CHAPTER 2
Polynomial and Rational Functions
Section 2.1 Quadratic Functions and Models
1. polynomial
2. nonnegative integer; real
3. quadratic; parabola
4. negative; maximum
5. () 2 2 fxx=− opens upward and has vertex () 0,2. Matches graph (b).
9.
6. ()() 2 12 fxx=+− opens upward and has vertex () 1,2. Matches graph (a).
7. ()() 2 4 fxx=−− opens downward and has vertex ()4,0. Matches graph (c).
8. ()()() 22 4224fxxx=−−=−−+ opens downward and has vertex ()2,4. Matches graph (d).
Horizontal shift one unit
Horizontal shrink
Horizontal stretch and Horizontal shift three to the right and a vertical shift a vertical shift three units to the left one unit upward units downward
(a) () 2 1 2 21 yx=−−+
Horizontal shift two units to the right, vertical shrink () 1 2 each -value is multiplied by , y reflection in the x-axis, and vertical shift one unit upward
() 2 1 2 13 yx
Horizontal shift two units to the left, vertical shrink () 1 2 each -value is multiplied by , y reflection in x-axis, and vertical shift one unit downward
2
Horizontal shift one unit to the right, horizontal stretch () each -value is multiplied by 2, x and vertical shift three units downward
Horizontal shift one unit to the left, horizontal shrink () 1 2 each -value is multiplied by , x and vertical shift four units upward
13. () () () 2 2 2 6 699 39 =− =−+− =−− fxxx xx x
Vertex: () 3,9
Axis of symmetry: 3 x =
Find x-intercepts: () 2 60 60 0 606 −= −= = −= = xx xx x xx
x-intercepts: ()() 0,0,6,0
14. () () () 2 2 2 8 81616 416 gxxx xx x =− =−+− =−−
Vertex: () 4,16
Axis of symmetry: 4 x =
Find x-intercepts: () 2 80 80 0 808 xx xx x xx −= −= = −= =
x-intercepts: ()() 0,0,8,0
15. ()() 2 2 8164hxxxx =−+=−
Vertex: () 4,0
Axis of symmetry: 4 x =
Find x-intercepts: () 2 40 40 4 x x x −= −= = x-intercept: () 4,0
16. ()() 2 2 211gxxxx=++=+
Vertex: () 1,0
Axis of symmetry: 1 x =−
Find x-intercepts: () 2 10 10 1 x x x += += =−
x-intercept: () 1,0
17. () () () () 2 2 2 2 62 6992 697 37 =−+ =−+−+ =−+− =−− fxxx xx xx x
Vertex: () 3,7
Axis of symmetry: 3 = x
Find x-intercepts: () 2 2 2 2 620 62 6929 37 37 −+= −=− −+=−+ −= =± xx xx xx x x x-intercepts: () 37, 0 ±
18. () () () 2 2 2 1661 16646461 83 =++ =++−+ =+− hxxx xx x
Vertex: () 8,3
Axis of symmetry: 8 =− x
Find x-intercepts: () 2 2 2 16610 16646164 83 83 83 ++= ++=−+ += +=± =−± xx xx x x x x-intercepts: () 83, 0 −±
19. () () () 2 2 2 821 8161621 45 =−+ =−+−+ =−+ fxxx xx x
Vertex: () 4, 5
Axis of symmetry: 4 = x
Find
Vertex: 1 ,1 2
Axis
Not a real number
No x-intercepts
20. () () () 2 2 2 1240 12363640 64 =++ =++−+ =++ fxxx xx x
Vertex: () 6, 4
Axis of symmetry: 6 =− x
Find
Not a real number
No
=++−+
22. () 2 2 2 1 3 4 991 3 444 3 2 2 fxxx xx x =++
=+−
Vertex: 3 ,2 2
Axis of symmetry: 3 2 x =−
Find x-intercepts: 2 1 30 4 391 2 3 2 2 xx x ++= −±− = =−±
x-intercepts: 33 2,0,2,0 22 −−−+
23. () ()() () 2 2 2 25 2115 16 fxxx xx x =−++ =−−+−−+ =−−+
Vertex: () 1,6
Axis of symmetry: 1 x =
Find x-intercepts: 2 2 250 250 2420 2 16 xx
Vertex: () 2,5
Axis of symmetry: 2 x =−
Find x-intercepts: 2 2 410 410 4164 2 25 xx xx x −−+= +−= −±+ = =−± x-intercepts: ()() 25,0,25,0 −−−+ 25. () 2 2 2 4421 11 4421 44 1 420 2 hxxx xx x =−+
Vertex: 1 ,20 2
Axis of symmetry: 1 2 x =
Find x-intercepts: () 2 44210 416336 24 xx x −+= ±− =
Not a real number
x-intercepts
=−+
26. () 2 2 2 2 21 1 21 2 11 221 416 17 2 48 fxxx xx x x =−+
=−−+
=−+
Vertex: 17 , 48
Axis of symmetry: 1 4 x =
Find x-intercepts: () 2 210 118 22 xx x −+= ±− =
Not a real number
No x-intercepts
27. ()()() 2 2 2314fxxxx=−+−=−++
Vertex: () 1,4
Axis of symmetry: 1 x =−
x-intercepts: ()() 3,0,1,0
28. ()() () () () 2 2 2 2 11 44 1121 24 30 30 30 fxxx xx xx x =−+− =−++ =−++++ =−++
Vertex: () 1121 24 ,
Axis of symmetry: 1 2 x =−
x-intercepts: ()() 6,0,5,0
29. ()() 2 2 81145gxxxx=++=+−
Vertex: () 4,5
Axis of symmetry: 4 x =−
x-intercepts: () 45,0−±
30. () () () 2 2 2 1014 10252514 511 fxxx xx x =++ =++−+ =+−
Vertex: () 5,11
Axis of symmetry: 5 x =−
x-intercepts: () 511,0−±
31. () () () () () 2 2 2 2 2 21218 269918 2691818 269 23 =−+− =−−+−− =−−++− =−−+ =−− fxxx xx xx xx x
Vertex: () 3, 0
Axis of symmetry: 3 = x x-intercept: () 3, 0
32. () () () () 2 2 2 2 42441 4641 4693641 435 fxxx xx xx x =−+− =−−− =−−++− =−−−
Vertex: () 3,5
Axis of symmetry: 3 x = No x-intercepts
33. ()()() 2 2 11 224223gxxxx=+−=+−
Vertex: () 2,3
Axis of symmetry: 2 x =−
x-intercepts: () 26,0−±
34. ()() () () 2 2 2 3 5 327 55 3 42 55 65 693 3 =+− =++−− =+− fxxx xx x
Vertex: () 42 5 3,
Axis of symmetry: 3 x =−
x-intercepts: () 314,0−±
35. () 2,1 is the vertex.
()() 2 21 fxax=+−
Because the graph passes through ()0,3, () 2 3021
341 44 1. =+− =− = = a a a a
So, () 2 21. yx=+−
36. () 2,2 is the vertex.
() 2 22 yax=++
Because the graph passes through ()1,0, () 2 0122 2. a a =−++ −=
So, () 2 222.yx=−++
37. () 2,5 is the vertex.
()() 2 25 fxax=++
Because the graph passes through ()0,9, () 2 9025 44 1. =++ = = a a a
So, ()()() 22 12525.fxxx=++=++
38. () 3,10 is the vertex.
()() 2 310=+−fxax
Because the graph passes through () 0, 8, () 2 80310 8910
189 2. =+− =− = = a a a a
So, ()() 2 2310.=+−fxx
39. () 1,2 is the vertex.
()() 2 12 fxax=−−
Because the graph passes through ()1,14, () 2 14112 1442 164 4. a a a a =−−− =− = = So, ()() 2 412.fxx=−−
40. () 2,3 is the vertex.
()() 2 23 fxax=−+
Because the graph passes through ()0,2, () 2 1 4 2023 243 14 a a a a =−+ =+ −= −=
So, ()() 2 1 4 23. fxx=−−+
41. ()5,12is the vertex.
()() 2 512 fxax=−+
Because the graph passes through () 7,15, () 2 3 4 157512
34. a aa =−+ = =
So, ()() 2 3 4 512. fxx=−+
42. () 2,2 is the vertex.
()() 2 22 fxax=+−
Because the graph passes through ()1,0, () 2 0122 02 2. a a a =−+− =− = So, ()() 2 222.fxx=+−
43. () 3 1 42 , is the vertex.
()() 2 3 1 42 fxax=++
Because the graph passes through ()2,0, () 2 3 1 42 349 24 21649 02 a aa =−++ −= =−
So, ()() 2 3 241 4942.fxx=−++
44. () 53 24 , is the vertex.
()() 2 53 24 fxax=−−
Because the graph passes through ()2,4, () 2 53 24 813 44 1981 44 19 81 42 4 . a a a a =−−− =− = =
So, ()() 2 1953 8124.fxx=−−
45. () 5 2 ,0 is the vertex.
()() 2 5 2 fxax=+
Because the graph passes through () 716 23,, () 2 1675 322 16 3 a a −=−+ −=
So, ()() 2 165 32fxx=−+
46. () 6,6 is the vertex.
()() 2 66 fxax=−+
Because the graph passes through () 613 102,, () 2 361 210 3 1 2100 9 1 2100 66 6 450. a a a a =−+ =+ −= −=
So, ()() 2 45066.fxx=−−+
47. 2 23=−−yxx
()() 2 023 031 =−− =−+ xx xx
3 = x or 1=− x
x-intercepts: ()() 3, 0, 1, 0
48. 2 45yxx=−−
()() 2 045 051 5or1 xx xx xx =−− =−+ ==−
x-intercepts: ()() 5,0,1,0
49. 2 253yxx=+− ()() 2 0253 0213 xx xx =+− =−+ 1 2 210 303 xx xx −= = += =−
x-intercepts: ()() 1 2 ,0,3,0
50. 2 253=−++yxx ()() 2 2 1 2 0253 0253 0213 210 303 =−++ =−− =+− += =− −= = xx xx xx xx xx
x-intercepts: ()() 1 2 , 0, 3, 0
51. () 2 4 f xxx =−
x-intercepts: ()() 0,0,4,0
() 2 04 04 0or4 xx xx xx =− =− ==
The x-intercepts and the solutions of () 0 fx = are the same.
52. () 2 210 f xxx =−+
x-intercepts: ()() 0,0,5,0
() 2 0210 025 x x xx =−+ =−− 200 505 xx xx −= = −= =
The x-intercepts and the solutions of () 0 fx = are the same. 48 4 4 16 6 14
53. () 2 918fxxx=−+
x-intercepts: ()() 3,0,6,0 ()() 2 0918 036 3or6 xx xx xx =−+ =−− ==
The x-intercepts and the solutions of () 0 fx = are the same.
54. () 2 820fxxx=−−
x-intercepts: ()() 2,0,10,0 ()() 2 0820 0210 202 10010 xx xx xx xx =−− =+− += =− −= =
The x-intercepts and the solutions of () 0 fx = are the same.
55. () 2 2730fxxx=−−
x-intercepts: ()() 5 2 ,0,6,0 ()() 2 5 2 02730 0256 or6 xx xx xx =−− =+− =−=
The x-intercepts and the solutions of () 0 fx = are the same.
56. ()() 2 7 10 1245fxxx=+−
x-intercepts: ()() 15,0,3,0 () ()()
2 7 10 01245 0153 15015 303 xx xx xx xx =+− =+− += =− −= =
The x-intercepts and the solutions of () 0 fx = are the same.
