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5.6: Find Rational Zeros, Part 1 (Really these answers are from 5.5)

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5.6: Find Rational Zeros, Part 1 (Really these answers are from 5.5)
5.6: Find Rational
Zeros, Part 1
(Really these answers
are from 5.5)
P. 366-367
Chapter 5,
15. 4
1
1
continued
25
0
22
4
24
216
21
24
218
21. 6
2
18
x24
0
24
6
23
9
215
23
5
29
1
1
x 3 2 4x 1 6
x13
1
22. 24
13
212
6
6
212
6
1
22
1
26
23. 8
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
28
12
1
22
240
264
8
48
64
8
0
6
235
f(x) 5 x 2 2x 2 40x 2 64
25
5
225
45
5 (x 2 8)(x 1 4)(x 1 2)
21
5
29
10
24. 210
1
10
x15
0
25
3
2
4
22
2
21
1
2
5 (x 2 8)(x 2 1 6x 1 8)
18
95
150
210
280
2150
8
15
0
1
f (x) 5 x 1 18x 1 95x 1 150
3
2
5 (x 1 10)(x2 1 8x 1 15)
5 (x 1 10)(x 1 3)(x 1 5)
25. 29
1
x22
20. The error is that a zero was not used as a place holder for
1
1
2
251
108
29
63
2108
27
12
0
f (x) 5 x 3 1 2x 2 2 51x 1 108
5 (x 1 9)(x 2 2 7x 1 12)
the missing x 2 term.
1
24
16
x 3 2 5x 1 3
x22
5 (x 1 9)(x 2 4)(x 2 3)
0
25
3
2
4
22
2
21
1
1
x 2 5x 1 3
} 5 x 2 1 2x 2 1 1 }
x22
x22
3
212
0
} 5 x 2 1 2x 2 1 1 }
1
5
4
error in how the quotient is written.
2
6
3
19. The division was done correctly, however, there is an
1
0
1
6
x26
x 4 1 4x 3 1 16x 2 35
x15
1
25
5 (x 1 4)(x 1 3)(x 2 1)
}} 5 x 3 2 x 2 1 5x 2 9 1 }
2
24
5 (x 1 4)(x2 1 2x 2 3)
x 4 2 5x 3 2 8x 2 1 13x 2 12
x26
1
230
1
2 23
0
f (x) 5 x 3 1 6x 2 1 5x 2 12
9
x13
28
1
224
1
}}} 5 x 3 1 x 2 2 2x 1 1 2 }
18. 25
6
5 (x 2 6)(x 2 5)(x 1 1)
25
1
30
5 (x 2 6)(x 2 2 4x 2 5)
} 5 x2 2 3x 1 5 2 }
17. 6
19
f (x) 5 x 3 2 10x 2 2 19x 1 30
} 5 x2 2 x 2 4 2 }
16. 23
210
1
x 2 5x 2 2
x24
3
1
26. 22
1
29
8
60
22
22
260
1 211 30
0
f (x) 5 x 3 2 9x 2 1 8x 1 60
5 (x 1 2)(x2 2 11x 1 30)
5 (x 1 2)(x 2 5)(x 2 6)
Algebra 2
Worked-Out Solution Key
279
Chapter 5,
2
215
34
221
2
213
21
32. 24
2 213
21
0
f (x) 5 2x 3 2 15x 2 1 34x 2 21
3
5 (x 2 1)(2x2 2 13x 1 21)
3
22
261
220
15
65
20
2
2
5 (x 2 5)(3x 1 1)(x 1 4)
22
221
218
23
15
18
225
2154
40
40
150
240
5 (x 2 10)(4x2 1 15x 2 4)
5 (x 2 10)(4x 2 1)(x 1 4)
71
42
70
277
242
26
0
f (x) 5 10x 2 81x 1 71x 1 42
2
5 (x 2 7)(10x2 2 11x 2 6)
5 (x 2 7)(5x 1 2)(2x 2 3)
2
3
The other zeros are 2}5 and }2.
280
Algebra 2
Worked-Out Solution Key
271
29
18
72
9
8
1
0
}
28 6 Ï64 2 4(2)
}
28x 6 Ï 56
x 5 }}
5}
4
2(2)
}
24 6 Ï 14
5
5
21
218
8
210
22
28
211
4
0
f (x) 5 5x 3 2 x 2 2 18x 1 8
5 (x 1 2)(5x 2 2 11x 1 4)
}}
11 6 Ï121 2 4(5)(4)
}
116 Ï41
x 5 }}
5}
10
10
35. D;
281
211
210
The other zeros are about 0.46 and 1.74.
1
The other zeros are }4 and 24.
3
0
The other zeros are about 20.13 and 23.87.
4
15
24
0
f (x) 5 4x 3 2 25x 2 2 154x 1 40
10
216
5 (x 2 9)(2x2 1 8x 1 1)
34. 22
The other zeros are 6 and 21.
10
22
5}
2
5 (x 1 3)(x 2 6)(x 1 1)
31. 7
64
f (x) 5 2x 3 2 10x 2 2 71x 2 9
0
5 (x 1 3)(x2 2 5x 2 6)
4
288
2
5 (x 2 5)(3x 1 13x 1 4)
30. 10
212
The other zeros are }3 and 28.
