.779 .
2
77
2
21
3
348
3
6
4
99
−4
4
−30
324
Chapter 3
Polynomial Functions
89. (a) Volume l w
h
(c)
height x
Box
Height
Box
Width
Box
Volume, V
length width 36 2x
1
36 21
136 212 1156
Thus, Vx 36 2x36 2xx x36 2x2.
2
36 22
236 222 2048
3
36 23
336 232 2700
4
36 24
436 242 3136
5
36 25
536 252 3380
6
36 26
636 262 3456
7
36 27
736 272 3388
(b) Domain: 0 < x < 18
The length and width must be positive.
(d)
3600
0
18
0
The volume is a maximum of 3456 cubic inches when the
height is 6 inches and the length and width are each
24 inches. So the dimensions are 6 24 24 inches.
The maximum point on the graph occurs at x 6.
This agrees with the maximum found in part (c).
90. (a) Volume w
h 24 2x24 4xx
(c)
212 x 46 xx
720
600
8x12 x6 x
(b) x > 0,
12 x > 0,
6x > 0
x < 12
x < 6
V
480
360
240
120
Domain: 0 < x < 6
x
1
2
3
4
5
6
x 2.6 corresponds to a maximum of about
665 cubic inches.
91. (a) A w 12 2xx 2x 2 12x square inches
(e)
4000
(b) 16 feet 192 inches
Vlw
h
12 2xx192
0
384x 2 2304x cubic inches
Maximum: 3, 3456
(c) Since x and 12 2x cannot be negative, we have
0 < x < 6 inches for the domain.
(d)
x
V
0
0
1
1920
2
3072
3
3456
4
3072
5
1920
6
0
When x 3, the volume is a
maximum with V 3456 in.3. The
dimensions of the gutter cross-section
are 3 inches 6 inches 3 inches.
6
0
The maximum value is the same.
(f) No. The volume is a product of the constant length and
the cross-sectional area. The value of x would remain the
same; only the value of V would change if the length was
changed.
Section 3.2
4
92. (a) V 3 r 3 r 24r
V
4
3
3 r
4r
325
(b) r ≥ 0
3
(d) V 120 ft 3 16
3 r
3
3
16
3 r
r 1.93 ft
length 4r 7.72 ft
150
(c)
Polynomial Functions of Higher Degree
0
2
0
93. y1 0.139t 3 4.42t 2 51.1t 39
94. y 0.056t 3 1.73t 2 23.8t 29
180
200
7
140
7
120
13
The data fit the model closely.
The model is a good fit to the actual data.
95. Midwest: y118 $259.368 thousand $259,368
13
96. Answers will vary.
South: y218 $223.472 thousand $223,472
Example: The median price of homes in the South are
all lower than those in the Midwest. The curves do
not intersect.
Since the models are both cubic functions with positive
leading coefficients, both will increase without bound
as t increases, thus should only be used for short term
projections.
97. G 0.003t 3 0.137t 2 0.458t 0.839, 2 ≤ t ≤ 34
(a)
y 0.009t 2 0.274t 0.458
(c)
60
− 10
45
−5
(b) The tree is growing most rapidly at t 15.
98. R 1
3
100,000 x
600x 2
The point of diminishing returns (where the graph
changes from curving upward to curving downward)
occurs when x 200. The point is 200, 160 which
corresponds to spending $2,000,000 on advertising to
obtain a revenue of $160 million.
b
0.274
15.222
2a 20.009
y15.222 2.543
Vertex 15.22, 2.54
(d) The x-value of the vertex in part (c) is approximately
equal to the value found in part (b).
99. False. A fifth degree polynomial can have at most four
turning points.
100. True. f x x 16 has one repeated solution.
101. True. A polynomial of degree 7 with a negative leading
coefficient rises to the left and falls to the right.
102. (a) Degree: 3
Leading coefficient: Positive
(c) Degree: 4
Leading coefficient: Positive
(b) Degree: 2
Leading coefficient: Positive
(d) Degree: 5
Leading coefficient: Positive
326
Chapter 3
Polynomial Functions
103. f x x4; f x is even.
y
(a) gx f x 2
5
4
Vertical shift two units upward
3
gx f x 2
2
1
f x 2
−3
gx
−2
−1
x
−1
1
2
3
Even
(b) gx f x 2
(c) gx f x x4 x 4
Horizontal shift two units to the left
Reflection in the y-axis. The graph looks the same.
Neither odd nor even
Even
1 4
(e) gx f 12 x 16
x
(d) gx f x x 4
Reflection in the x-axis
Horizontal stretch
Even
Even
(f) gx f x 1
2
(g) gx f x 34 x 344 x 3, x ≥ 0
1 4
2x
Vertical shrink
Neither odd nor even
Even
(h) gx f f x f f x f x 4 x 44 x16
Even
1
104. (a) y1 3x 25 1 is decreasing.
y2 35x 25 3 is increasing.
8
(c) Hx x5 3x3 2x 1
Since Hx is not always increasing or always decreasing,
Hx cannot be written in the form ax h5 k.
6
−12
12
y1
y2
−9
9
−8
(b) The graph is either always increasing or always
decreasing. The behavior is determined by a. If
a > 0, gx will always be increasing. If a < 0, gx
will always be decreasing.
105. 5x2 7x 24 5x 8x 3
−6
106. 6x3 61x2 10x x6x2 61x 10
x6x 1x 10
107. 4x 4 7x 3 15x 2 x 24x 2 7x 15
108. y3 216 y3 63
x24x 5x 3
109.
2x2 x 28 0
110. 3x2 22x 16 0
2x 7x 4 0
2x 7 0 ⇒ x y 6 y2 6y 36
3x 2x 8 0
72
x40 ⇒ x4
3x 2 0
x
23
or
x80
or
x8
Section 3.2
Polynomial Functions of Higher Degree
111. 12x2 11x 5 0
112. x2 24x 144 0
3x 14x 5 0
x 122 0
3x 1 0 ⇒ x 13
4x 5 0 ⇒ x x 12 0
54
x 12
114. x2 8x 2 0
x2 2x 21 0
113.
x2 2x 12 21 1 0
x2 8x 2
x 12 22 0
x2 8x 16 2 16
x 42 14
x 12 22
x 4 ± 14
x 1 ± 22
x 1 ± 22
x 4 ± 14
116. 3x2 4x 9 0
2x2 5x 20 0
115.
5
2 x2 x 20 0
2
4
x2 x 3 0
3
20 258 0
4
x2 x 3
3
5
5
2 x2 x 2
4
2
2 x
5
4
2
4
4
4
x2 x 3 3
9
9
185
0
8
x 45
x
2
x 32
185
16
185
5
±
4
4
x
x
5 ± 185
4
2
117. f x x 42
2 31
±
3
3
2 ± 31
3
118. f x 3 x 2
7
y
4
Reflection in the x-axis
and vertical shift of three
units upward of y x2
6
5
Transformation: Horizontal
shift four units to the left
319
x
y
x2
31
9
2
±
3
x
Common function: y 4
2
1
3
− 4 −3
2
− 7 − 6 − 5 −4 − 3 − 2 − 1
−1
119. f x x 1 5
−2
−1
x
−1
−2
−3
−5
2
4
−4
120. f x 7 x 6
1
3
−3
1
1
−3
1
−2
x
y
Common function: y x
x
−1
−1
1
Transformation: Horizontal
shift one unit to the left and
a vertical shift five units
downward
327
3
y
15
Horizontal shift of six units
to the right, reflection in the
x-axis, and vertical shift of
seven units upward
of y x
12
9
6
3
−3
x
−3
3
6
9
12
15
328
Chapter 3
Polynomial Functions
121. f x 2x 9
1
122. f x 10 3 x 3
y
Common function: y x
6
Transformation: Vertical
stretch each y-value is
multiplied by 2, then a
vertical shift nine units
upward
4
5
3
2
1
−6
−3 −2 −1
x
1
2
Horizontal shift of three units
to the left, vertical shrink
y-value is multiplied
each
by 13 , reflection in the x-axis
and vertical shift of ten
units upward of y x
−1
−2
Section 3.3
y
9
8
7
6
5
4
3
2
1
x
1 2 3 4 5 6 7 8 9
Polynomial and Synthetic Division
You should know the following basic techniques and principles of polynomial division.
■
The Division Algorithm (Long Division of Polynomials)
■
Synthetic Division
■
f k is equal to the remainder of f x divided by x k. (The Remainder Theorem)
■
f k 0 if and only if x k is a factor of f x.
Vocabulary Check
1. f x is the dividend; dx is the divisor; gx is the quotient; rx is the remainder
2. improper; proper
1. y1 3. synthetic division
x2
4
and y2 x 2 x2
x2
4. factor
2. y1 5. remainder
x4 3x2 1
39
and y2 x2 8 2
x2 5
x 5
x2
x2)
x2
x2 8
0x 0
x2
5)
x2 2x
3. y1 3x2 1
x4 5x2
2x 0
8x2 1
2x 4
8x2 40
4
39
x2
4
x2
and y1 y2.
x2
x2
Thus,
x4
x5 3x3
4x
and y2 x3 4x 2
x2 1
x 1
(a) and (b)
6
Thus,
x4 3x2 1
39
x2 8 2
and y1 y2.
x2 5
x 5
x3 4x
(c) x2 0x 1 ) x5 0x 4 3x3 0x2 0x 0
x5 0x 4 x3
4x3 0x2 0x
−9
4x3 0x2 4x
9
4x 0
−6
Thus,
x5 3x3
4x
and y1 y2.
x3 4x 2
x2 1
x 1
Section 3.3
4. y1 Polynomial and Synthetic Division
x 3 2x 2 5
2x 4
and y2 x 3 2
x2 x 1
x x1
329
x3
(c) x 2 x 1 ) x 3 2x 2 0x 5
(a) and (b)
x3 x2 x
8
3x 2 x 5
−12
12
3x 2 3x 3
2x 8
−8
Thus,
2x 4
5.
2x 4
x 3 2x 2 5
x3 2
and y1 y2.
x2 x 1
x x1
5x 3
6.
x 4 ) 5x2 17x 12
x 3 ) 2x2 10x 12
2x2 6x
5x2 20x
4x 12
3x 12
4x 12
3x 12
0
0
2x2 10x 12
2x 4
x3
5x2
x2 3x 1
7.
5x2
12x2
2x2 4x 3
8.
4x 5 ) 4x3 7x2 11x 5
4x3
17x 12
5x 3
x4
3x 2 )
6x3
6x3 4x2
11x
12x2 17x
12x2 15x
12x2 8x
4x 5
4x3
4x 5
9x 6
0
9x 6
11x 5
x2 3x 1
4x 5
7x2
x3 3x2
9.
x2)
x4
5x3
6x2
6x2
1
x2
x4 2x3
3x3
x2 7x 18
10.
x3)
x3
4x2 3x 12
7x2 21x
x2
18x 12
0
18x 54
x2
x3 3x2 1
x2
6x2
6x3 16x2 17x 6
2x2 4x 3
3x 2
7x2 3x
x2
5x3
0
x3 3x2
3x3 6x2
x4
16x2 17x 6
42
x3 4x2 3x 12
42
x2 7x 18 x3
x3
330
Chapter 3
Polynomial Functions
7
11.
12.
4
x 2 ) 7x 3
2x 1 ) 8x 5
7x 14
8x 4
11
9
11
7x 3
7
x2
x2
8x 5
9
4
2x 1
2x 1
3x 5
13.
14.
x
x2 0x1 ) x3 0x2 0x 9
2x2 0x 1 ) 6x3 10x2 x 8
6x3 0x2 3x
x3 0x2 x
10x2 2x 8
x 9
10x2 0x 5
x9
x 9
x 2
x2 1
x 1
3
2x 3
6x3 10x2 x 8
2x 3
3x 5 2
2x2 1
2x 1
x2 2x 4
15.
x2 2x 3 ) x4 0x3 3x2 0x 1
x4 3x2 1
2x 11
x2 2x 4 2
x2 2x 3
x 2x 3
⇒
x4 2x3 3x2
2x3 0x2 0x
2x3 4x2 6x
4x2 6x 1
4x2 8x 12
2x 11
x2
16.
x3
17.
x 3 0x2 0x 1 ) x5 0x4 0x3 0x2 0x 7
x3 3x2 3x 1 ) x4 0x3 0x2 0x 0
x4 3x3 3x2 x
x5 0x4 0x3 x2
x2
x5 7
x2 7
x2 3
3
x 1
x 1
x2
3x3 9x2 9x 3
6x2 8x 3
x4
6x2 8x 3
x3
x 13
x 13
2x
18.
3x3 3x2 x 0
7
19. 5
3
2x 1 ) 2x3 4x2 15x 5
2x3 4x2 2x
17x 5
17x 5
2x 3 4x 2 15x 5
2x 2
x 1 2
x 2x 1
3
17
15
2
15
10
5
25
25
0
3x 17x2 15x 25
3x2 2x 5
x5
3
Section 3.3
20. 3
5
18
15
7
9
6
6
5
3
2
0
21. 2
0 16
3 16
18
0 72
12 72
3
12
2
25. 4
5
6
8
20 52
29. 8
5 10
26 44
1
13
3
0
0 120
80
48 144
432 936
250
250
1
10
25
0
6
20
0
56
8
224
14
56
232
10
50
60
0
60
0
360
800
2160
10
60
360
1360
1
16
48 144
50x3 800
10x x6
4
10x3 10x2 60x 360 312 856
120x 80
856
x 4 16x 3 48x 2 144x 312 x3
x3
13x 4
1
0
8
0
64
512
512
1
8
64
0
30. 9
3
0
6
0
12
0
24
0
48
3
6
12
24
48
1
1
180x x6
x4
1
0
9
0
81
729
729
1
9
81
0
x3 729
x2 9x 81
x9
32. 2
48
3x4
3x3 6x2 12x 24 x2
x2
33. 6
75
100
10
x3 512
x2 8x 64
x8
31. 2
0
10
6x2
27. 6
44
5x 6x 8
5x 2 10x 26 x2
x2
1
8
232
5x 5x2 14x 56 x4
x4
3
3
x5
9
5
0
10
28. 3
0
5
0
16x 2
26. 2
4
18
18
0
x3 75x 250
x2 10x 25
x 10
72
3x 3x 2 2x 12
x6
3
9
0
0
9x 3 18x 2 16x 32
9x 2 16
x2
24. 6
8
8
23. 10
9 18 16 32
18
0 32
9
4
331
4x3 8x2 9x 18
4x2 9
x2
5x 3 18x 2 7x 6
5x 2 3x 2
x3
22. 2
Polynomial and Synthetic Division
0
6
0
36
180
216
0
216
6
36
36
216
x3 6x2 36x 36 216
x6
3
0
0
6 12
0
0
24 48
3
6 12
24 48
3x 4
48
3x 3 6x 2 12x 24 x2
x2
34. 1
1
1
2
1
3
5 3x 2x x1
2
3
3
5
6
6
11
x3
x2 3x 6 11
x1
1360
x6
332
Chapter 3
1
35. 2
4
4
4x3
Polynomial Functions
16
2
23
7
15
15
14
30
0
36.