57. ()()() ()() 2 33 33 9 =
=+− =− fxxx xx x opens upward ()()() ()() () 2 2 33 33 9 9 =− =−+− =−− =−+ gxxx xx x x opens downward
Note: ()()() 33=+−fxaxx has x-intercepts () 3,0 and ()3,0for all real numbers 0. ≠ a
58. ()()() ()() 2 55 55 25, opens upward fxxx xx x = =+− =− ()() () 2 , opens downward 25 gxfx gxx =− =−+
Note: ()() 2 25 fxax=− has x-intercepts () 5,0 and () 5,0 for all real numbers 0. a ≠
59. ()()() ()() 2 14 14 34
=−+− =−−− =−++ gxxx xx xx xx opens downward
=+− =−− fxxx xx xx opens upward ()()() ()() () 2 2 14 14 34 34
Note: ()()() 14=+−fxaxx has x-intercepts () 1,0 and ()4,0for all real numbers 0. ≠ a
60. ()()() ()() 2 23 23 6
=−+− =−−− =−++ gxxx xx xx xx opens downward
=+− =−− fxxx xx xx opens upward ()()() ()() () 2 2 23 23 6 6
Note: ()()() 23=+−fxaxx has x-intercepts () 2,0 and ()3,0for all real numbers 0. ≠ a
=++ =++ =++ fxxx xx xx xx
61. ()()()()
()()() ()() 2 1 2 1 2 32opens upward 32 321 273
()() 2 2 273opens downward =273 gxxx xx =−++
Note: ()()() 321fxaxx=++ has x-intercepts
() 3,0 and () 1 2 ,0 for all real numbers 0. a ≠
64. Let x = the first number and y = the second number. Then the sum is x ySySx+= =−
The product is ()() 2 P xxyxSxSxx ==−=− () 2 2 22 2 2 2 44 24 =− =−+ =−−+− =−−+ PxSxx xSx SS xSx SS x
()()() ()() 2 3 2 3 2 52 52 523 21315 =
62. ()()()()
=++ =++ =+− fxxx xx xx xx opens upward ()() 2 2 21315 21315 =−++ =−−− gxxx xx opens downward
Note: ()()() 523=++fxaxx has x-intercepts
() 5,0 and () 3 2 ,0 for all real numbers 0. ≠ a
63. Let x = the first number and y = the second number.
Then the sum is 110110. x yyx+= =−
The product is ()() 2 110110. P xxyxxxx ==−=−
() () 2 2 2 2 110 11030253025 553025 553025 =−+ =−−+−
The maximum value of the product occurs at the vertex of ()Px and is 3025. This happens when 55. xy==
The maximum value of the product occurs at the vertex of ()Px and is 2 4. S This happens when 2. x yS==
65. Let x = the first number and y = the second number.
Then the sum is 24 224. 2 x xyy += =
The product is () 24 . 2 x Pxxyx ==
() ()() 2 2 22 1 24 2 1 24144144 2 11 121441272 22 Pxxx xx xx =−+ =−−+− =−−−=−−+
The maximum value of the product occurs at the vertex of ()Px and is 72. This happens when 12 x = and () 241226. y =−= So, the numbers are 12 and 6.
66. Let x = the first number and y = the second number.
Then the sum is 42 342. 3 x xyy += =
The product is () 42 . 3 x Pxxyx == ()() () ()() 2 2 22 1 42 3 1 42441441 3 11 2144121147 33 Pxxx xx xx =−+ =−−+−
=−−−=−−+
The maximum value of the product occurs at the vertex of ()Px and is 147. This happens when 21 x = and 4221 7. 3 y == So, the numbers are 21 and 7.
a
67. 2 424 12 99 yxx=−++
The vertex occurs at () 249 3. 2249 b a −== The maximum height is ()()() 2 424 3331216 99 y =−++= feet.
68. 2 169 1.5 20255 yxx=−++
(a) The ball height when it is punted is the y-intercept. ()() 2 169 001.51.5 20255 y =−++= feet
(b) The vertex occurs at () 953645 2216202532 b x a =−=−= The maximum height is 2 364516364593645 1.5 32202532532 65616561656113,122966657 1.5 feet 104.02
(c) The length of the punt is the positive x-intercept.
The punt is about 228.64 feet.
69. 22 800100.250.2510800Cxxxx =−+=−+
The vertex occurs at () 10 20. 220.25 b x a =−=−=
The cost is minimum when 20 x = fixtures.
70. 2 230200.5 Pxx =+−
The
Because x is in hundreds of dollars, 201002000 ×= dollars is the amount spent on advertising that gives maximum profit.
71. () 2 251200 Rppp =−+
72. () 2 12150 Rppp =−+ (a) ()()() ()()() ()()() 2 2 2 41241504$408 61261506$468 81281508$432 R R R =−+= =−+= =−+=
(b) The vertex occurs at () 150 $6.25. 2212 b p a =−=−=
Revenue is maximum when price $6.25 = per pet. The maximum revenue is ()()() 2 6.25126.251506.25$468.75. R =−+=
30$13,500 thousand$13,500,000 R R R == == ==
25$14,375 thousand$14,375,000
(a) () () () 20$14,000 thousand$14,000,000
(b) The revenue is a maximum at the vertex. () () 1200 24 2225 2414,400 b a R −== =
The unit price that will yield a maximum revenue of $14,400,000 is $24.
(b) To find the dimensions that produce a maximum enclosed area, you can find the vertex.
To do this, either write the quadratic function in standard form or use , 2 =− b x a so the coordinates of the vertex are
74. (a)
Area of windowArea of rectangleArea of
To eliminate the y in the equation for area, introduce a secondary equation.
Perimeter perimeter of rectangleperimeter of semicircle
Substitute the secondary equation into the area equation.
(b) The area is maximum at the vertex.
The area will be at a maximum when the width is about 4.48 feet and the length is about 2.24 feet.
75. True. The equation 2 1210 x −−= has no real solution, so the graph has no x-intercepts.
76. True. The vertex of () f x is () 553 44 , and the vertex of () g x is () 571 44,.
77. () 2 75, fxxbx=−+− maximum value: 25
The maximum value, 25, is the y-coordinate of the vertex.
Find the x-coordinate of the vertex:
78. () 2 25 fxxbx=+− , minimum value: –50
The minimum value, –50, is the y-coordinate of the vertex.
Find the x-coordinate:
79. () 2
80. (a) Since the graph of P opens upward, the value of a is positive.
(b) Since the graph of P opens upward, the vertex of the parabola is a relative minimum at 2 = b t a
(c) Because of the symmetrical property of the graph of a parabola, and since the company made the same yearly profit in 2008 and 2016, the midpoint of the interval 816 ≤≤ t or 12 = t corresponds to the year 2012, when the company made the least profit.
So, the vertex occurs at 2
81. If () 2 f xaxbxc =++ has two real zeros, then by the Quadratic Formula they are 2 4 2 bbac x a −±− =
The average of the zeros of f is 22442 222
This is the x-coordinate of the vertex of the graph.
Section 2.2 Polynomial Functions of Higher Degree
1. continuous
2. Leading Coefficient Test
3. n; 1 n
4. (a) solution; (b) (); x a (c) x-intercept
5. touches; crosses
6. repeated zero; multiplicity
7. standard
8. Intermediate Value
9. () 2 25 f xxx =−− is a parabola with x-intercepts () 0,0 and () 5 2 ,0 and opens downward. Matches graph (c).
10. () 3 231fxxx=−+ has intercepts ()()() 11 22 0,1,1,0,3,0 and () 11 22 3,0. −+ Matches graph (f).
11. () 42 1 4 3 f xxx =−+ has intercepts () 0,0 and () 23,0.± Matches graph (a).
12. () 32 14 33fxxx=−+− has y-intercept Matches graph (e).
() 4 3 0,.
13. () 43 2 f xxx =+ has intercepts () 0,0 and ()2,0. Matches graph (d).
14. () 53 9 1 55 2 f xxxx =−+ has intercepts ()()()()() 0,0,1,0,1,0,3,0,3,0. Matches graph (b).
3 yx =
(a) ()() 3 4 fxx=−
Horizontal shift four units to the right
(b) () 3 4 fxx=−
Horizontal shift one unit to the left (b) () 5 1 fxx=+
Vertical shift four units downward (c) () 3
Vertical shift one unit upward (c) () 5 1 2 1 f xx =−
Reflection in the -axis and a vertical shrink each -value is multiplied by x y
(d) ()()3 44 fxx=−−
Reflection in the -axis, vertical shrink each -value is multiplied by , and vertical shift one unit upward x y (d) ()()5 1 2 1 fxx=−+
Horizontal shift four units to the right and vertical shift four units downward
Reflection in the -axis, vertical shrink each -value is multiplied by , and horizontal shift one unit to the left x y
Horizontal shift three Vertical shift three units
Reflection in the x-axis and then units to the left downward a vertical shift four units upward
Horizontal shift one unit to the right and a vertical shrink each -value is multiplied by y ( )
Vertical shift one unit upward and a horizontal shrink each -value is multiplied by 16 y ( ) 1 16
Vertical shift two units downward and a horizontal stretch each -value is multipied by y
18. 6 yx = (a) ()() 6 5 =−fxx
Horizontal shift to the right five units. ( ) 1 8
Vertical shrink each -value is multiplied by . y
Horizontal shift three units to the left and a vertical shift four units downward. (d) () 6 1 4 1 =−+
()() 6 21=−fxx ( ) 1 4
Reflection in the -axis, vertical shrink each -value is multiplied by , and vertical shift one unit upward x y ( ) Horizontal stretch each -value is multiplied by 4, and vertical shift two units downward x ( ) 1 2 Horizontal shrink each -value is multiplied by , and vertical shift one unit downward x
19. () 3 124=+ f xxx
Degree: 3
Leading coefficient: 12
The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right.
20. () 2 231fxxx=−+
Degree: 2
Leading coefficient: 2
The degree is even and the leading coefficient is positive. The graph rises to the left and rises to the right.
21. () 2 7 2 53 g xxx =−−
Degree: 2
Leading coefficient: 3
The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right.
22. () 6 1 hxx =−
Degree: 6
Leading coefficient: 1
The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right.
23. () 32 6 =−+ g xxxx
Degree: 3
Leading coefficient: 1
The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right.
24. () 54 1 4 8 =+− g xxx
Degree: 5
Leading coefficient: 1 4
The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right.
25. () 9.81.263=− f xxx
Degree: 6
Leading coefficient: 9.8
The degree is even and the leading coefficient is positive. The graph rises to the left and rises to the right.
26. () 10.52.753=−− f xxx
Degree: 5
Leading coefficient: 0.5
The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right.
27. ()() 32 7 8 571fssss=−+−+
Degree: 3
Leading coefficient: 7 8
The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right.
28. ()() 34 4 3 629=−−++htttt
Degree: 4
Leading coefficient: 8 3
The degree is even and the leading coefficient is negative.
The graph falls to the left and falls to the right.
29. ()()33 391;3 f xxxgxx =−+=
30. ()()()33 11 33 32, f xxxgxx =−−+=−
31. ()()()434 416; f xxxxgxx =−−+=−
32. ()()36,3424 f xxxgxx =−=
33. () 2 36 fxx=−
(a) ()() 2 036 066 x xx =− =+−
6060 66 xx xx +=−= =−=
Zeros: 6±
(b) Each zero has a multiplicity of one (odd multiplicity).
