277

Chapter 5,
continued
Lesson 5.5
6. 4
1
21
222
40
4
12
240
210
0
5.5 Guided Practice (pp. 362–365)
1.
2x2 2 3x 1 8
ww
2
4
3
x 1 2x 2 1q2x 1 x 1 0x2 1 x 2 1
1
2x 4 1 4x 3 2 2x 2
5 (x 2 4)(x 2 1 3x 2 10)
2
23x 2 6x 1 3x
8x 2 2 2x 2 1
8x 2 1 16x 2 8
5 (x 2 4)(x 1 5)(x 2 2)
7. 22
1
218x 1 7
218x 1 7
2x 4 1 x 3 1 x 2 1
}}
5 2x2 2 3x 1 8 1 }
x 2 1 2x 2 1
x 2 1 2x 2 1
0
18
29
0
0
2
5 (x 1 2)(x 2 2 9)
x 3 1 2x 2
5 (x 1 2)(x 2 3)(x 1 3)
The other zeros are 3 and 23.
8. 22
2
23x 2 6x
1
10x 2 10
10x 1 20
3. 23
1
4
21
23
12
1
24
11
1
4
1
23
7
4
5
2
5
2
214
22
212
14
27
0
6
2
f (x) 5 x 1 8x 1 5x 2 14
5 (x 1 2)(x 2 1 6x 2 7)
5 (x 1 2)(x 1 7)(x 2 1)
The other zeros are 27 and 1.
9. 25 5 215x 3 1 40x
0 5 215x3 1 40x 2 25
11
x3 1 4x2 2 x 2 1
}} 5 x 2 1 x 2 4 1 }
x13
x13
4
5
3
21
23
8
1
230
30
x 3 2 x 2 1 4x 2 10
}} 5 x 2 2 3x 1 10 2 }
x12
x12
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
22
f (x) 5 x 1 2x 2 9x 2 18
23x 1 4x
1
215
215
0
40
225
215
215
25
215
25
0
(x 2 1)(215x 2 2 15x 1 25) 5 0
9
x ø 0.9 is the other positive solution. The company
could still make the same profit producing about
900,000 shoes.
9
4x 3 1 x 2 2 3x 1 7
}} 5 4x 2 1 5x 1 2 1 }
x21
x21
1
218
ww
3
2
2
5. 4
29
3
x 1 2qx 2 x 1 4x 2 10
4. 1
2
1
x 2 2 3x 1 10
2.
2
f (x) 5 x 2 x 2 22x 1 40
23x 3 1 2x 2 1 x
3
3
3
26
5
12
4
28
212
22
23
0
5.5 Exercises (pp. 366–368)
1
Skill Practice
1. If a polynomial f (x) is divided by x 2 k, then the
remainder is r 5 f (k).
f (x) 5 x 3 2 6x 2 1 5x 1 12
5 (x 2 4)(x 2 2 2x 2 3)
5 (x 2 4)(x 2 3)(x 1 1)
2. The red numbers represent the coefficients of the
quotient and the blue number represents the remainder.
3.
x15
x 2 4qx 1 x 2 17
ww
2
x2 2 4x
5x 2 17
5x 2 20
3
3
x 2 1 x 2 17
}5x151}
x24
x24
Algebra 2
Worked-Out Solution Key
277
Chapter 5,
3x 1 4
5x 2 2 12x 1 37
9.
x 2 5q3x 2 11x 2 26
ww
2
x 1 2x 2 4q5x 2 2x 2
www
4
3
2
2
3x2 2 15x
4
5x 1 10x 2 20x 2
4x 2 26
212x 3 1 13x 2 1 0x
4x 2 20
212x2 2 24x 2 2 48x
37x2 1 48x 2 39
26
3x 2 2 11x 2 26
x25
37x 2 1 74x 2 148
6
x25
}} 5 3x 1 4 2 }
226x 1 109
x2 1 4x 1 7
5.
4
3
2
226x 1 109
x 1 2x 2 4
5x 2 2x 2 7x 2 39
x 1 2x 2 4
}}
5 5x 2 2 12x 1 37 1 }
2
2
x 3 1 3x 2 1 3x 1 2
x 2 1qww
x 3 2 x2
4x 2 1 12x 1 44
10.
2
x 2 3x 2 2q4x 1 0x 1 0x 1
www
4
3
2
2
4x 1 3x
2
4
4x 2 4x
3
4x 2 12x 2 8x
3
7x 2 7
44x 2 1 29x 2 4
x 3 1 3x 2 1 3x 1 2
x21
44x 2 2 132x 2 88
9
x21
}} 5 x 2 1 4x 1 7 1 }
2x 1 9
161x 1 84
4
8x2 2 2x
11. 5
36x 2 1
2
36x 2 9
8
8
4x 2 1
}} 5 2x 1 9 1 }
3x 1 8
3x 3 1 11x 2 1 4x 1 1
x 2 1 xqww
3
27
10
10
15
3
25
2
2
12. 2
4
213
25
8
210
25
215
2
8x 2 1 4x
8x2 1 8x
4
24x 1 1
24x 1 1
3x 3 1 11x 2 1 4x 1 1
}}
5 3x 1 8 1 }
x2 1 x
x2 1 x
8.