23x 15
4x2 14x 30
x 12
1
1
4
1
3
14
12
11
8
2
3
3
4
0
5
9
2
3
4
9
8
1
2
3
4
49
8
3
16x2
3x 3 4x 2 5
3
49
1
3x 2 x x 32
2
4 8x 12
38. f x x 3 5x2 11x 8, k 2
37. f x x3 x2 14x 11, k 4
4
3
2
2
1
5
2
1
7
11
14
8
6
3
2
f x x 4x 3x 2 3
f x x 2x 7x 3 2
f 4 4 4 144 11 3
f 2 23 522 112 8
2
2
3
2
8 20 22 8 2
40. f x 10x 3 22x2 3x 4, k 15
39. f x 15x4 10x3 6x2 14, k 23
23
15
6
10
0
14
10
0
4
8
3
0
6
4
34
3
15
1
5
10
f x x 23 15x3 6x 4 34
3
3
22
3
4
2
4
5
20
7
13
5
7
f x x 15 10x2 20x 7 13
5
f 15 1015 2215 315 4
f 23 15 23 10 23 6 23 14 34
3
4
10
2
3
2
2
3
65
13
25
22
25 5 4 25 5
42. f x x 3 2x2 5x 4, k 5
41. f x x3 3x2 2x 14, k 2
2
3
2
14
2
2 32
6
3 2
32
8
1
1
5
1
f x x 2x2 3 2x 32 8
2
5
4
5
25 5
10
2 5
25
6
f x x 5 x2 2 5 x 25 6
f 2 2 32 22 14 8
3
1
f 5 5 3 2 5 2 5 5 4
2
55 10 55 4 6
43. f x 4x3 6x2 12x 4, k 1 3
1 3
4
4
6
12
4
4 43
10 23
4
2 43
2 23
0
f x x 1 34x2 2 43x 2 23 0
f 1 3 41 3 61 3 121 3 4 0
3
2
f x x 2 2 3x2 2 32 x 8 42 0
44. f x 3x 3 8x2 10x 8, k 2 2
2 2
3
3
8
10
8
6 32
2 42
f 2 2 32 2 3 82 2 2 102 2 8
8
2 32
8 42
320 142 86 42 102 2 8
0
60 422 48 322 20 102 8
0
Section 3.3
45. f x 4x3 13x 10
(a) 1
4
4
(a) 2
10
9
1
f 1 1
(b) 2
4
13
16
3
0
8
8
10
6
4
4
f
10
6
4
4
10
1944
1954
10
42
32
5
9
14
3
(b)
3
(a) 1
1
96
97
3
3
10
1
1
2
6
8
83
5
3
3
3
1
1
0
2
2
3
6
14
1
0
4
4
16
0
48
3
0
192 780
2
3120
1
4
12
48
195 780
3122
1
0
3
4
9
0
15
3
45
0
144
2
432
1
3
5
15
48
144
434
1
0
1
4
1
0
3
3
3
0
0
2
0
1
1
3
3
0
0
2
0.4 1.6 0.7
0 2
0.4 1.2 0.5 0.5
0.4 1.6
0.8
0.7
0
4.8 11
2
22
0.4 2.4
5.5 11
20
f 2 20
5
6
1
10
2
8
1
16
17
(c) 5
5
15
10
7
4
3
10
50
40
1
200
199
0.7
0 2
2.0 13.5 67.5
0.4
2.7 13.5 65.5
0.4
(d) 10
0.4 1.6
4.0
0.7
0
56.0 567
2
5670
0.4 5.6
56.7 567
5668
f 10 5668
6
6
0
x 7x 6 x 2x2 2x 3
3
x 2x 3x 1
Zeros: 2, 3, 1
0.4 1.6
2.0
f 5 65.5
h5 199
49. 2
0
(b) 2
h2 17
(d) 5
0
0.4 1.2 0.5 0.5 2.5
h13 53
(c) 2
2
f 1 2.5
5
3
1
48. f x 0.4x 4 1.6x 3 0.7x 2 2
h3 97
1
3
2
12
g1 2
47. h x 3x3 5x2 10x 1
3
0
6
(d) 1
f 8 1954
(a) 3
3
0
g3 434
13
256
243
0
32
32
4
0
0
(c) 3
4
(d) 8
4
4
g4 3122
13
1
12
0
2
2
4
1
2
0
2
(b) 4
f 2 4
1
2
1
g2 14
4
(c)
333
46. g x x 6 4x 4 3x 2 2
13
4
9
0
4
4
Polynomial and Synthetic Division
50. 4
1
0 28 48
4
16
48
1
4 12
0
x 3 28x 48 x 4x 2 4x 12
x 4x 6x 2
Zeros: 4, 2, 6
334
51.
Chapter 3
1
2
Polynomial Functions
15
1
14
2
2
10
10
0
27
7
20
52.
x 2x2
48 80 41
32 32
48 48
2x3 15x2 27x 10
1
2
2
3
48x 3
80x 2
14x 20
6
6
9
0
41x 6 x 23 48x 2 48x 9
x 23 4x 312x 3
2x 1x 2x 5
3x 24x 34x 1
1
Zeros: 2, 2, 5
53. 3
Zeros:
1
3
3 23
23
2
3
2 3
1
3
2 3
3
2
1
1
6
6
0
54. 2
2 3 1
3, 4, 4
2
x3 2x2 3x 6 x 3 x 3 x 2
x3
Zeros: ± 3, 2
2
22 2
4
4
2 2
22
0
2
1
23
23
0
2
1
2x 2
1
2 2
2
22
22
1
2
0
2x 4 x 2 x 2x 2 Zeros: 2, 2, 2
55. 1 3
1
1
1 3
1
1
x3
3x2
3
1 3
2 3
0
1 3
1 3
2 3
1 3
1
1 3
1 3
0
2
2
0
2 x 1 3 x 1 3 x 1
x 1x 1 3 x 1 3 Zeros: 1, 1 ± 3
56. 2 5
2 5
1
1
2 5
13
7 35
3
3
1
1 5
6 35
0
1
1 5
2 5
6 35
6 35
1
3
0
x x 13x 3 x 2 5 x 2 5 x 3
3
2
Zeros: 2 5, 2 5, 3
57. f x 2x3 x2 5x 2; Factors: x 2, x 1
(a) 2
2
2
1
2
2
1
4
3
3
2
1
5
6
1
2
2
0
1
1
0
(b) The remaining factor of f x is 2x 1.
(c) f x 2x 1x 2x 1
(d) Zeros: 12, 2, 1
(e)
7
Both are factors of f x since the remainders are zero.
−6
6
−1
Section 3.3
Polynomial and Synthetic Division
58. f x 3x 3 2x2 19x 6; Factors: x 3, x 2
(a) 3
3
3
2
3
19
21
6
6
7
2
7
2
6 2
0
2
9
1
3
(c) f x 3x 3 2x2 19x 6
3x 1x 3x 2
(d) Zeros: 13, 3, 2
(e)
35
0
(b) The remaining factor is 3x 1.
−4
3
−10
59. f x x4 4x3 15x2 58x 40; Factors: x 5, x 4
(a) 5
4
5
1
1
1
4
1
1
15
5
10
10
12
2
1
4
3
(c) f x x 1x 2x 5x 4
40
40
0
58
50
8
(d) Zeros: 1, 2, 5, 4
(e)
8
8
0
20
−6
6
Both are factors of f x since the remainders are zero.
− 180
(b) x2 3x 2 x 1x 2
The remaining factors are x 1 and x 2.
60. f x 8x 4 14x 3 71x2 10x 24; Factors: x 2, x 4
(a) 2
8
8
4
14
16
71
60
10
22
24
24
30
11
12
0
8
30
32
11
8
12
12
8
2
3
0
(c) f x 4x 32x 1x 2x 4
(d) Zeros: 34, 12, 2, 4
(e)
40
−3
5
− 380
(b) 8x2 2x 3 4x 32x 1
The remaining factors are 4x 3 and 2x 1.
61. f x 6x3 41x2 9x 14; Factors: 2x 1, 3x 2
(a) 12
6
6
2
3
6
6
41
3
38
38
4
42
9
19
28
14
14
0
(b) 6x 42 6x 7
This shows that
28
28
0
so
f x
x 7.
2x 13x 2
The remaining factor is x 7.
Both are factors since the remainders are zero.
(c) f x x 72x 13x 2
f x
6x 7,
x 12 x 23 1 2
(d) Zeros: 7, ,
2 3
(e)
320
−9
3
− 40
335
336
Chapter 3
Polynomial Functions
63. f x 2x3 x2 10x 5;
62. f x 10x 3 11x2 72x 45;
Factors: 2x 1, x 5
Factors: 2x 5, 5x 3
(a) 52
3
5
10
11
25
72
90
45
45
10
36
18
0
36
6
10
(a)
5
1
1
0
10
0
10
2
0
25
10
25
0
2
10 30
0
(b) 10x 30 10x 3
f x
x x 5
2
3
5
10x 3,
10
(b) 2x 25 2x 5
This shows that
so
The remaining factor is x 3.
(c) f x x 32x 55x 3
5 3
(d) Zeros: 3, ,
2 5
f x
x 12 x 5
2x 5,
f x
x 5.
2x 1x 5
The remaining factor is x 5.
(c) f x x 5x 52x 1
(d) Zeros: 5, 5,
100
(e)
−4
5
5
0
Both are factors since the remainders are zero.
f x
so
x 3.
2x 55x 3
(e)
2
2
18
18
This shows that
1
2
1
2
14
4
−6
− 80
6
−6
64. f x x 3 3x2 48x 144; Factors: x 43 , x 3
(a) 3
3
3
48
0
1
0
43 1
48
0
1
1
144
144
0
48
43
48
43
0
(c) f x x 43 x 43 x 3
(d) Zeros: ± 43, 3
(e)
60
−8
8
(b) The remaining factor is x 43 .
65. f x x3 2x2 5x 10
−240
66. g x x 3 4x 2 2x 8
(a) The zeros of f are 2 and ± 2.236.
(a) The zeros of g are x 4, x 1.414, x 1.414.
(b) An exact zero is x 2.
(b) x 4 is an exact zero.
(c) 2
1
1
2
2
0
5
0
5
10
10
0
f x x 2x2 5
x 2x 5x 5
(c) 4
1
4
4
2
0
8
8
1
0
2
0
f x x 4x 2 2
x 4x 2 x 2 Section 3.3
Polynomial and Synthetic Division
3
2
68. f s s 12s 40s 24
67. ht t3 2t2 7t 2
(a) The zeros of h are t 2, t 3.732, t 0.268.
(a) The zeros of f are s 6, s 0.764, s 5.236
(b) An exact zero is t 2.
(b) s 6 is an exact zero.
(c) 2
1
1
2
2
4
7
8
1
337
1 12
40 24
6 36
24
(c) 6
2
2
0
6
1
ht t 2t2 4t 1
4
0
f s s 6s 2 6s 4
s 6 s 3 5 s 3 5 By the Quadratic Formula, the zeros of t2 4t 1
are 2 ± 3. Thus,
ht t 2t 2 3t 2 3 t 2t 2 3 t 2 3 .
69.
4x3 8x2 x 3
2x 3
3
2
8
6
2
4
4
4x3
8x2
Thus,
71.
1
3
2
x3
x 32
70.
8
3
3
0
1
1 64 64
8
56
64
1
7
8
0
64x 64
x 2 7x 8, x 8
x8
x2
3
4x3 8x2 x 3
2x2 x 1, x .
2x 3
2
1
1
2
x3
4x2 2x 2 22x 2 x 1
x 4 6x3 11x2 6x x 4 6x3 11x2 6x
x2 3x 2
x 1x 2
1
x3 x2 64x 64
x8
1
1
6
1
5
11
5
6
6
6
0
5
2
3
6
6
0
0
0
0
0
0
0
x4 6x3 11x2 6x
x2 3x, x 2, 1
x 1x 2
73. (a) and (b)
1800
3
1200
—CONTINUED—
13
72.
x 4 9x 3 5x 2 36x 4
x 4 9x 3 5x 2 36x 4
x2 4
x 2x 2
2
1
9
2
1
2
11
1
1
x 4
9x 3
5 36
22
34
4
4
2
0
17
11
17
2 18
2
2
1
0
9
36x 4
x 2 9x 1, x ± 2
x2 4
5x 2
338
Chapter 3
Polynomial Functions
73. —CONTINUED—
(c) M 0.242t 3 12.43t 2 173.4t 2118
Year, t
Military Personnel
M
3
1705
1703
4
1611
1608
5
1518
1532
6
1472
1473
7
1439
1430
8
1407
1402
9
1386
1388
10
1384
1385
11
1385
1393
12
1412
1409
13
1434
1433
(d) 18
0.242
12.43
4.356
173.4
145.332
2118
505.224
0.242
8.074
28.068
1612.776
M18 1613 thousand
No, this model should not be used to predict the
number of military personnel in the future. It
predicts an increase in military personnel until 2024
and then it decreases and will approach negative
infinity quickly.
The model is a good fit to the actual data.
75. False. If 7x 4 is a factor of f, then 47 is a zero of f.
74. (a) and (b)
40
2
12
0
(b) R 0.0026t 3 0.0292t 2 1.558t 15.632
(c) 18
0.0026
0.0026
0.0292
1.558
15.632
0.0468
0.3168
33.7464
0.0176
1.8748
49.3784
For the year 2008, the model predicts a monthly rate
of about $49.38.
76. True.
1
2
6
1
3
92
2
45
45
184
0
4
92
48
48
6
4
90
0
184
96
0
77. True. The degree of the numerator is greater than the
degree of the denominator.
f x 2x 1x 1x 2x 33x 2x 4
78. f x x kqx r
(a) k 2, r 5, qx any quadratic ax2 bx c
where a > 0. One example:
f x x 2x2 5 x3 2x2 5
(b) k 3, r 1, qx any quadratic ax2 bx c
where a < 0. One example:
f x x 3x2 1 x3 3x2 1
Section 3.3
x2n 6xn 9
79.
xn
3)
x3n
9x2n
27xn
27xn
27
xn
2 ) x 3x2n 5xn 6
3n
x3n 2x2n
x2n 5x n
6x2n 18xn
x2n 2xn
9xn 27
9xn 27
3x n 6
0
3xn 6
x3n 9x2n 27xn 27
x2n 6xn 9
xn 3
81. A divisor divides evenly into a dividend if the remainder
is zero.
83. 5
1
1
4
5
9
3
45
42
0
x3n 3x2n 5xn 6
x2n x n 3
xn 2
82. You can check polynomial division by multiplying the
quotient by the divisor. This should yield the original
dividend if the multiplication was performed correctly.
84. 2
c
210
c 210
To divide evenly, c 210 must equal zero. Thus, c must
equal 210.
85. f x x 32x 3x 13
The remainder when k 3 is zero since x 3
is a factor of f x.
87.
9x2 25 0
1
0
2
5
3
x2 21
16
x±
89. 5x2 3x 14 0
2116
21
4
90. 8x2 22x 15 0
5x 7x 2 0
4x 52x 3 0
57
x20 ⇒ x2
c
42
86. In this case it is easier to evaluate f 2 directly because
f x is in factored form. To evaluate using synthetic
division you would have to expand each factor and then
multiply it all out.
x±
4x 5 0 or 2x 3 0
x 54
91. 2x2 6x 3 0
b ± b2 4ac 6 ± 62 423 6 ± 12
2a
22
4
3 ± 3
2
1
20
1 2 4 10 21 c 42
To divide evenly, c 42 must equal zero. Thus, c must
equal 42.
3x 5 0 ⇒ x 35
2
8
16x2 21
3x 5 0 ⇒ x 5x 7 0 ⇒ x 0
4
88. 16x2 21 0
3x 53x 5 0
x
339
x2n x n 3
80.
x3n 3x2n
6x2n
Polynomial and Synthetic Division
or
3
x2
340
Chapter 3
Polynomial Functions
92. x2 3x 3 0
x
3 ± 32 413 3 ± 21
21
2
93. f x x 0x 3x 4
94. f x x 6x 1
xx 3x 4 xx 7x 12
x 6x 1
x 3 7x2 12x
x2 5x 6
2
Note: Any nonzero scalar multiple of f x would also
have these zeros.
Note: Any nonzero scalar multiple of f x would also
have these zeros.
95. f x x 3x 1 2 x 1 2 x 3x 1 2x 1 2
96. f x x 1x 2x 2 3x 2 3
x 3x 12 2 2
x 1x 2x 2 3x 2 3
x2 x 2x 22 3 2
x 3x2 2x 1
x2 x 2x2 4x 1
x 3 x2 7x 3
x4 3x3 5x2 9x 2
Note: Any nonzero scalar multiple of f x would also
have these zeros.
Section 3.4
Note: Any nonzero scalar multiple of f x would also have
these zeros.
Zeros of Polynomial Functions
■
You should know that if f is a polynomial of degree n > 0, then f has at least one zero in the complex number system.
■
You should know the Linear Factorization Theorem.
■
You should know the Rational Zero Test.
■
You should know shortcuts for the Rational Zero Test. Possible rational zeros factors of constant term
factors of leading coefficient
(a) Use a graphing or programmable calculator.
(b) Sketch a graph.
(c) After finding a root, use synthetic division to reduce the degree of the polynomial.
■
You should know that if a bi is a complex zero of a polynomial f, with real coefficients, then a bi
is also a complex zero of f.
■
You should know the difference between a factor that is irreducible over the rationals (such as x2 7) and a factor that is
irreducible over the reals (such as x2 9).
■
You should know Descartes’s Rule of Signs. (For a polynomial with real coefficients and a non-zero constant term.)
(a) The number of positive real zeros of f is either equal to the number of variations of sign of f or is less than
that number by an even integer.