(c) Turning points: 1 (the vertex of the parabola) (d)
34. () 2 81 f xx =− (a) ()() 2 081 099 x x x =− =−+
9090 99 xx xx −=+= ==−
Zeros: 9 ±
(b) Each zero has a multiplicity of one (odd multiplicity).
(c) Turning points: 1 (the vertex of the parabola) (d)
35. () 2 69httt=−+
(a) () 2 2 0693 ttt =−+=−
Zero: 3 t =
(c) Turning points: 1 (the vertex of the parabola) (d) 8 44 8 f
(b) 3 t = has a multiplicity of 2 (even multiplicity).
36. () 2 1025fxxx=++
(a) () 2 2 010255 xxx =++=+
Zero: 5 x =−
(b) 5 x =− has a multiplicity of 2 (even multiplicity).
(c) Turning points: 1 (the vertex of the parabola) (d)
37. () 2 112 333fxxx=+−
(a) () ()() 2 2 112 333 1 3 1 3 0 2 21 xx xx xx =+− =+− =+−
Zeros: 2,1xx=−=
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c) Turning points: 1 (the vertex of the parabola) (d)
38. () 2 153 222fxxx=+−
(a) For 2 153153 0,,,. 222222 xxabc +−====− 2 5513
= =−±
Zeros: 537 2 x −± =
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c) Turning points: 1 (the vertex of the parabola) (d)
39. ()() 2 521gxxxx=−− (a) () () 2 2 0521 021 xxx xxx =−−
For 2 210,1,2,1.xxabc −−===−=− ()()()() () 2 22411 21 28 2 12 x −−±−−− = ±
Zeros: 0,12xx==±
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c) Turning points: 2 (d) 40. ()() 223107=−+ftttt
Zeros: 7 3 0, , 1 ===ttt
(b) 0 = t has a multiplicity of 2 (even multiplicity).
7 3 = t and 1 = t each have a multiplicity of 1 (odd multiplicity).
(c) Turning points: 3 (d)
41. () 32 3123=−+ f xxxx
(a) ()322 03123341 =−+=−+ xxxxxx
Zeros: 0, 23==±xx (by the Quadratic Formula)
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c) Turning points: 2
(d)
42. () 432 30 f xxxx =−−
(a) () ()() 432 22 2 030 030 065 x xx xxx xxx =−− =−− =−+ 2 06050 065 xxx xxx =−=+= ===−
Zeros: 0,6,5xxx===−
(b) The multiplicity of 0 x = is 2 (even multiplicity).
The multiplicity of 6 x = is 1 (odd multiplicity).
The multiplicity of 5 x =− is 1 (odd multiplicity).
(c) Turning points: 3 (d)
44. (a) () () ()() 53 42 22 6 06 032 fxxxx xxx xxx =+− =+− =+−
Zeros: 0,2 x =±
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c) Turning points: 2 (d)
43. () 5369 g tttt =−+
(a) ()() ()() 2 53422 22 069693 33 tttttttt ttt =−+=−+=− =+−
Zeros: 0,3tt==±
(b) 0 t = has a multiplicity of 1 (odd multiplicity).
3 t =± each have a multiplicity of 2 (even multiplicity).
(c) Turning points: 4
(d)
45. () 42 396fxxx=++ (a) () ()() 42 42 22 0396 0332 0312 xx xx xx =++ =++ =++
No real zeros
(b) No multiplicity
(c) Turning points: 1 (d)
46. () 42 2240=−−fttt
(a) () ()() 42 42 22 02240 0220 0245 =−− =−− =+− tt tt tt
Zeros: 5 =± t
(c) Turning points: 3 (d) 6 24
(b) Each zero has a multiplicity of 1 (odd multiplicity).
47. () 323412gxxxx=+−−
(a) ()() ()()()()() 322 2 03412343 43223 xxxxxx xxxxx =+−−=+−+ =−+=−++
Zeros: 2,3xx=±=−
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c) Turning points: 2
(d)
48. () 32425100fxxxx=−−+
(a) ()() ()() ()()() 2 2 04254 0254 0554 xxx xx xxx =−−− =−− =+−−
Zeros: 5,4 x =±
(b) Each zero has a multiplicity of 1 (odd multiplicity).
(c) Turning points: 2 (d)
49. 32 42025 yxxx =−+
(a)
(b) x-intercepts: ()() 5 2 0,0,,0
(c) () () 32 2 2 5 2 042025 042025 025 0, x xx xxx xx x =−+ =−+ =− =
(d) The solutions are the same as the x-coordinates of the x-intercepts.
50. 32 4488yxxx=+−− (a)
(b) ()()() 1,0,1.41,0,1.41,0
(c) ()() ()() ()() 32 2 2 2 04488 04181 0481 0421 2,1 xxx xxx xx xx x =+−− =+−+ =−+ =−+ =±−
(d) The solutions are the same as the x-coordinates of the x-intercepts.
51. 5354 yxxx =−+ (a)
(b) x-intercepts: ()()() 0,0,1,0,2,0 ±± (c) ()() ()()()() 53 22 054 014 01122 0,1,2 xxx xxx xxxxx x =−+ =−− =+−+− =±±
(d) The solutions are the same as the x-coordinates of the x-intercepts.
52. 53 9 1 55 =− y xx (a)
(d) The solutions are the same as the x-coordinates of the x-intercepts. 8 16 7 4 9 20 9 140 2 4 6 12 3 11 3 2 66 4 4 1212 10 10
(b) x-intercepts: ()()() 0, 0, 3, 0, 3, 0 (c) () 53 9 1 55 32 1 5 3 1 5 2 0 09 0,0 90,3 =− =− = = −= =± xx xx xx xx
53. ()()() 2 07 7 =−− =− fxxx xx
Note: ()() 7 =−fxaxx has zeros 0 and 7 for all real numbers 0. a ≠
54. ()() ()() ()() 2 25 25 310 =−−− =+− =−− fxxx xx xx
Note: ()()() 25=+−fxaxx has zeros 2 and 5 for all real numbers 0. a ≠
55. ()()()() () 2 32 024 68 68 =−++ =++ =++ fxxxx xxx xxx
Note: ()()() 24=++fxaxxx has zeros 0,2, and 4 for all real numbers 0. ≠ a
56. ()()()() () 2 32 016 76 76 =−−− =−+ =−+ fxxxx xxx xxx
Note: ()()() 16=−−fxaxxx has zeros 0, 1, and 6 for all real numbers 0. ≠ a
57. ()()()()() ()() 2 432 4330 49 4936 fxxxxx xxx xxxx =−+−− =−− =−−+
Note: ()() 4324936 f xaxxxx =−−+ has zeros 4,3,3,and0 for all real numbers 0. a ≠
58. ()()()() ()()()() ()()()() ()() () 22 42 53 21012 2112 41 54 54 fxxxxxx xxxxx xxx xxx xxx =−−−−−−− =++−− =−− =−+ =−+
Note: ()()()()() 2112fxaxxxxx =++−− has zeros 2,1,0,1,and 2 for all real numbers 0. a ≠
59. ()()() ()() ()() 2 2 2 2 1212 1212 12 212 21
=−− =−+− =−− fxxx xx x xx xx
Note: ()() 2 21=−−fxaxx has zeros 12 + and 12 for all real numbers 0. ≠ a 60. ()()() ()() ()() 2 2 2 2 4343 4343 43 8163 813
Note: ()() 2 813=−+fxaxx has zeros 43 + and 43 for all real numbers 0. ≠ a
=−−+− =−−− =−++ fxxxx xxx xx xxx xxx xxx
=−−+−−
61. ()()()() ()()() ()() () ()() 2 2 2 32 22525 22525 225 2445 241 672
=−−−
Note: ()() 32672=−++fxaxxx has zeros 2, 25, + and 25 for all real numbers 0. ≠ a
=−−+− =−−− =−++ fxxxx xxx xx xxx xxx xxx
=−−+−−
62. ()()()() ()()() ()() () ()() 2 2 2 32 32727 32727 327 3447 343 799
=−−−−+
=−−−
Note: ()() 32799=−++fxaxxx has zeros 3, 27 + and 27 for all real numbers 0. ≠ a
63. ()()() 2 3369 =++=++ fxxxxx
Note: ()() 2 69,0,fxaxxa=++≠ has degree 2 and zero 3. x =−
64. ()()() 2 222 =−+=−fxxxx
Note: ()() 2 2, 0, =−≠fxaxa has a degree 2 and zeros 2 =± x
65. ()()()() () 2 32 051 45 45 fxxxx xxx xxx =−+− =+− =+−
Note: ()() 2 45,0,fxaxxxa=+−≠ has degree 3 and zeros 0,5, x =− and 1.
66. ()()()() ()() 2 32 226 26 22024 =++− =+− =−−− fxxxx xx xxx
Note: ()() 3222024, 0, =−−−≠ fxaxxxa has degree 3 and zeros 2 =− x and 6.
67. ()()()()() ()() ()()() ()() ()()() 2 432 2 432 2 42 512735550 or 512152310 or 512173620 fxxxxxxxx fxxxxxxxx fxxxxxxx =−−−−=+−−+ =−−−−=+−+− =−−−−=−+−
Note: Any nonzero scalar multiple of these functions would also have degree 4 and zeros 5,1,and 2. x =−
68. ()()()()() ()() 22 432 1144 14 10334016 =++++ =++ =++++ fxxxxx xx xxxx
Note: ()() 43210334016, 0, =++++≠ fxaxxxxa has degree 4 and zeros 1=− x and 4.
69. ()()()()()() () ()() () 3 32 53 00033 33 3 3 fxxxxxx xxx xx xx =−−−−−− =−+ =− =−
Note: ()() 533, 0, =−≠fxaxxa has degree 5 and zeros 0, 3, = x and 3.
70. ()()()()() ()()()()() ()()()()() ()()()()() 2 5432 2 5432 2 5432 2 5432 1478177911332224 or 147822169496208896 or 1478252237875321568 or 1478262418846401792 fxxxxxxxxxx fxxxxxxxxxx fxxxxxxxxxx fxxxxxxxxxx =+−−−=−+−−− =+−−−=−+−++ =+−−−=−+−++ =+−−−=−+−++
Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros 1,4,7, x =− and 8.
71. ()()() 2 2 7 11 442 2151ftttt=−+=−+
(a) Rises to the left; rises to the right (b) No real zeros (no x-intercepts) (c)
(d) The graph is a parabola with vertex () 7 2 1,.
