aat solutions - ch05

7/17/2019 Aat Solutions - Ch05

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Company.

Algebra 2

Worked-Out Solution Key

Prerequisite Skills (p. 328)

1. 23 and 1

2. 4

3. y5 ax21 bx1 c

4.

2

x

y

21

(1, 4)

x51

5.3

x

y

21

x521

2

(2 , 218 )1

2

3

4

6.

1

x

y

21

(22, 8)

x

5

2

2

7. x21 9x1 205 (x1 4)(x1 5)

8. 2x21 5x2 35 (2x2 1)(x1 3)

9. 9x22 645 (3x2 8)(3x1 8)

10. 2x21x1 65 0

x521 6 }}

(1)22 4(2)(6)}}

2(2) 5

21 6}

247}

4 5

21 6i}

47}

4

11. 10x21 13x5 3

10x21 13x2 35 0

(5x2 1)(2x1 3)5 0

5x2 15 0 or 2x1 35 0

5x5 1 or 2x5 23

x51

}

5

or x5 23}

2

12. x21 6x1 25 20

x21 6x2 185 0

x526 6 }}

(6)22 4(1)(218)}}

2(1)

526 6

}

108}

2

526 66

}

3}

2

5 23 63}

3

Lesson 5.1

5.1 Guided Practice (pp. 331333)

1. (42)35 46 Power of a power property

5 4096

2. (28)(28)35 (28)11 3Product of powers property

5 (28)4

5 4096

3. 12}9235

23

}

93 Power of a quotient property

58

}

729

4.6 p1024

}

9 p107 5

6

}

9p102427 Quotient of powers property

56

}

9

p10211

52

}

3 p1011

Negative exponent property

5.x26x5x35x261 51 3 Product of powers property

=x2

6. (7y2z5)(y24z21)5 7y21 (24)z51 (21) Product ofpowers property

5 7y22z4

57z4

}

y2


7. 1s3

}

t242

2

5

(s3)2

}

(t24)2

Power of a quotient property

5s6

}

t28

Power of a power property

5s6t8 Negative exponent property

8. 1x4y22

}

x3y6 2

3

5

(x4y22)3

}

(x3y6)3 Power of a quotient property

5x4 p3y22 p3

}

x3 p3y6 p3 Power of a product property

5x12y26

}

x9y18

5x1229y26218 Quotient of powers property

5x3

}

y24


5.1 Exercises (pp. 333335)

Skill Practice

1. a.Product of powers property

b.Negative exponent property

c.Power of a product property

2. No, 25.2 is not less then 10; the number should be

2.523 1022.

3. 33p325 3(31 2) Product of powers property

5 35

5 243 4. (422)35 426 Power of a power property

51}

46 Negative exponent property

51

}

4096

Chapter 5


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Algebra 2


5. (25)(25)45 (25)(11 4) Product of powers property

5 (25)5

5 23125

6. (24)25 28 Power of a power property

5 256

7.52

}

555 5(2 2 5) Quotient of powers property

5 523

51

}

53


51

}

125

8. 13}5245

34

}

54

Power of a quotient property

581

}

625

9. 12}7223

5223

}


5

73

}

23


5343

}

8

10. 93p9215 9(31 (21)) Product of powers property

5 92

5 81

11.34

}

322

5 342 (22) Quotient of powers property

5 36

5 729

12. 12}3225

12}3245 12}32

251 4 Product of powers property

5 12}3221

53

}

2


13. 63p60p6255 631 02 5 Product of powers property

5 622

51

}

62


51

}

36

14. 111}2225

22511}22210

Power of a power property

5

1210

}


5 210 Negative exponent property

5 1024

15. (4.23 103)(1.53 106)5 (4.23 1.5)(1033 106)

5 6.33 109

16. (1.23 1023)(6.73 1027)5 (1.23 6.7)(10233 1027)

5 8.043 10210

17. (6.33 105)(8.93 10212)5 (6.33 8.9)(1053 10212)

5 56.073 1027

5 5.6073 1013 102 7

5 5.6073 1026

18. (7.23 109)(9.43 108)5 (7.23 9.4)(1093 108)

5 67.683 1017

5 6.7683 1013 1017

5 6.7683 1018

19. (2.13 1024)35 2.133 (1024)3

5 9.2613 10212

20. (4.03 103)45 4.043 (103)4

5 2563 1012

5 2.563 1023 1012

5 2.563 1014

21.8.13 1012

}

5.43 109 5

8.1

}

5.43

1012

}

109

5 1.53 103

22. 1.13 1023

}

5.53 1028

5 1.1

}5.5

3 1023

}1028

5 0.23 105

5 2.03 10213 105

5 2.03 104

23.(7.53 108)(4.53 1024)

}}

1.53 107 5

(7.53 4.5)(1083 1024)

}}

1.53 107

533.753 104

}

1.53 107

53.3753 1013 104

}}

1.53 107

53.3753 105

}

1.53

107

53.375

}

1.5

3105

}

107

5 2.253 1022

24.w22

}

w6 5 w22 2 6 Quotient of powers property

5 w28

51

}

w8


25.(22y3)55 (22)5(y3)5 Power of a product property

5 210y15 Power of a power property

5 1024y15

26.(p3q2)215p23q22 Power of a product property

51

}

p3q2 Negative exponent property

27.(w3x22)(w6x21)5 w31 6x221 (21) Product of powersproperty

5 w9x23

5w9

}

x3

Negative exponent

property

Chapter 5, continued


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Algebra 2


28.(5s22t4)235 523(s22)23(t4)23 Power of a productproperty

51

}

125

ps22(23)pt4(23) Power of a power

property

51

}

125ps6pt212

5s6

}

125t12

Negative exponent

property

29. 13a3b52235 (3)23(a3)23(b5)23 Power of a productproperty

51

}

27

a3(23)b5(23) Power of a power

property

51

}

27

a29b215

51

}

27a9b15 Negative exponent property

30.x21y2

}

x2y215x2122y22 (21) Quotient of powers property

5x23y3

5y3

}x3


31.3c3d

}

9cd215

3

}

9

c32 1d12 (21) Quotient of powers property

5c2d2

}

3

32.4r4s5

}

24r4s255

4

}

24

r42 4s52 (25) Quotient of powers property

51

}

6

r0s10

51

}

6

p1 ps10 Zero exponent property

5

s10

}

6

33.2a3b24

}

3a5b225

2

}

3

a32 5b242 (22) Quotient of powers property

52

}

3

a22b22

52

}

3a2b2


34.y11

}

4z3p

8z7

}

y7 5

8y11z7

}

4y7z3

5 2y112 7z72 3 Quotient of powers property

5 2y4z4

35.x2y

2

3

}3y2 p y

2

}x245

x2y2

1

}

3x24y2

Product of powers property

5x22 (24)y2122

}}

3

Quotient of powers property

5x6y23

}

3

5x6

}

3y3


36. B;

2x2y

}

6xy215

x22 1y12 (21)

}

3

5xy2

}

3

37. The error is that the exponents were divided and not

subtracted.

x10

}

x2 5x102 2

5x8

38. The error is that the exponents were multiplied instead

of added.

x5px35x51 3

5x8

39. The error is that the bases were multiplied.

(23)2p(23)45 (23)6

40.A5

}

3}

4

s2 s5x

}

3

5 }

3}4 1x}322

5

}

3}

4 1x

2

}32

25

}

3x2

}

4 p9

5

}

3x2

}

36

41. v5 r2h r = x h5x

}

2

5 (x)21x}22

5 x2px

}

2

5x3

}

2

42. v5 lpwph l5 2x, w55x

}

3

, h5x

5(2x)15x}32(x)

510x3

}

3

43.x15y12z85x4y7z11p?

x15y12z8

}

x4y7z11

5 ?

x152 4y122 7z82 115 ?

x11y5z235x11y5

}

z3

5 ?

44. 3x3y2512x2y5

}

?

?512x2y5

}

3x3y2

5 4x22 3y52 2

5 4x21y3

54y3

}

x



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Algebra 2


45. (a5b4)25 a14b21p?

a5 p2b4 p25 a14b21p?

a10b85 a14b21p?

a10b8

}

a14b215 ?

a102 14p

b82 (21)5 a24b95b9

}

a

45 ?

46. Sample answer:

x12x165 (x4y4)(x8y12)5 (x2y15)(x10y1)5 (x25y7)(x17y9)

47.1}

am5

a0

}

am5 a02 m5 a2m

48.am

}

an5 amp

1}

an

a2np

am5 a2n1 m5 am2 n

Problem Solving

49. Pacific: V5 (1.563 1014)(4.033 103)

5 (1.563 4.03)(10143 103)

5 6.28683 1017meters3

Atlantic: V5 (7.683 1013)(3.933 103)

5 (7.683 3.93)(10133 103)

5 30.1824 3 1016

5 3.018243 1013 1016

5 3.018243 1017meters3

Indian: V5 (6.863 1013)(3.963 103)

5 (6.863 3.96)(10133 103)

5 27.16563 1016

5 2.716563 1013 1016

5 2.716563 1017meters3

Arctic: V5 (1.41 3 1013)(1.213 103)

5 (1.413 1.21)(10133 103)

5 1.70613 1016meters3

50. (0.000022)(125,000,000)5 (2.23 1025)(1.253 108)

5 (2.23 1.25)(10253 108)

5 2.753 103

The continent has moved 2.753 103miles.

51. Bead: d5 6 mm

r5 3 mm

V54

}

3

r354

}

3

(3)35 36

Pearl : d5 9 mm

r59

}

2

mm

V54

}

3r35

4

}

319}22

35

243}

2

Volume of pearl

}}

Volume of bead 5

243}

2

}

365

243

}

725 3.375

The volume of the pearl is about 3.375 times greater

than the volume of the bead.

52. a. V5 314}3r325 4r3

b.V5 r2h

c.h5 3(2r)5 6r

d.Volume of cylinder5 r2(6r)5 6r3

Occupied volume54r3

}

6r35

2

}

3

The tennis balls take up 2

}

3

of the cans volume.

53. a.V5 r2h5 (9.53)2(1.55) 442 mm35

4.4231027m3

b. Answers will vary.

c.pennies5Volume of classroom

}}

Volume of 1 penny

The answer is an overestimate because of the shape of

the pennies. There will be gaps between them.

54. a. Volume of inner core54

}

3

r3

Earths total volume54

}

3(5r)3

54

}

3

(125r3)

5 125 p4

}

3

r3

5 125 pVolume of inner core

ratio5 125:1

b.Volume of outer core54

}

3111}6r1 r2

32

4

}

3r3

54913

}

21614}3r322

4

}

3

r3

54697

}

21614}3r32

ratio5

4697

}216

p4

}

3r3

}4

}

3

r3

54697

}

216

Mixed Review

55.

1

x

y

21

56.1

x

y

21

57.

1

x

y

21

58.1

x

y

21



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Algebra 2


59.

1

x

y

21

60.

2

x

y

21

61. x1y5 2

7x1 8y5 21

A X B

F1 17 8GpFx

yG5F221G

A2151}

82 7F 8 2127 1G5F 8 2127 1G

X5A21B5F 8 2127 1G F221G5F25

7G5 FxyG

The solution is (25, 7).

62. 2x2 2y5 3

2x1 8y5 1

A X B

F21 22 2 8GpFx

yG5 F31G

A21

51

}

282 (24)F 8 222 21G5F22 21}21

}

2

1

}

4

G X5A21B5

F22 2

1

}

2

1

}

2

1

}

4

GF31G5F

213}2

7

}

4 G

5FxyG

The solution is 1213}2,7

}

42.

63. 4x1 3y5 6

6x2 2y5 10

A X B

F4 36 22G Fx

yG5F610G

A21

51

}

282 18 F22 2326 4G5

F

1

}

13

3

}

26

3

}

13

2

}

13

G X5A21B5F1}13 3}263

}

13 2

2

}

13

GF610G5F 21}1322}13

G5FxyG The solution is 121}13, 2

2

}

132.

64. (81 3i)2 (71 4i)5 81 3i2 72 4i5 12 i

65. (52 2i)2 (291 6i)5 52 2i1 92 6i5 142 8i

66. i(31 i)5 3i1 i2 5 3i1 (21)5 211 3i

67. (121 5i)2 (72 8i)5 121 5i2 71 8i5 51 13i

68. (51 4i)(2 + 3i)5 101 15i1 8i1 12i2

5 101 23i1 12(21)

5 221 23i

69. (82 4i)(11 6i)5 81 48i2 4i2 24i2

5 81 44i2 24(21)

5 321 44i

Lesson 5.2

Investigating Algebra Activity 5.2 (p. 336)

1. Asxapproaches 2`,

f(x) approaches 2`.