(b) The number of negative real zeros of f is either equal to the number of variations in sign of f x or is less than
that number by an even integer.
(c) When there is only one variation in sign, there is exactly one positive (or negative) real zero.
■
You should be able to observe the last row obtained from synthetic division in order to determine upper or lower bounds.
(a) If the test value is positive and all of the entries in the last row are positive or zero, then the test value is an upper
bound.
(b) If the test value is negative and the entries in the last row alternate from positive to negative, then the test value is a
lower bound. (Zero entries count as positive or negative.)
Section 3.4
Zeros of Polynomial Functions
Vocabulary Check
1. Fundamental Theorem of Algebra
2. Linear Factorization Theme
3. Rational Zero
4. conjugate
5. irreducible; reals
6. Descarte’s Rule of Signs
7. lower; upper
1. f x xx 62
2. f x x2x 3x2 1 x2x 3x 1x 1
The zeros are: x 0, x 6.
The five zeros are: 0, 0, 3, ± 1.
4. f x x 5x 82
3. gx x 2x 43
5. f x x 6x ix i
The three zeros are: 5, 8, 8.
The zeros are: x 2, x 4.
The three zeros are:
x 6, x i, x i.
6. ht t 3t 2t 3it 3i
The four zeros are: 3, 2, ± 3i.
7. f x x3 3x2 x 3
Possible rational zeros: ± 1, ± 3
Zeros shown on graph: 3, 1, 1
9. f x 2x4 17x3 35x2 9x 45
8. f x x 3 4x 2 4x 16
Possible rational zeros: ± 1, ± 2, ± 4, ± 8, ± 16
Zeros shown on graph: 2, 2, 4
3
Zeros shown on graph: 1, 2, 3, 5
10. f x 4x 5 8x 4 5x 3 10x 2 x 2
Possible rational zeros: ± 1, ± 2, ± 12, ± 14
Zeros shown on graph: 1,
12, 12,
Possible rational zeros: ± 1, ± 3, ± 5, ± 9, ± 15, ± 45,
1
3
5
9
15
45
± 2, ± 2, ± 2, ± 2, ± 2 , ± 2
1, 2
11. f x x3 6x2 11x 6
Possible rational zeros: ± 1, ± 2, ± 3, ± 6
1
1
1
6
1
5
11
5
6
6
6
0
x3 6x2 11x 6 x 1x2 5x 6
x 1x 2x 3
Thus, the rational zeros are 1, 2, and 3.
12. f x x 3 7x 6
13. gx x3 4x2 x 4 x2x 4 1x 4
Possible rational zeros: ± 1, ± 2, ± 3, ± 6
3
1
0
3
7
9
6
6
1
3
2
0
f x x 3x 2 3x 2 x 3x 2x 1
Thus, the rational zeros are 2, 1, 3.
x 4x2 1
x 4x 1x 1
Thus, the rational zeros of gx are 4 and ± 1.
341
342
Chapter 3
Polynomial Functions
14. hx x 3 9x 2 20x 12
15. ht t3 12t2 21t 10
Possible rational zeros: ± 1, ± 2, ± 3, ± 4, ± 6, ± 12
1
1
9
1
1
8
20 12
8 12
12
Possible rational zeros: ± 1, ± 2, ± 5, ± 10
1
1
12
1
11
1
0
hx x 1x 2 8x 12
t3
21
11
10
10
10
0
12t2 21t 10 t 1t2 11t 10
t 1t 1t 10
x 1x 2x 6
t 12t 10
Thus, the rational zeros are 1, 2, 6.
Thus, the rational zeros are 1 and 10.
17. Cx 2x3 3x2 1
16. px x 3 9x 2 27x 27
1
Possible rational zeros: ± 1, ± 2
Possible rational zeros: ± 1, ± 3, ± 9, ± 27
3
1
9 27 27
3 18 27
1
6
9
1
2
3
2
1
2
0
1
1
0
0
1
1
2x 3x2 1 x 12x2 x 1
f x x 3x 2 6x 9
3
x 1x 12x 1
x 3x 3x 3
x 122x 1
Thus, the rational zero is 3.
1
Thus, the rational zeros are 1 and 2.
18. f x 3x3 19x 2 33x 9
Possible rational zeros: ± 1, ± 3, ± 9, ± 13
3 19
33
9 30
3
3 10
3
9
9
0
f x x 33x 2 10x 3 x 33x 1x 3
Thus, the rational zeros are 3, 13.
19. f x 9x4 9x3 58x2 4x 24
20. f x 2x 4 15x 3 23x 2 15x 25
Possible rational zeros: ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24,
1
2
4
8
1
2
4
8
± 3, ± 3, ± 3, ± 3, ± 9, ± 9, ± 9, ± 9
2
9
9
3
9
9
9
18
27
58
54
4
4
8
12
27
27
0
4
0
4
12
12
0
24
24
0
Possible rational zeros: ± 1, ± 5, ± 25, ± 12, ± 52, ± 25
2
5
2 15
23
15 25
10 25 10
25
1
2
5
2
2
3
5
5
2
3
5
0
5
2
1
9x4 9x3 58x2 4x 24
x 2x 39x2 4
x 2x 33x 23x 2
2
Thus, the rational zeros are 2, 3, and ± 3.
2
2
3
2
5
5
2
5
0
5
0
f x x 5x 1x 12x 5
Thus, the rational zeros are 5, 1, 1, 52.
Section 3.4
xx 3 13x 12 0
Possible rational zeros: ± 1, ± 2, ± 4
1
1
2
1
1
343
22. x 4 13x 2 12x 0
21. z4 z3 2z 4 0
1
Zeros of Polynomial Functions
1
1
2
0
2
2
2
2
4
2
2
0
2
0
2
4
4
0
Possible rational zeros of x 3 13x 12:
± 1, ± 2, ± 3, ± 4, ± 6, ± 12.
4
4
0
1
0 13 12
1
1 12
1
1 12
1
xx 1
x2
0
x 12 0
z4 z3 2z 4 z 1z 2z2 2
xx 1x 4x 3 0
The only real zeros are 1 and 2.
The real zeros are 0, 1, 4, 3.
23. 2y4 7y3 26y2 23y 6 0
1
3
Possible rational zeros: ± 1, ± 2, ± 3, ± 6, ± 2, ± 2
1
2
2
6
2
6
6
0
23
17
6
17
18
1
9
12
3
2
2y4
26
9
17
7
2
9
6
6
0
7y3 26y2 23y 6 y 1y 62y2 3y 1 y 1y 62y 1y 1 y 12 y 62y 1
The only real zeros are 1, 6, and 12.
25. f x x3 x2 4x 4
24. x 5 x 4 3x 3 5x 2 2x 0
xx 4 x 3 3x 2 5x 2 0
Possible rational zeros of x ± 1, ± 2.
4
1
1
1
1
3
0
5
3
2
2
0
3
2
0
1
2
1
0
2
3
4
2
2
1
2
1
0
x3
(a) Possible rational zeros: ± 1, ± 2, ± 4
3x2
5x 2:
y
(b)
4
2
−6
x
−4
4
6
−4
−6
−8
xx 1x 2x 2 2x 1 0
(c) The zeros are: 2, 1, 2
xx 1x 2x 1x 1 0
The real zeros are 2, 0, 1.
26. f x 3x 3 20x 2 36x 16
(a) Possible rational zeros: ± 1, ± 2, ± 4, ± 8, ± 16, ± 13,
2
4
8
16
± 3, ± 3, ± 3, ± 3
27. f x 4x3 15x2 8x 3
1
3
1
3
(a) Possible rational zeros: ± 1, ± 3, ± 2, ± 2, ± 4, ± 4
y
(b)
y
(b)
10
8
6
4
4
2
2
−4 −2
−6 −4 −2
x
2
4
6
8 10
x
6
8 10 12
−4
−4
−6
−6
(c) Real zeros: 23, 2, 4
1
(c) The zeros are: 4, 1, 3
344
Chapter 3
Polynomial Functions
28. f x 4x 3 12x 2 x 15
29. f x 2x4 13x3 21x2 2x 8
(a) Possible rational zeros: ± 1, ± 3, ± 5, ± 15, ± 12, ± 32,
5
15
1
3
5
15
± 2, ± 2 , ± 4, ± 4, ± 4, ± 4
1
(a) Possible rational zeros: ± 1, ± 2, ± 4, ± 8, ± 2
(b)
16
y
(b)
15
−4
12
8
−8
1
(c) The zeros are: 2, 1, 2, 4
x
− 9 −6 − 3
6
9
12
(c) Real zeros: 1, 32, 52
31. f x 32x3 52x2 17x 3
30. f x 4x 4 17x 2 4
(a) Possible rational zeros: ± 1, ± 2, ± 4, ± 12, ± 14
(b)
9
1
3
1
3
(a) Possible rational zeros: ± 1, ± 3, ± 2, ± 2, ± 4, ± 4,
1
3
1
3
1
3
± 6 , ± 8 , ± 16 , ± 16 , ± 32 , ± 32
(b)
−8
6
8
−1
3
− 15
−2
(c) Real zeros: ± 2, ± 12
1 3
(c) The zeros are: 8, 4, 1
33. f x x4 3x2 2
32. f x 4x 3 7x 2 11x 18
(a) Possible rational zeros: ± 1, ± 2, ± 3, ± 6, ± 9, ± 18,
1
3
9
1
3
9
± 2, ± 2, ± 2, ± 4, ± 4, ± 4
(b)
8
(a) From the calculator we have x ± 1 and x ± 1.414.
(b) An exact zero is x 1.
1
−8
1
1
(c) 1
− 24
(c) Real zeros: 2,
1
1
1 145
±
8
8
3
1
2
0
1
1
8
1
1
0
0
2
2
2
0
2
2
2
0
2
2
0
f x x 1x 1x2 2
x 1x 1x 2 x 2 34. P t t 4 7t 2 12
35. hx x5 7x4 10x3 14x2 24x
(a) t ± 2, ± 1.732
(b) 2
2
(a) hx xx4 7x3 10x2 14x 24
1
0
2
7
4
0
6
12
12
1
2
3
6
0
1
2
2
3
0
6
6
1
0
3
0
(c) Pt t 2t 2t2 3
t 2t 2t 3 t 3 From the calculator we have x 0, 3, 4 and x ± 1.414.
(b) An exact zero is x 3.
3
1
1
(c) 4
1
1
7
3
4
10
12
2
4
4
0
2
0
2
14
6
8
24
24
0
8
8
0
h x xx 3x 4x2 2
xx 3x 4x 2 x 2 Section 3.4
Zeros of Polynomial Functions
345
36. gx 6x 4 11x 3 51x 2 99x 27
(a) x ± 3, 1.5, 0.333
(b) 3
(c) gx x 3x 36x 11x 3
2
x 3x 33x 12x 3
6 11 51
99 27
18
21 90
27
7 30
6
3
9
7 30
18
33
6
6 11
0
9
9
3
0
38. f x x 4x 3ix 3i 37. f x x 1x 5ix 5i
x 1x2 25
x 4x 2 9
x3 x2 25x 25
x 3 4x 2 9x 36
Note: f x ax3 x2 25x 25, where a is any
nonzero real number, has the zeros 1 and ± 5i.
39. f x x 6x 5 2ix 5 2i
Note: f x a x 3 4x 2 9x 36, where a is
any real number, has the zeros 4, 3i and 3i.
40. f x x 2x 4 ix 4 i
x 6x 5 2ix 5 2i
x 2x 2 8x 17
x 6x 52 2i2
x 3 10x 2 33x 34
Note: f x ax 3 10x 2 33x 34 where a is
any real number, has the zeros 2, 4 ± i.
x 6x2 10x 25 4
x 6x2 10x 29
x3 4x2 31x 174
Note: f x ax3 4x2 31x 174, where a is any
nonzero real number, has the zeros 6, and 5 ± 2i.
41. If 3 2i is a zero, so is its conjugate, 3 2i.
f x 3x 2x 1x 3 2ix 3 2i
3x 2x 1x 3 2ix 3 2i
3x2 x 2x 32 2i 2
3x2 x 2x2 6x 9 2
3x2 x 2x2 6x 11
42. If 1 3i is a zero, so is its conjugate, 1 3i.
f x x 5 2x 1 3ix 1 3i
x 2 10x 25x 2 2x 4
x 4 8x 3 9x 2 10x 100
Note: f x ax 4 8x 3 9x 2 10x 100,
where a is any real number, has the zeros 5, 5,
1 ± 3 i.
3x4 17x3 25x2 23x 22
Note: f x a3x4 17x3 25x2 23x 22,
where a is any nonzero real number, has the zeros
2
3 , 1, and 3 ± 2i.
43. f x x4 6x2 27
44. f x x 4 2x 3 3x 2 12x 18
(a) f x x2 9x2 3
(b) f x x2 9x 3x 3
(c) f x x 3ix 3ix 3x 3
x2 2x 3
x6)
x4
2x3
6x2
x4
x4
3x2 12x 18
2x3
2x3
3x2 12x 6
12x
2x3 3x2
3x 2
3x2 18
18
0
(a) f x x 2 6x 2 2x 3
(b) f x x 6 x 6 x 2 2x 3
(c) f x x 6 x 6 x 1 2ix 1 2 i
346
Chapter 3
Polynomial Functions
45. f x x 4 4x 3 5x 2 2x 6
x2 2x 3
x 2x 2 )x 4x 5x2 2x 6
2
4
3
x4 2x3 2x2
42x3
x
7x2
2x 6
2x 4x 4x
3
2
2x3 3x2 6x 6
(a) f x x2 2x 2x2 2x 3
(b) f x x 1 3 x 1 3 x2 2x 3
(c) f x x 1 3 x 1 3 x 1 2 i x 1 2 i Note: Use the Quadratic Formula for (b) and (c).
3x2 6x 6
3x2 6x 0
f x x2 2x 2x2 2x 3
46. f x x 4 3x 3 x 2 12x 20
x 2 3x 5
(a) f x x 2 4x 2 3x 5
x 2 4 ) x 4 3x 3 x 2 12x 20
x4
x 3 2 29 3 29
(c) f x x 2ix 2ix x 3 2 29
2
4x2
(b) f x x 2 4 x 3x 5x2 12x
3
3x3
12x
20
5x2
20
0
2
5i
2
2
3
10i
3 10i
50
50 15i
15i
3 10i
10i
3
15i
15i
0
The zero of 2x 3 is x 75
75
0
3x2 50x 75
3x2 50x 70
32.
Since 3i is a zero, so is 3i.
1
1
3i
9
9 3i
9
9
1
1 3i
3i
0
1
1 3i
3i
3i
3i
1
1
0
02x 3
x2 0x 25 )2x3 3x2 50x 75
2x3 3x2 50x 75
48. f x x 3 x 2 9x 9
3i
Since x ± 5i are zeros of f x, x 5ix 5i x2 25
is a factor of f x. By long division we have:
2x3 0x2 50x
3
The zeros of f x are x 2 and x ± 5i.
3i
Alternate Solution
Since 5i is a zero, so is 5i.
2
5x2
47. f x 2x3 3x2 50x 75
5i
3 29
2
The zero of x 1 is x 1.
The zeros of f are x 1 and x ± 3i.
Thus, f x x2 252x 3 and the zeros of f are
x ± 5i and x 32.
Section 3.4
49. f x 2x4 x3 7x2 4x 4
2
2
2i
2
2
347
Alternate Solution
Since 2i is a zero, so is 2i.
2i
Zeros of Polynomial Functions
1
4i
1 4i
7
8 2i
1 2i
4
4 2i
2i
1 4i
4i
1
1 2i
2i
1
2i
2i
0
4
4
0
Since x ± 2i are zeros of f x, x 2ix 2i x2 4
is a factor of f x. By long division we have:
2x2 4x 1
x2 0x 4 )2x4 x3 7x2 4x 4
2x4 0x3 8x2 2x4 0x3 8x2 4x
1
The zeros of 2x2 x 1 2x 1x 1 are x 2
and x 1.
1
The zeros of f x are x ± 2i, x 2, and x 1.
7x3 0x2 4x
0x30 x2 0x 4
0 x2 0x 4
0x2 0x 0
Thus, f x x2 42x2 x 1
f x x 2ix 2i2x 1x 1
and the zeros of f x are x ± 2i, x 2, and x 1.
1
50. gx x 3 7x 2 x 87
Since 5 2i is a zero, so is 5 2i.