2
1
25
26
f (x) 5 x 3 2 2x2 2 21x 2 18
264
5 (x 1 4)(3x 2 2)(x 1 8)
33. 9
1
72
5 (x 1 4)(3x 2 1 22x 2 16)
3
13
4
0
3
2
f (x) 5 3x 2 2x 2 61x 2 20
29. 23
34
f (x) 5 3x 3 1 34x 2 1 72x 2 64
5 (x 2 1)(2x 2 7)(x 2 3)
28. 5
3
26
4
4
15
263
254
224
54
54
29
29
0
f (x) 5 4x 3 1 15x 2 2 63x 2 54
5 (x 1 6)(4x 2 2 9x 2 9)
5 (x 1 6)(4x 1 3)(x 2 3)
3
The zeros are 26, 2}4 , and 3.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
27. 1
continued
Chapter 5,
continued
1
40. a. }
2
36. (x 1 4)(x 1 2) 5 x 2 1 6x 1 8
2x 3 1 17x 2 1 46x 1 40
x 1 6x 1 8
* 5 }}
2
b.
1
2
}
7
239
14
15
11
214
30
22
228
0
1
2
30
2x 1 5
x 2 1 6x 1 8qww
2x 3 1 17x 2 1 46x 1 40
2x 3 1 12x 2 1 16x
5x 1 30x 1 40
2
1
f(x) 5 x 2 }2 (30x 2 1 22x 2 28)
5x 2 1 30x 1 40
c. f (x) 5 1 x 2 }2 2(2)(15x 2 1 11x 2 14)
1
0
The missing dimension is 2x 1 5.
5 (2x 2 1)(3x 2 2)(5x 1 7)
37. (x 2 1)(x 1 6) 5 x 2 1 5x 2 6
Problem Solving
x 3 1 13x 2 1 34x 2 48
x 1 5x 2 6
w 5 }}
2
41. P 5 2x 3 1 4x 2 1 x
4 5 2x 3 1 4x 2 1 x
x1 8
0 5 2x 3 1 4x 2 1 x 2 4
x 3 1 13x 2 1 34x 2 48
x 2 1 5x 2 6qww
x3 1 5x 2 2 6x
4
21
8x 2 1 40x 2 48
8x 2 1 40x 2 48
The missing dimension is x 1 8.
25
1
1
212
26
236
23
218
0
f (x) 5 x 2 5x 2 12x 1 36
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3
2
5 (x 2 2)(x2 2 3x 2 18)
5 (x 2 2)(x 2 6)(x 1 3)
The other zeros are 6 and 23.
39. A;
5
1
21
k
230
5
20
30
1
4 k 1 20
0
For x 2 5 to be a factor, the remainder must be zero. In
the last column 30 must be added to 230 to get zero.
k 1 20 must equal 6 in order to have a product of 30 to
add to the 230. k 1 20 5 6 so, k 5 214.
24
0
4
1
0
0
P 5 (x 2 4)(2x 1 1)
5 (x 2 4)(1 2 x)(1 1 x)
x 5 1 is the only other positive solution. So, the company
could still make the same profit by producing only
1 million T-shirts.
42.
P 5 24x 3 1 12x 2 1 16x
48 5 24x 3 1 12x 2 1 16x
0 5 24x 3 1 12x 2 1 16x 2 48
3
24
b. The factors are (x 2 2), (x 2 6), and (x 1 3).
c. The solutions are x 5 2, x 5 6, x 5 23.
24
2
36
2
1
21
0
38. a. 2
4
12
16
248
212
0
48
24
0 16
0
2
(
)
P 5 (x 2 3) 24x 1 16 5 (x 2 3)(4 2 2x)(4 1 2x)
x 5 2 is the only other positive solution. The company
could produce 2 million MP3 players and still make the
same profit.
43. Let a 5 average attendance per team.
A
21.95x 3 1 70.1x 2 2 188x 1 2150
a5}
5 }}}
T
14.8x 1 725
20.132x 2 1 11.2x 2 561.35
14.8x 1 725q21.95x 1 70.1x 2 188x 1 2150
www
3
2
21.95x 3 2 95.7x 2
165.8x 2 2 188x
165.8x 2 1 8120x
28308x 1 2150
28308x 2 406,978.75
409,128.75
A function is
409,129
a 5 20.132x 2 1 11.2x 2 561 1 }
14.8x 1 725
Algebra 2
Worked-Out Solution Key
281
5.6: Find Rational
Zeros, Part 1
P. 374-377
Chapter 5,
continued
4. G; Let x 5 height.
Lesson 5.6
V 5 *wh
5.6 Guided Practice (pp. 371–373)
540 5 (x 1 15)(x 2 3)(x)
1. Factors of the constant term: 61, 63, 65, 615
0 5 x3 1 12x2 2 45x 2 540
Factors of the leading coefficient: 61
0 5 x2(x 1 12) 2 45(x 1 12)
0 5 (x2 2 45)(x 1 12)
3
1
5
15
Possible rational zeros: 6}1, 6}1, 6}1, 6}
1
}
The only positive solution is x 5 Ï45 , or x ø 6.7, so the
height of the box is about 6.7 inches.
5. A;
Simplified list: 61, 63, 65, 615
2. Factors of the constant term: 61, 62, 63, 66
Factors of the leading coefficient: 61, 62
Profit
Price per
(millions 5 camera
of dollars)
(dollars)
Number
+ of cameras
(millions)
Cost per
2 camera
(dollars)
Number
+ of cameras
(millions)
1
2
3
6
3
2
6
6}2, 6}2, 6}2
1
P 5 (100 2 10x2)(x) 2 30x
P 5 100x 2 10x3 2 30x
3
6
9
18
1
2
3. Possible rational zeros: 6}, 6}, 6}, 6}, 6}, 6}
1
1
1
1
1
1
Test x 5 1:
1
When P 5 60:
1
1
24
215
18
1
23
218
23
218
0
Because 1 is a zero of f, f (x) can be writen as:
0 5 10x 3 2 70x 1 60
You know that 2 is a solution, so x 2 2 is a factor of
10x 3 2 70x 1 60.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
10
10
3
Simplified list: 61, 62, 63, 66, 6}2 , 6}2
P 5 70x 2 10x 3
60 5 70x 2 10x 3
1
Possible rational zeros: 6}1, 6}1, 6}1, 6}1, 6}2,
0
270
60
20
40
260
20
230
0
So, (x 2 2)(10x2 1 20x 2 30) 5 0. use factoring to find
f(x) 5 (x 2 1)(x2 2 3x 2 18) 5 (x 2 1)(x 2 6)(x 1 3)
The zeros are 23, 1 and 6.