4x2 1 13x 2 5
x22
13. 24
1
7x 3 1 11x 2 1 7x 1 5
x 2 1 1qww
1 7x
1
11x 2 1 0x 1 5
11x 2 1 0x 1 11
26
2
15
x22
}} 5 4x 2 5 2 }
7x 1 11
3
25
x25
2x 2 7x 1 10
x25
}} 5 2x 1 3 1 }
3x 1 3x
7x 3
161x 1 84
x 2 3x 2 2
4x 1 5x 2 4
x 2 3x 2 2
}
5 4x 2 1 12x 1 44 1 }
2
2
8x 2 1 34x 2 1
4x 2 1qww
7.
8
1
24
216
4
215
2
x 1 8x 1 1
x14
15
x14
}5x142}
14. 3
1
6
7x 1 11x 1 7x 1 5
}}
5 7x 1 11 2 }
x2 1 1
x2 1 1
1
2
x 19
x23
0
9
3
9
3
18
18
x23
}5x131}
278
Algebra 2
Worked-Out Solution Key
5x
2
12x 2 36x 2 24x
9
8x 2 1 34x 2 1
4x 2 1
5x 2 4
2
12x 3 1 8x 2 1
7x 1 2
6.
7x 1 0x 2 39
3
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4.
continued
Chapter 5,
15. 4
1
1
3
continued
25
0
22
4
24
216
21
24
218
21. 6
18
x24
x 2 5x 2 2
x24
1
0
24
6
23
9
215
23
5
29
x 3 2 4x 1 6
x13
1
30
6
224
230
24
25
0
5 (x 2 6)(x 2 2 4x 2 5)
5 (x 2 6)(x 2 5)(x 1 1)
22. 24
25
28
13
212
6
6
212
6
1
22
1
26
1
6
5
212
24
28
12
1
2 23
0
f (x) 5 x 3 1 6x 2 1 5x 2 12
9
x13
} 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 }
1
210
1
2
16. 23
1
5 (x 1 4)(x2 1 2x 2 3)
5 (x 1 4)(x 1 3)(x 2 1)
1
23. 8
x 4 2 5x 3 2 8x 2 1 13x 2 12
x26
1
6
x26
}}} 5 x 3 1 x 2 2 2x 1 1 2 }
18. 25
1
1
1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
0
2
25
5
225
45
5 (x 2 8)(x 1 4)(x 1 2)
21
5
29
10
5 (x 2 8)(x 2 1 6x 1 8)
24. 210
1
10
x15
0
25
3
2
4
22
2
21
1
18
95
150
210
280
2150
1
8
15
0
3
2
f (x) 5 x 1 18x 1 95x 1 150
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
8
6
f(x) 5 x 2 2x 2 40x 2 64
x 3 2 5x 1 3
x22
3
64
235
} 5 x 2 1 2x 2 1 1 }
1
48
16
error in how the quotient is written.
2
8
0
19. The division was done correctly, however, there is an
1
264
4
x 4 1 4x 3 1 16x 2 35
x15
1
240
3
}} 5 x 3 2 x 2 1 5x 2 9 1 }
2
22
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
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
15
33. 9
2
5 (x 2 5)(3x 1 13x 1 4)
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
211
2
f (x) 5 10x 2 81x 1 71x 1 42
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
3
210
The other zeros are about 0.46 and 1.74.
1
The other zeros are }4 and 24.
10
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.
31. 7
22
5}
2
5 (x 1 3)(x 2 6)(x 1 1)
4
64
f (x) 5 2x 3 2 10x 2 2 71x 2 9
0
5 (x 1 3)(x2 2 5x 2 6)
30. 10
288
2
20
1
25
26
f (x) 5 x 3 2 2x2 2 21x 2 18
212
The other zeros are }3 and 28.
2
1
264
5 (x 1 4)(3x 2 2)(x 1 8)
3
13
4
0
3
2
f (x) 5 3x 2 2x 2 61x 2 20
29. 23
72
5 (x 1 4)(3x 2 1 22x 2 16)
220
65
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
30
}
7
239
14
15
11
214
30
22
228
0
1
2
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
1
f(x) 5 x 2 }2 (30x 2 1 22x 2 28)
2
5x 1 30x 1 40
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
8x 2 1 40x 2 48
0
1
25
1
212
26
236
23
218
0
3
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
f (x) 5 x 2 5x 2 12x 1 36
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
12
16
248
212
0
48
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
The missing dimension is x 1 8.