5 2i
5 2i
1
7
5 2i
1
14 6i
87
87
1
2 2i
15 6i
0
1
2 2i
5 2i
15 6i
15 6i
1
3
0
The zero of x 3 is x 3.
The zeros of f are x 3, 5 ± 2i.
51. gx 4x3 23x2 34x 10
Alternate Solution
Since 3 i is a zero, so is 3 i.
3 i
4
4
3 i
4
4
23
12 4i
11 4i
34
37 i
3 i
11 4i
12 4i
1
3 i
3i
0
The zero of 4x 1 is x 1
x 3 ± i and x 4.
1
4.
10
10
0
The zeros of gx are
Since 3 ± i are zeros of gx,
x 3 i x 3 i x 3 ix 3 i
x 32 i2
x2 6x 10
is a factor of gx. By long division we have:
34x 1
x2 6x 10 )4x3 23x2 34x 10
4x3 24x2 40x
4x324 x2 36x 10
x2 36x 10
0
Thus, gx x2 6x 104x 1 and the zeros of gx
are x 3 ± i and x 14.
348
Chapter 3
Polynomial Functions
52. hx 3x3 4x 2 8x 8
Since 1 3 i is a zero, so is 1 3 i.
3
4
3 33i
8
10 23i
8
8
3
1 33i
2 23i
0
3
1 33i
3 33i
2 23i
2 23i
3
2
0
The zero of 3x 2 is x 23.
1 3i
1 3i
2
The zeros of f are x 3, 1 ± 3i.
53. f x x 4 3x3 5x2 21x 22
Since 3 2 i is a zero, so is 3 2 i, and
x 3 2 i x 3 2 i x 3 2 ix 3 2 i
2
x 32 2 i
x 2 6x 11
is a factor of f x. By long division, we have:
x2 3x 2
x2 6x 11 ) x4 3x3 5x2 21x 22
x4 6x3 11x2
3x3 16x2 21x
3x3 18x2 33x
2x2 12x 22
2x2 12x 22
0
Thus,
f x x2 6x 11x2 3x 2
x 2 6x 11x 1x 2
and the zeros of f are x 3 ± 2 i, x 1, and x 2.
54. f x x 3 4x 2 14x 20
55. f x x2 25
Since 1 3i is a zero, so is 1 3i.
1 3i
1 3i
x 5ix 5i
1
4
1 3i
14
12 6i
20
20
1
3 3i
2 6i
0
1
3 3i
1 3i
2 6i
2 6i
1
2
0
The zero of x 2 is x 2.
The zeros of f are x 2, 1 ± 3i.
The zeros of f x are x ± 5i.
Section 3.4
56. f x x 2 x 56
1 ± 1 224 1 ± 223i
.
2
2
f x x 1 223i
2
By the Quadratic Formula, the zeros of hx are
x
x 1 2 223i
58. gx x 2 10x 23
4 ± 16 4
2 ± 3.
2
hx x 2 3 x 2 3 x 2 3 x 2 3 59. f x x4 81
By the Quadratic Formula, the zeros of f x are
x
349
57. hx x2 4x 1
By the Quadratic Formula, the zeros of f x are
x
Zeros of Polynomial Functions
10 ± 100 92 10 ± 8
5 ± 2.
2
2
gx x 5 2 x 5 2 60. f y y 4 625
x2 9x2 9
x 3x 3x 3ix 3i
The zeros of f x are x ± 3 and x ± 3i.
61. f z z2 2z 2
y2 25 y2 25
Zeros: y ± 5, ± 5i
f y y 5 y 5 y 5i y 5i
By the Quadratic Formula, the zeros of f z are
z
2 ± 4 8
1 ± i.
2
f z z 1 iz 1 i
z 1 iz 1 i
62. hx x 3 3x 2 4x 2
Possible rational zeros: ± 1, ± 2
1
1
3
1
4
2
2
2
1
2
2
0
Possible rational zeros: ± 1, ± 2, ± 5, ± 10
2
1
1
By the Quadratic Formula, the zeros of x2 2x 2
are x 63. gx x3 6x2 13x 10
2 ± 4 8
1 ± i.
2
6
2
4
13
8
5
10
10
0
By the Quadratic Formula, the zeros of x2 4x 5 are
x
4 ± 16 20
2 ± i.
2
Zeros: x 1, 1 ± i
The zeros of gx are x 2 and x 2 ± i.
hx x 1x 1 ix 1 i
gx x 2x 2 ix 2 i
x 2x 2 ix 2 i
64. f x x 3 2x 2 11x 52
Possible rational zeros: ± 1, ± 2, ± 4, ± 13, ± 26
4
1
2 11 52
4 24 52
1
6
13
0
By the Quadratic Formula, the zeros of x2 6x 13 are x Zeros: x 4, 3 ± 2i
f x x 4x 3 2ix 3 2i
6 ± 36 52
3 ± 2i.
2
350
Chapter 3
Polynomial Functions
65. hx x3 x 6
66. hx x 3 9x2 27x 35
Possible rational zeros: ± 1, ± 2, ± 3, ± 6
2
1
1
1
4
3
0
2
2
5
6
6
0
By the Quadratic Formula, the zeros of x2 2x 3 are
x
Possible rational zeros: ± 1, ± 5, ± 7, ± 35
2 ± 4 12
1 ± 2 i.
2
1
9
5
27
20
35
35
1
4
7
0
By the Quadratic Formula, the zeros of x2 4x 7
4 ± 16 28
2 ± 3i.
2
are x The zeros of hx are x 2 and x 1 ± 2 i.
Zeros: 5, 2 ± 3i
hx x 2x 1 2 i x 1 2 i h x x 5x 2 3ix 2 3i
x 2x 1 2 ix 1 2 i
67. f x 5x3 9x2 28x 6
68. gx 3x 3 4x 2 8x 8
Possible rational zeros:
1
2
3
6
± 1, ± 2, ± 3, ± 6, ± 5 , ± 5 , ± 5 , ± 5
15
5
5
9
1
10
28
2
30
Possible rational zeros:
1
2
4
8
± 1, ± 2, ± 4, ± 8, ± 3 , ± 3 , ± 3 , ± 3
23
6
6
0
By the Quadratic Formula, the zeros of
5x2 10x 30 5x2 2x 6 are
x
2 ± 4 24
1 ± 5 i.
2
3
4
2
8
4
8
8
3
6
12
0
By the Quadratic Formula, the zeros of
3x2 6x 12 3x2 2x 4 are
x
2 ± 4 16
1 ± 3i.
2
The zeros of f x are x 15 and x 1 ± 5 i.
Zeros: x 23, 1 ± 3i
f x x 15 5x 1 5 i x 1 5 i gx 3x 2x 1 3ix 1 3 i
5x 1x 1 5 ix 1 5 i
69. gx x4 4x3 8x2 16x 16
Possible rational zeros: ± 1, ± 2, ± 4, ± 8, ± 16
2
1
1
2
1
1
4
2
2
8
4
4
2
2
0
4
0
4
16
8
8
16
16
0
8
8
0
gx x 2x 2x 2 4 x 22x 2ix 2i
The zeros of gx are 2 and ± 2i.
70. hx x 4 6x 3 10x 2 6x 9
Possible rational zeros: ± 1, ± 3, ± 9
3
3
1
6
3
10
9
6
3
9
9
1
3
1
3
0
1
3
3
1
0
3
3
1
0
1
0
The zeros of
x2
1 are x ± i.
Zeros: x 3, ± i
hx x 3 2x ix i
71. f x x 4 10x 2 9
x 2 1x2 9
x ix ix 3ix 3i
The zeros of f x are x ± i and x ± 3i.
72. f x x 4 29x 2 100
x 2 25x 2 4
Zeros: x ± 2i, ± 5i
f x x 2ix 2ix 5ix 5i
Section 3.4
73. f x x3 24x2 214x 740
Zeros of Polynomial Functions
351
74. f s 2s 3 5s 2 12s 5
Possible rational zeros: ± 1, ± 2, ± 4, ± 5, ± 10, ± 20, ± 37,
± 74, ± 148, ± 185, ± 370, ± 740
Possible rational zeros: ± 1, ± 5, ± 12, ± 52
10
2000
−10
−20
10
10
−10
−1000
Based on the graph, try s 12 .
Based on the graph, try x 10.
10
1
24
10
14
1
214
140
74
1
2
740
740
0
By the Quadratic Formula, the zeros of x2 14x 74 are
x
14 ± 196 296
7 ± 5i.
2
The zeros of f x are x 10 and x 7 ± 5i.
75. f x 16x3 20x2 4x 15
Possible rational zeros:
1
3
5
15
1
3
± 1, ± 3, ± 5, ± 15, ± 2 , ± 2 , ± 2 , ± 2 , ± 4 , ± 4 ,
5
15
1
3
5
15
1
3
5
15
± 4 , ± 4 , ± 8 , ± 8 , ± 8 , ± 8 , ± 16 , ± 16 , ± 16 , ± 16
2
5
1
12
2
5
5
2
4
10
0
By the Quadratic Formula, the zeros of 2s 2 2s 5 are
s
2 ± 4 20
1 ± 2i.
2
1
The zeros of f s are s 2 and s 1 ± 2i.
76. f x 9x 3 15x 2 11x 5
Possible rational zeros: ± 1, ± 5, ± 13, ± 53, ± 19, ± 59
5
−5
20
5
−5
−3
Based on the graph, try x 1.
3
−5
1
Based on the graph, try x 34
16
16
20
12
32
4
24
20
34.
15
15
0
By the Quadratic Formula, the zeros of
16x2 32x 20 44x2 8x 5 are
8 ± 64 80
1
x
1 ± i.
8
2
9 15
9
11
6
5
5
6
5
0
9
By the Quadratic Formula, the zeros of 9x 2 6x 5 are
x
6 ± 36 180 1 2
± i.
18
3 3
1
2
The zeros of f x are x 1 and x 3 ± 3i.
The zeros of f x are x 34 and x 1 ± 12i.
77. f x 2x4 5x3 4x2 5x 2
1
Possible rational zeros: ± 1, ± 2, ± 2
Based on the graph, try x 2 and x 12.
2
2
20
2
12
−4
2
4
2
−5
5
4
1
4
2
2
1
1
0
2
0
2
5
4
1
2
2
0
1
1
0
The zeros of 2x2 2 2x2 1 are x ± i.
The zeros of f x are x 2, x 12, and x ± i.
352
Chapter 3
Polynomial Functions
78. gx x 5 8x 4 28x 3 56x 2 64x 32
2
1
8
2
28
12
56
32
64
48
32
32
1
6
16
24
16
0
1
6
2
16
8
24
16
16
16
1
4
8
8
0
Possible rational zeros: ± 1, ± 2, ± 4, ± 8, ± 16, ± 32
Based on the graph,
try x 2.
10
−10
2
10
−10
2
1
4
2
8
4
8
8
1
2
4
0
By the Quadratic Formula, the zeros of x2 2x 4 are
x
2 ± 4 16
1 ± 3i.
2
The zeros of gx are x 2 and x 1 ± 3 i.
79. gx 5x5 10x 5xx4 2
80. hx 4x 2 8x 3
Let f x x4 2.
Sign variations: 2, positive zeros: 2 or 0
Sign variations: 0, positive zeros: 0
hx 4x 2 8x 3
f x x4 2
Sign variations: 0, negative zeros: 0
Sign variations: 0, negative zeros: 0
81. hx 3x4 2x2 1
82. hx 2x 4 3x 2
Sign variations: 0, positive zeros: 0
Sign variations: 2, positive zeros: 2 or 0
hx 3x4 2x2 1
hx 2x 4 3x 2
Sign variations: 0, negative zeros: 0
Sign variations: 0, negative zeros: 0
83. gx 2x3 3x2 3
84. f x 4x 3 3x 2 2x 1
Sign variations: 1, positive zeros: 1
gx 2x3
3x2
f x 4x 3 3x 2 2x 1
3
Sign variations: 0, negative zeros: 0
Sign variations: 3, positive zeros: 3 or 1
2
Sign variations: 0, negative zeros: 0
4
4
0
1
0
0
0
0
0
0
1
1
4
1
5
0
5
5
1 is a lower bound.
0
5
5
3 12
8
20
8
32
2
5
8
4 is an upper bound.
40
(a) 4
15
0
15
4 is an upper bound.
(b) 1
Sign variations: 3, negative zeros: 3 or 1
88. f x 2x 3 3x 2 12x 8
87. f x x4 4x3 15
1
Sign variations: 0, positive zeros: 0
f x 3x 3 2x 2 x 3
f x 5x x x 5
(a) 4
Sign variations: 0, negative zeros: 0
86. f x 3x 3 2x 2 x 3
85. f x 5x3 x2 x 5
3
Sign variations: 3, positive zeros: 3 or 1
15
5
20
(b) 3
2
2
3 12
8
6
27 45
2 9
15 37
3 is a lower bound.
Section 3.4
4
5
1
1
1
0
5
5
16
205
189
16
25
41
4
3
7
1
1
0
21
21
16
63
47
4
2
6
18
3 is an upper bound.
46
141
2
0
8
0
8
32 128
3
544
32 136
547
92. f z 12z 3 4z 2 27z 9
1
1
Possible rational zeros: ± 1, ± 2, ± 4
0
4
4
3
138
0
6
2
8
3 is a lower bound.
91. f x 4x3 3x 1
4
8
54
2
(b) 4
16
141
125
3 is a lower bound.
1
0
18
(a) 3
5 is an upper bound.
(b) 3
353
90. f x 2x 4 8x 3
89. f x x 4 4x 3 16x 16
(a) 5
Zeros of Polynomial Functions
3
4
1
1
3
9
1
Possible rational zeros: ± 1, ± 3, ± 9, ± 2, ± 2, ± 2, ± 3,
1
3
9
1
1
± 4 , ± 4 , ± 4 , ± 6 , ± 12
1
1
0
3
2
4x3 3x 1 x 14x2 4x 1
12
12
4 27
18
21
9
9
6
0
14
f z 2z 32 6z 2 7z 3
x 12x 12
1
Thus, the zeros are 1 and 2.
2z 33z 12z 3
3 1 3
Real zeros: 2, 3, 2
94. gx 3x3 2x 2 15x 10
93. f y 4y3 3y2 8y 6
1
3
1
3
Possible rational zeros: ± 1, ± 2, ± 3, ± 6, ± 2, ± 2, ± 4, ± 4
Possible rational zeros: ± 1, ± 2, ± 5, ± 10, ± 13, ± 23,± 53, ± 10
3
34
2
3
4
4
4y3
3
3
0
8
0
8
6
6
0
3y2 8y 6 y 34 4y2 8
y 34 4y2 2
3
2
2
3
0
15 10
0 10
15
0
gx x 23 3x 2 15 3x 2x2 5
2
Thus, the only real zero is 3.
4y 3 y2 2
3
Thus, the only real zero is 4.
25
95. Px x4 4 x2 9
1
96. f x 22x 3 3x 2 23x 12
144x4 25x2 36
Possible rational zeros: ± 1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 2, ± 2
144x2 9x2 4
4
1
4 2x
1
2
32x 3x 2x 2
3
The rational zeros are ± 2 and ± 2.
2
f x 1
2 x
3 23
12
8
20 12
3
5
4
2x 2
0
5x 3 12x 42x 1x 3
1
The rational zeros are 3, 2, and 4.
97. f x x3 14x2 x 14
144x3 x2 4x 1
14x24x 1 14x 1
144x 1x2 1
144x 1x 1x 1
The rational zeros are 14 and ± 1.
3
354
Chapter 3
Polynomial Functions
99. f x x3 1 x 1x2 x 1
98. f z 166z 3 11z 2 3z 2
Possible rational zeros: ± 1, ± 2, ± 12, ± 13, ± 23, ± 16
Rational zeros: 1 x 1
2
Irrational zeros: 0
6
6
11
12
3
2
2
2
1
1
0
Matches (d).
f x 16x 26x 2 x 1
16x 23x 12x 1
1 1
Rational zeros: 2, 3, 2
100. f x x 3 2
101. f x x3 x xx 1x 1
3
3
3
x 2 x2 2x 4
Rational zeros: 3 x 0, ± 1
Rational zeros: 0
Irrational zeros: 0
3
2
Irrational zeros: 1 x Matches (b).
Matches (a).
102. f x x 3 2x
xx 2 2
x x 2 x 2 Rational zeros: 1 x 0
x ± 2 Irrational zeros: 2
Matches (c).