72. ()()() 2 101628gxxxxx =−+−=−−−
(a) Falls to the left; falls to the right (b) Zeros: 2,8 (c)
(d)
73. ()()() 3 2555fxxxxxx =−=+−
(a) Falls to the left; rises to the right (b) Zeros: 0,5,5 (c) (d)
74. ()()() 422933gxxxxxx =−=+−
(a) Rises to the left; rises to the right (b) Zeros: 3,0,3 (c) (d)
(a) Rises to the left; falls to the right (b) Zero: 2 (c) (d) t 1 0 1 2 3 () f t 4.5 3.75 3.5 3.75 4.5 4224 2 6 8 t y x 1 3 5 7 9 () g x 7 5 9 5 7 4610 2 4 6 8 10 x (2,0)(8,0) y x 2 1 0 1 2 () f x 42 24 0 24 42 y x 62 84268 24 36 48 48 (5,0) (0,(5,0) 0) x 2 1 0 1 2 () f x 24 8 0 8 24 y x 21 4124 15 20 25 5 10 15 (3,0)(0,0)(3,0) x 2 1 0 1 2 () f x 0 15 2 8 15 2 0 y x 4 848 10 4 8 (2,0)(2,0) x 2 1 0 1 2 () f x 16 9 8 7 0 y x 321 4134 6 4 2 12 10 14 (2,0)
75. ()() ()()() 44 2 11 22 1 2 816 422 =−+=− =+−+ fxxx xxx
(a) Rises to the left; rises to the right (b) Zeros 2:=± x (c) (d)
76. ()()() 824232 f xxxxx =−=−++
77. ()()() 32 31518323fxxxxxxx =−+=−− (d)
(a) Falls to the left; rises to the right (b) Zeros: 0,2,3 (c)
78. () () ()() 32 2 4415 4415 2523 f xxxx xxx xxx =−++ =−−− =−−+
(a) Rises to the left; falls to the right (b) Zeros: 35 22 ,0, (c)
79. ()() 232 55 f xxxxx =−−=−+ (d)
(a) Rises to the left; falls to the right (b) Zeros: 0,5 (c)
80. () () 24 22 483 316 f xxx xx =−+ =− (d)
(a) Rises to the left; rises to the right (b) Zeros; 0,4 ±
(c)
81. ()() 2 2 92=+fxxx
(a) Falls to the left, rises to the right (b) Zeros: 0, 2 =− x (c)
82. ()() 2 3 1 3 4 hxxx=−
(a) Falls to the left; rises to the right (b) Zeros: 0,4 (c) (d)
83. ()()() 22 1 4 22gttt=−−+
(a) Falls to the left; falls to the right (b) Zeros: 2,2 (c) (d)
84. ()()() 23 1 10 13gxxx=+−
(a) Falls to the left; rises to the right (b) Zeros: 1,3 (c) (d)
85. ()()() 3 1644fxxxxxx =−=−+
Zeros: 0 of multiplicity 1;4 of multiplicity 1; and 4 of multiplicity 1
86. () 42 1 4 2 f xxx =−
Zeros: 22 and 22 of multiplicity 1; 0 of multiplicity 2 87. ()()()() 2 1 5 1329gxxxx =+−−
Zeros: 1 of multiplicity 2; 3 of multiplicity 1; 9 2 of multiplicity 1 88. ()()() 22 1 5 235hxxx=+−
Zeros: 5 3 2,, both with multiplicity 2
89. () 3233=−+fxxx
92. () 42103hxxx=−+
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.±
The function has three zeros. They are in the intervals [][] 1,0,1,2, and [] 2,3. They are 0.879,1.347,2.532. x ≈−
90. () 32 0.112.079.816.88 =−+− fxxxx
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.
91. () 43 343gxxx=+−
93. (a) Volume height lengthwidth362 lwh x x =⋅⋅ = ==−
So, ()()()()() 2 362362362. Vxxxxxx =−−=− (b) Domain: 018 x <<
(c)
The length and width must be positive.
The maximum point on the graph occurs at 6. x = This agrees with the maximum found in part (c). 5
The function has two zeros. They are in the intervals [] 2,1 and []0,1. They are 1.585,0.779. x ≈−
(d)
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 62424 ×× inches.
(a)
(b) 0,120,60 126 xxx xx >−>−> <<
Domain: 06 x <<
(c)
95. (a)
2.5 x ≈ corresponds to a maximum of 665 cubic inches.
Using trace and zoom features, the relative maximum is approximately () 4.44, 1512.60 and the relative minimum is approximately () 11.97, 189.37.
(b) The revenue is increasing on () 3, 4.44 and () 11.97, 16 and decreasing on () 4.44, 11.97 .
(c) Answers will vary. Sample answer: The revenue for the software company was increasing from 2003 to midway through 2004 when it reached a maximum of approximately $1.5 trillion. Then from 2004 to 2012 the revenue was decreasing. It decreased to $189 million. From 2012 to 2016 the revenue has been increasing.
96. (a)
Using trace and zoom features, the relative maximum is approximately () 11.85, 136.84and the relative minimum is approximately () 8.60, 125.93 and () 14.55, 130.16.
(b) The revenue is increasing on () 8.60, 11.85and () 14.55, 16and decreasing on () 7, 8.60and () 11.85, 14.55.
(c) Answers will vary. Sample answer: The revenue for the construction company was decreasing from 2007 to midway through 2008 when it reached a minimum of approximately $125 million. Then from 2008 to late 2011 the revenue was increasing to a maximum of $137 million. The revenue once again was decreasing from 2012 to 2015. The lower revenue was about $130 million. Since 2015 to 2016, revenue has been increasing.
97. () 32 1 600 100,000 R xx=−+
The point of diminishing returns (where the graph changes from curving upward to curving downward) occurs when 200. x = The point is () 200,160 which corresponds to spending $2,000,000 on advertising to obtain a revenue of $160 million.
98. 32 0.0030.1370.4580.839,234Gtttt =−++−≤≤ (a)
(b) The tree is growing most rapidly at 15. t ≈ (c) () () 2 0.0090.2740.458 0.274 15.222 220.009 15.2222.543 ytt b a y =−++ −=≈ ≈ () Vertex15.22,2.54 ≈
(d) The x-value of the vertex in part (c) is approximately equal to the value found in part (b).
99. True. A polynomial function only falls to the right when the leading coefficient is negative.
100. False. A fifth-degree polynomial can have at most four turning points.
101. False. The range of an even function cannot be (),. −∞∞
An even function’s graph will fall to the left and right or rise to the left and right.
102. False. f has at least one real zero between 2 x = and 6. x =
103. Answers will vary. Sample answers: 4 0 a < 4 0 a >
104. Answers will vary. Sample answers: 5 0 a < 5 0 a >
105. ()() 4; f xxfx = is even.
(a) ()() 2 gxfx=+
Vertical shift two units upward ()() () () 2 2 gxfx fx gx −=−+ =+ = Even
(b) ()() 2 gxfx=+ (c) ()()() 4 4 g xfxxx =−=−=
Horizontal shift two units to the left Reflection in the y-axis. The graph looks the same. Neither odd nor even Even
(d) ()() 4 g xfxx =−=− (e) ()() 4 11 216 g xfxx ==
Reflection in the x-axis Horizontal stretch Even Even (f) ()() 4 11 22 g xfxx ==
()()() 4 3/43/43,0gxfxxxx
Vertical shrink Neither odd nor even Even (h) ()()()()()()() 4 4416 g xffxffxfxxx
106. (a) Degree: 3
Leading coefficient: Positive (b) Degree: 2
Leading coefficient: Positive
(c) Degree: 4
Leading coefficient: Positive
(d) Degree: 5
Leading coefficient: Positive
(a) ()5 1 1 3 21=−−+yx is decreasing and ()5 2 3 5 23=+−yx is increasing.
Section 2.3 Polynomial and Synthetic Division 147
(b) It is possible for ()()5 =−+ g xaxhk to be strictly increasing if 0 > a and strictly decreasing if 0. < a
(c) f cannot be written in the form ()()5 f xaxhk =−+ because f is not strictly increasing or strictly decreasing.
Section 2.3 Polynomial and Synthetic Division
1. () f x is the dividend; ()dx is the divisor; ()qx is the quotient; ()rx is the remainder
2. proper
3. improper
4. synthetic division
5. Factor
6. Remainder
9. 2 12 212 ,1 33 xx yyx xx +− ==−+ ++
(a) and (b)
(c) 2 2 1 321 3 1 3 2 x xxx xx x x ++− +
So, 2 212 1 33 xx x xx +− =−+ ++ and 12.yy =
10. 42 2 1222 11 , 11 xx yyx xx +− ==− ++
(a) and (b)
(c) 2 2432 432 01001 0 1 x xxxxxx xxx +++++− ++
So, 42 2 22 11 11 xx x xx +− =− ++ and 12.yy =
11. 2 2 24 321012 26 412 412 0 + +++ + + + x xxx xx x x 2 21012 24,3 3 xx xx x ++ =+≠ +
12. 2 2 53 451712 520 312 312 0 x xxx xx x x + 2 51712 53,4 4 xx xx x =+≠
17. 6 165 66 1 ++ + xx x 651 6 11 + =− ++ x xx 9 9 9 3 6 2 6 6
13. 2 32 32 2 2 31 4547115 45 1211 1215 45 45 0 xx xxxx xx xx xx x x −+ +−−+ + + + 32 2 471155 31, 454 xxx xxx x −−+ =−+≠− +
14. 2 32 32 2 2 243 32616176 64 1217 128 96 96 0 xx xxxx xx xx xx x x −+ −−+− −+ −+ 32 2 6161762 243, 323 xxx xxx x −+− =−+≠
15. 32 432 43 32 32 31 2562 2 36 36 2 2 0 xx xxxxx xx xx xx x x +− +++−− + + + 432 32 562 31,2 2 xxxx xxx x ++−− =+−≠− +
16. 2 32 32 2 2 718 34312 3 73 721 1812 1854 42 xx xxxx xx xx xx x x ++ −+−− 32 2 431242 718 33 xxx xx xx +−− =+++
18. 3 3294 96 10 +− + xx x 9410 3 3232 =− ++ x xx 19. 232 32 01009 0 9 x xxxxx xxx x ++++− ++ 3 22 99 11 xx x xx −+ =− ++
20. 4325432 5432 000100007 000 7 +++−+++++ +++− + x xxxxxxxxx xxxxx x 5 44 77 11 ++ =+ xx x xx
21. 232 32 2 2 28 012839 202 89 808 1 x xxxxx xxx xx xx x ++−+− ++ −+− 32 22 28391 28 11 xxxx x xx −+−− =−+ ++
22. 2 2432 432 32 32 2 2 69 3502016 3 6320 6618 9216 9927 711 xx xxxxxx xxx xxx xxx xx xx x ++ −−++−− +− + 43 2 22 52016711 69 33 xxxx xx xxxx +−−+ =+++
23. 32432 432 32 32 2 3 3310000 33 330 3993 683 x xxxxxxx xxxx xxx xxx xx + −+−++++ −+− −++ −+− −+ ()() 42 33 683 3 11 xxx x xx −+ =++
24. 232 32 2 2124155 242 175 −+−−+ −+ −+ x xxxxx xxx x () 32 2 2 24155175 2 21 1 xxxx x xx x −−+− =− −+ 25. 32 2 2101424 226, 4 4 −+− =−+≠ xxx xxx x 26. 32 2 51876 532,3 3 xxx xxx x ++− =+−≠− + 27.