Asxapproaches 1`,

f(x) approaches 1`.


f(x) approaches1`.

Asxapproaches1`,

f(x) approaches 2`.


f(x) approaches1`.

Asxapproaches1`,

f(x) approaches1`.


f(x) approaches 2`.

Asxapproaches1`,

f(x) approaches 2`.

5. a.f (x)5xn, wherexis odd: Asxapproaches1`,

f (x) approaches1`. Asxapproaches 2`,

f (x) approaches 2`.

b.f (x)5 2xn, where nis odd: Asxapproaches2`,

f (x) approaches1`. Asxapproaches1 `,

f (x) approaches 2`.

c.f (x)5xn, where nis even : Asxapproaches2`,


f (x) approaches1`.

d.f (x)5 2xn, wherexis odd : Asxapproaches2`,


f (x) approaches2`.

6. Asxapproaches 2`,f (x) approaches1`. Asx

approaches1`,f (x) approaches 1`.f(x) is going to

resemble its highest power. Becausex6is the highest

power, the end behavior will be the same asf(x)5x6.



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Algebra 2



1. The function is a polynomial function written as

f (x)5 22x 1 13 in standard form. It has degree 1 (linear)

and a leading coefficient of 22.

2. The function is not a polynomial function because the term

5x22has an exponent that is not a whole number.


h(x)5 6x22 3x1 in standard form. It has a degree

of 2 (quadratic) and a leading coefficient of 6.

4. f (x)5x41 2x31 3x227

f (22)5 (22)41 2(22)31 3(22)22 7

5 162 161 122 7

5 5

5. g(x)5x32 5x21 6x1 1

g(4)5(4)32 5(4)21 6(4)1 1

5 642 801 241 1

5 9

6. 2 5 3 21 7

10 26 50

5 13 25 57

f (2)5 57

7. 21 22 21 0 4 25

2 21 1 25

22 1 21 5 210

f (21)5 210

8. The degree is odd and the leading coefficient is negative.

9. x 23 22 21 0 1 2 3

y 132 37 4 23 4 37 132

x

y

21

1

10. x 23 22 21 0 1 2 3

y 32 9 0 21 0 23 216

x

y

22

1

11. x 23 22 21 0 1 2 3

y 58 20 6 4 2 212 250

x

y

21

1

12. s 0 5 10 15 20 25 30

E 0 3.2 51 258.2 816 1992.2 4131

The wind speed needed to generate the wave is about 25

miles per hour.

Wind speed

(miles per hour)

Energy

pe

rsquare

foot

(foot-pounds)

0 10 20 30 s

E

0

1000

2000

3000


Skill Practice

1.f (x) is a degree 4 (quartic) polynomial function. The

leading coefficient is 25 while the constant term is 6.

2. The end behavior of a polynomial is the behavior the

function demonstrates as it approaches 6`.

3. The function is a polynomial function written asf (x)5 2x21 8 in standard form. It has a degree 2

(quadratic) and a leading coefficient of 21.


f (x)5 8x41 6x2 3 in standard form. It has a degree

4 (quartic) and a leading coefficient of 8.

5. The function is a polynomial that is already written in

standard form. It has degree 4 (quartic) and a leading

coefficient of .

6. The function is not a polynomial function because the term

5x22has an exponent that is not a whole number.

7. The function is a polynomial function already written

in standard form. It has degree 3 (cubic) and a leading

coefficient of 25}2

.

8. The function is not a polynomial function because the

term2

}

x has an exponent that is not a whole number.

9. f (x)5 5x32 2x21 10x2 15

f (21)5 5(21)32 2(21)21 10(21)2 15

5 252 22 102 15

5 232



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Algebra 2


10. f (x)5 5x42x32 3x21 8x

f (2)5 5(2)42 (2)32 3(2)21 8(2)

5 802 82 121 16

5 76

11. g(x)5 22x51 4x3

g(23)5 22(23)51 4(23)35 4862 1085 378

12. h(x)5 6x32 25x1 20

h(5)5 6(5)32 25(5)1 205 7502 1251 205 645

13. h(x)51

}

2

x423

}

4

x31x1 10

h(24)51

}

2(24)42

3

}

4(24)31 (24)1 10

5 1281 482 41 10

5 182

14. g(x)5 4x51 6x31x2210x1 5

g(22)5 4(22)51 6(22)31 (22)22 10(22)1 5

5 21282 481 41 201 5

5 2147

15. 3 5 22 28 16

15 39 93

5 13 31 109

f (3)5 109

16. 22 8 12 6 25 9

216 8 228 66

8 24 14 233 75

f (22)5 75

17. 26 1 8 27 35

26 212 114

1 2 219 149

f (26)5 149

18. 4 28 0 14 235

232 2128 2456

28 232 2114 2491

f (4)5 2491

19. 2 22 3 0 28 13

24 22 24 224

22 21 22 212 211

f(2)5 211

20. 23 6 0 10 0 0 227

218 54 2192 576 21728

6 218 64 2192 576 21755

f (23)5 21755

21. 32

7 11 4 0

221 230 278

27 210 226 278

f (3)5 278

22. 4 1 0 0 3 220

4 16 64 268

1 4 16 67 248

f (4)5 248

23. The error is that a zero was not placed as the coefficient ofthe missingx3term.

22 24 0 9 221 7

8 216 14 14

24 8 27 27 21

f (22)5 21 117

24. A; The ends approach opposite directions so the function

must be odd. It imitates an odd function whose coefficient

is positive.

25. The degree is even and the leading coefficient is positive.

26. The degree is odd and the leading coefficient is negative.27. The degree is even and the leading coefficient is negative.

28. f (x) 1àsx 2`

f (x) 1àsx 1`

29. f (x) 2àsx 2`

f (x) 2àsx 1`

30. f (x) 1àsx 2`

f (x) 2àsx 1`

31. f (x) 2àsx 2`

f (x) 1àsx 1`

32. f (x) 1àsx 2`

f (x) 1àsx 1`

33. f (x) 1àsx 2`

f (x) 2àsx 1`

34. f (x) 2àsx 2`

f (x) 1àsx 1`

35. f (x) 1àsx 2`

f (x) 1àsx 1`

36. f (x) 1àsx 2`

f (x) 2àsx 1`



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Algebra 2


37. Sample answer:f (x)5 2x52 4x21 3x2 1

x

y

211

38. x 23 22 21 0 1 2 3

y 227 28 21 0 1 8 27

2

x

y

22

39. x 23 22 21 0 1 2 3

y 281 216 21 0 21 216 281

2

x22

y

40. x 23 22 21 0 1 2 3

y 2240 229 2 3 4 35 246

1

x

y

22

41. x 23 22 21 0 1 2 3

y79 14

2

12

22

1 14 79

x

y

4

22

42. x 23 22 21 0 1 2 3

y 32 13 6 5 4 23 222

1

x

y

21

43. x 23 22 21 0 1 2 3

y 212 2 4 0 24 22 12

1

x

y

21

44. x 23 22 21 0 1 2 3

y 2105 232 29 0 7 0 257

1

x

y

21

45. x 23 22 21 0 1 2 3

y 2246 234 22 0 2 34 246

1

x

y

21

46. x 23 22 21 0 1 2 3

y 65 29 11 5 5 5 21

x

y

21

1



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Algebra 2


47. x 23 22 21 0 1 2 3

y 2238 232 24 24 22 32 248

x

y

21

1

48. x 23 22 21 0 1 2 3

y 42 2 2 6 2 2 42

x

y

21

1

49. x 23 22 21 0 1 2 3

y 2158 237 22 1 2 7 22

x

y

22

2

50. B;f(x) 5 21

}

3

x31 1

From the graph,f(x) 1` asx2`;and

f(x) 2` asx1`. So, the degree is odd

and the leading coefficient is negative. The

constant term 5f(0)5 1.

51. Wheng(x)5 2f(x):

g(x) 2` asx2` andg(x) 1` asx1`.

52.f(x)5 a3x31 a

2x21 a

1x1 a

0

When a35 2 and a

05 25:

f(x)5 2x31 a2x21 a

1x2 5

Whenf(1)5 0:

2(1)31 a2(1)21 a

1(1)2 55 0

a21 a

15 3

Whenf(2)5 3:

2(2)31 a2(2)21 a

1(2)2 55 3

4a21 2a

15 28

Solve the system of equations:

a21 a

15 3 324 24a

22 4a

15 212

4a21 2a

15 28 4a

21 2a

15 28

22a15 220

a15 10

Substitute a1into one of the original equations. Solve for a

2.

a21 105 3 a

25 27

The cubic function isf(x)5 2x32 7x21 10x2 5.

So,f(25)5 2(25)32 7(25)21 10(25) 2 55 2480.

53. a.x f(x) g(x)

f(x)}

g(x)

10 1000 840 1.19

20 8000 7280 1.10

50 125,000 120,200 1.04

100 1,000,000 980,400 1.02

200 8,000,000 7,920,800 1.01

b.Asx1`,f(x)

}

g(x)

1.

c.For the graph of a polynomial function, the end behavior

is determined by the functions degree and the sign of its

leading coefficient.

Problem Solving

54. w5 0.00071d32 0.090d21 0.48d

w5 0.0071(15)32 0.090(15)21 0.48(15)

5 23.96252 20.251 7.2 5 10.9125 carats

55. t 0 2 4 6 8

S 5.5 5.1 4.7 6.7 10.5

Skateboarding

participants(millions)

02 4 6 810 3 5 7 109

2

4

6

8

10

14

12

Years since 1992

t

S

There were 8 million skateboarders 6 years after

1992 (1998).



10/73

CopyrightbyM


Company.

296

Algebra 2


56. a.The function is degree 3 (cubic).

b. t 0 2 4 6 8

M 21,600 21,396 22,800 25,284 28,320

t 10 12 14 16

M 31,380 33,936 35,460 35,424

c.

Indoormoviescreens

(thousands)

04 8 12 160

40

20

10

30

Years since 1987

t

M

57. t 0 2 4 6 8 10 12

s 1.2 1.46 1.68 1.59 1.44 1.97 4.41

Snowboarding

participants(millions)

02 4 6 810 3 5 7 109

2

1

Years since 1992

t

S

The number of snowboarders was more than 2 million

in 2002.

58. a. As t1`,P(x) 1`.

As t2`,P(x) 1`.

b.

06 12 18 240

4000

2000

1000

3000

t

P

Quarterlyperiodicals

Years since 1980

c. p(t)5 0.138t42 6.24t31 86.8t22 239t1 1450

p(30)5 0.138(30)42 6.24(30)31 86.8(30)2

2 239(30)1 1450

5 111,7802 168,4801 78,1202 71701 1450

5 15,700 periodicals

It is not appropriate to use this model to predict the

number of periodicals in 2010 beacuse the model was

made for 1980 to 2002.

59. a. S5 20.122t31 3.49t22 14.6t1 136

S(5)5 20.122(5)31 3.49(5)22 14.6(5)1 136

5 215.251 87.252 731 136

5 135 grams

H5 20.115t31 3.71t22 20.6t1 124

H(5)5 20.115(5)31 3.71(5)22 20.6(5)1 124

5 214.3751 92.752 1031 124

5 99.375 grams

Difference5 S(5)2H(5)

5 1352 99.375

5 35.625 grams

b.

Weight(grams)

0 2 4 6 80

60

120

240

180

Days after hatching

t

W

S

H

c. The chick is more likely to be a sarus chick than a

hooded chick. At 3 days the sarus chicks weight is about

120 grams, higher than the hooded chicks. Even though

neither chicks average weight at 3 days equals 130

grams, the sarus chick is closer.

60. a. y5 0.000304x3

2.20y5 0.000304(0.394x)3

y50.000304(0.394x)3

}}

2.2

(8.4523 1026)x3

b.

100 20 30 40 50

1

0

2

3y50.000304x3

Length (cm and in.)

Width(kgandlb)

x

y

y 5 (8.45231026)x3

The function from part (a) is a vertical shrink of the

original function.