4. Possible rational zeros: 61, 62, 67, 614
Test x 5 1:
1
1
28
5
14
1
27
22
22
212
28
5
14
21
9
214
29
14
0
that x 5 1 and x 5 23 are the other solutions.
1
27
1 is not a zero.
The company could produce 1 million cameras and make
the same profit.
Test x 5 21:
6. In 2001, t 5 6.
21
1
C 5 20.027(6)4 1 0.32(6)3 2 0.25(6)2 2 4.9(6) 1 51
5 234.992 1 69.12 2 9 2 29.4 1 51
ø 46.73
The average monthly cell phone bill in 2001 was about
$46.73.
1
7. V 5 } s2h
3
1
Because 21 is a zero of f, f(x) can be written as:
f (x) 5 (x 1 1)(x2 2 9x 1 14) 5 (x 1 1)(x 2 7)(x 2 2)
The zeros of f are 21, 2 and 7.
1
48 5 }3 (3x 2 6)2(x)
1
48 5 }3 x(9x2 2 36x 1 36)
0 5 3x 3 2 12x 2 1 12x 2 48
0 5 3x 2(x 2 4) 1 12(x 2 4)
0 5 (3x 2 1 12)(x 2 4)
The only positive solution is x 5 4, so the height of the
sculpture is 4 feet.
Algebra 2
Worked-Out Solution Key
283
Chapter 5,
continued
3
1
1
1
5. Possible rational zeros: 61, 63, 6}, 6}, 6}, 6},
2
2
3
4
3
3
3
1
1
1
1
1
1
}
}
}
}
}
}
}
}
}
6 4, 66 , 68, 68 , 612, 6 16, 616, 624, 648
3
2
}
2
2
1
220
21
3
6
221
4
214
0
f(x) 5 (x 1 2) + g (x)
5 (x 1 2)1 x 2 }2 2(2x2 1 4x 2 14)
3
2x2 1 4x 2 14 5 0
3
3
1
}}
Reasonable values: x 5 2}4, x 5 }
, x 5 }2
16
24 6 Ï(4)2 2 4(2)(214)
2(2)
3
Check: x 5 2}4 :
4
220
3
236
24
23
232
4
0
48
48
}
5 21 6 2Ï2
7.
}
1
V 5 }3 (x2)(x 2 1)
1
6 5 }3 x2(x 2 1)
3
2}4 is a zero.
1
}
3
The real zeros of f are: 22, }2, 21 1 2Ï2 , and 21 2 2Ï2 .
18 5 x3 2 x2
2
3
f (x) 5 x 1 }4 (48x2 2 32x 1 4)
0 5 x3 2 x2 2 18
Possible rational zeros: 61, 62, 63, 66, 69, 618
5 1 x 1 }4 2(4)(12x2 2 8x 1 1)
3
2
21
0
218
2
2
4
1
1
2
214
1
21
0
218
3
6
18
1
5 (4x 1 3)(6x 2 1)(2x 2 1)
3 1
4 6
1
2
The real zeros of f are 2}, }, and }.
6. Possible rational zeros: 61, 62, 63, 66, 67, 614,
3
7
1
21
621, 642, 6}2 , 6}2, 6}2, 6}
2
3
1
2 6
0
The other solutions,
which satisfy x 2 1 2x 1 6, are
}
x 5 21 6 iÏ 5 and can be discarded because they are
imaginary. The only real solution is x 5 3. The base is
3 3 3 feet. The height is x 2 1 or 2 feet.
7
3
Reasonable zeros: x 5 2}2 , x 5 22, x 5 }2 , x 5 2
7
2}2
22
218
219
27
7
}
2
22
2
5
2
2
5
42
2273
4
2105
}
4
}
211
77
2
39
}
2
218
219
42
22
40
242
1
220
21
0
f (x) 5 (x 1 2)(2x 1 x 2 20x 1 21) 5 (x 1 2) + g(x)
2
1
Possible rational zeros of g(x): 61, 63, 67, 621, 6}2,
3
7
21
6}2, 6}2 , 6}
2
3
Reasonable zeros: x 5 }2 , x 5 2
284
Algebra 2
Worked-Out Solution Key
Skill Practice
1. If a polynomial function has integer coefficients, then
p
every rational zero of the function has the form }q where
p is a factor of the constant term and q is a factor of the
leading coefficient.
2. To shorten the list, draw the graph to see approximately
24
3
5.6 Exercises (pp. 374–377)
where the zeros are.