32 2 6726248 62574 33 xxx xx xx +−+ =+++ 28. 32 2 2121435 242 44 ++− =+−+ ++ xxx xx xx 29. 32 2 48918 49, 2 2 +−− =−≠− + xxx xx x 4 2 10 14 24 8 8 24 2 2 6 0 3 5 18 7 6 15 9 6 5 3 2 0 3 6 7 1 26 18 75 222 6 25 74 248 4 2 12 14 3 8 16 8 2 4 2 5 2 4 8 9 18 8 0 18 4 0 9 0
41. 4 32 180216 63636 66 xx xxx xx =−−−−−
42. 23 2 53211 36 11 xxx xx xx −+− =−+−+ ++ 43. 32 2 41623151 41430, 1 2 2 xxx xxx x +−− =+−≠−
1 0 0 180 0 6 36 216 216 1 6 36 36 216 13 4 6 12 4 443−+ 1023−+ 4 4 243 + 223 + 0
45. () 32 107, 3 =−−+=fxxxxk ()()() ()() 2 32 3245 33310375 =−+−− =−−+=− fxxxx f
46. () 324108, 2 =−−+=−fxxxxk ()()() ()()()() 2 32 2624 224210284 =+−++ −=−−−−−+= fxxxx f 47. () 432 2 3 1510614, fxxxxk=+−+=−
() ()()()() 3 432 34 2 33 34 2222 33333 1564 1510614 fxxxx f =+−++ −=−+−−−+= 48. () 32 1 5 102234, fxxxxk =−−+= ()()() ()()()() 2 32 13 1 55 13 1111 55555 10207 102234 fxxxx f =−−−+ =−−+=
49. () 32 46124,13fxxxxk=−+++=− ()()()() ()()()() 2 32 134243223 13413613121340 fxxxx f
50. () 32 38108,22fxxxxk=−++−=+
51. () 3 273fxxx=−+
(a) Using the Remainder Theorem:
(b) Using the Remainder Theorem: ()()() 3 1217132 f =−+=−()()() 3 2227231 f −=−−−+=
Using synthetic division:
Verify using long division:
(c) Using the Remainder Theorem: ()()() 3 32373336 =−+= f
Using synthetic division:
Using synthetic division:
Verify using long division:
(d) Using the Remainder Theorem: ()()() 3 2227235 f =−+=
Using synthetic division:
(a) Using the Remainder Theorem:
Using
Using the Remainder Theorem:
53. () 32574hxxxx=−−+
(a) Using the Remainder Theorem:
Using synthetic division:
Using the Remainder Theorem:
Using synthetic division:
Verify
Verify
(c) Using the Remainder Theorem: (d) Using the Remainder Theorem:
Using synthetic division:
Using synthetic division:
Verify
Verify using
54.
(a) Using the Remainder Theorem:
Using the Remainder Theorem: ()()()()
Using synthetic division:
(c) Using the Remainder Theorem:
Using
Using the Remainder Theorem:
58.
Zeros: 1 2 ,2,5
1 2 2 15 27 10 1 7 10 2 14 20 0 2 3 48 80 41 6
Zeros:
Zeros: 1,13,13
3 1 15 635 0
+ 1 15 635 25 + 635 + 1 3 0
Zeros: 25,25,3 −+−
63. () 32 252;fxxxx=+−+ Factors: ()() 2,1xx+−
(a)
Zeros: 3,3,2 60.
Zeros: 2,2,2
Both are factors of () f x because the remainders are zero.
(b) The remaining factor of () f x is () 21. x (c) ()()()() 2121fxxxx=−+−
(d) Zeros: 1 2 ,2,1 (e)
64. () 32 384;=−−−fxxxx
Factors: ()() 1, 2 +−xx
(a)
Both are factors of () f x because the remainders are zero.
(b) The remaining factor is () 32. + x
(c) () ()()() 32 384 3212 =−−− =++− fxxxx xxx
(d) Zeros: 2 3 ,1,2
(e)
65. () 432893840;=−++− fxxxxx
Factors: ()() 5, 2 −+xx
(a)
Both are factors of () f x because the remainders are zero.
(b) ()() 2 5414 −+=−− xxxx
The remaining factors are () 1 x and ()2. x
(c) ()()()()() 5214 =−+−− fxxxxx
(d) Zeros: 2,1,4,5 (e)
66. () 432 814711024;fxxxxx =−−−+
Factors: ()() 2,4xx+−
(a)
Both are factors of () f x because the remainders are zero.
(b) ()() 2 8234321 xxxx +−=+−
The remaining factors are () 43 x + and () 21. x (c) ()()()()() 432124fxxxxx =+−+−
(d) Zeros: 3 1 42,,2,4 (e)
67. () 32 641914;fxxxx=+−−
Factors: ()() 21,32 xx+−
(a)
Both are factors of () f x because the remainders are zero.
68. () 32 10117245;fxxxx =−−+
(b) ()64267 xx+=+
This shows that ()() 67, 12 23 fx x xx =+
+−
so () ()() 7. 2132 fx x xx =+ +−
The remaining factor is ()7. x +
(c) ()()()() 72132fxxxx =++−
(d) Zeros: 12 7,, 23 (e)
Factors: ()() 25,53 xx+− (a) (b) () 1030103 xx−=−
This shows that ()() 103, 53 25 fx x xx =−
so () ()() 3. 2553 fx x xx =− +−
The remaining factor is ()3. x
Both are factors of () f x because the remainders are zero.
(c) ()()()() 32553fxxxx =−+−
(e)
(d) Zeros: 53 3,, 25
69. () 32 2105;fxxxx=−−+
Factors: ()() 21,5 xx−+ (a)
1 2 2 1 10 5 1 0 5
2 0 10 0
(d) Zeros: 43,3±− (e)
71. () 322510fxxxx=−−+
(a) The zeros of f are 2 x = and 2.236. x ≈±
Both are factors of () f x because the remainders are zero.
(b) () 22525 xx−=−
−+
This shows that () ()() 25, 1 5 2 fx x xx =−
so () ()() 5. 215 fx x xx =− −+
The remaining factor is () 5 x
(c) ()()()() 5521fxxxx =+−−
(d) Zeros: 1 5,5, 2 (e)
70. () 32348144;fxxxx=+−−
Factors: ()() 43,3xx++ (a)
5 2 0 10 25 10 2 25 0 6 66 14 3 1 3 48 144 3 0 144 1 0 48 0 43 1 0 48 43 48 1 43 0
Both are factors of () f x because the remainders are zero.
(b) The remaining factor is () 43 x (c) ()()()() 43433fxxxx =−++
(b) An exact zero is 2. x = (c) ()()() ()()() 2 25 255 fxxx xxx =−− =−−+
72. () 32326=+−−gxxxx
(a) The zeros of g are 3,1.414,1.414.=−≈−≈xxx
(b) 3 =− x is an exact zero. (c) ()()() ()()() 2 32 322 =+− =+−+ gxxx xxx
73. () 32272htttt=−−+
(a) The zeros of h are 2,3.732,0.268.ttt =−≈≈
(b) An exact zero is 2. t =− (c) ()()() 2 241htttt=+−+
By the Quadratic Formula, the zeros of 2 41tt−+ are 23. ± Thus, ()()()() 22323.htttt =+−+−−
74. () 32124024fssss=−+−
(a) The zeros of f are 6,0.764,5.236sss=≈≈
(b) 6 s = is an exact zero.
(c) ()()()
2 664 63535 fssss sss =−−+
75. () 54327101424 hxxxxxx =−++−
(a) The zeros of h are 0,3,4,xxx=== 1.414,1.414.xx≈≈−
(b) An exact zero is 4. x =
(c) ()()() ()()()() 432 4326 4322 hxxxxxx xxxxx =−−−+ =−−+−
76. () 432 611519927gxxxxx =−−+−
(a) The zeros of a are 3,3,1.5,xxx==−= 0.333. x ≈
(b) An exact zero is 3. x =−
(c) ()()()
32 3629369 332331 axxxxx xxxx =+−+− =+−−−
64 1 7 8 0 2 1 9 5 36 4 2 22 34 4 1 11 17 2 0
2 1 11 17 2 2 18 2 1 9 1 0
80. ()() 432432 2 9536495364 422 xxxxxxxx xxx +−−++−−+ = −+− 432 2 2 95364 91,2 4 xxxx xxx x +−−+ =+−≠± 6 1 12 40 24 6 36 24
32 6464 8 xxx x +−− + 32 2 6464 78,8 8 xxx xxx x +−− =−−≠− +
32 483 23 xxx x −++
483
3 2 xxx xxxx x
So, 32 2 4833 21,. 232 xxx xxx x −++ =−−≠ 79. ()() 432432 2 61166116 3212 x xxxxxxx xxxx ++++++ = ++++ ()() 432 2 6116 3,2,1 12 xxxx xxx xx +++ =+≠−− ++
81. (a)
3,200,000
(c) 25 x = So, an advertising expense of $250,000 yields a profit of $2,171,000, which is close to $2,174,375.
82. (a) and (b)
(c)
432 2.6439119.7882044.3315,679.546,891 =−+−+ Ntttt
The data fits the model well.
The estimated values are close to the original data values. (d) Because the remainder is 936, ≈ r you can conclude that ()14936. ≈ N This confirms the estimated value.
83. False. If () 74 x + is a factor of , f then 4 7 is a zero of f 84. True. ()()()()()()()
(b) Using the trace and zoom features, when 25, x = an advertising expense of about $250,000 would produce the same profit of $2,174,375. 1 2 6 1
85. True. The degree of the numerator is greater than the degree of the denominator. 86. False. The equation 32 2 34 44 1 xx xx x −+ =−+ + is not true for 1 x =− since this value would result in division by zero in the original equation. So, the equation should be written as 32 2 34 44, 1 xx xx x −+ =−+ + 1. x ≠−
89. To divide 2 35+−xx by 1 + x using synthetic division, the value of k is 1=− k not 1 = k as shown.
90. (a) An advertising expense of $200,000 when 20, x = also appears to yield a profit of about $936,660. (b) Evaluate l at 20. x =
()20936,760 P =
So, an advertising expense of $20,000 will yield a profit of $936,760.