38. a. 2
4
21
8x 2 1 40x 2 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
Chapter 5,
continued
44. a. Total revenue
46.
5 (number of radios sold)(price per radio)
5 (x)(40 2 4x 2) 5 40x 2 4x 3
P 5 26x 3 1 72x
96 5 26x 3 1 72x
0 5 26x 3 1 72x 2 96
3
An expression is 40x 2 4x .
2
A function is P 5 25x 2 4x 3.
c. When P 5 24:
26
24 5 25x 2 4x 3
0
225
24
6
9
224
6
216
0
20.00722x 4 1 0.176x 3 2 1.40x 2 1 3.39x 1 17.6
2444,215.872
5379.39x 1 444,233.472
5379.39x 1 17.6
265.1x 2 2 5376x
265.1x 2 1 3.39x
0.772x 3 1 63.7x 2
P5}
5 }}}}
V
3.10x 1 256
0.772x 3 2 1.40x 2
0
Mixed Review for TAKS
of visits to the national park that were overnight stays.
2 0.00722x 4 2 0.596x 3
48
5 26(x 2 2)(x 1 4)(x 2 2)
47. C;
Let t 5 time it takes James to walk to school in minutes.
Total distance 5
rate + time (James) 1 rate + time (friend)
3
So,
444,216
P 5 20.00233x 3 1 0.249x 2 2 21x 1 1735 2 }
.
3.10x 1 256
The function for the percent of visits that were overnight
stays is the overnight stays divided by the total visits.
Algebra 2
Worked-Out Solution Key
10
+t1}
+ (t 2 4)
85}
60
60
13t 2 40
85}
60
45. Answers may vary due to rounding. Let P 5 the precent
2 0.00233x 3 1 0.249x 2 2 21x 1 1735.287
3.10x 1 256q 20.00722x 1 0.176x 2 1.40x 1 3.39x 1 17.6
212
The zeros are at 2 and 24. Production levels cannot be
negative so x 5 2 is the only solution.
because you cannot produce a negative number of
radios.
wwwww
4
3
2
96
5 23(x 2 2)(2)(x 1 4)(x 2 2)
d. No. The solution x ø 22.88 does not make sense
282
224
5 (23x 1 6)(2x 1 8)(x 2 2)
So, (x 2 1.5)(4x2 1 6x 2 16) 5 0. Use the quadratic
formula to find x ø 1.39, the other positive solution.
So, about 1.39 million radios also make a profit of
$24,000,000.
S
212
480 5 13t 2 40
520 5 13t
40 5 t
It took James 40 minutes to walk to school.
48. G; 2x 1 3y 5 15
2x 1 3(0) 5 15
15
x5}
2
,0 .
The x-intercept is 1 }
2 2
15
Mixed Review for TEKS (p. 369)
1. B;
1.64 3 1011
1.2 3 10
1.64
1.2
1011
10
}
5}3}
ø 1.37 3109
2
2
It would take 1.37 3 109 football fields stretched
end-to-end to reach from Earth to the sun.
2. H;
C(x) 5 (4x 2 2)(x 2 2)(2x 2 4)
5 (4x2 2 10x 1 4)(2x 2 4)
5 8x3 2 36x 2 1 48x 2 16
The function C(x) 5 8x3 2 36x2 1 48x 2 16 gives the
volume of the inside of the cooler in terms of the width x.
3. C; If f (x) l 2` as x l 2`, f (x) l 2` as x l 1`,
and the degree is even (4), then the leading coefficient
must be negative.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4
296
P 5 (26x 2 12x 1 48)(x 2 2)
You know that 1.5 is a solution, so x 2 1.5 is a factor
of 4x3 2 25x 1 24.
4
72
2
0 5 4x3 2 25x 1 24
1.5
0
26
b. P 5 (40x 2 4x 3) 2 15x 5 25x 2 4x 3
Chapter 5,
continued
Lesson 5.6
4. G; Let x 5 height.
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)
3
1
5
15
Possible rational zeros: 6}1, 6}1, 6}1, 6}
1
0 5 (x2 2 45)(x 1 12)
}
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.
2
10
3
Simplified list: 61, 62, 63, 66, 6}2 , 6}2
P 5 70x 2 10x3
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
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
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Test x 5 1:
10
So, (x 2 2)(10x2 1 20x 2 30) 5 0. use factoring to find
that x 5 1 and x 5 23 are the other solutions.
The company could produce 1 million cameras and make
the same profit.
6. In 2001, t 5 6.
1
1
ø 46.73
The average monthly cell phone bill in 2001 was about
$46.73.
1
7. V 5 } s2h
3
5
14
1
27
22
22
212
28
5
14
21
9
214
29
14
0
1
27
1 is not a zero.
Test x 5 21:
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
28
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