103. (a)
(b) V l w
15
x
9
9−
x
−
15
2x
2x
x
(c)
x9 2x15 2x
Since length, width, and height must be positive,
we have 0 < x < 92 for the domain.
(d) 56 x9 2x15 2x
V
125
Volume of box
h 15 2x9 2xx
56 135x 48x2 4x3
100
50 4x3 48x2 135x 56
75
50
25
x
1
2
3
4
5
Length of sides of
squares removed
The volume is maximum when x 1.82.
The dimensions are: length 15 21.82 11.36
width 9 21.82 5.36
height x 1.82
1.82 cm 5.36 cm 11.36 cm
1 7
The zeros of this polynomial are 2, 2, 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 817 56 so it
showed up as an extraneous solution.]
Thus, the volume is 56 cubic centimeters when
x 12 centimeter or x 72 centimeters.
Section 3.4
104. (a) Combined length and width:
4x y 120 ⇒ y 120 4x
4x 3 120x 2 13,500 0
x 3 30x 2 3375 0
15
1
30
15
0
3375
225 3375
1
15
225
4x 230 x
(b)
18,000
355
13,500 4x 230 x
(c)
Volume l w h x 2 y
x 2120 4x
Zeros of Polynomial Functions
x 15
x2
0
15x 225 0
Using the Quadratic Formula,
0
x 15,
30
0
Dimensions with maximum volume:
20 in. 20 in. 40 in.
15 ± 155
.
2
The value of
it is negative.
15 155
is not possible because
2
P 76x3 4830x2 320,000, 0 ≤ x ≤ 60
105.
2,500,000 76x3 4830x2 320,000
76x3 4830x2 2,820,000 0
The zeros of this equation are x 46.1, x 38.4, and x 21.0. Since 0 ≤ x ≤ 60, we
disregard x 21.0. The smaller remaining solution is x 38.4. The advertising expense is $384,000.
106.
P 45x 3 2500x 2 275,000
800,000 0
45x 3
45x 3
2500x 2
2500x 2
107. (a) Current bin: V 2 3 4 24 cubic feet
New bin: V 524 120 cubic feet
275,000
V 2 x3 x4 x 120
1,075,000
0 9x 3 500x 2 215,000
(b) x3 9x2 26x 24 120
The zeros of this equation are x 18.0, x 31.5, and
x 42.0. Because 0 ≤ x ≤ 50, disregard x 18.02.
The smaller remaining solution is x 31.5, or an
advertising expense of $315,000.
108. (a) A 250 x160 x 1.5160250
x3 9x2 26x 96 0
The only real zero of this polynomial is x 2. All the
dimensions should be increased by 2 feet, so the new
bin will have dimensions of 4 feet by 5 feet by 6 feet.
(c) A 250 2x160 x 60,000
60,000
(b) 60,000 x2 410x 40,000
0 x2 410x 20,000
x
410 ± 4102 4120,000
21
410 ± 248,100
2
x must be positive, so
x
410 248,100
2
44.05.
The new length is 250 44.05 294.05 ft and the
new width is 160 44.05 204.05 ft, so the new
dimensions are 204.05 ft 294.05 ft.
2x2 570x 20,000 0
x
570 ± 5702 4220,000
22
x must be positive, so
x
570 484,900
31.6.
4
The new length is 250 231.6 313.2 ft and the new
width is 160 31.6 191.6 ft, so the new dimensions
are 191.6 ft 313.2 ft.
356
Chapter 3
109. C 100
x
200
2
Polynomial Functions
x
,x ≥ 1
x 30
C is minimum when 3x3 40x2 2400x 36000 0.
The only real zero is x 40 or 4000 units.
110. h(t 16t2 48t 6
Let h 64 and solve for t
64 16t2 48t 6
16t2 48t 58 0
By the Quadratic Formula we have t 48 ± i1408
.
32
Since the equation yields only imaginary zeros, it is not
possible for the ball to have reached a height of 64 feet.
P R C xp C
111.
x140 0.0001x 80x 150,000
112. (a) A 0.0167t3 0.508t2 5.60t 13.4
(b)
The model is a good fit to
the actual data.
12
0.0001x 2 60x 150,000
9,000,000 0.0001x2 60x 150,000
7
Thus, 0 0.0001x 2 60x 9,150,000.
x
60 ± 60
300,000 ± 10,00015i
0.0002
Since the solutions are both complex, it is not possible
to determine a price p that would yield a profit of
9 million dollars.
13
0
(c) A 8.5 when t 10 which corresponds to the
year 2000.
(d) A 9 when t 11 which corresponds to the
year 2001.
(e) Yes. The degree of A is odd and the leading
coefficient is positive, so as x increases, A will increase.
This implies that attendance will continue to grow.
113. False. The most nonreal complex zeros it can have is
two and the Linear Factorization Theorem guarantees
that there are 3 linear factors, so one zero must be real.
114. False. f does not have real coefficients.
115. gx f x. This function would have the same zeros
as f x so r1, r2, and r3 are also zeros of gx.
116. gx 3 f x. This function has the same zeros as f
because it is a vertical stretch of f. The zeros of g are
r1, r2, and r3.
117. gx f x 5. The graph of gx is a horizontal shift
of the graph of f x five units to the right so the zeros
of gx are 5 r1, 5 r2, and 5 r3.
118. gx f 2x. Note that x is a zero of g if and only if 2x
r3
r1 r2
is a zero of f. The zeros of g are , , and .
2 2
2
119. gx 3 f x. Since gx is a vertical shift of the graph
of f x, the zeros of gx cannot be determined.
120. gx f x. Note that x is a zero of g if and only if
x is a zero of f. The zeros of g are r1, r2, and r3.
121. f x x4 4x2 k
x2 4 ± 42 41k 4 ± 24 k
2 ± 4 k
21
2
x ± 2 ± 4 k
(a) For there to be four distinct real roots, both 4 k and 2 ± 4 k must be positive. This occurs
when 0 < k < 4. Thus, some possible k-values are k 1, k 2, k 3, k 12, k 2, etc.
(b) For there to be two real roots, each of multiplicity 2, 4 k must equal zero. Thus, k 4.
—CONTINUED—
Section 3.4
Zeros of Polynomial Functions
357
121. —CONTINUED—
(c) For there to be two real zeros and two complex zeros, 2 4 k must be positive and 2 4 k
1
must be negative. This occurs when k < 0. Thus, some possible k-values are k 1, k 2, k 2, etc.
(d) For there to be four complex zeros, 2 ± 4 k must be nonreal. This occurs when k > 4. Some possible
k-values are k 5, k 6, k 7.4, etc.
122. (a) gx f x 2
No. This function is a horizontal shift of f x.
Note that x is a zero of g if and only if x 2 is a
zero of f; the number of real and complex zeros is
not affected by a horizontal shift.
123. Zeros: 2, 12, 3
(b) gx f 2x
No. Since x is a zero of g if and only if 2x is a zero of
f, the number of real and complex zeros of g is the
same as the number of real and complex zeros of f.
y
124.
f x x 22x 1x 3
50
(− 1, 0)
2x3 3x2 11x 6
y
10
(1, 0)
(4, 0)
x
8
(−2, 0)
−8
−4
4
(3, 0) 4
5
( (
1
,0
2
(3, 0)
x
4
8
12
Any nonzero scalar multiple of f would have the same
three zeros. Let gx af x, a > 0. There are infinitely
many possible functions for f.
125. Answers will vary. Some of the factoring techniques are:
126. (a) Zeros of f x: 2, 1, 4
1. Factor out the greatest common factor.
(b) The graph touches the x-axis at x 1
2. Use special product formulas.
(c) The least possible degree of the function is 4 because
there are at least 4 real zeros (1 is repeated) and a
function can have at most the number of real zeros
equal to the degree of the function. The degree
cannot be odd by the definition of multiplicity.
a2 b2 a ba b
a2 2ab b2 a b2
a2 2ab b2 a b2
a3 b3 a ba2 ab b2
a3 b3 a ba2 ab b2
3. Factor by grouping, if possible.
(d) The leading coefficient of f is positive. From the
information in the table, you can conclude that the
graph will eventually rise to the left and to the right.
(e) Answers may vary. One possibility is:
4. Factor general trinomials with binomial factors
by “guess-and-test” or by the grouping method.
f x x 12x 2x 4
x 12x 2x 4
5. Use the Rational Zero Test together with synthetic
division to factor a polynomial.
6. Use Descartes’s Rule of Signs to determine the
number of real zeros. Then find any zeros and
use them to factor the polynomial.
7. Find any upper and lower bounds for the real zeros
to eliminate some of the possible rational zeros.
Then test the remaining candidates by synthetic
division and use any zeros to factor the polynomial.
x2 2x 1x2 2x 8
x 4 4x3 3x2 14x 8
(f)
y
(−2, 0)
−3
2
(1, 0)
−1
−4
−6
−8
− 10
2
(4, 0)
x
3
5
358
Chapter 3
Polynomial Functions
127. (a) f x x b ix b i x2 b
128. (a) f x cannot have this graph since it also has a zero at
x 0.
(b) f x x a bi x a bi
(b) g x cannot have this graph since it is a quadratic
function. Its graph is a parabola.
x a bi x a bi
(c) h x is the correct function. It has two real zeros,
x 2 and x 3.5, and it has a degree of four,
needed to yield three turning points.
x a2 bi2
x 2ax 2
a2
b2
(d) k x cannot have this graph since it also has a zero at
x 1. In addition, since it is only of degree three,
it would have at most two turning points.
129. 3 6i 8 3i 3 6i 8 3i 11 9i
130. 12 5i 16i 12 11i
131. 6 2i1 7i 6 42i 2i 14i2 20 40i
132. 9 5i9 5i 81 25i 2 81 25 106
133. gx f x 2
134. gx f x 2
135. gx 2f x
y
y
y
(6, 4)
4
3
3
2
(2, 2)
2
x
−1
(0, 0) 2
3
4
5
6
(2, 0)
(0, 4)
x
−2 − 1
6
1
2
3
4
(2, 4)
−2
Horizontal shift two units
to the right
(− 2, 0)
Vertical shift two units downward
Vertical stretch (each y-value is
multiplied by 2)
y
(−4, 4)
8
3
6
(−2, 2)
1
4
(0, 2)
(2, 0)
(1, 2)
x
1
(0, 2)
(8, 4)
(4, 2)
2
x
−2
(− 4, 0)
(− 1, 0)
−2
Reflection in the y-axis
−1
x
1
−2
2
4
6
8
2
Horizontal shrink each x-value is
1
multiplied by 2 Mathematical Modeling and Variation
You should know the following the following terms and formulas.
■ Direct variation (varies directly, directly proportional)
■
(a) y kx
(b) y kxn as nth power
Inverse variation (varies inversely, inversely proportional)
■
(a) y kx
(b) y kxn as nth power
Joint variation (varies jointly, jointly proportional)
■
(a) z kxy
(b) z kxnym as nth power of x and mth power of y
k is called the constant of proportionality.
■
8
10
4
(0, 2)
Section 3.5
6
y
(2, 4)
3
− 4 − 3 − 2 −1
−1
4
1
138. gx f 2 x
y
4
2
−2
137. gx f 2x
136. gx f x
x
(−2, − 2)
−3
−2
(4, 8)
8
(0, 0)
(4, 2)
1
10
(4, 2)
Least Squares Regression Line y ax b. Use your calculator or computer to enter
the data points and to find the “best-fitting”linear model.
Horizontal stretch each x-value
is multiplied by 2
Section 3.5
Mathematical Modeling and Variation
359
Vocabulary Check
1. variation; regression
2. sum of square differences
3. correlation coefficient
4. directly proportional
5. constant of variation
6. directly proportional
7. inverse
8. combined
9. jointly proportional
1. y 1767.0t 123,916
Actual Number
(in thousands)
Model
(in thousands)
1992
128,105
127,450
1993
129,200
129,217
145,000
Number of employees
(in thousands)
Year
y
140,000
135,000
130,000
125,000
t
2
1994
131,056
130,984
1995
132,304
132,751
1996
133,943
134,518
1997
136,297
136,285
1998
137,673
138,052
1999
139,368
139,819
2000
142,583
141,586
2001
143,734
143,353
2002
144,863
145,120
y
Winning time (in minutes)
2.
5.0
4.8
4.6
4.4
4.2
4.0
3.8
6
8 10 12
Year (2 ↔ 1992)
The model is a good fit for the actual data.
y
3.
The model is not a “good
fit” for the actual data. It
appears that another type
of model may be a better
fit.
5.4
4
5
4
2
1
t
0
8 16 24 32 40 48 56
x
Year (0 ↔ 1950)
1
2
3
4
5
Using the points 0, 3 and 4, 4,
we have y 14x 3.
4.
y
5.
y
y
6.
5
5
5
4
4
4
3
3
2
2
1
1
x
1
2
3
4
5
The line appears to pass through
2, 5.5 and 6, 0.5, so its equation
is y 54x 8.
1
x
1
2
3
4
5
Using the points 2, 2 and 4, 1,
we have y 12x 3.
x
1
2
3
4
5
The line appears to pass through
0, 2 and 3, 3 so its equation is
y 13x 2.
360
Chapter 3
Polynomial Functions
(c) y 1.03t 130.27
7. (a) and (b)
y
(d) The models are similar.
240
(e) When t 108, we have:
Length (in feet)
220
200
Model in part (b): 238 feet
180
160
Model in part (c): 241.51 feet
140
(f) Answers will vary.
t
12
36
60
84
108
Year (12 ↔ 1912)
y t 130
(b) The line appears to pass through 7, 1151.6 and
10, 1906.0, so the equation is about y 251.5x 609.
8. (a) and (b)
Total revenues
(in millions of dollars)
y
(c) y 251.56x 608.79
3000
2500
2000
(d) Answers will vary.
1500
(e) Using the model in (b), y 251.515 609 $3164.6 million.
1000
Using the model in (c), y 251.5615 608.79 $3165.2 million.
500
t
5
7
9
11
13
(f) Answers will vary.
Year (5 ↔ 1995)
10. (a) y 0.4306x 67.708
9. (a) and (c)
800
(b)
800
5
5
14
0
110
60
14
0
90
90
The model is a good fit to the actual data. r 0.98
The model is a good fit to the data. r 0.97
(b) S 38.4t 224
(c) y 0.430690 67.708 106.5 million
(d) For 2005, use t 15: S $800.4 million
(d) For every increase of one million households with
cable TV, there is a 0.43 million increase in the
number of households with color TV.
For 2007, use t 17: S $877.3 million
(e) Each year the annual gross ticket sales for Broadway
shows in New York City increase by approximately
$38.4 million.
11. The graph appears to represent y 4x, so y varies
inversely as x.
12. The graph appears to represent y 2x which is a direct
variation.
13. k 1
14. k 2
x
2
y
kx2
4
4
6
16
8
36
64
3
10
x
100
y kx
2
y
2
4
6
8
10
8
32
72
128
200
y
100
200
80
160
60
120
40
80
20
40
x
2
4
6
8
10
x
2
4
6
8
10
Section 3.5
Mathematical Modeling and Variation
1
16. k 4
15. k 12
x
2
4
6
8
10
x
2
4
6
8
10
y kx2
2
8
18
32
50
y kx2
1
4
9
16
25
y
y
50
25
40
20
30
15
20
10
10
5
x
2
4
6
8
x
10
2
17. k 2
6
8
10
18. k 5
x
k
x2
y
4
2
4
6
8
10
x
1
2
1
8
1
18
1
32
1
50
y
y
k
x2
2
4
6
8
10
5
4
5
16
5
36
5
64
1
20
y
5
10
5
4
4
10
1
3
10
3
4
2
10
2
4
1
10
1
4
x
2
4
6
8
x
10
2
19. k 10
6
8
10
20. k 20
x
y
4
k
x2
2
4
6
8
10
x
5
2
5
8
5
18
5
32
1
10
y
y
k
x2
2
4
6
8
10
5
5
4
5
9
5
16
1
5
y
5
2
5
2
4
3
2
3
1
2
1
2
1
x
x
2
4
6
8
10
21. The table represents the equation y 5x.
2
4
6
8
10
22. The table represents the equation y 25 x.
361
362
Chapter 3
23.
y kx
Polynomial Functions
y
k
x
24 k
5
24.
7 k10
7
k
10
12
k
5
Thus, y 120x. This equation
checks with the other points given
in the table.