91. To divide evenly, 210 c + must equal zero. So, c must equal 210.
92. To divide evenly, 42 c must equal zero. So, c must equal 42.
93. If 4 x is a factor of () 32 28,fxxkxkx=−+− then ()40. f = ()()()() 32 444248 0641688 568 7 fkk kk k k =−+− =−+− −=− =
Section 2.4 Complex Numbers
1. real
2. imaginary
3. pure imaginary 4. 1;1
principal
2165
2347
16. 2733−= i
17. 23
18. 50
19. () 2 661 16 −+=−+− =−− iii i
20. () 2 24214 24 −+=−−+ =+ iii i
21. 0.040.04 0.2 −= = i i
22. 0.00250.0025 0.05 −= = i i
23. ()() 523523 74 +++=+++ =+ iiii i
24. ()() 1325684 −+−+=+iii
25. ()() 981 ii −−−=
26. ()() 3261332613 311 iiii i +−+=+−− =−−
27. ()() 28550222552332 iii−+−+−−=−++−=−
28. ()() 8184328324324 iii +−−+=+−−=
29. () 1314713147 1420 iiii i −−=−+ =−+
30. () 251011151526 iii+−++=+
31. ()() 2 1323232 325 iiiii ii +−=−+− =++=+
32. ()() 2 72352135610 214110 1141 iiiii i i −−=−−+ =−− =−
33. () 2 121912108 12108 10812 iiii i i −=− =+ =+
34. () 2 8947232 3272 iiii i −+=−− =−
35. ()() 2 232329 2911 +−=− =+= iit
36. ()() 2 4747167 16723 +−=− =+= iii
37. () 2 2 67368449 368449 1384 iii i i +=++ =+− =−+
38. () 2 2 54254016 254016 940 iii i i −=−+ =−− =−
39. The complex conjugate of 92i + is 92. i ()() 2 9292814 814 85 iii+−=− =+ =
40. The complex conjugate of 810i is 810. i + ()() 2 81081064100 64100 164 iii−+=− =+ =
41. The complex conjugate of 15i is 15. i −+ ()() 2 151515 156 iii−−−+=− =+=
42. The complex conjugate of 32i −+ is 32. i ()() 2 323292 92 11 iii−+−−=− =+ =
43. The complex conjugate of 2025i −= is 25i ()() 2 25252020 iii−=−=
44. The complex conjugate of 1515i −= is 15. i ()() 2 15151515 iii−=−=
45. The complex conjugate of 6 is 6. ()() 666 =
46. The complex conjugate of 18 + is 18. + ()() 18181288 942 ++=++ =+
47. () 2245 454545 245 810810 1625414141 i iii i i i + =⋅ −−+ ++ ===+ +
48. () () 2 1 13131313131313 111222 i ii i iii +++ ⋅===+ −+−
49. () () 2 2 5 52510 5525 2410125 261313 i iii iii i i + +++ ⋅= −+− + ==+
50. 2 2 6712612714 121214 205 4 5 iiiii iii i i −++−− ⋅= −+− + ==+
51. 2 2 9494 49 iiii i iii −−−+ ⋅==−−
52. 2 2 81621632 84 224 iiii i iii +−−− ⋅==−
53. () 2 2 2 333940 164025940940 45 2712012027 8116001681 12027 16811681 −+ ==⋅ −+−−−+ −+−− == + =−− iiii iiii i iii i
54. () 2 2 2 2 55 4129 23 5512 512512 2560 25144 60256025 169169169 = +++ =⋅ −+−− = ==− ii ii i ii ii ii i i i
55. ()() ()() 2131 23 1111 2233 11 15 2 15 22 ii iiii ii i i −−+ −= +−+− = + = =−−
56. ()() ()() 2 2 2252 25 2222 42105 4 129 5 129 55 iii i iiii iii i i i −++ += +−+− −++ = + = =+
57. ()() ()() 22 2 2 2 38232 2 32383238 3864 924616 49 91816 492518 25182518 10072225162 625324 6229762297 949949949 iiii ii iiii iiii iii ii i ii ii iii i i ++− += −+−+ ++− = +−− + = ++ −+− =⋅ +− −++− = + + ==+
58. ()() () 2 2 2 143 13 44 443 4 514 1414 520 116 520 1717 iii i iiii iiii ii i ii i i i ++−− −= −+−− = =⋅ +− = =−
59. ()()()() 2 626212231 23 −⋅−===− =− iii
60. ()() () 2 510510 5052152 ii i −⋅−= ==−=−
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61. ()() 22 2 15151515 ii −===−
62. ()() 22 2 75757575 ii −===−
63. 850850 2252 72 −+−=+ =+ = ii ii i 64. 455455 355 25 −−−=− =− = ii ii i
65. ()()()()
3571035710 213107550 215075310 215275310 +−−−=+−
66. ()()() () 2 2 262626 426266 4262661 4646 246 −−=−− =−−+ =−−+− =−− =−− ii iii ii i i
67. 2 220; 1, 2, 2 xxabc −+===−= ()()()() () 2 22412 21 24 2 22 2 1 x i i −−±−− = ±− = ± = =±
68. 2 6100; 1, 6, 10 xxabc ++==== ()() () 2 664110 21 64 2 62 2 3 −±− = −±− = −+ = =−± x i i
69. 2 416170; 4, 16, 17 xxabc ++==== ()()() () 2 16164417 24 1616 8 164 8 1 2 2 x i i −±− = −±− = −± = =−±
70. 2 96370; 9, 6, 37 xxabc −+===−= ()()()() () 2 664937 29 61296 18 6361 2 183 x i i −−±−− = ±− = ± ==±
71. 2 416210; 4, 16, 21 ++==== xxabc ()()() () 2 16164421 24 1680 8 1680 8 1645 8 5 2 2 −±− = −±− = −± = −± = =−± x i i i
72. 2 16430; 16, 4, 3 ttabc −+===−=
()()()() () 2 444163 216 4176 32 4411 32 111 88 t i i −−±−− = ±− = ± = =±
73. 2 2 3 690 Multiply both sides by 2. 2 312180; 3, 12, 18 −+= −+===−= xx xxabc
()()()() () 2 12124318 23 1272 6 1262 6 22 x i i −−±−− = ±− = ± = =±
74. 2 735 0 8416 xx−+= Multiply both sides by 16. 2 141250; 14, 12, 5 xxabc −+===−= ()()()() () 2 12124145 214 12136 28 12234 28 334 714 x i i −−±−− = ±− = ± = =±
75. 22 1.4210014201000; 14, 20, 100 −+= −+= ==−= xxxx abc
()()()() () 2 2020414100 214 205200 28 202013 28 202013 2828 5513 77 −−±−− = ±− = ± = =± =± x i i i
76. 2 4.53120; 4.5, 3, 12 xxabc −+===−= ()()()() () 2 3344.512 24.5 3207 9 3323 9 123 33 −−±−− = ±− = ± = =± x i i
77. ()() 663222 611 61 16 iiiii i i i −+=−+ =−−+− =− =−+
78. ()() 2322 4242412142 iiiiiii −=−=−−−=−+
79. ()()() 522 1414141114 iiiiii −=−=−−−=−
80. ()()()()()() 3 32 1111 iiiiii −=−=−=−−=
81. ()() () () () 33 3 33 2 7262 62 21622 43221 4322 −= = = =− =− i i ii i i
82. ()()()()() 66 6222 228881118 iiiii −====−−−=−
83. 322 1111 ii i iiiiiii ===⋅==
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84. ()3 322 11111881 88888648 2 ii i iiiiiii i ====⋅==
85. ()()() 4 422 38181811181 iiii===−−=
86. ()()()() 6 6222 1111 iiiii −===−−−=−
87. (a) 12 916, 2010 zizi =+=− (b)
12 111112010916296 91620109162010340230 34023029611,240463011,2404630 296296877877877 iii zzziiiii iii zi ii −+++ =+=+==
88. (a)
89. False.
Sample answer: ()() 1342 +++=+iii which is not a real number.
90. False.
If 0 b = then abiabia +=−=
That is, if the complex number is real, the number equals its conjugate.
91. True.
92. False.
The pattern ,1,, 1 ii repeats. Divide the exponent by 4.
If the remainder is 1, the result is i
If the remainder is 2, the result is 1.
If the remainder is 3, the result is i If the remainder is 0, the result is 1.
97. ()()()() 11221212 abiabiaabbi +++=+++
The complex conjugate of this sum is ()() 1212 .aabbi +−+
94. (i) D (ii) F (iii) B (iv) E (v) A (vi) C
95. 2 666666 iii −−===−
96. ()() ()() 2 112212122112 12121221 abiabiaaabiabibbi aabbababi ++=+++ =−++
The complex conjugate of this product is ()() 12121221 aabbababi −−+
The product of the complex conjugates is ()() ()() 2 112212122112 12121221 abiabiaaabiabibbi aabbababi −−=−−− =−−+
So, the complex conjugate of the product of two complex numbers is the product of their complex conjugates.
The sum of the complex conjugates is ()()()() 11221212 abiabiaabbi −+−=+−+
So, the complex conjugate of the sum of two complex numbers is the sum of their complex conjugates.
Section 2.5 Zeros of Polynomial Functions
1. Fundamental Theorem of Algebra
2. Linear Factorization Theorem
3. Rational Zero
4. complex conjugate
5. linear; quadratic; quadratic
6. irreducible; reals
7. Descartes’s Rule of Signs
8. lower; upper
9. f is a 3rd degree polynomial function, so there are three zeros.
10. f is a 4th degree polynomial function, so there are four zeros.
11. f is a 5th degree polynomial function, so there are five zeros.
12. f is a 6th degree polynomial function, so there are six zeros.
13. f is a 2nd degree polynomial function, so there are two zeros.
14. ()()() ()() 22 22 11 2121 4 httt tttt t =−−+ =−+−++ =−
h is a 1st degree polynomial function, so there is one real zero.
15. () 3222fxxxx=+−−
Possible rational zeros: 1,2±±
Zeros shown on graph: 2,1,1,2
16. () 324416fxxxx=−−+
Possible rational zeros: 1,2,4,8,16 ±±±±±
Zeros shown on graph: 2,2,4
17. () 432 21735945fxxxxx =−++−
Possible rational zeros: 3591545 1 222222 1,3,5,9,15,45, ,,,,, ±±±±±± ±±±±±±
Zeros shown on graph: 3 2 1,,3,5
18. () 5432 485102fxxxxxx =−−++−
Possible rational zeros: 11 1,2,,24±±±±
Zeros shown on graph: 11 22 1,,,1,2
19. () 3 76fxxx=−−
Possible rational zeros: 1,2,3,6±±±±
3 1 0 7 6 3 9 6 1 3 2 0
()()() ()()() 2 332 321 fxxxx xxx =−++ =−++
So, the rational zeros are 2,1,and 3.
20. () 3 1312fxxx=−+
Possible rational zeros: 1,2,3,4,6,12 ±±±±±±
3 1 0 13 12 3 9 12
3 4 0
()()() ()()() 2 334 341 fxxxx xxx =−+− =−+−
So, the rational zeros are 3,4,and 1.
21. () 3244=−+gttt
Possible rational zeros: 1,2,4±±±
After testing all six possible rational zeros by synthetic division, you can conclude there are no rational zeros.
22. () 3 1930=−+pxxx
Possible rational zeros: 1,2,3,5,6,10,15,30 ±±±±±±±± ()()() ()()() 2 3310 352 =−+− =−+− pxxxx xxx
So, the rational zeros are 3,5,2.
23. () 328136htttt=+++
Possible rational zeros: 1,2,3,6±±±± ()() ()()() 3228136621 611 tttttt ttt +++=+++ =+++
So, the rational zeros are 1 and6.
24. () 3281218=+++gxxxx
Possible rational zeros: 1,2,3,6,9,18. ±±±±±±
After testing all six possible rational zeros by synthetic division, you can conclude there are no rational zeros.
25. () 32 231Cxxx=+−
3 1 0 19 30 3 9 30 1 3 10 0 6 1 8 13 6 6 12 6 1210 1 2 3 0 1 2 1 1 2 1 1 0
Possible rational zeros: 1 2 1, ±± ()() ()()() ()() 322 2 231121 1121 121 xxxxx xxx xx +−=++− =++− =+−
So, the rational zeros are 1 2 1 and .
26. () 32 319339fxxxx=−+−
Possible rational zeros: 1 3 1,3,9, ±±±± ()()() ()()() 2 33103 3313 fxxxx xxx =−−+ =−−−
3 3 19 33 9 9 30 9 3 10 3 0
So, the rational zeros are 1 3 3 and .