3. f(x) 5 x3 2 3x 1 28
Factors of constant term: 61, 62, 64, 67, 614, 628
Factors of leading coefficient: 61
1
2
4
7
14
28
, 6}
Possible rational zeros: 6}1, 6}1, 6}1, 6}1 , 6}
1
1
Simplified list: 61, 62, 64, 67, 614, 628
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3
2}4
}
24 6 Ï128
4
x 5 }} 5 }
Chapter 5,
continued
Simplified list: 61, 63, 65, 615, 6}2 , 6}2 ,
Factors of constant term: 61, 62, 65, 610
5
1
1
15
3
1
5
10. h(x) 5 6x 3 2 3x 2 1 12
Simplified list: 61, 62, 65, 610
Factors of the constant term: 61, 62, 63, 64, 66, 612
5. f(x) 5 2x4 1 6x3 2 7x 1 9
Factors of the leading coefficient: 61, 62, 63, 66
Factors of constant term: 61, 63, 69
1
9
3
1
9
12
Possible rational zeros: 6}1 , 6}1, 6}1, 6}2 , 6}2, 6}2
3
1
3
2
4
6
Possible rational zeros: 6}1, 6}1, 6}1, 6}1, 6}1 ,
Factors of leading coefficient: 61, 62
3
5
6}
8
10
1
Possible rational zeros: 6}, 6}, 6}, 6}
1
3
1
15
5
1
2
1
15
6}2 , 6}
, 6}4 , 6}4, 6}4, 6}
, 6}8, 6}8 , 6}8 ,
7
2
Factors of leading coefficient: 61
2
1
3
6
4
12
, 6}2 , 6}
, 6}2, 6}2 , 6}2 , 6}
,
6}
2
1
2
9
Simplified list: 61, 63, 69, 6}2, 6}2, 6}2
1
2
3
4
6
12
6
4
12
1
2
3
6}3 , 6}3, 6}3, 6}3 , 6}3, 6}
, 6}6, 6}6, 6}6,
3
6. h(x) 5 2x 3 1 x 2 2 x 2 18
Factors of constant term: 61, 62, 63, 66, 69, 618
6}6 , 6}6, 6}
6
Factors of leading coefficient: 61, 62
Simplified list: 61, 6 2, 63, 64, 66, 612,
3
6
9
18
1
2
1
, 6}2,
Possible rational zeros: 6}1, 6}1 , 6}1, 6}1 , 6}1 , 6}
1
3
6
9
18
2
6}2 , 6}2 , 6}2, 6}2 , 6}
2
3
9
1
Simplified list: 61, 62, 63, 66, 69, 618, 6}2, 6}2, 6}2
6}2 , 6}2, 6}3, 6}
, 6}3 , 6}6
3
7. g (x) 5 4x5 1 3x3 2 2x 2 14
3
1
7
2
14
1
624
1
212
35
224
1
211
24
24
0
14
1
2
7
14
6}
, 6}4, 6}4 , 6}4, 6}
2
4
7
1
1
8. f(x) 5 3x4 1 5x3 2 3x 1 42
Factors of the constant term: 61, 62, 63, 66, 67,
614, 621, 642
3
2
6
7
Possible rational zeros: 6 }1, 6}1, 6}1, 6}1 , 6}1,
14
21
42
1
3
2
6
7
21
42
, 6}
6}
3
3
Simplified list: 61, 62, 63, 66, 67, 614,
2
7
Real zeros are 1, 3, and 8.
12. Possible rational zeros: 61, 62, 64, 67, 68, 614, 628,
656
25
222
56
1
24
226
1 24 226
Test x 5 21:
30
21
1
1
14
, 6}
, 6}
, 6}3 , 6}3 , 6}3, 6}3 , 6}3 , 6}
,
6}
3
1
1
1
1
5 (x 2 1)(x 2 3)(x 2 8)
Test x 5 1:
Factors of the leading coefficient: 61, 63
1
f(x) 5 (x 2 1)(x 2 11x 1 24)
2
7
Simplified list: 61, 6 2, 6 7, 6 14, 6}2 , 6}2, 6}4, 6}4
211
1
7
2
1
14
621, 642, 6}3 , 6}3, 6}3, 6}
3
1
9. h(x) 5 8x4 1 4x3 2 10x 1 15
25
222
56
21
6
16
26
216
72
Factors of the constant term: 61, 63, 65, 615
Test x 5 2:
Factors of the leading coefficient: 61, 62, 64, 68
2
1
3
5
1
Test x 5 1:
Factors of the leading coefficient: 61, 62, 64
1
4
11. Possible rational zeros: 61, 62, 63, 64, 66, 68, 612,
1
Possible rational zeros: 6}1, 6}1 , 6}1, 6}
, 6}2, 6}2, 6}2,
1
2
1
Factors of the constant term: 61, 62, 67, 614
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3
1
4. g (x) 5 x 3 2 4x 2 1 x 2 10
15
1
1
, 6}2,
Possible rational zeros: 6}1 , 6}1 , 6}1, 6}
1
3
5
15
3
5
15
3
1
1
6}2 , 6}2, 6}
, 6}4 , 6}4 , 6}4, 6}
, 6}8, 6}8,
2
4
5
15
6}8 , 6}
8
1
25
222
56
2
26
256
23
228
0
f(x) 5 (x 2 2)(x2 2 3x 2 28)
5 (x 2 2)(x 2 7)(x 1 4)
Real zeros are 2, 7, and 24.
Algebra 2
Worked-Out Solution Key
285
Chapter 5,
continued
13. Possible rational zeros: 61, 62, 63, 65, 66, 610, 615,
630
Test x 5 1:
1
1
0
231
230
1
1
230
1
230
260
1
1
231
230
21
1
30
21
230
0
5 (x 1 1)(x 2 6)(x 1 5)
Real zeros are 21, 6, and 25.