Mixed Review

61. 2b1 115 152 6b

2b5 42 6b

8b5 4

b51

}

2

62. 2.7n1 4.35 12.94 63. 27< 6y2 1 < 5

2.7n5 8.64 26 < 6y< 6

n5 3.2 21


11/73

CopyrightbyM


Company.

Algebra 2


64. x22 14x1 485 0

(x2 6)(x2 8)5 0

x2 65 0 or x2 85 0

x5 6 or x5 8

65. 224q22 90q5 21

224q22 90q2 215 0

8q21 30q1 75 0

(4q1 1)(2q1 7)5 0

4q1 15 0 or 2q1 75 0

4q5 21 or 2q5 27

q5 21

}

4

or q5 27

}

2

66. z21 5z< 36

z21 5z2 36 < 0

z21 5z2 365 0

(z1 9)(z2 4)5 0

z1 95 0 (z2 4)5 0

z5 29 z5 4

Part ofz-axis Chosen pt.,x z21 5z 2 36

z< 29 210 (210)21 5(210)2 365 14

29 4 5 (5)21 5(5)2 365 14

Is z215z236


12/73

CopyrightbyM


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298

Algebra 2


Graphing Calculator Activity 5.2 (p. 345)

1. 210x 10, 210y 50

2. 220x 50, 22000y 8000

3. 210x 10, 210x 10

4. 25x 5, 210x 30

5. 25x 10, 210x 50

6. 25x 10, 240x 40

7. 25x 10, 220x 50

8. 25x 5, 220x 20

9. Sample answer:The window forg(x) should be the same

x-interval, but they-interval will be shifted by cunits.

10. Sample answer: 210x 20, 0y 3000

Lesson 5.3


1. (t22 6t1 2)1 (5t22 t2 8)

5 t21 5t22 6t2 t1 22 8

5 6t22 7t2 6 2. (8d2 31 9d3)2 (d32 13d22 4)

5 8d2 31 9d32 d31 13d21 4

5 9d32 d31 13d21 8d2 31 4

5 8d31 13d21 8d1 1

3. (x1 2)(3x22x2 5)

5 (x1 2)3x22 (x1 2)(x)2 (x1 2)(5)

5 3x31 6x22x22 2x2 5x2 10

5 3x31 5x22 7x2 10

4. (a2 5)(a1 2)(a1 6)5 (a22 3a2 10)(a1 6)

5 (a22 3a2 10)a1 (a22 3a2 10)6

5 a32 3a22 10a1 6a22 18a2 60

5 a31 3a22 28a2 60

5. (xy2 4)35 (xy)32 3(xy)2(4)1 3(xy)(4)22 (4)3

5x3y32 12x2y21 48xy2 64

6. T5 CpD

5 0.542t2 2 7.16t 1 79.4

3 109t 1 4010

2173.42t22 28711.6t 1 318,394

5.9078t32 780.44t21 8654.6t

59.078t31 1392.98t22 20,057t1 318,394

T5 59.078t31 1392.98t22 20,057t1 318,394


Skill Practice

1. When you add or subtract polynomials, you add or subtract

the coefficients of like terms.

2. Subtracting two polynomials is simply adding the opposite

of the subtracted polynomial to the first polynomial.

3. (3x22 5)1 (7x22 3)5 3x21 7x22 52 3

5 10x22 8

4. (x22 3x1 5)2 (24x21 8x1 9)

5x22 3x1 51 4x22 8x2 9

5x21 4x22 3x2 8x1 52 9

5 5x22 11x2 4

5. (4y21 9y2 5)2 (4y22 5y1 3) 5 4y21 9y2 52 4y21 5y2 3

5 4y22 4y21 9y1 5y2 52 3

5 14y2 8

6. (z21 5z2 7)1 (5z22 11z2 6)

5z21 5z21 5z2 11z2 72 6

5 6z22 6z2 13

7. (3s31s)1 (4s32 2s21 7s1 10)

5 3s31 4s32 2s21s1 7s1 10

5 7s32 2s21 8s1 10

8. (2a22 8)2 (a31 4a22 12a1 4)

5 2a22 82 a32 4a21 12a2 4

5 2a31 2a22 4a21 12a2 82 4

5 2a32 2a21 12a2 12

9. (5c21 7c1 1)1 (2c32 6c1 8)

5 2c31 5c21 7c2 6c1 11 8

5 2c31 5c21 c1 9

10. (4t32 11t21 4t)2 (2 7t22 5t1 8)

5 4t32 11t21 4t1 7t21 5t2 8

5 4t32 11t21 7t21 4t1 5t2 8

5 4t32 4t21 9t2 8

11. (5b2 6b31 2b4)2 (9b31 4b42 7)

5 5b2 6b31 2b42 9b32 4b41 7

5 2b42 4b42 6b32 9b31 5b1 7

5 22b42 15b31 5b1 7

12. (3y22 6y41 52 6y)1 (5y42 6y31 4y)

5 26y41 5y42 6y31 3y22 6y1 4y1 5

5 2y42 6y31 3y22 2y1 5

13. (x42x31x22x1 1)1 (x1x42 12x2)

5x41x42x31x22x22x1x1 12 1

5 2x42x3

14. (8v42 2v21 v2 4)2 (3v32 12v21 8v)

5 8v42 2v21 v2 42 3v31 12v22 8v

5 8v42 3v32 2v21 12v21 v2 8v2 4

5 8v42 3v31 10v22 7v2 4

15. B;

(8x42 4x32x1 2)2 (2x42 8x22x1 10)

5 8x42 4x32x1 22 2x41 8x21x2 10

5 8x42 2x42 4x31 8x22x1x1 22 10

5 6x42 4x31 8x22 8



13/73

CopyrightbyM


Company.

Algebra 2



16. x(2x22 5x1 7)5 2x2(x)2 5x(x)1 7(x)

5 2x32 5x21 7x

17. 5x2(6x1 2)5 6x(5x2)1 2(5x2)

5 30x31 10x2

18. (y2 7)(y1 6)5 (y2 7)(y)1 (y2 7)(6)

5y22 7y1 6y2 42

5y22y2 42

19. (3z1 1)(z2 3)5 (3z1 1)(z)2 (3z1 1)(3)

5 3z21z2 9z2 3

5 3z22 8z2 3

20. (w1 4)(w21 6w2 11)

5 (w1 4)(w2)1 (w1 4)(6w)2 (w1 4)(11)

5 w31 4w21 6w21 24w2 11w2 44

5 w31 10w21 13w2 44

21. (2a2 3)(a22 10a2 2)

5 (2a2 3)(a2)2 (2a2 3)(10a)2 (2a2 3)(2)

5 2a32 3a22 20a21 30a2 4a1 6

5 2a32 23a21 26a1 6

22. (5c22 4)(2c21 c2 3)

5 (5c22 4)(2c2)1 (5c22 4)(c)2 (5c22 4)(3)

5 10c42 8c21 5c32 4c2 15c21 12

5 10c41 5c32 23c22 4c1 12

23. (2x21 4x1 1)(x22 8x1 3)

5 (2x21 4x1 1)(x2)2 (2x21 4x1 1)(8x)

1 (2x21 4x1 1)(3)

5 2x41 4x31x21 8x32 32x22 8x2 3x2

1 12x1 3

5 2x41 12x32 34x21 4x1 3

24. (2d21 4d1 3)(3d22 7d1 6)

5 (2d21 4d1 3)(3d2)2 (2d21 4d1 3)(7d)

1 (2d21 4d1 3)(6)

5 23d41 12d31 9d21 7d32 28d22 21d

2 6d21 24d1 18

5 23d41 19d32 25d21 3d1 18

25. (3y21 6y2 1)(4y22 11y2 5)

5 (3y21 6y2 1)(4y2)2 (3y21 6y2 1)(11y)

2 (3y21 6y2 1)(5)

5 12y41 24y32 4y22 33y32 66y21 11y

2 15y22 30y1 5

5 12y42 9y32 85y22 19y1 5

26. The error is that the negative sign was not carried through

the subtracted polynomial (x31 7x2 2).

(x22 3x1 4)2 (x31 7x2 2)

5x22 3x1 42x32 7x1 2

5 2x31x22 10x1 6

27. (2x2 7)3is not the difference of the cubes of (2x) and (7).

It is the quantity (2x2 7) multiplied together three times.

(2x2 7)35 (2x)32 3(2x)2(7)1 3(2x)(7)22 (7)3

5 8x32 21(4x2)1 6x(49)2 343

5 8x32 84x21 294x2 343

28. (x1 4)(x2 6)(x2 5)5 (x22 2x2 24)(x2 5)

5 (x2

2 2x2 24)(x)2 (x2

2 2x2 24)(5) 5x32 2x22 24x2 5x21 10x1 120

5x32 7x22 14x1 120

29. (x1 1)(x2 7)(x1 3)5 (x22 6x2 7)(x1 3)

5 (x22 6x2 7)(x)2 (x22 6x2 7)(3)

5x32 6x22 7x1 3x22 18x2 21

5x32 3x22 25x2 21

30. (z2 4)(2z1 2)(z1 8)5 (2z21 6z2 8)(z1 8)

5 (2z21 6z2 8)(z)1 (2z21 6z2 8)(8)

5 2z31 6z22 8z2 8z21 48z2 64

5 2z32 2z21 40z2 64

31. (a2 6)(2a1 5)(a1 1)5 (2a22 7a2 30)(a1 1) 5 (2a22 7a2 30)(a)1 (2a22 7a2 30)(1)

5 2a32 7a22 30a1 2a22 7a2 30

5 2a32 5a22 37a2 30

32. (3p1 1)(p1 3)(p1 1)5 (3p21 10p1 3)(p1 1)

5 (3p21 10p1 3)(p)1 (3p21 10p1 3)(1)

5 3p31 10p21 3p1 3p21 10p1 3

5 3p31 13p21 13p1 3

33. (b2 2)(2b2 1)(2b1 1)5 (2b22 5b1 2)(2b1 1)

5 2(2b22 5b1 2)(b)1 (2b22 5b1 2)(1)

5 22b31 5b22 2b1 2b22 5b1 2

5 22b31 7b22 7b1 2

34. (2s1 1)(3s2 2)(4s2 3)5 (6s22s2 2)(4s2 3)

5 (6s22s2 2)(4s)2 (6s22s2 2)(3)

5 24s32 4s22 8s2 18s21 3s1 6

5 24s32 22s22 5s1 6

35. (w2 6)(4w2 1)(23w1 5)

5 (4w22 25w1 6)(23w1 5)

5 2(4w22 25w1 6)(3w)1 (4w22 25w1 6)(5)

5 212w31 75w22 18w1 20w22 125w1 30

5 212w31 95w22 143w1 30

36. (4x2 1)(22x2 7)(25x2 4)

5 (28x22 26x1 7)(25x2 4) 5 2(28x22 26x1 7)(5x)2 (28x22 26x1 7)(4)

5 40x31 130x22 35x1 32x21 104x2 28

5 40x31 162x21 69x2 28


14/73

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300

Algebra 2


37. (3q2 8)(29q1 2)(q2 2)

5 (227q21 78q2 16)(q2 2)

5 (227q21 78q2 16)(q)2 (227q21 78q2 16)(2)

5 227q31 78q22 16q1 54q22 156q1 32

5 227q31 132q22 172q1 32

38. (x1 5)(x2 5)5 (x)22 (5)25x22 25

39. (w2 9)25 (w)22 2(wp9)1 (9)2

5 w22 18w1 81

40. (y1 4)35 (y)31 3(y)2(4)1 3(y)(4)21 (4)3

5y31 12y21 48y1 64

41. (2c1 5)25 (2c)21 2(2cp5)1 (5)2

5 4c21 20c1 25

42. (3t2 4)35 (3t)32 3(3t)2(4)1 3(3t)(4)22 (4)3

5 27t32 108t21 144t2 64

43. (5p2 3)(5p1 3)5 (5p)22 (3)2

5 25p22 9

44. (7x2y)35 (7x)32 3(7x)2(y)1 3(7x)(y)22 (y)3

5 343x32 147x2y1 21xy22y3

45. (2a1 9b)(2a2 9b)5 (2a)22 (9b)2

5 4a22 81b2

46. (3z1 7y)35 (3z)31 3(3z)2(7y)1 3(3z)(7y)21 (7y)3

5 27z31 189z2y1 441zy21 343y3

47. D;

(3x2 2y)25 (3x)22 2(3xp2y)1 (2y)2

5 9x22 12xy1 4y2

48. V5 lwh l5 3x1 1; w5x; h5x1 3

5 (3x1 1)(x)(x1 3)