27. () 432 9958424fxxxxx =−−++
Possible rational zeros: 12412488 33339999 1,2,3,4,6,8,12,24, ,,,,,,, ±±±±±±±± ±±±±±±±±
2 9 9 58 4 24 18 54 8 24 9 27 4 12 0
3 9 27 4 12 27 0 12 9 0 4 0
()()()() ()()()() 2 2394 233232 fxxxx xxxx =+−− =+−−+
So, the rational zeros are 2 3 2,3,, and 2 3
28. () 432 215231525fxxxxx =−++−
Possible rational zeros: 525 1 1,5,25,,,222 ±±±±±±
5 2 15 23 15 25 10 25 10 25 2 5 2 5 0 1 2 5 2 5 2 3 5 2 3
()()()()() 51125fxxxxx =−−+−
So, the rational zeros are 5,1,1 and 5 2
29. 32 511420 −+−−= xxx
Possible rational zeros: 1,212,,1,2 1,555 ±± =±±±± ±± ()() ()()()() () () 2 2 2 2 2 15620 5620 5620 4 2 66452 25 676 10 2319 10 319 5 xxx xx xx bbac x a x x x x −−++= −++= −−= −±− = −−±−−− = ± = ± = ± =
So, the real zeros are 319 1, . 55 ==±xx
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30. 32 8101560 +−−=xxx
Possible
31. 432631660 ++−+=xxxx
Possible
32. 43281417420 ++−−=xxxx
Possible rational zeros: 1,2,3,6,7,14,21,42 ±±±±±±±±
33. () 32 44fxxxx=+−−
(a) Possible rational zeros: 1,2,4±±± (b)
35. () 32 41583fxxxx=−+−−
(a) Possible rational zeros: 1133 1,3,,,,2244 ±±±±±± (b)
(c) Real zeros: 2,1,2
34. () 32 3203616fxxxx =−+−+
(a) Possible rational zeros:
(b)
(c) Real zeros: 1 4 ,1,3
36. () 32 41215fxxxx=−−+
(a) Possible rational zeros: 3 1 22 5153515 1 224444 1,3,5,15,,, ,,,,, ±±±±±± ±±±±±±
(b)
(c) Real zeros: 2 3 ,2,4
(c) Real zeros: 35 1,,22
37. () 432 2132128fxxxxx =−+−++
(a) Possible rational zeros: 1 2 1,2,4,8, ±±±±± (b)
(c) Real zeros: 1 2 ,1,2,4
38. () 42 4174fxxx=−+
(a) Possible rational zeros: 11 1,2,4,,24 ±±±±± (b)
(c) Real zeros: 11 22 2,,,2
39. () 32 3252173fxxxx =−++
(a) Possible rational zeros: 1133 2244 111333 8816163232 1,3,,,,, ,,,,, ±±±±±± ±±±±±± (b)
(b)
(c) Real zeros: 3 1 84,,1
40. () 32 471118fxxxx=+−−
(a) Possible rational zeros: 1,2,3,6,9,18, 139139 ,,,,, 222444 ±±±±±± ±±±±±±
45. If 32i + is a zero, so is its conjugate, 32. i
(c) Real zeros: 1145 2, 88 −±
41. ()()()() ()() 2 32 155 125 2525 f xxxixi xx xxx =−−+ =−+ =−+−
Note: ()() 32 2525,fxaxxx=−+− where a is any nonzero real number, has the zeros 1 and 5. i ±
42. ()()()() ()() 2 32 433 49 4936 f xxxixi xx xxx =−−+ =−+ =−+−
Note: ()() 324936,fxaxxx=−+− where a is any real number, has the zeros 4,3, i and 3. i
43. If 1 + i is a zero, so is its conjugate, 1. i ()()()()()() () ()() 22 432 2211 4422 614168 =−−−+−− =−+−+ =−+−+ f xxxxixi xxxx xxxx
Note: ()() 432614168,=−+−+ fxaxxxx where a is any nonzero real number, has the zeros 2, 2 and 1. ± i
44. If 32i is a zero, so is its conjugate, 32. + i ()()()()()() () ()() 22 432 153232 45613 10322265 =+−−−−+ =−−−+ =−+−− f xxxxixi xxxx xxxx
Note: ()() 43210322265,=−+−− fxaxxxx where a is any nonzero real number, has the zeros 1,5 and 32. ± i
Note: ()() 432 317252322,fxaxxxx =−++− where a is any nonzero real number, has the zeros 2 3 ,1, and 32. i ±
46. If 13 + i is a zero, so is its conjugate, 13. i ()()()()() ()() 22 432 2551313 2152524 211310100 =++−−−+ =++−+ =++++ f xxxxixi xxxx xxxx
Note: ()() 432 211310100,=++++ fxaxxxx
where a is any real number, has the zeros 5 2 ,5, and 13. ± i
47. ()()()()() ()() () 22 432 21 21 2 =+−−+ =+−+ =+−+− f xaxxxixi axxx axxxx
Since ()04 =− f ()()()() () 432 400002 42 2 −=+−+− −=− = a a a
So, ()() 432 432 22 22224. =+−+− =+−+− fxxxxx xxxx
48. ()()()()() ()() () 22 43 1222 22 24 =+−−+ =−−+ =−−− f xaxxxixi axxx axxx
Since ()112 f = ()()() () 43 1211214 126 2 =−−− =− =− a a a
So, ()() 43 43 224 2248. fxxxx xxx =−−−− =−+++
53. () 432231218fxxxxx =−−+− 2 2432 42 32 3 2 2 23 6231218 6 2312 212 318 318 0 xx xxxxx xx xxx xx x x −+ −−−+− −++ −+
49. ()()()()() () ()() () 2 32 31313 324 212 f xaxxixi axxx axxx =+−+−− =+−+ =+−+
Since ()212 f −= ()()() () 32 12222212 1212 1 =−+−−−+ = = a a a
So, ()()() 43 32 124 212. fxxxx xxx =−−− =+−+
50. ()()()()() () ()() () 2 3 21212 223 6 f xaxxixi axxx axx =+−+−− =+−+ =−+
Since ()112 f −=− ()() () 3 12116 126 2 −=−−−+ −= =− a a a
So, ()()() 3 3 26 2212. fxxx xx =−+ =−+−
51. () 4228fxxx=+−
(a) ()()() 2242=+−fxxx
(b) ()()()() 2 422=++− fxxxx
(c) ()()()()()2222=+−+− fxxixixx
52. () 42627fxxx=+−
(a) ()()() 2293fxxx=+−
(b) ()()()() 2 933fxxxx =++−
(c) ()()()()() 3333=+−+− fxxixixx
(a) ()()() 22623fxxxx=−−+
(b) ()()()() 2 6623fxxxxx =+−−+
(c) ()()()()()661212=+−−−−+ f xxxxixi
Note: Use the Quadratic Formula for (c).
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54. () 43231220fxxxxx =−−−− 2 2432 42 32 3 2 2 35 431220 4 3512 3 12 520 520 0 xx xxxxx xx xxx xx x x +−−−− +
(a) ()()() 22435fxxxx=+−−
(b) ()() 2 329329 4 22fxxxx
(c) ()()() 329329 22 22fxxixixx
Note: Use the Quadratic Formula for (b).
55. () 32 44fxxxx=−+−
Because 2i is a zero, so is 2i.
2i 1 1 4 4
2i 42i 4
1 21 i 2i 0
2i 1 21 i 2i 2i 2i 1 1 0
()()()() 221fxxixix=−+−
The zeros of () f x are 1,2. x i =±
56. () 32 231827fxxxx=+++
Because 3i is a zero, so is 3. i
3i 2 3 18 27 6i 918 i 27
2 36i + 9i 0
3i 2 36i + 9i 6i 9i 2 3 0
()()()() 3323fxxixix =−++
The zeros of () f x are 3 2 3,.xi=±−
Alternate Solution:
Because 2 x i =± are zeros of (), f x ()() 2 224xixix+−=+ is a factor of () f x By long division, you have: 232 32 2 2 1 0444 04 04 04 0 x xxxxx xxx xx xx ++−+− ++ −+− −+−
()()() 2 41fxxx=+−
The zeros of () f x are 1,2. x i =±
Alternate Solution:
Because 3 x i =± are zeros of (), f x ()() 2 339xixix−+=+ is a factor of () f x By long division, you have: 232 32 2 2 23 09231827 2018 3027 3027 0 x xxxxx xxx xx xx + +++++ ++ ++ ++
()()() 2 923fxxx=++
The zeros of () f x are 3 2 3,.xi=±−
57. () 3282526=−+−fxxxx
Because 32 + i is a zero, so is 32. i
58. () 3292517=+++fxxxx
Because 4 −+ i is a zero, so is 4 i
()()()() ()() 32322 =−+−−− fxxixix
The zeros of () f x are 32,2.=± x i
Alternate Solution:
Because 32=± x i are zeros of ()()()() () 2 ,3232613 −+−−=−+ fxxixixx is a factor of () f x
By long division, you have: 232 32 2 22 2 61382526 613 21226 21226 0 −+−+− −+ −+− −+− x xxxxx xxx xx xx
()()() 2 6132=−+−fxxxx
The zeros of () f x are 32,2.=± x i
59. () 432614189 =−+−+ fxxxxx
Because 12 i is a zero, so is 12, +
()()()() ()() 441 =−−+−−−+ fxxixix
The zeros of () f x are 4,1.=−±−xi
Alternate Solution:
Because 4 =−± x i are zeros of ()()()() () 2 ,44817 −−+−−−=++ fxxixixx is a factor of () f x
By long division, you have:
()()() 2 8171=+++fxxxx
The zeros of () f x are 4,1.=−±−xi
is a factor of (). f x By
The zeros of () f x are 12,1,3.=± xi
60. () 43231314=+−−+ fxxxxx
Because 23−+ i is a zero, so is 23, i and
is a factor of () f x By long division, you
()()() 22 2 4732 4712 =++−+ =++−− fxxxxx xxxx
The zeros of () f x are 23,1,2.=−± xi
61. () ()() 2 36 66 fxx x ixi =+ =+−
The zeros of () f x are 6. x i =±
62. () ()() 2 49 77 =+ =−+ fxx x ixi
The zeros of () f x are 7. =± x i
63. () 2 217hxxx=−+
By the Quadratic Formula, the zeros of () f x are 264 2468 14. 22 x i ±−±− ===±
()()()() () ()() 1414 1414 =−+−− =−−−+ f xxixi xixi
64. () 2 1017gxxx=++ By the Quadratic Formula, the zeros of () f x are: 10100681032 522. 22 x −±−−± ===−± ()()()() () ()() 522522 522522 =−−+−−− =+−++ fxxx xx
65. () ()() ()()()() 4 22 16 44 2222 fxx xx x xxixi =− =−+ =−+−+
Zeros: 2,2i ±±
66. () ()() ()()()() 4 22 256 1616 4444 fyy yy yyyiyi =− =−+ =−+−+
Zeros: 4,4i ±±
67. () 2 22fzzz=−+
By the Quadratic Formula, the zeros of () f z are 248 1. 2 zi ±− ==±
()()() ()() 11 11 f zzizi zizi = −+
=−−−+
68. () 32342hxxxx=−+−
Possible rational zeros: 1,2±±
By the Quadratic Formula, the zeros of 2 22xx−+ are 248 1. 2 x i ±− ==±
Zeros: 1,1 i ±
()()()() 111 hxxxixi =−−−−+
69. () 3235gxxxx=−++
Possible rational zeros: 1,5±±
By the Quadratic Formula, the zeros of 2 45xx−+
are: 41620 2 2 x i ±− ==±
Zeros: 1,2 i −±
()()()() 122 g xxxixi =+−−−+
70. () 32 39 fxxxx=−++
Possible rational zeros: 1,3,13,39 ±±±±
By the Quadratic Formula, the zeros of 2 413xx−+ are: 41652 23 2 x i ±− ==±
Zeros: 3,23i −±
()()()() 32323 f xxxixi =+−−−+
71. () 432481616gxxxxx =−+−+
Possible rational zeros: 1,2,4,8,16 ±±±±±
()()()()
()()() 2 2 224 222 gxxxx x xixi =−−+ =−+−
Zeros: 2,2i ±
72. () 43261069hxxxxx =++++
Possible rational zeros: 1,3,9±±±
The zeros of () f x are 10 x =− and 75. x i =−± 1 1 3 4 2
The zeros of 2 1 x + are x i =±
Zeros: 3, i −±
()()()() 2 3 hxxxixi =++−
73. () 3224214740fxxxx=+++