14. Possible rational zeros: 61, 62, 63, 64, 66, 68, 69,
612, 618, 624, 636, 672
21
1
44
24
1
8
34
78
8
34
78
102
1
8
29
272
1
9
0
9
2
1
29
272
21
27
16
7
216
256
1
1
1
8
29
272
2
20
22
10
11
250
20
24
0
6
20
24
2
16
52
8
26
76
1
6
20
24
22
28
224
4
12
0
h(x) 5 (x 1 1)(x 1 2)(x2 1 4x 1 12)
0 5 x2 1 4x 1 12
}}
24 6 Ï(4)2 2 4(1)(12)
x 5 }}
2(1)
16. Possible rational zeros: 61, 62, 63, 64, 66, 68, 612,
624
272
22
212
42
6
221
230
1
1
1
22
29
10
224
1
21
210
0
21
210
0
224
22
29
10
224
21
3
6
216
23
26
16
240
Test x 5 21:
29
272
3
33
72
11
24
0
h(x) 5 (x 2 3)(x2 1 11x 1 24)
5 (x 2 3)(x 1 8)(x 1 3)
The real zeros are 3, 28, and 23.
286
6
Test x 5 1:
Test x 5 3:
3 1
8
1
224
The only real solutions are 21 and 22.
29
1
220
}
8
1
26
5 22 6 2iÏ 2 , which are not real.
Test x 5 22:
22
21
1
Test x 5 2:
2
24
Test: x 5 22:
22
8
1
44
Test: x 5 2:
Test x 5 21:
21
26
Possible rational zeros of g (x): 61, 62, 63, 64, 66, 68,
612, 624
1
272
0
7
1
Test x 5 1:
1
26
h(x) 5 (x 1 1)(x3 1 6x2 1 20x 1 24) 5 (x 1 1) + g (x)
g(x) 5 (x 1 1)(x2 2 x 2 30)
1
7
Test x 5 21:
0
1
1
1
Test x 5 21:
21
624
Test x 5 1:
Algebra 2
Worked-Out Solution Key
21
1
1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
15. Possible rational zeros: 61, 62, 63, 64, 66, 68, 612,
Chapter 5,
continued
f (x) 5 (x 2 1)(x 1 1)(x2 1 2x 2 8)
Test x 5 2:
2
1
22
29
10
224
2
0
218
216
0
29
28
240
1
5 (x 2 1)(x 1 1)(x 1 4)(x 2 2)
The real zeros are 1, 21, 24, and 2.
18. Possible rational zeros: 61, 65, 625
Test x 5 1:
1
Test x 5 22:
22
1
1
22
29
10
224
22
8
2
224
24
21
12
248
Test x 5 3:
29
10
224
3
3
218
224
1
1
Test x 5 23:
26
28
248
1
23
1
1
240
225
1
1
215
255
1
215
255
280
0
216
240
225
21
1
15
25
215
225
0
Test x 5 21:
1
21
1
g (x) 5 (x 1 1)(x 2 x 2 15x 2 25) 5 (x 1 1) + f(x)
3
2
Possible zeros of f(x): 61, 65, 625
22
29
10
224
23
15
218
24
25
6
28
0
Test x 5 5:
5
1
21
215
225
5
20
25
4
5
1
f(x) 5 (x 1 3)(x3 2 5x2 1 6x 2 8) 5 (x 1 3) + g(x)
0
Possible rational zeros of g (x): 61, 62, 64, 68
g(x) 5 (x 1 1)(x 2 5)(x 1 4x 1 5)
Test x 5 4:
x2 1 4x 1 5 5 0
4
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
216
1
21
22
3
0
1
1
1
25
6
28
4
24
8
21
2
0
2
}}
24 6 Ï(4)2 2 4(1)(5)
x 5 }}
2(1)
5 22 6 i, which are not real.
The only real zeros are 21 and 5.
f(x) 5 (x 1 3)(x 2 4)(x2 2 x 1 2)
x2 2 x 1 2 5 0
1
1
19. Possible rational zeros: 61, 62, 64, 68, 616, 6}, 6}
2
4
Reasonable rational zeros: x 5 1
}
1 6 Ï1 2 4(1)(2)
2
Test x 5 1:
x 5 }}
1
}
1 6 iÏ7
4
5}
which are not real.
2
The real solutions are 23 and 4.
17. Possible rational zeros: 61, 62, 64, 68
Test x 5 1:
1
1
1
2
29
22
8
1
3
26
28
3
26
28
0
f(x) 5 (x 2 1)(x 1 3x 2 6x 2 8) 5 (x 2 1) + g (x)
2
Possible rational zeros of g (x): 61, 62, 64, 68
1
16
4
4
216
4
216
0
5 (x 2 1)(4)(x2 1 x 2 4)
0 5 x2 1 x 2 4
}}
21 6 Ï(1) 2 2 4(1)(24)
x 5 }}
2(1)
}
21 6 Ï 17
5}
2
}
}
21 1 Ï17 21 2 Ï17
Test x 5 21:
1
220
f (x) 5 (x 2 1)(4x 2 1 4x 2 16)
3
21
4
0
3
26
28
21
22
8
2
28
0
,}
Real zeros: 1, }
2
2
Algebra 2
Worked-Out Solution Key
287
Chapter 5,
continued
3
1
20. Possible rational zeros: 61, 63, 65, 615, 6}, 6},
2
2
5
15
3
5
15
1
}
}
}
}
}
}
62, 6 2 , 64, 64, 64, 6 4
3
5
Reasonable zeros: x 5 21, x 5 }2, x 5 }2
Test x 5 21:
21
4
4
212
21
15
24
16
215
216
15
0
28
24
4
8
2
22
24
0
The real zeros are 22, 21, and 2.
5
The real zeros are 21, }2 , and }2 .