5 (3x21x)(x1 3)

5 (3x21x)(x)1 (3x21x)(3)

5 3x31x21 9x21 3x

5 3x31 10x21 3x

49. V5 r2h r5x2 4 ; h5 2x1 3

5 (x2 4)2(2x1 3)

5 (x22 8x1 16)(2x1 3)

5 ((x22 8x1 16)(2x)1 (x22 8x1 16)(3))

5 (2x32 16x21 32x1 3x22 24x1 48)

5 (2x32 13x21 8x1 48)

5 2x32 13x21 8x1 48

50. V5s3 s5x2 5

5 (x2 5)3

5(x)32 3(x)2(5)1 3(x)(5)22 (5)3

5x32 15x21 75x2 125

51. V51

}

3

Bh B5 (2x2 3)2; h5 (3x1 4)

51

}

3(2x2 3)2(3x1 4)

51

}

3

(4x22 12x1 9)(3x1 4)

51

}

3

((4x22 12x1 9)(3x)1 (4x22 12x1 9)(4))

51

}

3

(12x32 36x21 27x1 16x22 48x1 36)

51

}

3

(12x32 20x22 21x1 36)

5 4x3220

}

3

x22 7x1 12

52. (a1 b)(a2 b)0 a22 b2

(a1 b)(a)2 (a1 b)(b)0 a22 b2

a21 ab2 ab2 b20 a22 b2

a22 b25 a22 b2

53. (a1 b)20 a21 2ab1 b2

(a1 b)(a1 b)0 a21 2ab1 b2

(a1 b)(a)1 (a1 b)(b)0 a21 2ab1 b2

a21 ab1 ab1 b20 a21 2ab1 b2

a21 2ab1 b25 a21 2ab1 b2

54. (a1 b)30 a31 3a2b1 3ab21 b3

(a1 b)(a1 b)(a1 b)0 a31 3a2b1 3ab21 b3

(a21 2ab1 b2)(a1 b)0 a31 3a2b1 3ab21 b3

(a21 2ab1 b2)(a)1 (a21 2ab1 b2)(b)

0 a31 3a2b1 3ab21 b3

a31 2a2b1 ab21 a2b1 2ab21 b3

0 a31 3a2b1 3ab21 b3

a31 3a2b1 3ab21 b35 a31 3a2b1 3ab21 b3

55. (a2 b)30 a32 3a2b1 3ab22 b3

(a2 b)(a2 b)(a2 b)0 a32 3a2b1 3ab22 b3

(a22 2ab1 b2)(a2 b)0 a32 3a2b1 3ab22 b3

(a22 2ab1 b2)(a)2 (a22 2ab1 b2)(b)

0 a32 3a2b1 3ab22 b3

a32 2a2b1 ab22 a2b1 2ab21 b3

0 a32 3a2b1 3ab22 b3

a32 3a2b1 3ab21 b35 a32 3a2b1 3ab22 b3

56. a.p(x)1 q(x)5 (x42 7x1 14)1 (x22 5)

5x41x22 7x1 9

degree 4 b.p(x)2 q(x)5 (x42 7x1 14)2 (x22 5)

5x42x22 7x1 19

degree 4

c.p(x) pq(x)5 (x42 7x1 14)(x22 5)

5 (x42 7x1 14)(x2)2 (x42 7x1 14)(5)

5x62 7x31 14x22 5x41 35x2 70

5x62 5x42 7x31 14x21 35x2 70

degree 6



15/73

CopyrightbyM


Company.

Algebra 2


d.p(x)1 q(x) : degree m

p(x)2 q(x):degree m

p(x) pq(x) : degree (m1 n)

57. a.x52 15 (x2 1)(x41x31x21x1 1)

Check: (x41x31x21x1 1)(x)

2 (x41x31x21x1 1)

5x51x41x31x21x2x42x3

2x22x2 1

5x52 1

x62 15 (x2 1)(x51x41x31x21x1 1)

Check: (x51x41x31x21x1 1)(x)

2 (x51x41x31x21x1 1)

5x61x51x41x31x21x

2x52x42x32x22x2 1

5x62 1

b.xn2 15 (x2 1)(xn2 11xn2 21xn2 31 . . .1xn2 n)

5 (x2 1)(xn2 11xn2 21xn2 31 . . .1 1)

Check: (xn2 11xn2 21xn2 31 . . .1 1)(x) 2 (xn2 11xn2 21xn2 31 . . .1 1)

5xn2 11 11xn 2 21 11xn2 31 11 . . .1x

2xn2 12xn2 22xn2 32 . . .2 1

5xn1xn2 11xn2 21 . . .xn2 (n2 1)

2xn2 12xn2 22xn2 3. . .21

5xn2 1

58.f(x)5 (x1 a)(x1 b)(x1 c)(x1 d)

5 (x21 ax1 bx1 ab)(x1 c)(x1 d)

5 ((x21 ax1 bx1 ab)(x)

1 (x21 ax1 bx1 ab)(c))(x1 d)

5 (x31 ax21 bx21 abx1 cx2

1 acx1 bcx1 abc)(x1 d)

5 (x31 ax21 bx21 abx1 cx21 acx

1 bcx1 abc)(x)1 (x31 ax21 bx21 abx 1 cx21 acx1 bcx1 abc)(d)

5x41 ax31 bx31 abx21 cx31 acx2

1 bcx21 abcx1 dx31 adx21 bdx21 abdx

1 cdx21 acdx1 bcdx1 abcd

5x41 (a1 b1 c1 d )x3

1 (ab1 ac1 bc1 ad1 bd1 cd )x2

1 (abc1 abd1 acd1 bcd )x1 abcd

coefficient ofx35 (a1 b1 c1 d )

constant term5 abcd

Problem Solving

59. T5M1F

5 (0.091t32 4.8t21 110t1 5000)1 (0.19t32 12t2

1 350t1 3600)

5 0.281t32 16.8t21 460t1 8600

60. R5PpD

5 (6.82t22 61.7t1 265)(4.11t1 4.44)

5 (6.82t22 61.7t1 265)(4.11t)1 (6.82t22 61.7t

1 265)(4.44)

5 28.0302t32 253.587t21 1089.15t1 30.2808t2

2 273.948t1 1176.6

5 28.0302t32 223.3062t21 815.202t1 1176.6

R(3)5 28.0302(3)32 223.3062(3)21 815.202(3)

1 1176.6

5 756.81542 2009.75581 2445.6061 1176.6

5 2369.2656

The total revenue in 2002 from DVD sales was 2369.2656

million dollars, or $2,369,265,600.

61. F5 0.0116s21 0.789

P5 0.00267sF

5 0.00267s(0.0116s21 0.789)

5 0.00003s31 0.00211s

The model isP5 0.00003s31 0.00211s.

P(10)5 0.00003(10)31 0.00211(10)

5 0.031 0.0211

5 0.0511

So, you need 0.0511 horsepower to keep the bicycle

moving 10 miles per hour.

62.

r11

r

0.5 cm

1 cm

M(r)5 volume of cookie 1 volume of marshmallow

5 (r1 1)2h11

}

214}3r32

5 (r1 1)211}221}

2

p4

}

3

r3

5}

2

(r21 2r1 1)1

}

2

p4

}

3

r3

5}

214}3r31 r21 2r1 12

52}

3r31

}2r21 r1

}

2

D(r)51

}

214}3(r1 1)321 (r1 1)21

1

}

22

5

}21

4

}3

(r1 1)321}2(r1 1)2

5}

2 p

4

}

3

(r31 3r21 3r1 1)1}2(r21 2r 1 1)

5}

214}3r31 4r21 4r1

4

}

31 r21 2r 1 12

52}

3r31

5}2r21 3r1

7}

6



16/73

CopyrightbyM


Company.

302

Algebra 2


C(r)5D(r)2M(r)

52}

3r31

5}

2r21 3r1

7}

6

2 12}3r31}2r21 r1

}22

52}

3r31

5}

2r21 3r1

7}

62

2}

3r3

2

}2r22 r2

}2

5 2r21 2r12}

3

63.N5 total mens players1 total womens players

5 (Lm

pSm

)1 (LwpS

w)

Lm

pSm

5 (20.127t31 0.822t22 1.02t1 31.5)p

(5.57t1 182)

5 (20.127t31 0.822t22 1.02t1 31.5)(5.57t)

1 (20.127t31 0.822t22 1.02t1 31.5)(182)

5 20.70739t42 18.53546t31 143.9226t2

2 10.185t1 5733

LwpS

w5 (20.0662t31 0.437t22 0.725t1 22.3)p

(12.2t1 185)

5 (20.0662t31 0.437t22 0.725t1 22.3)p

(12.2t)1 (20.0662t31 0.437t2

2 0.725t1 22.3)(185)

5 20.80764t42 6.9156t31 72t21 137.935t

1 4125.5

N5 20.70739t42 18.53546t31 143.9226t2

2 10.185t1 57332 0.80764t4

2 6.9156t31 72t21 137.935t1 4125.5

5 21.51503t42 25.45106t31 215.9226t2

1 127.751 9858.5

The model for the total number of people is written by

adding the number of men players to the number of

women players. The number of men and women players

is written by multiplying the number of teams by the

average team size for each group.

64.s5 t2 5

t5s1 5

C5 20.00105t31 0.0281t21 0.465t1 48.8

C5 20.00105(s1 5)31 0.0281(s1 5)21

0.465(s1 5)1 48.8

5 20.00105(s31 3s2(5)1 3s(25)1 125)

1 0.0281(s21 10s1 25)1 0.465s1 2.3251 48.8

5 20.00105s32 0.01575s22 0.07875s2 0.13125

1 0.0281s21 0.281s1 0.70251 0.465s1 2.325

1 48.8

5 20.00105s31 0.01235s21 0.66725s1 51.69625

Mixed Review

65. 2x2 75 11 66. 102 3x5 25

2x5 18 23x5 15

x5 9 x5 25

67. 4t2 75 2t

275 22t

7

}

25 t

68. y22 2y2 485 0

(y1 6)(y2 8)5 0

y1 65 0 or y2 85 0

y5 26 or y5 8

69. w22 15w1 545 0

(w2 6)(w2 9)5 0

w2 65 0 or w2 95 0

w5 6 or w5 9

70. x21 9x1 145 0

(x1 2)(x1 7)5 0 x1 25 0 or x1 75 0

x5 22 or x5 27

71. 4z21 21z2 185 0

(4z2 3)(z1 6)5 0

4z2 35 0 or z1 65 0

4z5 3 or z5 26

z53

}

4

72. 9a22 30a1 255 0

(3a2 5)25 0

3a2

55

0

3a5 5

a55

}

3

73. x1y2 2z5 24 y5 2x1 2z2 4

3x2y1z5 22

2x1 2y1 3z5 29

3x 2 (2x1 2z2 4)1z5 22

3x1x2 2z1 41z5 22

4x2z5 18

2x1 2(2x1 2z2 4)1 3z5 29

2x2 2x1 4z2 81 3z5 29 23x1 7z5 21

4x2z5 18 33 12x2 3z5 54

23x1 7z5 21 34 212x1 28z5 24

25z5 50

z5 2

23x1 7(2)5 21 x5 5

y5 251 2(2)2 45 25

The solution is (5, 25, 2).



17/73

CopyrightbyM


Company.