Possible rational zeros: 1,2,4,5,10,20,37, 74,148,185,370,740 ±±±±±±± ±±±±±
Based on the graph, try 10. x =−
3 1 6 10 6 9 3 9 3 9 1 3 1 3 0 3 1 3 1 3 3 0 3 1 0 1 0 10 1 24 214 740 10 140 740 1 14 74 0 20 1000 10 2000
By the Quadratic Formula, the zeros of 2 1474xx++ are 14196296 75. 2 x i −±− ==−±
74. () 32 25125fssss=−+−
Possible rational zeros: 15 1,5,, 22 ±±±±
76. () 32 915115fxxxx=−+−
Possible rational zeros: 1515 1,5,,,, 3399 ±±±±±±
Based on the graph, try 1 . 2 s =
By the Quadratic Formula, the zeros of () 2 225 ss−+ are 2420 12. 2 s i ±− ==±
The zeros of () f s are 1 2 s = and 12. s i =±
75. () 32 1620415fxxxx=−−+
±±±±±±±±±±
±±±±±±±±±±
Possible rational zeros: 1351513 1,3,5,15,,,,,,, 222244 5151351513515 ,,,,,,,,, 44888816161616
Based on the graph, try 3 4 x =−
Based on the graph, try 1. x =
The zeros of () f x are 1 2,, 2 xx=−=− and . x i =± 1 2 2 5 12 5 1 2 5 2 4 10 0
By the Quadratic Formula, the zeros of ()22 1632204485 xxxx −+=−+ are 864801 1. 82 x i ±− ==±
The zeros of () f x are 3 4 x =− and 1 1. 2 x i =±
By the Quadratic Formula, the zeros of 2 965 xx−+ are 63618012 1833 x i ±− ==±
The zeros of () f x are 1 x = and 12 33 x i =±
77. () 432 25452fxxxxx =++++
Possible rational zeros: 1 1,2, 2 ±±±
Based on the graph, try 2 x =− and 1 . 2 x =−
The zeros of ()22 2221 xx+=+ are x i =±
or duplicated, or posted to a publicly accessible website, in whole or in
78. () 5432828566432gxxxxxx =−+−+−
Possible rational zeros: 1,2,4,8,16,32 ±±±±±±
Based on the graph, try 2. x =
By the Quadratic Formula, the zeros of 2 24xx−+ are 2416 13. 2 x i ±− ==±
The zeros of () g x are 2 x = and 13. x i =±
79. () 32 233gxxx=−−
Sign variations: 1, positive zeros: 1 () 32 233gxxx−=−−−
Sign variations: 0, negative zeros: 0
80. () 2 483=−+hxxx
Sign variations: 2, positive zeros: 2 or 0 () 2 483hxxx−=++
Sign variations: 0, negative zeros: 0
81. () 32 231hxxx=++
Sign variations: 0, positive zeros: 0 () 32 231hxxx−=−++
Sign variations: 1, negative zeros: 1
82. () 4 232=−−hxxx
83. () 432 6232=+−+gxxxx
Sign variations: 2, positive zeros: 2 or 0 () 432 6232−=−−+ gxxxx
Sign variations: 2, negative zeros: 2 or 0
84. () 32 4321 =−−−fxxxx
Sign variations: 1, positive zeros: 1 () 32 4321 −=−−−− fxxxx
Sign variations: 2, negative zeros: 2 or 0
85. () 32 55=+−+fxxxx
88. () 3241fxxx=−+ (a) 4 is an upper bound. (b) 1 is a lower bound. 2 1 6 16 24 16 2 8 16 16
Sign variations: 1, positive zeros: 1 () 4 232−=+−hxxx
Sign variations: 1, negative zeros: 1
Sign variations: 2, positive zeros: 2 or 0 () 32 55−=−+++ fxxxx
Sign variations: 1, negative zeros: 1.
86. () 32 323=−−+fxxxx
Sign variations: 2, positive zeros: 2 or 0 () 32 323−=−−++ fxxxx
Sign variations: 1, negative zeros: 1
87. () 32321fxxxx=+−+ (a) 1 is an upper bound. (b) 4 is a lower bound.
89. () 4341616fxxxx=−+−
(a) 5 is an upper bound. (b) 3 is a lower bound.
91. () 32 161243 =−−+fxxxx
Possible
However, the function factors by grouping.
90. () 4 283fxxx=−+ (a) 3 is an upper bound. (b) 4 is a lower bound
So,
zeros
92. () 32 124279fzzzz=−−+
Possible
So, the real zeros are 3 1 23,, and 3 2
93. () 32 4386fyyyy=+++ Possible rational zeros: 1133 1,2,3,6,,,,2244 ±±±±±±±±
94. () 32 321510gxxxx=−+−
Possible rational zeros: 510 12 1,2,5,10,,,,3333 ±±±±±±±± ()()()()()22 2 3 315325gxxxxx =−+=−+
2 3 3 2 15 10 2 0 10 3 0 15 0
So, the only real zero is 2 3
95.
The rational zeros are 3 2 ± and 2. ±
96. ()() 32 1 2 232312fxxxx =−−+
Possible rational zeros: 3 1 1,2,3,4,6,12,,22 ±±±±±±±
98. ()() 32 1 6 61132fzzzz=+−−
Possible rational zeros: 1121 1,2,,,,2336 ±±±±±± ()()() ()()() 2 1 6 1 6 261 23121 fxxxx xxx =+−− =++−
4 2 3 23 12 8 20 12 2 5 3 0 2 6 11 3 2 12 2 2 6 1 1 0
The rational zeros are 1 2 3,, and 4.
97. () () ()() ()() ()()() 32 32 2 2 11 44 1 4 1 4 1 4 1 4 441 41141 411 4111 fxxxx xxx xxx xx xxx =−−+ =−−+ =−−− =−− =−+−
The rational zeros are 1 4 and 1.±
The rational zeros are 1 3 2,, and 1 2
99. ()()() 32111fxxxxx =−=−++
Rational zeros: () 11 x = Irrational zeros: 0 Matches (d).
100. () ()() 3 333 2 2 224 fxx xxx =− =−++
Rational zeros: 0
Irrational zeros: () 3 12 x = Matches (a).
101. ()()() 3 11fxxxxxx =−=+−
Rational zeros: () 30,1 x =± Irrational zeros: 0 Matches (b).
102. () () ()() 3 2 2 2 22 fxxx xx xxx =− =− =+−
Rational zeros: ()10 x =
Irrational zeros: () 22 x =± Matches (c).
103. (a) (b) ()() ()() 15292 92152 Vlwhxxx xxx =⋅⋅=−− =−−
Because length, width, and height must be positive, you have 9 2 0 x << for the domain.
(c)
The volume is maximum when 1.82. x ≈
The dimensions are: () 32 ,. rakbkckdfkr =+++= 1.82 cm5.36 cm 11.36 cm ××
104. (a) Combined length and width: 41201204 x yyx+= =− () () 2 2 2 Volume 1204 430 lwhxy x x xx =⋅⋅= =− =−
(b) Dimensions with maximum volume: 20 in. 20 in. 40 in. ××
(c) () 2 32 32 13,500430 412013,5000 3033750 x x xx xx =− −+= −+= ()() 2 15152250xxx−−−=
Using the Quadratic Formula, 15155 15,. 2 x ± =
The value of 15155 2 is not possible because it is negative.
(d) ()() 23 32 5692152 56135484 044813556 xxx
The zeros of this polynomial are 7 1 22,, and 8. x cannot equal 8 because it is not in the domain of V [The length cannot equal 1 and the width cannot equal 7. The product of ()()() 81756 −−= so it showed up as an extraneous solution.]
So, the volume is 56 cubic centimeters when 1 2 x = centimeter or 7 2 x = centimeters.
105. (a) Current bin: 23424 V =××= cubic feet
New bin: () ()()()() 524120 cubic feet 234120 V Vxxxx == =+++=
(b) 32 32 92624120 926960 xxx xxx +++= ++−=
The only real zero of this polynomial is 2. x = 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.
106. 2 200 100,1 30 x Cx xx =+≥ + C is minimum when 32 3402400360000. xxx−−−=
The only real zero is 40 x ≈ or 4000 units.
107. False. The most complex zeros it can have is two, and the Linear Factorization Theorem guarantees that there are three linear factors, so one zero must be real.
108. False. f does not have real coefficients.
109. ()(). g xfx =− This function would have the same zeros as (), f x so 12,,rr and 3r are also zeros of () g x
110. ()() 3. g xfx = This function has the same zeros as f because it is a vertical stretch of f. The zeros of g are 12,,rr and 3.r
111. ()()5. gxfx=− The graph of () g x is a horizontal shift of the graph of () f x five units of the right, so the zeros of () g x are 12 5,5, rr++ and 3 5. r +
112. ()()2. g xfx = Note that x is a zero of g if and only if 2x is a zero of f. The zeros of g are 12 22,, rr and 3 2 r
113. ()() 3. g xfx =+ Because () g x is a vertical shift of the graph of (), f x the zeros of () g x cannot be determined.
114. ()() g xfx =− Note that x is a zero of g if and only if x is a zero of f. The zeros of g are 12,, rr and 3r
115. Zeros: 1 2 2,,3 ()()()() 32 2213 23116 fxxxx xxx =−+−− =−++−
Any nonzero scalar multiple of f would have the same three zeros. Let ()(),0.gxafxa=> There are infinitely many possible functions for f 116.
10 (1,0) (3,0) (1,0)(4,0) y
x 45
117. Because 1 i + is a zero of f, so is 1. i From the graph, 1 is also a zero.
()()()() ()() ()() 2 32 111 221 342 fxxixix xxx xxx =−+−−− =−+− =−+−
118. Because 1 i + is a zero of f, so is 1. i From the graph, 1 is also a zero.
()()()() ()() ()() 2 32 111 221 2 fxxixix xxx xx =−+−−+ =−++ =−+
Because the graph rises to the left and falls to the right, () 32 1, and 2. afxxx =−=−+−
119. If = x i is a zero, then =− x i is also a zero. So the function is ()()()()() 23.5.=−−−+ f xxxxixi
120. (a) Zeros of () :2,1,4 fx
(b) The graph touches the x-axis at 1. x =
(c) The least possible degree of the function is 4 because there are at least four 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.
(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)
f y x x 1 32 (2,0) (1,0)(4,0) 4 6 8 2 35 10 y
121. Because ()()20,fifi== then i and 2i are zeros of f. Because i and 2i are zeros of f, so are i and –2. i ()()()()() ()() 22 42 22 14 54 f xxixixixi xx xx =−+−+ =++ =++
122. Because ()20, f = 2 is a zero of f. Because () 0, fi = i is a zero of f. Because i is a zero of f, so is i ()()()() ()() 2 32 12 121 22 f xxxixi xx xxx =−−+− =−−+ =−+−+
123. (a) ()()() 2 f xxbixbixb =−+=+ (b) ()()() ()() ()() 22 222 2 f xxabixabi x abixabi xabi xaxab