3
1
21. Possible rational zeros: 61, 63, 6 5, 615, 6}, 6},
2
2
5
15
5
5
1
1
6}2, 6}
, 6}3, 6}3 , 6}6 , 6}6
2
5
1
Reasonable zeros: x 5 23, x 5 2}3, x 5 }2
Test x 5 23:
3
9
1
25. Possible rational zeros: 61, 63, 69, 6}, 6}, 6}
2
2
2
3
Reasonable zeros: x 5 21, x 5 }2, x 5 3
Test x 5 21:
21
2
2
27
0
9
22
9
29
29
9
0
g(x) 5 (x 1 1)(2x2 2 9x 1 9) 5 (x 1 1)(2x 2 3)(x 2 3)
3
25
16
215
218
221
15
7
25
0
6
28
5 (x 1 2)(2x 1 2)(x 2 2)
5 (x 1 1)(2x 2 3)(2x 2 5)
23
2
2
f(x) 5 (x 1 2)(2x2 2 2x 2 4)
f (x) 5 (x 1 1)(4x2 2 16x 1 15)
3
22
6
The real zeros are 21, }2 , and 3.
3
1
26. Possible rational zeros: 61, 63, 65, 6 15, 6 }, 6},
2
2
5
15
6}2, 6}
2
5
Reasonable zeros: x 5 2}2, x 5 1, and x 5 3
5 (x 1 3)(2x 2 1)(3x 1 5)
5
1
5
Test x 5 2}2:
5
Real zeros are 23, }2, and 2}3.
2}2
23
214
15
25
20
215
2
28
6
0
1
2
2
22. Possible rational zeros: 61, 62, 6 4, 68, 616,
8
16
1
2
4
6 }3, 6 }3 , 6 }3, 6 }3 , 6 }
3
2
Reasonable zeros: x 5 }
, x 5 2, x 5 4
3
2
Test x 5 }3 :
2
}
23
20 236
16
3
22
12
216
18
224
0
23
f (x) 5 1 x 2 }3 2(23x2 1 18x 2 24)
2
5 1 x 2 }3 2(3)(2x2 1 6x 2 8)
2
5 (3x 2 2)(2x 1 2)(x 2 4)
2
Real zeros are }3, 2, and 4.
3
9
1
23. C; Possible zeros: 61, 63, 69, 6}, 6}, 6}
2
2
2
1
24. Possible zeros: 61, 62, 64, 68, 6}
2
Reasonable zeros: x 5 22, x 5 21, and x 5 2
Test x 5 22:
288
Algebra 2
Worked-Out Solution Key
5
h(x) 5 x 1 }2 (2x2 2 8x 1 6)
5 1 x 1 }2 2(2)(x2 2 4x 1 3)
5
5 (2x 1 5)(x 2 3)(x 2 1)
5
The real zeros are 2}2 , 3, and 1.
1
27. Possible rational zeros: 61, 62, 63, 64, 66, 612, 6},
3
2
4
6}3 , 6}3
1
Reasonable zeros: x 5 24, x 5 2}3, and x 5 3
Test x 5 24:
24
3
3
4
235
212
212
32
12
23
0
28
f (x) 5 (x 1 4)(3x 2 8x 2 3) 5 (x 1 4)(3x 1 1)(x 2 3)
2
1
The real zeros are 24, 2}3 , and 3.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
f (x) 5 (x 1 3)(6x 1 7x 2 5)
2
Chapter 5,
continued
1
28. Possible real zeros: 61, 62, 63, 64, 66, 612, 6},
3
2
4
}
}
63, 63
2
Reasonable zeros: x 5 26, x 5 21, x 5 }3
Test x 5 26:
26
4
212
218
26
12
22
0
3
1
5 (x 1 4)1 x 2 }2 2(2)(x2 1 x 1 1)
1
5 (x 1 4)(2x 2 1)(x2 1 x 1 1)
}}
21 6 Ï(1)2 2 4(1)(1)
2(1)
x 5 }}
}
21 6 iÏ3
, which are imaginary numbers
5}
2
1
The only real solutions are 24 and }2 .
f (x) 5 (x 1 6)(3x 1 x 2 2)
2
5 (x 1 6)(3x 2 2)(x 1 1)
1
31. Possible rational zeros: 61, 62, 64, 6}
2
2
The real zeros are 26, 21, and }3 .
Reasonable zeros: x 5 22 and x 5 2
7
1
29. Possible real zeros: 61, 62, 67, 614, 6}, 6}
2
2
7
Reasonable zeros: x 5 2}2 , x 5 21, and x 5 2
7
Test x 5 2}2 :
7
2}2 2
5 211 214
2
1
x2 1 x 1 1 5 0
19
3
g(x) 5 (x 1 4)1 x 2 }2 2 (2x2 1 2x 1 2)
27
7
14
22
24
0
1
Test x 5 22:
22
2
2
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
5 (2x 1 7)(x 2 2)(x 1 1)
9
5
3
24
28
24
24
4
1
1
21
0
g(x) 5 (x 1 4)(2x 1 x 1 x 2 1) 5 (x 1 4) + f(x)
2
1
Possible real solutions of f(x): 61, 6}2
1
Test x 5 }2 :
1
}
2
1
1 21
2
2
0
3
22
0
25
3
22
4
22
2
21
1
0
}}
}
1 6 iÏ 7
The real solutions are x 5 22 and x 5 2.
Test x 5 24:
2
25
5}
, which are not real.
4
1
Resonable zeros: x 5 24 and x 5 }2
2
4
1 6 Ï(21)2 2 4(2)(1)
x 5 }}
2(2)
1
30. Possible real solutions: 61, 62, 64, 6}
2
1
26
2x2 2 x 1 1 5 0
7
1
10
h(x) 5 (x 1 2)(x 2 2)(2x2 2 x 1 1)
The real zeros are 2}2 , 2, and 21.