Algebra 2


74. x2 2y1z5 213 x5 2y2z2 13

2x1 4y1z5 35

3x1 2y1 4z5 28

2(2y 2z 2 13)1 4y1z5 35

22y1z1 131 4y1z5 35

2y1 2z5 22

y1z5 11

3(2y2z2 13)1 2y1 4z5 28

6y2 3z2 391 2y1 4z5 28

8y1z5 67

y1z5 11 321 2y2z5 211

8y1z5 67 8y1z5 67

7y 5 56

y 5 8

81z5 11 z5 3

x5 2(8)2 (3) 2 135 0


75. 3x2y2 2z5 20 y5 3x2 2z2 20

2x1 3y2z5 216

22x2y1 3z5 25

2x1 3(3x2 2z2 20)2z5 216

2x1 9x2 6z2 602z5 216

8x2 7z5 44

22x2(3x2 2z2 20)1 3z5 25

22x2 3x1 2z1 201 3z5 25

25x1 5z5 225

2x1z5 25

8x2 7z5 44 8x2 7z5 44

2x1z5 25 38 28x1 8z5 240

z5 4

2x1 45 25 x5 9

y5 3(9)2 2(4)2 205 21


76. detF3 243 1 G5 3(1)2 3(24)5 31 125 15

77. det

F

5 7

24 9G

5 5(9)2 (24)(7)5 451 285 73

78. detF21 8 0 3 4 2325 2 1G5 (241 1201 0)2 (01 61 24) 5 1162 305 86

79. detF 2 3 2426 1 523 21 22

G5 (242 452 24)2 (122 101 36) 5 2732 385 2111

Quiz 5.15.3 (p. 352)

1. 35p3215 351 (21)5 345 81

2. (24)25 24 p25 285 256

3. 1 2}322225

22

}

(322)25

22

}

3245 22p345 4 p815 324

4. 13}5222

5 15}3225

52

}32

525

}

9

5. (x4y22)(x23y8)5x41 (23)y(22)1 85xy6

6. (a2b25)235 (a2)23(b25)235 a26pb155b15

}

a6

7.x3y7

}

x24y05x32(24)y72 05x7y7

8.

c3d22

}

c5d215

c

32 5

pd

222 (21)5

c

22

d

215

1

}

c2d

9.x 23 22 21 0 1 2 3

y 244 29 2 1 0 11 46

4

x

y

22

10.

x 23 22 21 0 1 2 3

y 95 26 7 2 21 10 71

1

x

y

21

11.x 23 22 21 0 1 2 3

y 58 15 22 25 26 217 250

1

x

y

22



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304

Algebra 2


12. (x31x22 6)2 (2x21 4x2 8)

5x31x22 62 2x22 4x1 8

5x32x22 4x1 2

13. (23x21 4x2 10)1 (x22 9x1 15)

5 23x21x21 4x2 9x2 101 15

5 22x22 5x1 5

14. (x2 5)(x22 5x1 7)

5 (x22 5x1 7)(x)2 (x22 5x1 7)(5)

5x32 5x21 7x2 5x21 25x2 35

5x32 10x21 32x2 35

15. (x1 3)(x2 6)(3x2 1)5 (x22 3x2 18)(3x2 1)

5 (x22 3x2 18)(3x)2 (x22 3x2 18)

5 3x32 9x22 54x2x21 3x1 18

5 3x32 10x22 51x1 18

16. (7,282,000,000,000)4 (294,000,000)5

7.282 3 1012

}

2.94 3 108 5

7.282

}

2.94 3

102

}

108

2.48 3104

5 $24,800

Lesson 5.4

5.4 Guided Practice (pp. 354-356)

1. x32 7x21 10x5x(x22 7x1 10)5x(x2 5)(x2 2)

2. 3y52 75y35 3y3(y22 25)5 3y3(y2 5)(y1 5)

3. 16b51 686b25 2b2(8b31 343)

5 2b2(2b1 7)(4b22 14b1 49)

4. w32 275 (w2 3)(w21 3w1 9)

5. x31 7x22 9x2 635x2(x1 7) 2 9(x1 7)

5 (x1 7)(x22 9) 5 (x1 7)(x2 3)(x1 3)

6. 16g42 6255 (4g2)22 (25)2

5 (4g22 25)(4g21 25)

5 (2g2 5)(2g1 5)(4g21 25)

7. 4t62 20t41 24t25 4t2(t42 5t21 6)

5 4t2(t22 2)(t22 3)

8. 4x52 40x31 36x5 0

4x(x42 10x21 9)5 0

4x(x22 9)(x22 1)5 0

4x(x2 3)(x1 3)(x2 1)(x1 1)5 0

x5 0,x5 3,x5 23,x5 1, orx5 21

9. 2x51 24x5 14x3

2x52 14x31 24x5 0

2x(x42 7x21 12)5 0

2x(x22 3)(x22 4)5 0

2x(x22 3)(x2 2)(x1 2)5 0

x5 0,x5 2}

3 ,x5 }

3 ,x5 2, orx5 22

10. 227x31 15x25 26x4

6x42 27x31 15x25 0

3x2(2x22 9x1 5)5 0

3x25 0 or 2x22 9x1 55 0

x5 0 orx59 6 }}

922 4 p2 p5}}

2 p2 5

9 6 }

41}

4

11. Volume

(ft3)5

Interiorlength

(ft)

p

Interiorwidth

(ft)

p

Interiorheight

(ft)

40 5 (6x2 2) p (3x2 2) p (x2 1)

405 18x32 36x21 22x2 4

05 18x32 36x21 22x2 44

05 18x2(x2 2)1 22(x2 2)

05 (18x21 22)(x2 2)

The only real solutions isx5 2. The basin is 12 ft long,

6 ft wide, and 2 ft high.


Skill Practice

1. The expression 8x61 10x32 3 is in quadraticform

because it can be written as 2u21 5u2 3 where u5 2x3.

2. The factorization of a polynomial is factored completely if

it is written as a product of unfactorable polynomials with

integer coefficients.

3. 14x22 21x5 7x(2x2 3)

4. 30b32 54b25 6b2(5b2 9)

5. c31 9c21 18c5 c(c21 9c1 18)5 c(c1 3)(c1 6)

6.z32 6z22 72z5z (z22 6z2 72)5z(z2 12)(z1 6)

7. 3y52 48y35 3y3(y22 16)5 3y3(y2 4)(y1 4)

8. 54m51 18m41 9m35 9m3(6m21 2m1 1) 9. A;

(2x72 32x3)5 2x3(x42 16)

5 2x3(x22 4)(x21 4)

5 2x3(x1 2)(x2 2)(x21 4)

10.x31 85x31 235 (x1 2)(x22 2x1 4)

11.y32 645y32 435 (y2 4)(y21 4y1 16)

12. 27m31 15 (3m)31 135 (3m1 1)(9m22 3m1 1)

13. 125n31 2165 (5n)31 635 (5n1 6)(25n22 30n1 36)

14. 27a32 10005 (3a)32 103

5 (3a2 10)(9a21 30a1 100)

15. 8c31 3435 (2c)31 735 (2c1 7)(4c22 14c1 49)16. 192w32 35 3(64w32 1)

5 3((4w)32 13)

5 3(4w2 1)(16w21 4w1 1)

17. 25z31 3205 25(z32 64)

5 25(z32 43)

5 25(z2 4)(z21 4z1 16)



19/73

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Algebra 2


18. x31x21x1 15x2(x1 1)1 (x1 1)5 (x1 1)(x21 1)

19.y32 7y21 4y2 285y2(y2 7)1 4(y2 7)

5 (y2 7)(y21 4)

20. n31 5n22 9n2 455 n2(n1 5)2 9(n1 5)

5 (n1 5)(n22 9)

5 (n1 5)(n2 3)(n1 3)

21. 3m32 m21 9m2 35 3m31 9m2 m22 3

5 3m(m21 3)2 (m21 3)

5 (m21 3)(3m2 1)

22. 25s32 100s22s1 45 25s2(s2 4)2 (s2 4)

5 (s2 4)(25s22 1)

5 (s2 4)(5s2 1)(5s1 1)

23. 4c31 8c22 9c2 185 4c2(c1 2)2 9(c1 2)

5 (c1 2)(4c22 9)

5 (c1 2)(2c2 3)(2c1 3)

24.x42 255 (x2)22 525 (x22 5)(x21 5)

25. a41 7a21 65 (a21 1)(a21 6)

26. 3s42s22 245 (3s21 8)(s22 3)

27. 32z52 2z5 2z (16z42 1)

5 2z (4z21 1)(4z22 1)

5 2z (4z21 1)(2z1 1)(2z2 1)

28. 36m61 12m41 m25 m2(36m41 12m21 1)

5 m2(6m21 1)(6m21 1)

5 m2(6m21 1)2

29. 15x52 72x32 108x5 3x(5x42 24x22 36)

5 3x(5x21 6)(x22 6)

30. The error is that when the binomial was factored, it was

not factored properly.

8x32 275 0

(2x2 3)(4x21 6x1 9)5 0

2x2 35 0

x53

}

2

23

}

2

31. The error is thatx5 0 is not included in the solutions,

andx22 16 was not factored completely before finding

the zeros.

3x(x22 16)5 0

3x(x2 4)(x1 4)5 0

x5 0,x5 4, orx5 24

32. y32 5y25 0

y2(y2 5)5 0

y5 0 ory5 5

33. 18s35 50s

18s32 50s5 0

2s(9s22 25)5 0

2s(3s2 5)(3s1 5)5 0

s5 0,s55

}

3, ors5 2

5

}

3

34. g31 3g22g2 35 0

g2(g1 3)2 (g1 3)5 0

(g1 3)(g22 1)5 0

(g1 3)(g1 1)(g2 1)5 0

g5 23,g5 21, org5 1

35. m31 6m22 4m2 245 0

m2(m1 6)2 4(m1 6)5 0

(m1 6)(m22 4)5 0

(m1 6)(m1 2)(m2 2)5 0

m5 26, m5 22, or m5 2

36. 4w41 40w22 445 0

4(w41 10w22 11)5 0

4(w21 11)(w22 1)5 0

4(w21 11)(w1 1)(w2 1)5 0

w5 21 or w5 1

37. 4z55 84z3

4z52 84z35 0

4z3(z22 21)5 0 z5 0,z5

}

21 , orz5 2}

21

38. 5b31 15b21 12b5 236

5b31 15b21 12b1 365 0

5b2(b1 3)1 12(b1 3)5 0

(b1 3)(5b21 12)5 0

b5 23

39. x62 4x42 9x21 365 0

x4(x22 4)2 9(x22 4)5 0

(x22 4)(x42 9)5 0

(x2 2)(x1 2)(x22 3)(x21 3)5 0

x5 2,x5 22,x5 }

3 , orx5 2}

3

40. 48p55 27p3

48p52 27p35 0

3p3(16p22 9)5 0

(3p3)(4p1 3)(4p2 3)5 0

p5 0,p5 23

}

4

, orp53

}

4

41. C;

3x42 27x21 9x5x3

3x42x32 27x21 9x5 0

x(3x32 27x2x21 9)5 0

x(3x(x22 9)2 (x22 9))5 0 x(x22 9)(3x2 1)5 0

x(x2 3)(x1 3)(3x2 1)5 0

x5 0,x5 3,x5 23, orx51

}

3

42. 16x32 44x22 42x5 2x(8x22 22x2 21)

5 2x(4x1 3)(2x2 7)

43. n42 4n22 605 (n21 6)(n22 10)



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Algebra 2


44. 24b42 500b5 24b(b31 125)

5 24b(b1 5)(b22 5b1 25)

45. 36a32 15a21 84a2 355 36a31 84a2 15a22 35

5 12a(3a21 7)2 5(3a21 7)

5 (3a21 7)(12a2 5)

46. 18c41 57c32 10c25 c2(18c21 57c2 10)

5 c2(6c2 1)(3c1 10)

47. 2d42 13d 22 455 (2d 21 5)(d 22 9)

5 (2d 21 5)(d1 3)(d2 3)

48. 32x52 108x25 4x2(8x32 27)

5 4x2(2x2 3)(4x21 6x1 9)

49. 8y62 38y42 10y25 2y2(4y42 19y22 5)

5 2y2(4y21 1)(y22 5)

50.z52 3z42 16z1 485z4(z2 3)2 16(z2 3)

5 (z2 3)(z42 16)

5 (z2 3)(z21 4)(z22 4)

5 (z2 3)(z21 4)(z2 2)(z1 2)

51.A5 lpw l5 3x1 2, w5x1 4

485 (3x1 2)(x1 4)

485 3x21 14x1 8

05 3x21 14x2 40

05 (3x1 20)(x2 2)

x5 220

}

3 orx5 2, but a side length cannot be negative

sox5 2.

52. V5 lpwph l5 2x, w5x2 1, h5x2 4

405 (2x)(x2 1)(x2 4)

405 (2x22 2x)(x2 4)

405 2x32 10x21 8x

05 2x32 10x21 8x2 40

05 2x2(x2 5)1 8(x2 5)

05 (x2 5)(2x21 8)

The only real solution isx5 5.