This equation checks with the other
points given in the table.
y kx
y kx
27.
14 k2
290
3
y 205x
I kP
31.
187.50 k5000
87.50 k2500
0.035 k
k
y
290
3 x
y kx
33
13
I 0.035P
32.
53
14
33
13 x
53
5 gallons: y 145 18.9 liters
25 gallons: y 1425 94.6 liters
53
When x 20 inches,
y 50.8 centimeters.
y kx
34.
y kx
10.22 k145.99
5520 k150,000
0.0368 k
0.07 k
k
y 53
14 x
When x 10 inches,
y 25.4 centimeters.
33.
y kx
53 k14
k
y
I 0.0375P
I kP
29.
33 k13
0.0375 k
12
x
5
580 k6
205 k
y 7x
y
y kx
28.
2050 k10
7k
30.
12 k5
120 k
7
y x
10
26.
y kx
25.
d kF
35.
0.15 k265
3
5300
k
y 0.0368x
y 0.07x
y 0.0368200,000
y 0.07540.50
3
(a) d 5300
90 0.05 meter
y 37.84
3
(b) 0.1 5300
F
$7360
3
d 5300
F
530
3
The sales tax is $37.84.
The property tax is $7360.
F
F 17623 newtons
36.
d kF
37.
1.9 k25 ⇒ k 0.076
0.12 k220
3
5500
k
d 0.076F
3
d 5500
F
When the distance compressed is
3 inches, we have
3
0.16 5500
F
880
3
3 0.076F
F
The required force is
d kF
1
2933
newtons.
F 39.47.
No child over 39.47 pounds should
use the toy.
38. d kF
1 k15
1
k 15
1
d 15
F
8
2
1
15
F
F 60 lb per spring
Combined lifting force 2F
120 lbs
39. A kr2
40. V ke3
44. z kx2y3
45. P Mathematical Modeling and Variation
k
x2
42. h 46. R kT Te
47. F 41. y k
V
50. S 4 r 2
49. A 12bh
52. V r 2h
53. r The volume of a right circular
cylinder is jointly proportional to
the height and the square of the
radius.
A kr2
56.
k
A r 2
k
x
3
k
25
75
x
F 14rs3
61.
z
kx2
y
6
k62
4
81
24
k
36
2 27
k
3
2
k
3
18 k
P
18x
y2
k
4
28
x
4158 k1133
z 2xy
42 k
7
F krs3
k 14
28
3
k
x
y
2k
28 k42
2
3
9
y
28 k
59.
kx
y2
kgW
varies directly as the square root
of g and inversely as the square
root of W. Note: The constant of
proportionality is k.
75 k
64 k48
P
The volume of a sphere varies
directly as the cube of its radius.
57.
z kxy
60.
kg
r2
48. R kSS L
54. y
z
363
51. V 43r3
d
t
y
43. F km1m2
r2
Average speed is directly
proportional to the distance and
inversely proportional to the time.
9 k32
58.
k
s
The surface area of a sphere
varies directly as the square of
the radius r.
The area of a triangle is jointly
proportional to its base and height.
55.
Section 3.5
23x 2 2x2
y
3y
62.
v
1.5 k pq
s2
k4.16.3
1.22
1.51.44
k
4.16.3
2.16
k
25.83
24
k
287
24pq
v
287s 2
364
Chapter 3
63.
d kv2
0.02 k
4
1
Polynomial Functions
64. d kv 2
If the velocity is doubled:
2
d k2v2
k 0.32
d k 4v 2
d 0.32v2
4kv 2
4
kv 2
0.12 0.32v2
v2 v
0.12 3
0.32 8
3
22
d increases by a factor of 4 when velocity is doubled.
6
4
0.61 mihr
d 2
kl
, A r 2 A
4
4kl
r
d 2
r
65.
66.17 66. From Exercise 65:
k 5.73 108.
41000k
0.0126
12 33.5 45.73 108l
d 2
d
45.73 r 10
d
10 14
45.730.05
2
k 5.73 108
r
r
45.73 108l
0.0126
12 2
33.5 l
45.73 108
8
l
8
d 0.0045 feet 0.054 inch
45.73 108l
0.0126 2
12
0.0126 2
12
l 506 feet
W kmh
67.
68.
2116.8 k1201.8
k
2116.8
9.8
1201.8
W 9.8mh
When m 100 kilograms and h 1.5 meters, we have
W 9.81001.5 1470 joules.
P kA k r 2 k
8.78 k
2
9
2
d
2
2
48.78
k
81
k 0.138
However, we do not obtain $11.78 when d 12 inches.
P 0.138
2
Instead, k 12
2
$15.61
11.78
0.104.
36
For the 15-inch pizza, we have k 414.18
0.080.
225
The price is not directly proportional to the surface area.
The best buy is the 15-inch pizza.
69. v v
k
A
k
4 k
0.75A 3 A
The velocity is increased by one-third.
Section 3.5
70. Load 365
kwd 2
l
(a) load k 2wd 2 kwd 2
2l
l
(b) load (c) load k 2w2d 2 8kwd 2
l
l
The safe load is eight times as great.
The safe load is unchanged.
k 2w2d 2 4kwd 2
2l
l
(d) load kwd22 14kwd2
l
l
The safe load is one-fourth as great.
The safe load is four times as great.
71. (a)
Mathematical Modeling and Variation
Temperature (in °C)
C
5
4
3
2
1
d
2000
4000
Depth (in meters)
(b) Yes, the data appears to be modeled (approximately) by the inverse proportion model.
4.2 k1
1000
1.9 4200 k1
(c) Mean: k (d)
k2
2000
3800 k2
1.4 k3
3000
1.2 4200 k3
k4
4000
4800 k4
0.9 k5
5000
4500 k5
4200 3800 4200 4800 4500
4300
4300, Model: C 5
d
(e) 3 6
d
0
4300
d
1
4300
1433 meters
3
3
6000
0
72. (a)
73. y y
262.76
x2.12
74. I k
d2
Length
(in centimeters)
7
(a)
6
When the distance is doubled:
0.2
5
4
I
3
2
1
F
2
4
6
8
10 12
25
55
0
Force (in pounds)
(b) It appears to fit Hooke’s Law.
k
6.9
0.575
12
(c) y kF
9 0.575F
F 15.7 pounds
(b) y 262.76
252.12
0.2857 microwatts per sq.cm.
k
k
.
2d2 4d 2
The illumination is one-fourth as
great. The model given in Exercise
73 is very close to I kd 2.
The difference is probably due to
measurement error.
366
Chapter 3
Polynomial Functions
75. False. y will increase if k is
positive and y will decrease if k
is negative.
77. False. The closer the value of r is
to 1, the better the fit.
76. False. E is jointly proportional (not
“directly proportional”) to the
mass of an object and the square
of its velocity.
78. (a) The data shown could be represented by a linear model which would be a good approximation.
(b) The points do not follow a linear pattern. A linear model would be a poor approximation.
A quadratic model would be better.
(c) The points do not follow a linear pattern. A linear model would be a poor approximation.
(d) The data shown could be represented by a linear model which would be a good approximation.
79. The accuracy of the model in predicting prize winnings is
questionable because the model is based on limited data.
80. Answers will vary.
81. 3x 2 > 17
82. 7x 10 ≤ 1 x
8x ≤ 11
3x > 15
x ≥
x > 5
11
8
x
3
4
5
6
7
8
11
8
9
x
−1
83.
2x 1
< 9
4 < x < 5
x
1
2
3
3
4
5
4
4 3x
≥ 5
4 3x ≤ 5
8 < 2x < 10
0
2
84. 4 3x 7 ≥ 12
9 < 2x 1 < 9
−4 −3 −2 −1
1
0
or 4 3x ≥ 5
3x ≤ 9
or
x ≥ 3
or
3x ≥ 1
x ≤ 13
5
−1
3
x
−2 −1
85. f x x2 5
x3
(a) f 0 02 5
5
03
3
32 5
14
7
(b) f 3 3 3
6
3
42 5
(c) f 4 4 3 21
86. f x 0
1
2
3
10,
x
6x 1,
2
2
4
x ≥ 2
x < 2
(a) f 2 22 10 4 10 6
(b) f 1 12 10 1 10 9
(c) f 8 682 1 384 1 383
Review Exercises for Chapter 3
Review Exercises for Chapter 3
1. (a) y 2x2
(b) y 2x2
Vertical stretch
Vertical stretch and a reflection in the x-axis
y
y
4
4
3
3
2
2
1
x
− 4 − 3 −2 −1
−1
1
2
3
− 4 −3 − 2 − 1
4
x
1
2
3
4
−2
−3
−3
−4
−4
(d) y x 22
(c) y x2 2
Vertical shift two units upward
Horizontal shift two units to the left
y
(a)
y
4
4
3
1
1
x
− 4 − 3 −2 − 1
−1
1
2
3
4
−2
−2
−3
−3
−4
−4
2. (a) y x 2 4
1
2
3
4
(b) y 4 x 2
Vertical shift four units downward
y
Reflection in the x-axis and a vertical shift
four units upward
y
3
2
−4 −3
x
−4 −3 −2 −1
−1
5
x
−1
−1
1
3
3
4
2
−2
1
−4 −3
−5
x
−1
−1
1
3
4
−2
−3
(c) y x 3 2
(d) y 12 x 2 1
Horizontal shift three units to the right
y
Vertical shrink (each y-value is multiplied by 2 ,
and a vertical shift one unit downward
1
5
y
4
3
4
2
3
2
1
− 3 −2 − 1
−1
−2
−3
1
x
1
2
3
4
5
x
−4 −3 −2
2
−2
−3
−4
3
4
367
368
Chapter 3
Polynomial Functions
3. gx x2 2x
4. f x 6x x2
y
x2 2x 1 1
7
x 12 1
5
x2 6x 9 9
6
6
Vertex: 3, 9
3
Axis of symmetry: x 1
x
−3 −2 −1
−1
0 x 2 2x xx 2
8
x 32 9
4
Vertex: 1, 1
y
10
1
2
3
4
5
6
2
Axis of symmetry: x 3
0 6x x2 x6 x
−2
4
x
−2
2
4
8
10
−2
x-intercepts: 0, 0, 6, 0
x-intercepts: 0, 0, 2, 0
6. hx 3 4x x2
5. f x x2 8x 10
x2 4x 3
x2 8x 16 16 10
x 42 6
x2 4x 4 4 3
y
y
x 22 7
Vertex: 4, 6
Axis of symmetry: x 4
0 x 42 6
−8
x
−4
2
−2
x 42 6
8
Vertex: 2, 7
6
4
Axis of symmetry: x 2
−4
x 4 ± 6
10
x 22 7
2
0 3 4x x2
−6
2
x
−2
2
4
6
8
10
0 x2 4x 3
x 4 ± 6
x-intercepts: 4 ± 6, 0
x
4 ± 42 413
21
4 ± 28
2 ± 7
2
x-intercepts: 2 ± 7, 0
7. f t 2t2 4t 1
8. f x x2 8x 12
2t2 2t 1 1 1
x2 8x 16 16 12
2t 12 1 1
x 42 4
2t 12 3
Vertex: 1, 3
6
Axis of symmetry: t 1
4
2t 12 3
2
32
t1 ±
t-intercepts:
1
4
t
1
2
3
4
5
6
x-intercepts: 2, 0, 6, 0
2
x
−2
4
−2
−4
3
6
±
6
0 x 2x 6
1
− 3 −2 − 1
8
0 x2 8x 12
3
0 2t 1 3
y
Axis of symmetry: x 4
5
2
t1±
Vertex: 4, 4
y
6
2
,0
8
10
Review Exercises for Chapter 3
9. hx 4x2 4x 13
10. f x x2 6x 1
4x2 x 13
4 x2 x 4
x2
Vertex:
x2 6x 9 9 1
x 32 8
1 1
13
4 4
Vertex: 3, 8
1
x
1 13
4
4 x
1
2
2
Axis of symmetry: x 3
12
21, 12
2
x
15
1
Axis of symmetry: x 2
x 21
0 x2 6x 1
y
20
1
04 x
2
369
10
12
−3
−2
6 ± 32
3 ± 22
2
y
x-intercepts: 3 ± 22, 0
5
2
6 ± 62 411
21
x
−1
1
2
2
x
−2
3
2
4
8
10
−2
−4
3
−6
No real zeros
−8
x-intercepts: none
12. f x 4x2 4x 5
11. hx x2 5x 4
x2 5x 5
2
5
2
x
x
Vertex:
2
2
25 25
4
4
4
4 x2 x 25 16
4
4
−8
41
4
x 2 −6
−4
2
4 x
−2
−4
25, 414
Vertex:
5
Axis of symmetry: x 2
− 10
5 ±2
41
,0
5 ± 41
.
2
Vertex:
1
5
x
3
2
2
4
10
8
2
−8 −6 −4 −2
The equation has no real zeros.
No x-intercepts
y
4
2
By the Quadratic Formula, x 1
25 25
x2 5x 4
3
4
4
1
2
21, 4
1
13. f x x2 5x 4
3
y
x
−2
2
4
6
0 4x 2 4x 5
By the Quadratic Formula, x 1
5
x
3
2
1
1
Axis of symmetry: x 2
0 x 2 5x 4
x-intercepts:
2
12
x
−2
1
4
y
1 1 5
4 4 4
2
41
4
5 41
,
2 12
2
−8
−4
x
−2
2
−4
41
12
−6
0 x 2 5x 4
Axis of symmetry: x −6
By the Quadratic Formula, x 5
2
x-intercepts:
5 ±2
41
,0
5 ± 41
.
2
4 ± 8i
1
± i.
8
2
370
Chapter 3
Polynomial Functions
1
14. f x 6x2 24x 22
2
y
14
3x2 12x 11
12
10
3x2 4x 4 4 11
8
6
3x 22 34 11
4
2
3x 2 1
2
x
–6 –4 –2
Vertex: 2, 1
4
6
8 10
Axis of symmetry: x 2
0 3x2 12x 11
x
3
12 ± 122 4311 12 ± 12
2 ±
23
6
3
x-intercepts:
2 ±
3
3
,0
15. Vertex: 4, 1 ⇒ f x ax 42 1
16. Vertex: 2, 2 ⇒ f x ax 2 2 2
Point: 0, 3 ⇒ 3 a0 22 2
Point: 2, 1 ⇒ 1 a2 42 1
2 4a
3 4a 2
12 a
1 4a
1
4
1
Thus, f x 2x 42 1.
a
f x 14x 2 2 2
17. Vertex: 1, 4 ⇒ f x ax 12 4
18. Vertex: 2, 3 ⇒ f x ax 2 2 3
Point: 1, 6 ⇒ 6 a1 2 2 3
Point: 2, 3 ⇒ 3 a2 12 4
6 9a 3
1a
3 9a
Thus, f x x 12 4.
1
3
f x 19. (a) A xy x
8x
since
2
x 2y 8 0 ⇒ y (c)
8x
.
2
(b) Since the figure is in the first quadrant and x and y
must be positive, the domain of
Ax
(d)
8x
2
is 0 < x < 8.
9
0
8
0
The maximum area of 8 occurs at the vertex when
x 4 and y —CONTINUED—
84
2.
2
x
1
3 x
a
2 2 3
y
Area
1
4
1
2 1
14 121 72
2
4 122
24 122 6
3
4 123
34 123 152
4
4 124
44 124 8
5
4 125
54 125 152
6
4 126
64 126 6
Review Exercises for Chapter 3
371
19. —CONTINUED—
(e) A x
1
1
8x
8x x 2 x 2 8x
2
2
2
1
1
1
x2 8x 16 16 x 42 16 x 42 8
2
2
2
84
The maximum area of 8 occurs when x 4 and y 2.
2
(b) 2x 2y 200
20. (a)
(c) Area 100x x2
x2 100x 2500 2500
x y 100
y
x 502 2500
y 100 x
x
Area xy
x 502 2500
x100 x
The maximum area occurs at the vertex when x 50
and y 100 50 50. The dimensions with the
maximum area are x 50 meters and y 50 meters.
100x x2
21. R 10p2 800p
(a) R20 $12,000
22. Let x the number of $30 increases in rent. Then,
Rent 540 30x
R25 $13,750
Number of occupied units 50 x
R30 $15,000
Revenue Number of occupied unitsrent
(b) The maximum revenue occurs at the vertex of the
parabola.
b
800
$40
2a 210
R40 $16,000
R 50 x540 30x
Cost Number of occupied units$18
C 50 x18
Profit Revenue Cost
P 50 x540 30x 50 x18
The revenue is maximum when the price is $40 per
unit. The maximum revenue is $16,000.