1
24
Test x 5 2:
2
7
3
24
h(x) 5 (x 1 2)(2x3 2 5x2 1 3x 2 2) 5 (x 1 2) + f(x)
5 1 x 1 }2 2(2)(x2 2 x 2 2)
2
4
1
2
2
27
Possible rational zeros of f: 61, 62, 6}2
7
g(x) 5 x 1 }2 (2x2 2 2x 2 4)
24
21
32. Possible rational zeros: 61, 62, 63, 64, 66, 612,
1
2
4
6}3, 6}3, 6}3
Reasonable zeros: x 5 23, and x 5 4
Test x 5 23:
23
3
3
26
232
35
212
29
45
239
12
215
13
24
0
h(x) 5 (x 1 3)(3x3 2 15x2 1 13x 2 4) 5 (x 1 3) + g(x)
1
2
4
Possible rational zeros of g: 61, 62, 64, 6}3 , 6}3, 6}3
Algebra 2
Worked-Out Solution Key
289
Chapter 5,
continued
Test x 5 4:
3
3
Test x 5 21:
215
13
24
12
212
4
23
1
21
0
1
h(x) 5 (x 1 3)(x 2 4)(3x 2 3x 1 1)
2
}}
3 6 Ï 32 2 4(3)(1)
3 6 iÏ3
x 5 }}
5}
, which are not real.
6
2(3)
}
The only real solutions are 23 and 4.
33. Possible rational zeros: 61, 62, 63, 65, 66, 610,
3
5
0
37
230
24
26
252
30
1
1
213
26
215
0
0
26
11
26
1
25
6
25
6
0
Test x 5 23:
23
15
2
213
26
215
2
211
15
211
15
0
5
23
22
25
3
26
3
0
6
23
21
0
22
1
0
2
h(x) 5 (x 1 3)(2x4 2 x3 2 2x 1 1) 5 (x 1 3) + f(x)
1
Possible rational zeros of f: 61, 6}2
1
Test x 5 }2:
1
2
}
2
5 (x 1 2)(x 2 1)(2x 2 5)(x 2 3)
5
The real zeros are 22, 1, }2 , and 3.
2
34. Possible rational zeros: 61, 62, 63, 64, 66, 612
Reasonable zeros: x 5 22, x 5 21, x 5 1, x 5 2,
and x 5 3
Test x 5 22:
1
26
1
f (x) 5 (x 1 2)(x 2 1)(2x2 2 11x 1 15)
1
11
Reasonable zeros: x 5 23, x 5 }2, and x 5 1
Test x 5 1:
22
26
f(x) 5 (x 1 2)(x 1 1)(x 2 1)(x2 2 5x 1 6)
6}2, 6}2, 6}
2
2
6
3
1
35. Possible rational zeros: 61, 63, 6}, 6}
2
2
1
2
211
The real zeros of f are 22, 21, 1, 2, and 3.
Possible rational zeros of g: 61, 63, 65, 615, 6}2,
1
6
5 (x 1 2)(x 1 1)(x 2 1)(x 2 2)(x 2 3)
29
f (x) 5 (x 1 2)(2x3 2 13x2 1 26x 2 15) 5 (x 1 2) + g(x)
5
21
Test x 5 1:
1
Test x 5 22:
3
26
5 (x 1 2)(x 1 1) + h(x)
5
2
5
15
Reasonable zeros: x 5 22, x 5 1, x 5 }2 , and x 5 3
2
5
Possible rational zeros of h: 61, 62, 63, 66
615, 630, 6}2 , 6}2, 6}2, 6}
.
2
22
25
f (x) 5 (x 1 2)(x 1 1)(x3 2 6x2 1 11x 2 6)
3x2 2 3x 1 1 5 0
1
1
23
25
15
4
212
22
10
210
210
12
25
5
5
26
0
f (x) 5 (x 1 2)(x4 2 5x3 1 5x2 1 5x 2 6)
5 (x 1 2) + g(x)
Possible rational zeros of g: 61, 62, 63, 66
21
0
22
1
1
0
0
21
0
0
22
0
1
2
1
h(x) 5 (x 1 3) x 2 }2 (2x3 2 2)
5 (x 1 3)1 x 2 }2 2(2)(x3 2 1)
1
5 (x 1 3)(2x 2 1)(x 2 1)(x2 1 x 1 1)
x2 1 x 1 1 5 0
}}
21 6 Ï(1)2 2 4(1)(1)
x 5 }}
2(1)
}
21 6 iÏ3
5}
, which are not real.
2
1
The only real solutions of h are 23, }2, and 1.
36. The error is that only the positive factors were listed as
possible rational zeros.
Possible zeros: 61, 62, 67, 614
290
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4
Chapter 5,
continued
37. The error is that the factors of the coefficient were
divided by the factors of the constant term. The possible
5
5
5
1
1
1
zeros are: 61, 65, 6}2 , 6}2, 6}3, 6}3, 6}6, 6}6
38. Sample answer: f(x) 5 4x 2 1 3
3
2
1
2
1
4
3
4
h(x) 5 (x 1 2)(x2 2 x 1 1)
x2 2 x 1 1 5 0
}}
1 6 Ï(21)2 2 4(1)(1)
x 5 }}
2(1)
}
1 6 i Ï3
, which are imaginary numbers.
5}
2
Possible zeros: 61, 63, 6}, 6}, 6}, 6}
39. B;
y
The only real zero is 22, which matches graph A.
44. It is not possible for a cubic function to have more than
3 real zeros. At most a cubic function can have 3 real
zeros. The number of possible zeros is determined by
degree. The cubic function will either have 1 real zero or
3 real zeros, and never no zeros because the graphs of the
functions cross the x-axis at least once.
( (
4
,
5
20
0
2 x
( (
3
,
4
(2 , 0(
3
2
0
Problem Solving
40. sr factors. Each factor of a0 (there are s) is divided by
each factor of an (there are r), then at most there are sr
possible rational zeros.