53. V51

}

3r2h r5 2x2 5, h5 3x

12551

}

3(2x2 5)2(3x)

3755 (4x22 20x1 25)(3x)

3755 12x32 60x21 75x

05

12x32

60x21

75x2

375 05 4x32 20x21 25x2 125

05 4x2(x2 5)1 25(x2 5)

05 (x2 5)(4x21 25)

The only real solution isx5 5.

54.x3y62 275 (xy2)32 33

5 (xy22 3)(x2y41 3xy21 9)

55. 7ac21 bc22 7ad22 bd25 c2(7a1 b)2d2(7a1 b)

5 (7a1 b)(c22 d2)

5 (7a1 b)(c2 d )(c1 d)

56.x2n2 2xn1 15 (xn2 1)(xn2 1)

5 (xn2 1)2

57. a5b22 a2b41 2a4b2 2ab31 a32 b2

5 a2b2(a32 b2)1 2ab(a32 b2)1 (a32 b2) 5 (a32 b2)(a2b21 2ab1 1)

5 (a32 b2)(ab1 1)(ab1 1)

5 (a32 b2)(ab1 1)2

Problem Solving

58. V5 lpwph l5 12x2 15, w5 12x2 21, h5x

9455 (12x2 15)(12x2 21)(x)

9455 (144x22 432x1 315)x

05 144x32 432x21 315x2 945

05 16x32 48x21 35x2 105

05

16x2

(x2

3)1

35(x2

3) 05 (x2 3)(16x21 35)

The only real solution isx5 3. The block is

3 meters high.

59. V5 lpwph l5x2 2, w5 3x2 2, h5 6x2 2

1125 (x2 2)(3x2 2)(6x2 2)

1125 (3x22 8x1 4)(6x2 2)

1125 18x32 48x21 24x2 6x21 16x2 8

1125 18x32 54x21 40x2 8

05 18x32 54x21 40x2 120

05 9x32 27x21 20x2 60

05 9x2(x2 3)1 20(x2 3)

05 (9x21 20)(x2 3)

x5 3 cm

3x5 9 cm

6x5 18 cm

The outer dimensions of the mold should be

3 cm3 9 cm3 18 cm.

60. a. V5 lpwph

Volume of platform 15 (8x)(6x)(x) 5 48x3

Volume of platform 25 (6x)(4x)(x) 5 24x3

Volume of platform 35 (4x)(2x)(x)5 8x3

b. 12505 48x31 24x31 8x3

12505 80x3

c. 12505 80x3

15.6255x3

2.5 ft5x

Platform 1: 20 ft3 15 ft3 2.5 ft





21/73

CopyrightbyM


Company.

Algebra 2


61. l5x

x

x25

x25

h5x2 5

w5x2 5

V5 lpwph

2505 (x)(x2 5)(x2 5)

2505x(x22 10x1 25)

2505x32 10x21 25x

05x32 10x21 25x2 250

05x2(x2 10)1 25(x2 10)

05 (x21 25)(x2 10)

x5 10

x2 55 5

The dimensions of the prism should be

10 in.3 5 in.3 5 in.

62. Volume of steel 5 volume of outside 2 volume of inside

6.425 (x)(x2 2)(x1 8)2 89.58

6.425x(x21 6x2 16)2 89.58

6.425x31 6x22 16x2 89.58

05x31 6x22 16x2 96

05x2(x1 6)2 16(x1 6)

05 (x1 6)(x22 16)

05 (x1 6)(x2 4)(x1 4)

x5 4 feet

The outer dimensions of the tank should be

4 ft3 2 ft3 12 ft.

63. V5 lpwph l5x2 2, w5 32 2x, h5 3x1 4

7

}

35 (x2 2)(32 2x)(3x1 4)

7

}3

5 (x2 2)(9x1 122 6x22 8x)

7

}

35 9x21 12x2 6x32 8x22 18x2 241 12x21 16x

7

}

35 26x31 13x21 10x2 24

75 218x31 39x21 30x2 72

05 218x31 39x21 30x2 79

From the graph, the only real real solution is 21.37. This

result would yield a negative dimension of the platform.

So, the volume cannot be7

}

3

cubic feet.

x

y

10

1(2

1.37, 0)

64. a.y y31y2

1 11 15 2

2 81 45 12

3 271 95 36

4 641 165 80

5 1251 255 150

6 2161 365 252

7 3431 495 392

8 5121 645 576

9 7291 815 810

10 10001 1005 1100

b.x31 2x25 96

x3

}8

1x2

}4

5 12 1x}2231 1x}22

25 12

y5x

}

2 25

x

}

2

x5 4

c.3x31 2x25 512

27x3

}

8 1

9x2

}

4 5 576 13x}22

31 13x}22

25 576

y53x

}

2 85

3x

}

2

x516

}

3

d. ax41 bx35 c Original equation

a4x4

}

b4 1

a3x3

}

b3 5

a3c

}

b4 Multiply each side by

a3

}

b4.

1ax}b241 1ax}b2

35

a3c

}

b4 Rewrite equation.

Then, create a table with values fory41y3.

65. a. The three solids form a cube with a smaller cube

taken out of the corner. If the solids were connected,

V5 a32 b3because the side of the large cube5

aand the side of the cube taken out5 b.

b. I: V5 apap(a2 b)5 a2(a2 b)

II: V5 ap(b)(a2 b) 5 ab(a2 b)

III: V5 bpb(a2 b)5 b2(a2 b)

c. V5 I1 II1 II

5 a2(a2 b)1 ab(a2 b)1 b2(a2 b)

5 (a2 b)(a21 ab1 b2)

5 a32 b3

Mixed Review

66.

1

x

y

21

67.

1

x

y

21



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308

Algebra 2


68.

2

x

y

21

69.1

x

y

21

70.

1

x

y

21

71.1

x

y

21

72.

1

x

y

21

73.

1

x

y

21

74.1

x

y

21

75.

1

x

y

21

76. 2 5 23 4 21 10

10 14 36 70

5 7 18 35 80

f (2)5 80

77. 23 23 0 1 26 2 4

9 227 78 2216 642

23 9 226 72 2214 646

f (23)5 646

78. 22 5 0 24 12 0 20

210 20 232 40 280

5 210 16 220 40 260

f (22)5 260

79. 4 26 0 0 9 215

224 296 2384 21500

26 224 296 2375 21515

f (4)5 21515

Problem Solving Workshop 5.4 (p. 361)

1.y5x31 4x22 8x

x y

1 23

2 8

3 39

4 96

x5 4

2.y5x32 9x22 14x1 7

y5 233

Intersections : (22.56, 233)

(1.56, 233)

(10,233)

x5 22.56, 1.56, or 10

3.y5 2x32 11x21 3x1 5

x y

1 21

2 217

3 231

4 231

5 25

6 59

x5 6

4.y5x41x32 15x22 8x1 6

y5 245 Intersections: (23.43, 245)

(22.53, 245)

(1.96, 245)

(3, 245)

x5 23.43, 22.53, 1.96, or 3

5.y5 2x41 2x3 1 6x21 17x2 4

y5 32

Intersections: (1.45, 32)

(4, 32)

x5 1.45 or 4



23/73

CopyrightbyM


Company.

Algebra 2


6. y5 23x41 4x31 8x21 4x2 11

y5 13

Intersections: (1.64, 13)

(2, 13)

x5 1.64 or 2

7. y5 4x42 16x31 29x22 95x

x y

0 0

0.5 242

1 278

1.5 2111

2 2138

2.52

150

x5 2.5

8. y5 (2x2 2)(x2 2)(x2 1)

x y

2 0

3 8

4 36

5 96

6 200

The volume is 200 cubic

feet whenx5 6.

Length5 2x5 12 ft

Width5x5 6 ft

Height5x5 6 ft

Intersection:x5 6,y5 200

Length5 2x5 12 ft

Width5x5 6 ft

Height5x5 6 ft

9. x5 height

x2 45 width

x1 65 length

V5 lpwph

y5 (x16)(x2 4)(x)

x y

6 144

7 273

8 448

9 675

10 960

11 1309

12 1728

Intersection at:x5 12,

y5 1728

x5 12

height5x5 12 ft

width5x2 45 8 ft

length5x1 65 18 ft

10. y5 0.0000984x42 0.00712x31 0.162x22 1.11x1 12.3

x y

1 11.35

2 10.67

3 10.24

4 10.02

5 9.97

x5 5 Intersection:x5 5,y5 9.97

In 1975, pineapple consumption was about 9.97 pounds

per person.

Lesson 5.5


1. 2x22 3x1 8

x21 2x2 1qww2x41x31 0x21x2 1

2x41

4x32

2x2

23x31 2x21x

23x32 6x21 3x

8x22 2x2 1

8x21 16x2 8

218x1 7

2x41x31x2 1

}}

x21 2x2 1

5 2x22 3x1 81218x1 7

}

x21 2x2 1

2. x22 3x1 10

x1 2qwwx32x21 4x2 10

x31 2x2

2

3x

21

4x 23x22 6x

10x2 10

10x1 20

230

x32x21 4x2 10

}}

x1 2

5x22 3x1 10230

}

x1 2

3. 23 1 4 21 21

23 23 12

1 1 24 11

x31 4x22x2 1

}}

x1 3

5x2

1x2 4111

}

x1 3

4. 1 4 1 23 7

4 5 2

4 5 2 9

4x31x22 3x1 7

}}

x2 1 5 4x21 5x1 21

9

}

x2 1



24/73

CopyrightbyM


Company.

310

Algebra 2


5. 4 1 26 5 12

4 28 212

1 22 23 0

f (x)5x32 6x21 5x1 12

5 (x2 4)(x22 2x2 3)

5 (x2 4)(x2 3)(x1 1)

6. 4 1 21 222 40

4 12 240

1 3 210 0

f (x)5x32x22 22x1 40

5 (x2 4)(x21 3x2 10)

5 (x2 4)(x1 5)(x2 2)

7. 22 1 2 29 218

22 0 18

1 0 29 0

f (x)5x31 2x22 9x2 18

5 (x1 2)(x22 9)

5 (x 1 2)(x2 3)(x1 3)

The other zeros are 3 and 23.

8. 22 1 8 5 214

22 212 14

1 6 27 0

f (x)5x3

1 8x2

1 5x2 14 5 (x1 2)(x21 6x2 7)

5 (x1 2)(x1 7)(x2 1)


9. 255 215x31 40x

05 215x31 40x2 25

1 215 0 40 225

215 215 25

215 215 25 0

(x2 1)(215x22 15x1 25)5 0

x 0.9 is the other positive solution. The companycould still make the same profit producing about

900,000 shoes.


Skill Practice

1. If a polynomialf (x) is divided byx2 k,then the

remainder is r5f (k).

2. The red numbers represent the coefficients of the

quotient and the blue number represents the remainder.

3. x1 5

x2 4qwwx21x2 17

x22 4x

5x2 17

5x2 20 3

x21x2 17

}

x2 4 5x1 51

3

}

x2 4

4. 3x1 4

x2 5qww3x22 11x2 26

3x22 15x

4x2 26

4x2 20

26

3x22 11x2 26

}}

x2 5 5 3x1 42

6

}

x2 5

5. x21 4x1 7

x2 1qww

x31 3x21 3x1 2

x32x2

4x21 3x

4x22 4x

7x1 2

7x2 7

9

x31 3x21 3x1 2

}}

x2 1 5x21 4x1 71

9

}

x2 1

6. 2x1 9

4x2 1qww

8x21 34x2 1

8x22 2x

36x2 1

36x2 9

8

8x21 34x2 1

}}

4x2 1 5 2x1 91

8

}

4x2 1

7. 3x1 8

x21xqww3x31 11x21 4x1 1

3x31 3x2

8x21 4x

8x21 8x

24x1 1

3x31 11x21 4x1 1

}}

x21x 5 3x1 81

24x1 1

}

x21x



25/73

CopyrightbyM


Company.