27,000 960x 30x2 900 18x
30x2 978x 26,100
The maximum profit occurs at the vertex.
b
978
16.3 16 increases
2a 230
The corresponding rent is 540 3016 $1020.
The minimum cost occurs at the vertex of the parabola.
Vertex: b
120
1091 units
2a
20.055
Approximately 1091 units should be produced each day to
yield a minimum cost.
24. 26 0.107x2 5.68x 48.5
0 0.107x2 5.68x 74.5
x
5.68 ± 5.682 40.10774.5
20.107
x 23.7, 29.4
The age of the bride is
approximately 24 years
when the age of the groom
is 26 years.
y
27
Age of groom
23. C 70,000 120x 0.055x2
26
25
24
23
22
x
20 21 22 23 24 25
Age of bride
372
Chapter 3
Polynomial Functions
27. y x 4, f x 2 x 4
26. y x3, f x 4x3
25. y x3, f x x 43
y
y
y
3
5
4
3
2
1
3
2
1
x
−2
1 2 3 4
−3
6 7
−2
−1
−3
−4
1
x
1
2
−3
3
−2
−2
−3
−3
y
5
4
3
2
1
6
5
4
3
8
6
4
2
x
−2
1
3 4 5 6 7
1
x
−3 −2 −1
1
2
3
4
5
3
1
30. y x5, f x 2x5 3
y
y
2
Transformation: Reflection in
the x-axis and a vertical shift
two units upward
29. y x5, f x x 35
28. y x4, f x 2x 24
1
−1
f x is a reflection in the x-axis
and a vertical stretch of the graph
of y x3.
Transformation: Reflection in
the x-axis and a horizontal shift
four units to the right
x
−2
−1
−6
6
−4
x
−2
2
4
6
−2
−5
−3
f x is a vertical stretch and a
shift to the right two units of the
graph of y x4.
32. f x 12x3 2x
31. f x x2 6x 9
The degree is even and the leading coefficient is negative.
The graph falls to the left and falls to the right.
The degree is even and the leading coefficient is positive.
The graph rises to the left and rises to the right.
35. f x 2x2 11x 21
0 2x 11x 21
−9
9
−40
−6
6
−5
x 3 of multiplicity 2
(even multiplicity)
Turning points: 1
Turning points: 2
37. f t t 3 3t
38. f x x3 8x2
3
0
0 t 3 3t
Turning points: 2
0 xx 3
3
2
Zeros: x 0 of multiplicity 1
(odd multiplicity)
3
Zeros: x 2, 7, all of
multiplicity 1 (odd multiplicity)
Zeros: t 0, ± 3 all of
multiplicity 1 (odd multiplicity)
The degree is odd and the leading coefficient is negative.
The graph rises to the left and falls to the right.
36. f x xx 3 2
20
2x 3x 7
0 t t 2 3
The degree is odd and the leading coefficient is positive.
The graph falls to the left and rises to the right.
34. hx x5 7x2 10x
33. g x 34x4 3x2 2
2
f x is a vertical shrink and a
vertical shift three units upward
of the graph of y x5.
Transformation: Horizontal shift
three units to the right
−5
4
−3
x3
8x2
10
−10
10
0 x 2x 8
Zeros: x 0 of multiplicity 2
(even multiplicity)
x 8 of multiplicity 1
(odd multiplicity)
Turning points: 2
−80
Review Exercises for Chapter 3
39. f x 12x 3 20x2
40. gx x4 x3 2x2
10
0 4x23x 5
−5
−4
0 x2x2 x 2
5
Zeros: x 0 of multiplicity 2
(even multiplicity)
5
x2x 1x 2
−5
−3
Zeros: x 0 of multiplicity 2 (even multiplicity)
5
x 3 of multiplicity 1
(odd multiplicity)
x 1 of multiplicity 1 (odd multiplicity)
x 2 of multiplicity 1 (odd multiplicity)
Turning points: 2
Turning points: 3
41. f x x3 x2 2
42. g x 2x3 4x2
(a) The degree is odd and the leading coefficient, 2, is
positive. The graph rises to the right and falls to the left.
(a) The degree is odd and the leading coefficient is
negative. The graph rises to the left and falls to
the right.
(b) gx 2x3 4x2
(b) Zero: x 1
0 2x3 4x2
x
3
2
1
0
1
2
0 2x2x 2
f x
34
10
0
2
2
6
0 x2x 2
The zeros are 0 and 2.
y
(d)
3
0 x4 x3 2x2
0 12x 3 20x2
(c)
373
(c)
4
3
2
x
3
2
1
0
1
gx
18
0
2
0
6
1
(−1, 0)
1
2
3
y
(d)
x
−4 −3 −2
4
4
3
−3
2
−4
(− 2, 0)
− 4 −3
(0, 0)
−1
−1
1
2
x
3
4
−2
−3
−4
44. h x 3x2 x4
43. f x xx3 x2 5x 3
(a) The degree is even and the leading coefficient, 1 , is
negative. The graph falls to the left and falls to the right.
(a) The degree is even and the leading coefficient is
positive. The graph rises to the left and rises to
the right.
(b) gx 3x2 x4
(b) Zeros: x 0, 1, 3
(c)
0 3x2 x4
x
4
3
2
1
0
1
2
3
f x
100
0
18
8
0
0
10
72
(−3, 0)
−4
3
−2 −1
The zeros are 0, 3, and 3.
(c)
y
(d)
0 x23 x2
(1, 0)
x
1
2
(0, 0)
3
x
2
1
0
1
2
hx
4
2
0
2
4
4
y
(d)
4
3
−15
−18
−21
(0, 0)
2
(− 3, 0(
−4 −3
( 3, 0(
−1
−1
−2
−3
−4
x
1
3
4
374
Chapter 3
Polynomial Functions
45. f x 3x3 x2 3
46. f x 0.25x3 3.65x 6.12
x
3
2
1
0
1
2
3
x
6
5
4
3
2
f x
87
25
1
3
5
23
75
f x
25.98
6.88
4.72
10.32
11.42
The zero is in the interval 1, 0.
x
1
0
1
2
3
4
Zero: x 0.900
f x
9.52
6.12
2.72
0.82
1.92
7.52
The only zero is in the interval 5, 4.
It is x 4.479.
47. f x x4 5x 1
48. f x 7x4 3x3 8x2 2
x
3
2
1
0
1
2
3
x
3
2
1
0
1
2
f x
95
25
5
1
5
5
65
f x
416
58
2
2
4
106
There are zeros in the intervals 2, 1 and 1, 0.
They are x 1.211 and x 0.509.
There are two zeros, one in the interval 1, 0 and one
in the interval 1, 2
Zeros: x 0.200, x 1.772
8x 5
49.
3x 2 ) 24x2 4
3
50.
x 8
3x 2 ) 4x 7
24x2 16x
4x 83
15x 8
29
3
15x 10
29
4x 7 4
3x 2 3 33x 2
2
Thus,
2
24x2 x 8
8x 5 .
3x 2
3x 2
5x 2
51.
x2
3x 1 )
5x3
13x2
3x2
52.
5x3 15x2 5x
4
3x4
3
3x2
2x2
6x 2
3x2
0
2x2
6x 2
3x2
3
3
0
Thus,
5x3 13x2 x 2
5x 2.
x2 3x 1
x2 3x 2
53.
x2 0x 2 ) x4 3x3 4x2 6x 3
3
3x4
3x2 3 2
x2 1
x 1
3x2 5x 8
54.
2x2 0x 1 ) 6x4 10x3 13x2 5x 2
6x4 0x3 3x2
x4 0x3 2x2
3x 3 2x2 6x
10x3 16x2 5x
3x3 0x2 6x
10x3 0x2 5x
2x2 0x 3
16x2 0x 2
2x2 0x 4
16x2 0x 8
1
1
x 3x 4x 6x 3
x2 3x 2 2
.
x2 2
x 2
4
Thus,
3
3
x 1 ) 3x 0x 0x2 0x 0
x2
2
2
10
6x4
10x 13x 5x 2
10
3x2 5x 8 2
2x2 1
2x 1
3
2
Review Exercises for Chapter 3
55. 2
6
4
12
27
16
18
22
0
8
6
8
11
4
8
56. 5
0.1
0.3
0.5
0.8
0.1
0.5
20
19.5
0
4
4
0.1x3 0.3x2 0.5
19.5
0.1x 2 0.8x 4 x5
x5
Thus,
6x4 4x3 27x2 18x
8
6x3 8x2 11x 4 .
x2
x2
57. 4
2
19
8
38
44
24
24
2
11
6
0
58. 3
3
20
9
11
3
12
12
0
29
33
4
3x3 20x2 29x 12
3x2 11x 4
x3
2x3 19x2 38x 24
Thus,
2x2 11x 6.
x4
59. f x 20x 4 9x 3 14x2 3x
(a) 1
20
9
20
14
11
3
3
0
0
20
11
3
0
0
3
4
20
9
15
14
18
3
3
0
0
20
24
4
0
0
3
Yes, x 1 is a zero of f.
(c) 0
(b)
Yes, x 4 is a zero of f.
20
9
0
14
0
3
0
0
0
20
9
14
3
0
(d) 1
Yes, x 0 is a zero of f.
20
9
20
14
29
3
15
0
12
20
29
15
12
12
No, x 1 is not a zero of f.
60. f x 3x3 8x2 20x 16
(a) 4
8
12
4
3
3
20
16
4
(b) 4
16
16
0
3
(c)
8
2
6
3
3
20
4
24
20
80
60
16
240
224
No, x 4 is not a zero of f.
Yes, x 4 is a zero of f.
2
3
8
12
20
3
(d) 1
16
16
0
8
3
11
3
3
20
11
9
16
9
25
No, x 1 is not a zero of f.
2
3
Yes, x is a zero of f.
61. f x x4 10x3 24x2 20x 44
(a) 3
1
10
3
24
21
20
135
44
465
1
7
45
155
421
Thus, f 3 421.
(b) 1
1
10
1
24
9
20
33
44
53
1
9
33
53
9
f 1 9
375
376
Chapter 3
Polynomial Functions
62. gt 2t5 5t4 8t 20
(a) 4
5
8
13
2
2
0
52
52
0
208
208
8
832
824
20
3296
3276
Thus, g4 3276.
(b) 2
2
5
22
0
52 4
0
10 42
8
102 8
20
20
2
5 22
5 2 4
10 42
102
0
Thus, g2 0.
63. f x x 3 4x2 25x 28; Factor: x 4
(a) 4
1
4
4
25
32
28
28
1
8
7
0
64. f x 2x3 11x2 21x 90
(a) 6
11
12
1
2
21
6
15
(b) 2x2 x 15 2x 5x 3
(b) x2 8x 7 x 7x 1
The remaining factors are 2x 5 and x 3.
The remaining factors of f are x 7 and x 1.
(c) f x 2x 5x 3x 6
(c) f x x 3 4x2 25x 28
5
(d) Zeros: x 2, 3, 6
x 7x 1x 4
(d) Zeros: 7, 1, 4
(e)
50
−7
80
−8
90
90
0
Yes, x 6 is a factor of f x.
Yes, x 4 is a factor of f x.
(e)
2
5
5
− 100
− 60
65. f x x 4 4x 3 7x2 22x 24
66. f x x4 11x3 41x2 61x 30
Factors: x 2, x 3
(a) 2
1
1
3
(a) 2
4
2
7
12
22
10
24
24
6
5
12
0
1
6
3
5
9
12
12
1
3
4
0
Both are factors since the remainders are zero.
(b) x2 3x 4 x 1x 4
The remaining factors are x 1 and x 4.
(c) f x x 1x 4x 2x 3
11
2
41
18
61
46
30
30
1
9
23
15
0
1
9
5
23
20
15
15
1
4
3
0
Yes, x 2 and x 5 are both factors of f x.
(b) x2 4x 3 x 1x 3
The remaining factors are x 1 and x 3.
(c) f x x 1x 3x 2x 5
(d) Zeros: x 1, 2, 3, 5
(e)
(d) Zeros: 2, 1, 3, 4
(e)
5
1
4
−6
40
12
−8
−3
5
− 10
Review Exercises for Chapter 3
67. f x 3xx 22
68. f x x 4x 92
Zeros: x 0, x 2
69. f x x2 9x 8
70. f x x 3 6x
x 1x 8
Zeros: x 9, 4
xx2 6
Zeros: x 1, x 8
71. f x x 4x 6x 2ix 2i
Zeros: x 4, x 6, x 2i, x 2i
73. f x 4x3 8x2 3x 15
Possible rational zeros: ± 1, ± 2, ± 3, ± 6, ± 9, ± 18
2
1
3
1
21
3
18
72. f x x 8x 52x 3 ix 3 i
Zeros: x 5, 8, 3 ± i
1
2
4
18
18
0
Possible rational zeros: ± 1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15,
1
2
5
10
± 30, ± 3 , ± 3 , ± 3 , ± 3
1
x3 2x2 21x 18 x 1x2 3x 18
x 1x 6x 3
3
20
3
7
23
30
30
3
23
30
0
So, f x 3x 3 20x2 7x 30
x 13x2 23x 30
The zeros of f x are x 1, x 6, and x 3.
x 13x 5x 6
0 x 13x 5x 6
5
Zeros: x 1, 3, 6
77. f x x3 10x2 17x 8
78. f x x 3 9x2 24x 20
Possible rational zeros: ± 1, ± 2, ± 4, ± 8
1
10
1
9
1
1
x3
17
9
8
8
8
0
4
1
1
4
4
0
20
20
1
4
4
0
So, f x x 3 9x2 24x 20
x 5x 22.
11
12
1
1
3
4
1
0
1
4
4
0
12
12
0
x x 11x2 x 12 x 3x 4x2 1
4
24
20
x 12x 8
Possible rational zeros: ± 1, ± 2, ± 3, ± 4, ± 6, ± 12
1
9
5
x 5x2 4x 4
79. f x x4 x3 11x2 x 12
1
3
4
1
x 1x 1x 8
The zeros of f x are x 1 and x 8.
1
Possible rational zeros: ± 1, ± 2, ± 4, ± 5, ± 10, ± 20
5
10x2 17x 8 x 1x2 9x 8
3
8
Possible rational zeros: ± 1, ± 2, ± 4, ± 8, ± 3, ± 3, ± 3, ± 3
76. f x 3x 3 20x2 7x 30
75. f x x3 2x2 21x 18
1
Zeros: x 0, ± 6i
74. f x 3x4 4x3 5x2 8
1
3
5
15
Possible rational zeros: ± 1, ± 3, ± 5, ± 15, ± 2, ± 2, ± 2, ± 2 ,
1
3
5
15
± 4, ± 4, ± 4, ± 4
1
377
3
The real zeros of f x are x 3, and x 4.
Zeros: x 5, 2
378
Chapter 3
Polynomial Functions
80. f x 25x 4 25x 3 154x2 4x 24
Possible rational zeros: ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24, ± 15, ± 25, ± 35, ± 45, ± 65, ± 85, ± 12
5,
24
1
2
3
4
6
8
12
24
± 5 , ± 25 , ± 25 , ± 25 , ± 25 , ± 25 , ± 25 , ± 25 , ± 25
3
25
25
75
154
150
4
12
24
24
25
25
50
50
50
4
8
4
8
0 8
0
2
25
0 4
0
So, f x 25x 4 25x 3 154x2 4x 24
x 3x 225x2 4
x 3x 25x 25x 2.
Zeros: x 3, 2, ± 25
81. f x 3x 23 x 4x 3 ix 3 i
3x 2x 4
x2
3
Multiply by 3 to clear the fraction.
Since 3i is a zero, so is 3i.
3x2 14x 8x2 3
3x4 14x3 17x2 42x 24
2
Note: f x a3x4 14x3 17x2 42x 24, where a is any real nonzero number, has zeros 3, 4, and ± 3 i.
82. Since 1 2i is a zero and the coefficients are real,
1 2i must also be a zero.
f x x 2x 3x 1 2ix 1 2i
83. f x x3 4x2 x 4, Zero: i
Since i is a zero, so is i.
i
1
4
i
1
1 4i
4
4
1
4 i
4i
0
x2 x 6x 12 4
x2 x 6x2 2x 5
x4
x 3x 17x 30
3
2
i
1
4 i
i
4i
4i
1
4
0
f x x ix ix 4, Zeros: x ± i, 4
84. hx x3 2x2 16x 32
85. g x 2x 4 3x 3 13x2 37x 15, Zero: 2 i
Since 4i is a zero, so is 4i.