41. f(x) 5 x 2 2x 2 x 1 2
3
2
h5*14
63 5 *2(* 1 4)
63 5 *3 1 4*2
0 5 *3 1 4*2 2 63
Possible rational zeros: 61, 62
Possible zeros: 1, 3, 7, 9, 21, 63
Test x 5 1:
Test * 5 1:
1
1
1
22
21
2
1
21
22
21
22
0
1
1
1
f (x) 5 (x 2 1)(x 2 2 x 2 2)
Real zeros are 21, 1, and 2, which matches graph B.
3
1
42. g (x) 5 x 2 3x 1 2
3
2
Possible rational zeros: 61, 62
1
Test x 5 1:
1
1
23
0
2
1
22
22
22
1
22
5
5
5
5
258
4
0
263
3
21
63
7
21
0
Dimensions: 3 in. 3 3 in. 3 7 in.
0
w5d15
* 5 d 1 35
2000 5 (d 1 35)(d 1 5)(d)
2000 5 (d 2 1 40d 1 175)d
0 5 d 3 1 40d 2 1 175d 2 2000
}}
Reasonable zero: d 5 5
Test d 5 5:
}
}
}
Real zeros: 1, 1 1 Ï3 , and 1 2 Ï 3 , which matches
graph C.
5
1
43. h(x) 5 x 3 1 x 2 2 x 1 2
Possible rational zeros: 61, 62
1
40
175
22000
5
225
2000
400
0
45
V 5 (d 2 5)(d 1 45d 1 400)
2
Test x 5 22:
1
1
46. V 5 * + w + d
2 6 Ï (2)2 2 4(1)(22)
x 5 }}
2(1)
1
263
* 5 3 is the only real zero.
x2 2 2x 2 2 5 0
22
0
V 5 (* 2 3)(*2 1 7* 1 21)
g(x) 5 (x 2 1)(x2 2 2x 2 2)
5 1 6 Ï3
4
Test * 5 3:
5 (x 2 1)(x 2 2)(x 1 1)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
45. V 5 *2h
1
21
2
22
2
22
21
1
0
The only positive real value for d is 5.
Dimensions: Depth 5 d 5 5 feet
Width 5 d 1 5 5 10 feet
Length 5 d 1 35 5 40 feet
Algebra 2
Worked-Out Solution Key
291
Chapter 5,
47.
continued
V 5 x(x 2 1)(x 2 2)
Test t 5 8:
24 5 x(x2 2 3x 1 2)
8
12
2264
2028
23924
212,384
96
21344
5472
12,384
2168
684
1548
0
0 5 x 2 3x 1 2x 2 24
3
2
Possible rational solutions: 61, 62, 63, 64, 66, 68,
612, 624
12
1
48. V 5 } (x2(2x 2 5))
3
1
}
3 5 3 (2x3 2 5x2)
t 5 8 is a solution.
d.
9 5 2x3 2 5x2
0 5 2x3 2 5x2 2 9
1
3
9
Possible rational solutions: 61, 63, 69, 6}2, 6}2, 6}2
The only reasonable solution is 8 because the other
solutions are either negative or imaginary. The year
1998 corresponds to the solution.
49. a. 20,300 5 210t3 1 140t 2 2 20t 1 18,150
0 5 210t 1 140t 2 20t 2 2150
2
b. Possible solutions: 1, 2, 5
c. 1
210
51.
140
220
22150
210
130
110
130
110
22040
3x
x
l
210
d
21x 1 6
Not drawn to scale
Ramp length 5 (21x 1 6) 2 3x 5 18x 1 6
1
2
210
210
5
210
210
140
220
22150
220
240
440
120
220
21710
140
220
22150
250
450
2150
90
430
0
F
1
G
150 5 3x[3x 1 x 1 6x 1 2x]
2
2
50 5 x[9x2 1 3x]
0 5 9x3 1 3x2 2 50
The possible rational solutions: 61, 62, 65, 610, 625,
1
2
5
10
25
50
1
2
5
25
50
6}
, 6}
9
9
5
Test x 5 }3:
t 5 5 corresponds to 1999.
5
3
}
9
0 5 12t 4 2 264t 3 1 2028t 2 2 3924t 2 12,384
b. Possible solutions: 1, 2, 3, 4, 6, 8, 9
9
c. Test t 5 4:
2264
2028
23924
212,384
48
2864
4656
2928
2216
1164
732
29456
3
0
250
15
30
50
18
30
0
(9x 2 1 18x 1 30)1 x 2 }53 2 5 0
5
12
Test t 5 6:
6
12
12
2264
2028
23924
212,384
72
21152
5256
7992
2192
876
1332
24392
x 5 }3 is the only positive, real solution.
right ramp: * 5 12 ft
Algebra 2
Worked-Out Solution Key
left ramp: * 5 24 ft
5
h 5 }3 ft
h 5 }3 ft
w 5 5 ft
w 5 5 ft
5
5.6 Mixed Review for TAKS
52. B; The sale price is dependent on the original price.
132s8t17
A
53. F; w 5 } 5 }
5 11s3t8
*
12s5t9
The width is 11s 3t 8 units.
292
10
, 6}
, 6}
, 6}9, 6}9 , 6}9, 6}
,
650, 6}3, 6}3, 6}3, 6}
3
3
3
9
50. a. 56,300 5 12t 4 2 264t 3 1 2028t 2 2 3924t 1 43,916
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* 5 }3 (18x 1 6) 5 12x 1 4
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V 5 3x }2(x)(6x 1 2) 1 }2(x)(12x 1 4)
t 5 5 is an actual solution.
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d 5 }3 (18x 1 6) 5 6x 1 2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
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