Algebra 2


8. 7x111

x21 1qww7x31 11x21 7x1 5

7x3 1 7x

11x21 0x1 5

11x21 0x1 11

26

7x31 11x21 7x1 5

}}

x21 1 5 7x1 112 6

}

x21 1

9. 5x22 12x1 37

x21 2x2 4qwww5x42 2x32 7x21 0x2 39

5x41 10x32 20x2

212x31 13x21 0x

212x22 24x22 48x

37x21 48x2 39

37x21 74x2 148

226x1 109

5x42 2x32 7x22 39

}}

x

21

2x2

4

5 5x22 12x1 371226x1 109

}

x21 2x2 4

10. 4x21 12x1 44

x22 3x2 2qwww

4x41 0x31 0x21 5x2 4

4x42 12x32 8x2

12x31 8x21 5x

12x32 36x22 24x

44x21 29x2 4

44x22 132x2 88

161x1 84

4x41 5x2 4

}

x22 3x2 2 5 4x21 12x1 441

161x1 84

}

x22 3x2 2

11. 5 2 27 10

10 15

2 3 25

2x22 7x1 10

}}

x2 5 5 2x1 31

25

}

x2 5

12. 2 4 213 25

8 210

4 25 215

4x21 13x2 5

}}

x2 2 5 4x2 52

15

}

x2 2

13. 24 1 8 1

24 216

1 4 215

x21 8x1 1

}

x1 4 5x1 42

15

}

x1 4

14. 3 1 0 9

3 9

1 3 18

x21 9

}

x2 3 5x1 31

18

}

x2 3

15. 4 1 25 0 22

4 24 216

1 21 24 218

x32 5x22 2

}

x2 4

5x22x2 4218

}

x2 4

16. 23 1 0 24 6

23 9 215

1 23 5 29

x32 4x1 6

}

x1 3

5x22 3x1 52 9

}

x1 3

17. 6 1 25 28 13 212

6 6 212 6

1 1 22 1 26

x42 5x32 8x21 13x2 12

}}}

x2 6 5x31x22 2x1 12

6

}

x2 6

18. 25 1 4 0 16 235

25 5 225 45

1 21 5 29 10

x41 4x31 16x2 35

}}

x1 5 5x32x21 5x2 91

10

}

x1 5

19. The division was done correctly, however, there is an

error in how the quotient is written.

2 1 0 25 3

2 4 22

1 2 21 1

x32 5x1 3

}

x2 2 5x21 2x2 11

1

}

x2 2

20. The error is that a zero was not used as a place holder for

the missingx2term.

2 1 0 25 3

2 4 22

1 2 21 1

x32 5x1 3

}

x2 2 5x21 2x2 11

1

}

x2 2



26/73

CopyrightbyM


Company.

312

Algebra 2


21. 6 1 210 19 30

6 224 230

1 24 25 0

f (x)5x32 10x22 19x1 30

5 (x2 6)(x22 4x2 5)

5 (x2 6)(x2 5)(x1 1)

22. 24 1 6 5 212

24 28 12

1 2 23 0

f (x)5x31 6x21 5x2 12

5 (x1 4)(x21 2x2 3)

5 (x1 4)(x1 3)(x2 1)

23. 8 1 22 240 264

8 48 64

1 6 8 0

f(x)5x32 2x22 40x2 64

5 (x2 8)(x21 6x1 8)

5 (x2 8)(x1 4)(x1 2)

24. 210 1 18 95 150

210 280 2150

1 8 15 0

f (x)5x31 18x21 95x1 150

5 (x1 10)(x21 8x1 15)

5 (x1 10)(x1 3)(x1 5)

25. 29 1 2 251 108

29 63 2108

1 27 12 0

f (x)5x31 2x22 51x1 108

5 (x1 9)(x22 7x1 12)

5 (x1 9)(x2 4)(x2 3)

26. 22 1 29 8 60

22 22 260

1 211 30 0 f (x)5x32 9x21 8x1 60

5 (x1 2)(x22 11x1 30)

5 (x1 2)(x2 5)(x2 6)

27. 1 2 215 34 221

2 213 21

2 213 21 0

f (x)5 2x32 15x21 34x2 21

5 (x2 1)(2x22 13x1 21)

5 (x2 1)(2x2 7)(x2 3)

28. 5 3 22 261 220

15 65 20

3 13 4 0

f(x)5 3x32 2x22 61x2 20

5 (x2 5)(3x21 13x1 4)

5 (x2 5)(3x1 1)(x1 4)

29. 23 1 22 221 218

23 15 18

1 25 26 0

f (x)5x32 2x22 21x2 18

5 (x1 3)(x22 5x2 6)

5 (x1 3)(x2 6)(x1 1)


30. 10 4 225 2154 40

40 150 240

4 15 24 0

f (x)5 4x32 25x22 154x1 40

5 (x2 10)(4x21 15x2 4)

5 (x2 10)(4x2 1)(x1 4)

The other zeros are1

}

4

and 24.

31. 7 10 281 71 42

70 277 242

10 211 26 0

f (x)5 10x32 81x21 71x1 42

5 (x2 7)(10x22 11x2 6)

5 (x2 7)(5x1 2)(2x2 3)


2

}

5

and

3

}

2

.



27/73

CopyrightbyM


Company.

Algebra 2


32. 24 3 34 72 264

212 288 64

3 22 216 0

f (x)5 3x31 34x21 72x2 64

5 (x1 4)(3x21 22x2 16)

5 (x1 4)(3x2 2)(x1 8)


}

3

and 28.

33. 9 2 210 271 29

18 72 9

2 8 1 0

f (x)5 2x32 10x22 71x2 9

5 (x2 9)(2x21 8x1 1)

x528 6 }

642 4(2)}}

2(2) 5

28x 6 }

56}

4

524 6

}

14}2

The other zeros are about 20.13 and 23.87.

34. 22 5 21 218 8

210 22 28

5 211 4 0

f (x)5 5x32x22 18x1 8

5 (x1 2)(5x22 11x1 4)

x5

11 6 }}

1212 4(5)(4)

}}

10

5

116 }

41

}

10

The other zeros are about 0.46 and 1.74.

35. D;

26 4 15 263 254

224 54 54

4 29 29 0

f (x)5 4x31 15x22 63x2 54

5 (x1 6)(4x22 9x2 9)

5 (x1 6)(4x1 3)(x2 3)

The zeros are 26, 23

}4

, and 3.

36. (x1 4)(x1 2)5x21 6x1 8

l52x31 17x21 46x1 40

}}

x21 6x1 8

2x1 5

x21 6x1 8qww2x31 17x21 46x1 40

2x31 12x21 16x

5x21 30x1 40

5x21 30x1 40

0

The missing dimension is 2x1 5.

37. (x2 1)(x1 6)5x21 5x2 6

w5x31 13x21 34x2 48

}}

x21 5x2 6

x1 8

x21 5x2 6qwwx31 13x21 34x2 48

x31 5x22 6x

8x21 40x2 48

8x21 40x2 48 0

The missing dimension isx1 8.

38. a. 2 1 25 212 36

2 26 236

1 23 218 0

f (x)5x32 5x22 12x1 36

5 (x2 2)(x22 3x2 18)

5 (x2 2)(x2 6)(x1 3)


b. The factors are (x2 2), (x2 6), and (x1 3).

c. The solutions arex5 2,x5 6,x5 23.

39. A;

5 1 21 k 230

5 20 30

1 4 k1 20 0

Forx2 5 to be a factor, the remainder must be zero. In

the last column 30 must be added to 230 to get zero.

k1 20must equal 6 in order to have a product of 30 to

add to the 230. k1 205 6 so, k5 214.



28/73

CopyrightbyM


Company.

314

Algebra 2


40. a.1}

2

b.1

}

2

30 7 239 14

15 11 214

30 22 228 0

f(x)5 1x2 1}22(30x21 22x2 28)

c.f(x)51x2 1}22(2)(15x21 11x2 14) 5 (2x2 1)(3x2 2)(5x1 7)

Problem Solving

41. P5 2x31 4x21x

45 2x31 4x21x

05 2x31 4x21x2 4

4 21 4 1 24

24 0 4

21 0 1 0

P5 (x2 4)(2x21 1)

5 (x2 4)(12x)(11x)

x5 1 is the only other positive solution. So, the company

could still make the same profit by producing only

1 million T-shirts.

42. P5 24x31 12x21 16x

485 24x31 12x21 16x

05 24x31 12x21 16x2 48

3 24 12 16 248

212 0 48

24 0 16 0

P5 (x2 3)(24x21 16)5 (x2 3)(42 2x)(41 2x)

x5 2 is the only other positive solution. The company

could produce 2 million MP3 players and still make the

same profit.

43. Let a5 average attendance per team.

a5A

}

T5

21.95x31 70.1x22 188x1 2150

}}}

14.8x1 725

20.132x21 11.2x2 561.35

14.8x1 725qwww21.95x31 70.1x22 188x1 2150

21.95x32 95.7x2

165.8x22 188x

165.8x21 8120x

28308x1 2150

28308x2 406,978.75

409,128.75

A function is

a5 20.132x21 11.2x2 5611409,129

}

14.8x1 725

44. a. Total revenue

5 (number of radios sold)(price per radio)

5 (x)(402 4x2)5 40x2 4x3

An expression is 40x2 4x3.

b.P5 (40x2 4x3)2 15x5 25x2 4x3

A function isP5 25x2 4x3.

c. WhenP5 24:

245 25x2 4x3

05 4x32 25x1 24

You know that 1.5 is a solution, sox2 1.5 is a factor

of 4x32 25x1 24.

1.5 4 0 225 24

6 9 224

4 6 216 0

So, (x2 1.5)(4x21 6x2 16)5 0. Use the quadraticformula to findx1.39, the other positive solution.

So, about 1.39 million radios also make a profit of$24,000,000.

d.No. The solutionx22.88 does not make sense

because you cannot produce a negative number of

radios.

45. LetP5 the precent of visits to the national park that

were overnight stays.

P5S

}

V5

20.00722x41 0.176x32 1.40x21 3.39x1 17.6

}}}}

3.10x1 256

20.

00233x

310.

249x

222

1x11

73

5.

287

3.

10x12

56qw

wwww

20.0

0

722x

410.

176x

321.

40x

213.3

9x

11

7.6

20.0

0722x

420.

596x

3

0.

772x

321.

40x

2

0.

772x

316

3.

7x

2

265.

1x

213.

39x

265.

1x

225

376x

5379.

39x11

7.

6

5379.

39x14

4,

233.

472

2

44215.

872

So,

P5 20.00233x31 0.249x22 21x1 17352444,216

}

3.10x1 256

.

The function for the percent of visits that were overnight

stays is the overnight stays divided by the total visits.



29/73

CopyrightbyM


Company.

Algebra 2


46. P5 26x31 72x

965 26x31 72x

05 26x31 72x2 96

2 26 0 72 296

212 224 96

26 212 48 0

P5 (26x22 12x1 48)(x2 2)

5 (23x1 6)(2x1 8)(x2 2)

5 23(x2 2)(2)(x1 4)(x2 2)

5 26(x2 2)(x1 4)(x2 2)

The zeros are at 2 and 24. Production levels cannot be

negative sox5 2 is the only solution.

Mixed Review

47. x2 4y< 5 x2 4y< 5

12 4(4) < 5 4 2 4(21) < 5

12

16 < 5 41

4 < 5 215 < 5 yes 8 < 5 no

(1, 4) is a solution.

48. 3x1 2y 1 3x1 2y 1

3(22)1 2(4) 1 3(1)1 2(23) 1

261 8 1 32 6 1

2 1 yes 23 1 no


49. 5x2 2y> 10 5x2 2y> 10

5(4)2 2(6) > 10 5(8)2 2(10) > 10

202 12 > 10 402 20 > 10

8 > 10 no 20 > 10 yes (8, 10) is a solution.

50. 6x1 5y 15 6x1 5y 15

6(25)1 5(10) 15 6(21)1 5(4) 15

2301 50 15 261 20 15

20 15 no 14 15 yes


51. x21 3x2 405 0

(x2 5)(x1 8)5 0

x2 55 0 or x1 85 0

x5 5 or x5 28

52. 5x2

1 13x1 65 0 (5x1 3)(x1 2)5 0

5x1 35 0 or x1 25 0

5x5 23

x5 23

}

5

or x5 22

53.x21 7x1 25 0

x527 6 }}

(7)22 4(1)(2)}}

2(1)

527 6

}

41}

2

20.30 and 26.70

54. 4x21 15x1 105 0

x5215 6 }}

(15)22 4(4)(10)}}

2(4) 5

215 6 }

65}

8

22.88 and 20.87

55. 2x21 15x1 315 0

x5215 6 }}

(15)22 4(2)(31)}}

2(2) 5

215 6 }

223}}

4 5

215 6 i}

23}

4

The solutions are215 1 i

}

23}

4 and

215 2 i}

23}

4 .