4i
4i
Since 2 i is a zero, so is 2 i
1
2
4i
16
16 8i
32
32
1
2 4i
8i
0
1
2 4i
4i
8i
8i
0
2
1
hx x 4ix 4ix 2
Zeros: x ± 4i, 2
2i
2i
2
3
4 2i
13
5i
37
31 3i
15
15
2
1 2i
13 5i
6 3i
0
2
1 2i
4 2i
13 5i
10 5i
6 3i
6 3i
2
5
3
0
gx x 2 ix 2 i2x2 5x 3
x 2 ix 2 i2x 1x 3
Zeros: x 2 ± i, 12, 3
Review Exercises for Chapter 3
379
4
3
2
86. f x 4x 11x 14x 6x
x4x3 11x2 14x 6
One zero is x 0. Since 1 i is a zero, so is 1 i.
1i
1i
4
11
4 4i
14
11 3i
6
6
4
7 4i
3 3i
0
4
7 4i
4 4i
4
3
3 3i
3 3i
f x xx 1 ix 1 i4x 3
xx 1 ix 1 i4x 3
Zeros: 0, 34, 1 i, 1 i
87. f x x3 4x2 5x
88. gx x3 7x2 36
xx2 4x 5
2
1
7
2
0
18
36
36
1
9
18
0
xx 5x 1
Zeros: x 0, 5, 1
The zeros of x2 9x 18 x 3x 6 are
x 3, 6. The zeros of gx are 2, 3, 6.
gx x 2x 3x 6
89. g x x 4 4x 3 3x2 40x 208, Zero: x 4
4
4
1
4
4
3
0
40
12
208
208
1
0
3
52
0
1
0
4
3
16
52
52
1
4
13
0
90. f x x4 8x3 8x2 72x 153
3
1
8
3
8
33
72
123
153
153
1
11
41
51
0
3
1
11
3
41
24
1
8
17
51
51
gx x 42x2 4x 13
By the Quadratic Formula, the zeros of x2 8x 17 are
By the Quadratic Formula the zeros of x2 4x 13 are
x 2 ± 3i. The zeros of gx are x 4 of multiplicity
2, and x 2 ± 3i.
x
gx x 42x 2 3ix 2 3i
x 42x 2 3ix 2 3i
91. f x x4 2x 1
(a)
8 ± 82 4117 8 ± 4
4 ± i.
21
2
The zeros of f x are 3, 3, 4 i, 4 i.
f x x 3x 3x 4 ix 4 i
92. gx x3 3x2 3x 2
(a)
7
6
−6
−6
6
6
−1
−2
(b) The graph has two x-intercepts, so there are
two real zeros.
(b) One real zero because the graph has only one
x-intercept.
(c) The zeros are x 1 and x 0.54.
(c) The zero is x 0.44.
380
Chapter 3
Polynomial Functions
93. hx x3 6x2 12x 10
(a)
94. f x x5 2x3 3x 20
(a)
4
−6
10
−3
12
5
−8
− 30
(b) The graph has one x-intercept, so there is one
real zero.
(b) One real zero because the graph has only one
x-intercept.
(c) x 3.26
(c) The zero is x 1.72.
95. g x 5x3 3x2 6x 9
96. hx 2x5 4x3 2x2 5
g x has two variations in sign, so g has either two or no
positive real zeros.
hx has three variations in sign, so h has either three or
one positive real zeros.
g x 5x3 3x2 6x 9
hx 2x5 4x3 2x2 5
g x has one variation in sign, so g has one negative
real zero.
3
4
4
3
5
4
1
hx has two variations in sign, so h has either two or
no negative real zeros.
98. gx 2x3 5x2 14x 8
97. f x 4x3 3x2 4x 3
(a) 1
2x5 4x3 2x2 5
(a) 8
4
3
1
4
1
3
54
4
4
5
17
4
(b) 4
Median income
(in thousands of dollars)
2
14
52
5
8
8
152
100. (a) and (b)
y
(a)
8
592
2 13
38 144
Since the last row entries alternate in sign, x 4 is a
lower bound.
Since the last row entries alternate in sign,
x 41 is a lower bound.
99. I 2.09t 37.2
14
88
2
11
74 600
Since the last row has all positive entries, x 8 is an
upper bound.
4
1 5
2
Since the last row has all positive entries,
x 1 is an upper bound.
(b) 14
5
16
2
8000
65
60
55
5
2000
50
45
t
5 6 7 8 9 10 11 12
Year (5 ↔ 1995)
(b) The model is a good fit to the actual data. r 0.992
13
(b) S 627.02t 346
The model is a good fit. r 0.995
(c) S 627.0218 346 $10,940.36 million
(d) The factory sales of electronic gaming software in
the U.S. increases by $627.02 million each year.
Problem Solving for Chapter 3
101. D km
102.
4 k2.5
8
5
103. F ks2
P kS 3
750 k273
k
D 85m
If the speed is doubled,
k 0.03810
F k2s2
P 0.03810403
F 4ks2.
In 2 miles,
D 852 3.2 kilometers.
381
2438.7 kilowatts
The force will be changed by a
factor of 4.
In 10 miles,
D 8510 16 kilometers.
104.
k
p
105. T k
800 5
3
x
C khw2
106.
28.80 k1662
k
65
k 0.05
k 365 195
k 4000
x
k
r
4000
667 boxes
6
T
C 0.051482
195
r
$44.80
When t 80 mph,
T
195
2.4375 hours
80
2 hours, 26 minutes.
107. False. A fourth-degree polynomial can have at most four
zeros and complex zeros occur in conjugate pairs.
108. True. If y kx, then
109. The maximum (or minimum) value of a quadratic
function is located at its graph’s vertex. To find the
vertex, either write the equation in standard form or
use the formula
110. Answers will vary. Sample answer:
2ab , f 2ab .
If the leading coefficient is positive, the vertex is a
minimum. If the leading coefficient is negative, the
vertex is a maximum.
1
x y.
k
Polynomials of degree n > 0 with real coefficients can
be written as the product of linear and quadratic factors
with real coefficients, where the quadratic factors have
no real zeros.
Setting the factors equal to zero and solving for the
variable can find the zeros of a polynomial function.
To solve an equation is to find all the values of the
variable for which the equation is true.
Problem Solving for Chapter 3
1. (a) (i)
gx x2 4x 12
0 x 6x 2
Zeros: 6, 2
(iii) gx x2 3x 10
0 x 5x 2
Zeros: 5, 2
(ii)
gx x2 5x
0 xx 5
Zeros: 0, 5
(iv) gx x2 4x 4
0 x 2x 2
Zeros: 2, 2
2
(v) gx x 2x 6
(vi) gx x2 3x 4
0 x2 2x 6
0 x2 3x 4
By the Quadratic Formula, x 1 ± 7.
By the Quadratic Formula, x Zeros: 1 ± 7
—CONTINUED—
Zeros:
3 ± 7i
2
3 ± 7i
.
2
382
Chapter 3
Polynomial Functions
1. —CONTINUED—
(b) (i)
f x x 2x2 4x 12
(ii)
f x x 2x2 5x
30
30
−10
10
−10
10
−30
−15
(iii) f x x 2x2 3x 10
(iv)
f x x 2x2 4x 4
60
4
−4
−10
−30
(v)
8
10
−4
f x x 2x2 2x 6
(vi)
f x x 2x2 3x 4
20
2
−9
−10
9
10
−10
−10
x 2 is an x-intercept of f x in all six graphs. All the graphs, except (iii), cross the x-axis at x 2.
(c) (i) other x-intercepts: 2, 0, 6, 0
(ii) other x-intercepts: 5, 0, 0, 0
(iii) other x-intercepts: 5, 0
(iv) other x-intercepts: No additional x-intercepts
(v) other x-intercepts: 1.6, 0, 3.6, 0
(vi) other x-intercepts: No additional x-intercepts
(d) The x-intercepts of f x x 2gx are the same as the zeros of gx plus x 2.
2.
r
l
(a) r 2 rl 600
(c)
2000
rl 600 r 2
l
(b) V 600 r 2
r
1 2
1
600 r 2
r l r 2
2
2
r
1
r 600 r 2
2
1
300r r 3
2
1
r3 300r
2
0
20
0
The maximum volume is V 1595.8 cubic feet. This
occurs when r 8 feet and l 16 feet.
Problem Solving for Chapter 3
383
3. f x ax3 bx2 cx d
x k)
ax3
ax2 ak bx ak2 bk c
bx2
cx
d
ax3 akx2
ak bx2 cx
ak bx2 ak2 bkx
ak2 bk cx d
ak2 bk cx ak3 bk2 ck
ak3 bk2 ck d
Thus, f x ax3 bx2 cx d x kax2 ak bx ak2 bx c ak3 bk2 ck d and
f k ak3 bk2 ck d. Since the remainder r ak3 bk2 ck d, f k r.
4. (a)
(d) 3x3 x2 90; a 3, b 1 ⇒
y3 y2
y
1
2
2
12
3
36
4
80
5
150
6
252
7
392
8
576
9
810
10
1100
93 x 3 9x2 990
3x3 3x2 810 ⇒ 3x 9 ⇒ x 3
(e) 2x3 5x2 2500; a 2, b 5 ⇒
x2
(c)
x3
2x2
2x5 2x5 3
x
2
7x6 7x6 3
5. V l w
2x
4 ⇒ x 10
5
36 ⇒
2
392 ⇒
49
a2
b3 216
7x
7 ⇒ x6
6
(g) 10x3 3x2 297; a 10, b 3 ⇒
a2 100
b3
27
100 2
100
100
10x3 3x 297
27
27
27
10x3 10x3 3
2
80 ⇒
49
49
49
7 x 3 6x2 1728
216
216
216
a2 1
288; a 1, b 2 ⇒ 3 b
8
3
2
(f) 7x3 6x2 1728; a 7, b 6 ⇒
1
1 3 1 2
x 2x 288
8
8
8
x
2
a2
4
b3 125
4
4
4
2 x 3 5x2 2500
125
125
125
252 ⇒ x 6
(b)
x3
a2
9
b3
x
3 ⇒ x6
2
2
1100 ⇒
10x
10 ⇒ x 3
3
h x2x 3
x2x 3 20
x3 3x2 20 0
Possible rational zeros: ± 1, ± 2, ± 4, ± 5, ± 10, ± 20
2
1
3
2
0
10
20
20
1
5
10
0
x 2x2 5x 10 0
x 2 or x 5 ± 15i
2
x
x+3
x
Choosing the real positive value for x we have: x 2 and x 3 5.
The dimensions of the mold are 2 inches 2 inches 5 inches.
384
Chapter 3
6. (a)
Polynomial Functions
Function
Zeros
Sum of Zeros
Product of Zeros
f1x x2 5x 6
2, 3
5
6
f2x x3 7x 6
3, 1, 2
0
6
3, 1, ± 2i
2
12
3, 0, 2, 2 ± 3
3
0
f3x x4 2x3 x2 8x 12
f4x x5 3x4 9x3 25x2 6x
(b) Conjecture: Sum of Zeros an1
(c) Conjecture: Product of Zeros:
aa, ,
0
0
if f x anx n an1x n1 . . . a1x a0
if n is even
if n is odd
if f x anx n an1x n1 . . . a1x a0
7. False. Since f x dxqx rx, we have
f x
rx
qx .
dx
dx
The statement should be corrected to read f 1 2 since
f 1
f x
qx .
x1
x1
8. (a) y ax 2 bx c
0, 4: 4 a0 b0 c
2
(b) Enter the data points 0, 4, 1, 0, 2, 2, 4, 0,
6, 10 and use the regression feature to obtain
y x 2 5x 4.
4 c
4, 0: 0 a42 b4 4
0 16a 4b 4 44a b 1
0 4a b 1 or b 1 4a
1, 0: 0 a12 b1 4
4ab
4 a 1 4a
4 1 3a
3 3a
a 1
b 1 41 5
y x 2 5x 4
9. (a) Slope 94
5
32
(b) Slope Slope of tangent line is less than 5.
(c) Slope 4.41 4
4.1
2.1 2
Slope of tangent line is less than 4.1.
(e)
Slope 4 h,
h0
4 1 3
415
4 0.1 4.1
The results are the same as in (a)–(c).
41
3
21
Slope of tangent line is greater than 3.
(d) Slope f 2 h f 2
2 h 2
2 h2 4
h
4h h2
h
4 h, h 0
(f) Letting h get closer and closer to 0, the slope approaches
4. Hence, the slope at 2, 4 is 4.
Problem Solving for Chapter 3
10. (a) x 2y 100
y
(b)
1500
100 x
2
Ax xy x
Area of pasture
(in square meters)
y
1002 x
100x x2
2
50x 1250
1000
750
500
250
x
20 40 60 80 100
Length of pasture (in meters)
x2
The maximum occurs at an area of 1250 square meters
when x 50 meters and
2
Domain: 0 < x < 100
y
100 50
25 meters.
2
11. Let x length of the wire used to form the square. Then 100 x length of wire used to form the circle.
(a) Let s the side of the square. Then 4s x ⇒ s x
4 and the area of the square is s2 x
42.
Let r the radius of the circle. Then
2r 100 x ⇒ r 100 x
100 x 2
.
and the area of the circle is r2 2
2
The combined area is:
Ax 4x 2
1002 x
2
x2
10,000 200x x2
16
4 2
2500 50x
x2
x2
16
4
161 41 x
164x
50
2500
x
2
2
50x 2500
(b) Domain: Since the wire is 100 cm, 0 ≤ x ≤ 100.
(c) Ax 385
164 x
2
50
2500
x
164x
164x
4
400
164x 400
4 16 4 164x 400
4 10,000
2500
4
164x 400
4 2500
4
2
2
800
2500
x 4
800
400
x
4
4
2500
400
4 2
2
2
2
2
2
2500
The minimum occurs at the vertex when x 400
4 56 cm and Ax 350 cm2. The maximum
occurs at one of the endpoints of the domain. When x 0, Ax 796 cm2. When x 100, Ax 625 cm2.
Thus, the area is maximum when x 0 cm.
(d) Answers will vary. Graph Ax to see where the minimum and maximum values occur.
386
12.
Chapter 3
Polynomial Functions
xn 1
x1
xn1 xn2 xn3 . . . x1
n1
n2
n3
2
.
.
.
x 1 ) x 0x
0x
0x
0x 0x 1
n
xn xn1
xn1 0xn2
xn1 xn2
xn2 0xn3
xn2 xn3
xn3 0xn4
x 2 0x
x2 x
x1
x1
0
xn 1
xn1 xn2 xn3 . . . x 1
x1
Practice Test for Chapter 3
Practice Test for Chapter 3
1. Sketch the graph of f x x2 6x 5 and identify the vertex and the intercepts.
2. Find the number of units x that produce a minimum cost C if
C 0.01x2 90x 15,000.
3. Find the quadratic function that has a maximum at 1, 7 and passes through
the point 2, 5.
4. Find two quadratic functions that have x-intercepts 2, 0 and 3, 0.
4
5. Use the leading coefficient test to determine the right and left end behavior of
the graph of the polynomial function f x 3x5 2x3 17.
6. Find all the real zeros of f x x5 5x3 4x.
7. Find a polynomial function with 0, 3, and 2 as zeros.
8. Sketch f x x3 12x.
9. Divide 3x4 7x2 2x 10 by x 3 using long division.
10. Divide x3 11 by x2 2x 1.
11. Use synthetic division to divide 3x5 13x4 12x 1 by x 5.
12. Use synthetic division to find f 6 given f x 7x 3 40x 2 12x 15.
13. Find the real zeros of f x x3 19x 30.
14. Find the real zeros of f x x4 x3 8x2 9x 9.
15. List all possible rational zeros of the function f x 6x3 5x2 4x 15.
10
2
16. Find the rational zeros of the polynomial f x x3 20
3 x 9x 3 .
17. Write f x x4 x3 3x2 5x 10 as a product of linear factors.
18. Find a polynomial with real coefficients that has 2, 3 i, and 3 2i as zeros.
19. Use synthetic division to show that 3i is a zero of f x x3 4x2 9x 36.
20. Find a mathematical model for the statement, “z varies directly as the square of
x and inversely as the square root of y.”
387