56.x21 2x1 105 0

x522 6

}}

(2)22 4(1)(10)

}}

2(1)

522 6

}

236

}

2

5 21 63i

The solutions are 21 1 3iand 21 2 3i.

57. (x22 4x1 15)1 (23x21 6x2 12)

5x22 4x1 152 3x21 6x2 12

5 22x21 2x1 3

58. (2x22 5x1 8)2 (5x22 7x2 7)

5 2x22 5x1 82 5x21 7x1 7

5 23x21 2x1 15

59. (3x2 4)(3x31 2x22 8)

5 3x(3x31 2x22 8)2 4(3x31 2x22 8)

5 9x41 6x

32 24x2 12x

32 8x

21 32

5 9x42 6x32 8x22 24x1 32

60. (3x2 5)35 (3x)32 3(3x)2(5)1 3(3x)(5)22 (5)3

5 27x32 135x21 225x2 125

Mixed Review of Problem Solving (p. 369)

1. a. 164,000,000,0005 1.643 1011yards

b.1.643 1011

}

1.203 102

51.64

}

1.20

31011

}

102

5 1.3673 109football fields

2. a. T(x)5 lpwph

5 (4x)(x)(2x)

5 8x3

b. C(x)5 (4x2 2)(x2 2)(2x2 4)

5 (4x22 10x1 4)(2x2 4)

5 8x32 36x21 48x2 16

c.I(x)5 T(x)2 C(x)



30/73

CopyrightbyM


Company.

316

Algebra 2


d.I(x)5 8x32 (8x32 36x21 48x2 16)

5 36x22 48x1 16

I(8)5 36(8)22 48(8)1 16

5 23042 3841 16

5 1936

The volume of the insulation is 1936 cubic inches.

3. Cube: V5x3A}V

56x2

}

x3 5

6

}x

A5 6x2

Sphere: V54

}

3

1x}223

A}V

5x2

}}

6x3

56

}x

5}

6

x3

A5 41x}222

5 x2

The cells exchange at the same rate.

4. f(x)5 2x41 2x21 1

x

y

211

5. a.Cis a degree 4 (quartic).

b.t, year 0 1 2 3 4

C, cost 51 46.14 42.33 40.50 40.97

t, year 5 6 7 8

C, cost 43.38 46.73 49.38 49.05

c.

1 2 3 4 5 6 7 8 90

20

10

0

40

60

50

30

Years since 1995

Cost(dollars)

t

C

No, the model will not accurately predict cell phone

bills beyond 2003. The model was not intended to be

used past 2003. From the graph you can see that it drops

sharply past 2003.

6. a.R5x(1002 10x)5 100x2 10x3

A function for the total revenue isR(x)5 100x2 10x3.

b.P5 (100x2 10x3)2 30x5 70x2 10x3

A function for the profit isP(x)5 70x2 10x3.

c. WhenP5 60:

605 70x2 10x3

05 10x32 70x1 60

You know that 2 is a solution, sox2 2 is a factor of

10x32 70x1 60.

2 10 0 270 60

20 40 260

10 20 230 0

So, (x2 2)(10x21 20x2 30)5 0. Use factoring tofind thatx5 1 andx5 23 are the other solutions.

d.No. The solutionx5 23 does not make sense because

you cannot produce a negative number of cameras.

7. V51

}

3

h(s2)

4851

}

3(x)(3x2 6)2

485x

}

3

(9x22 36x1 36)

485 3x32 12x21 12x

05 3x32 12x21 12x2 48

05 3x2(x2 4)1 12(x2 4)

05 (3x21 12)(x2 4)

x5 4

The height is 4 feet.

Lesson 5.6


1. Factors of the constant term: 61, 63, 65, 615

Factors of the leading coefficient: 61

Possible rational zeros: 61

}

1, 6

3

}

1, 6

5

}

1, 6

15

}

1

Simplified list: 61, 63, 65, 615

2. Factors of the constant term: 61, 62, 63, 66

Factors of the leading coefficient: 61, 62


}

1

, 62

}

1

, 63

}

1

, 66

}

1

, 61

}

2

,

62

}

2

, 63

}

2

, 66

}

2

Simplified list: 61, 62, 63, 66, 61

}

2

, 63

}

2

3. Possible rational zeros: 61

}

1

, 62

}

1

, 63

}

1

, 66

}

1

, 69

}

1

, 618

}

1

Testx5 1:

1 1 24 215 18

1 23 218

1 23 218 0

Because 1 is a zero off, f(x) can be writen as:

f(x)5 (x2 1)(x22 3x2 18)5 (x2 1)(x2 6)(x1 3)

The zeros are 23, 1 and 6.



31/73

CopyrightbyM


Company.

Algebra 2


4. Possible rational zeros: 61, 62, 67, 614

Testx5 1:

1 1 28 5 14

1 27 22

1 27 22 212

1 is not a zero. Testx5 21:

21 1 28 5 14

21 9 214

1 29 14 0

Because 21 is a zero off,f(x) can be written as:

f(x)5 (x1 1)(x22 9x1 14)5 (x1 1)(x2 7)(x2 2)

The zeros offare 21, 2 and 7.

5. Possible rational zeros: 61, 63, 61

}

2, 6

3

}

2, 6

1

}

3, 6

1

}

4,

63

}4

, 61

}6

, 61

}8

, 63

}8

, 61

}12

, 61

}16

, 63

}16

, 61

}24

, 61

}48

Reasonable values:x 5 23

}

4,x5

3

}

16,x5

1

}

2

Check:x5 23

}

4:

23

}

4 48 4 220 3

236 24 23

48 232 4 0

23

}

4

is a zero.

f(x)5 1x1 3}42(48x22 32x1 4)

5 1x1 3}42(4)(12x22 8x1 1) 5 (4x1 3)(6x2 1)(2x2 1)

The real zeros offare 23

}

4

,1

}

6

, and1

}

2

.

6. Possible rational zeros: 61, 62, 63, 66, 67, 614,

621, 642, 61

}

2

, 63

}

2

, 67

}

2

, 621

}

2

Reasonable zeros:x5 27

}

2,x5 22,x5

3

}

2,x5 2

27

}

2

2 5 218 219 42

27 777

}

2

2273

}

4

2 22 21139

}

2

2105

}

4

22 2 5 218 219 42

24 22 40 242

2 1 220 21 0

f(x)5 (x1 2)(2x31x22 20x1 21)5 (x1 2) pg(x)

Possible rational zeros ofg(x): 61, 63, 67, 621, 61

}

2

,

63

}

2

, 67

}

2

, 621

}

2

Reasonable zeros:x53

}

2

,x5 2

3

}

2

2 1 220 21

3 6 221

2 4 214 0

f(x)5 (x1 2) pg (x)

5 (x1 2)1x2 3}22(2x21 4x2 14) 2x21 4x2 145 0

x524 6 }}

(4)22 4(2)(214)}}

2(2) 5

24 6 }

128}

4

5 21 6 2}

2

The real zeros offare: 22,3

}

2

, 211 2

}

2 , and 212 2

}

2 .

7. V 51

}

3

(x2)(x2 1)

651

}

3

x2(x2 1)

185x32x2

05x32x2218

Possible rational zeros: 61, 62, 63, 66, 69, 618

2 1 21 0 218

2 2 4

1 1 2 214

3 1 21 0 218

3 6 18

1 2 6 0

The other solutions, which satisfyx21 2x1 6, are

x5 21 6 i}

5 and can be discarded because they are

imaginary. The only real solution isx5 3. The base is

3 33 feet. The height isx2 1 or 2 feet.



32/73

CopyrightbyM


Company.

318

Algebra 2



Skill Practice

1. If a polynomial function has integer coefficients, then

every rational zero of the function has the formp

}qwhere

pis a factor of the constant termand qis a factor of the

leading coefficient.

2. To shorten the list, draw the graph to see approximatelywhere the zeros are.

3. f(x)5x32 3x1 28

Factors of constant term: 61, 62, 64, 67, 614, 628

Factors of leading coefficient: 61


}

1, 6

2

}

1, 6

4

}

1, 6

7

}

1, 6

14

}

1, 6

28

}

1

Simplified list: 61, 62, 64, 67, 614, 628

4.g (x)5x32 4x21x2 10

Factors of constant term: 61, 62, 65, 610

Factors of leading coefficient: 61


}

1

, 62

}

1

, 65

}

1

, 610

}1


5.f(x)5 2x41 6x32 7x1 9

Factors of constant term: 61, 63, 69

Factors of leading coefficient: 61, 62


}

1

, 63

}

1

, 69

}

1

, 61

}

2

, 63

}

2

, 69

}

2


}

2

, 63

}

2

, 69

}

2

6. h(x)5 2x31x22x2 18

Factors of constant term: 61, 62, 63, 66, 69, 618

Factors of leading coefficient: 61, 62


}

1

, 62

}

1

, 63

}

1

, 66

}

1

, 69

}

1

, 618

}

1

, 61

}

2

,

62

}

2

, 63

}

2

, 66

}

2

, 69

}

2

, 618

}

2

Simplified list: 61, 62, 63, 66, 69, 618, 61

}

2

, 63

}

2

, 69

}

2

7.g (x)5 4x51 3x32 2x2 14

Factors of the constant term: 61, 62, 67, 614

Factors of the leading coefficient: 61, 62, 64


}

1, 6

2

}

1, 6

7

}

1, 6

14

}

1, 6

1

}

2, 6

2

}

2, 6

7

}

2,

614

}

2, 6

1

}

4, 6

2

}

4, 6

7

}

4, 6

14

}

4

Simplified list: 61, 62, 67, 614, 61

}2

, 67

}2

, 61

}4

, 67

}4

8.f(x)5 3x41 5x32 3x1 42

Factors of the constant term: 61, 62, 63, 66, 67,

614, 621, 642

Factors of the leading coefficient: 61, 63


}

1

, 62

}

1

, 63

}

1

, 66

}

1

, 67

}

1

,

614

}1, 621

}1, 642

}1, 61}3, 62}3, 63}3, 66}3, 67}3, 614

}3

,

621

}

3

, 642

}

3

Simplified list: 61, 62, 63, 66, 67, 614,

621, 642, 61

}

3

, 62

}

3

, 67

}

3

, 614

}

3

9. h(x)5 8x41 4x32 10x1 15

Factors of the constant term: 61, 63, 65, 615

Factors of the leading coefficient: 61, 62, 64, 68


}

1, 6

3

}

1, 6

5

}

1, 6

15

}

1, 6

1

}

2,

6

3

}

2

,6

5

}

2

,6

15

}

2

,6

1

}

4

,6

3

}

4

,6

5

}

4

,6

15

}

4

,6

1

}

8

,6

3

}

8

,

65

}

8

, 615

}

8

Simplified list: 61, 63, 65, 615, 61

}

2

, 63

}

2

,

65

}

2

, 615

}

2

, 61

}

4

, 63

}

4

, 65

}

4

, 615

}

7

, 61

}

8

, 63

}

8

, 65

}

8

,

615

}

8

10. h(x)5 6x32 3x21 12

Factors of the constant term: 61, 62, 63, 64, 66, 612

Factors of the leading coefficient: 61, 62, 63, 66


}

1

, 62

}

1

, 63

}

1

, 64

}

1

, 66

}

1

,

612

}

1, 6

1

}

2, 6

2

}

2, 6

3

}

2, 6

4

}

2, 6

6

}

2, 6

12

}

2,

61

}

3

, 62

}

3

, 63

}

3

, 64

}

3

, 66

}

3

, 612

}

3

, 61

}

6

, 62

}

6

, 63

}

6

,

64

}

6, 6

6

}

6, 6

12

}

6

Simplified list: 61, 62, 63, 64, 66, 612,

61

}

2

, 63

}

2

, 61

}

3

, 62

}

3

, 64

}

3

, 61

}

6

11. Possible rational zeros: 61, 62, 63, 64, 66, 68, 612,

624

Testx5 1:

1 1 212 35 224

1 211 24

1 211 24 0

f(x)5 (x2 1)(x22 11x1 24)

5 (x2 1)(x2 3)(x2 8)

Real zeros are 1, 3, and 8.



33/73

CopyrightbyM


Company.

Algebra 2


12.Possible rational zeros: 61, 62, 64, 67, 68, 614, 628,

656

Testx5 1:

1 1 25 222 56

1 24 226

1 24 226 30 Testx5 21:

21 1 25 222

aat solutions - ch05

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