Chapter 5 Homework The Constant Term And The Middle Term

Chapter 5

Polynomials: Factoring

Exercise Set 5.1

75 = 3 ·5·5

4. 90 = 2 ·3·3·5

6. 12 = 2 ·2·3

8. x2=x2

10. 8x4=2·2·2·x4

12. 8x2=2·2·2·x2

14. 16p6q4=2·2·2·2·p6·q4

16. −x2=−1·x2

−6x=−1·2·3·x

18. −x9y6=−1·x9·y6

x3y3=x3·y3

20. x(x+5)

28. 4(2x2−x−5)

36. 4(2y3−5y2+3y−4)

52. 3x3+2x2+3x+2=x2(3x+2)+(3x+2)=

102 Chapter 5: Polynomials: Factoring

58. 7x3−14x2−x+2=7x2(x−2) −(x−2) =

64. y2+14y+49

72. y3−3y2+5y

78. x3−x2−2x+5=x2(x−1) −(2x−5)

Exercise Set 5.2

2. (x+ 2)(x+3)

12. (t+4)

2

22. x3−x2−42x=x(x2−x−42) = x(x+ 6)(x−7)

30. (x2−3)(x2+2)

38. 5w4−20w3−25w2=5w2(w2−4w−5)=5w2(w−5)(w+1)

44. 4x2+40x+ 100 = 4(x2+10x+25)=4(x+5)

2

56. −z2+36−9z=−z2−9z+36

=−1(z2+9z−36)

58. (d+ 6)(d+ 16)

62. (t−0.5)(t+0.2)

68. (x+3y)(x+8y)

Exercise Set 5.3 103

74. −2y+11y= 108

78. 10 −(x−7)=4x−(1+5x)

82. A=p+w

2

84. a−c+r=2

88. a2+ba −50

=1

Exercise Set 5.3

2. (3x−4)(x+1)

8. (7x+ 1)(x+2)

20. (7x+ 4)(5x+2)

(0, ⫺3)

104 Chapter 5: Polynomials: Factoring

32. 6x2+33x+ 15 = 3(2x2+11x+ 5) = 3(2x+ 1)(x+5)

38. (2t+ 3)(3t+2)

42. (5x−3)(3x−2)

48. 15x3+19x2−10x=x(15x2+19x−10) =

50. 33t−15 −6t2=−6t2+33t−15

52. 1+p−2p2=−2p2+p+1

54. 70x4+68x3+16x2=2x2(35x2+34x+8)=

58. (3x2+ 2)(3x2+4)

74. 30a+87ab +30b2= 3(10a2+29ab +10b2)=

78.

80.

82.

Exercise Set 5.4 105

Exercise Set 5.4

4. a2+5a−2a−10 = a(a+5)−2(a+5)=

10. 8x2−6x−28x+21=2x(4x−3) −7(4x−3) =

18. 3x2+x−4=3x2+4x−3x−4=

28. 6−13x+6x2=6−9x−4x+6x2=

32. 2x2−5x+2=2x2−4x−x+2=

36. 15x2−19x−10 = 15x2−25x+6x−10 =

5x(3x−5) + 2(3x−5) = (3x−5)(5x+2)

42. 17x−4x2+15=−4x2+17x+15=

−1(4x2−17x−15)=−1(4x2−20x+3x−15) =

46. 6x2+33x+ 15 = 3(2x2+11x+5)=

52. 6t2+t−15 = 6t2+10t−9t−15 =

2t(3t+5)−3(3t+5)=(3t+ 5)(2t−3)

60. 18x2+3x−10 = 18x2+15x−12x−10 =

64. 1−p−2p2=−2p2−p+1=−1(2p2+p−1) =

106 Chapter 5: Polynomials: Factoring

66. −15x2−19x−6=−1(15x2+19x+6)=

2x2(35x2+14x+20x+8)=

70. 144x5+ 168x4+48x3=24x3(6x2+7x+2)=

72. 9x4+18x2+8=9x4+12x2+6x2+8=

(3x−7)2

and whose sum is 3. Thus, 4x2+6x+ 3 cannot be factored

into a product of binomial factors. It is prime.

3p(p−6q)+2q(p−6q)=(p−6q)(3p+2q)

88. 30a2+87ab +30b2= 3(10a2+29ab +10b2)=

3(10a2+25ab +4ab +10b2)=

90. 15a2−5ab −20b2= 5(3a2−ab −4b2)=

92. 60x+4x2−8x3=−8x3+4x2+60x=

94. 15x3+33x4+6x5=6x5+33x4+15x3=

−14 =1

102. Solve: x+(2x−10)+(4x+ 15) = 180

106. (a+4)

2−2(a+4)+1=

Chapter 5 Mid-Chapter Review

5. 10y3−18y2+12y=2y·5y2−2y·9y+2y·6

2x2−x−6=2x2−4x+3x−6

7. x3=x3

Chapter 5 Mid-Chapter Review 107

8. 5x4=5·x4

9. 6x5=2·3·x5

10. −8x=−1·2·2·2·x

11. 15x3y2=3·5·x3·y2

Each coefficient has a factor of 5. The GCF of the powers

12. x2y4=x2·y4

−x3y3=−1·x3·y3

16. 3t6−5t4−2t3=t3·3t3−t3·5t−t3·2=t3(3t3−5t−2)

17. x2+4x+3

20. 8y5−48y3=8y3·y2−8y3·6=8y3(y2−6)

22. 6−11t+4t2,or4t2−11t+6

(4t+)(t+ ) or (2t+ )(2t+).

(4) Look for combinations of factors from steps (2) and

(3) such that the sum of their products is the middle

23. z2+4z−5

z2+4z−5=(z+ 5)(z−1)

24. 2z3+8z3+5z+20=2z2(z+4)+5(z+4)

29. 6y2+7y−10

We will use the ac-method.

30. 3x2−3x−18 = 3(x2−x−6)

Consider x2−x−6. Look for a pair of factors of −6 whose

31. 6x3+4x2+3x+2=2x2(3x+2)+(3x+2)

32. 15 −8w+w2,orw2−8w+15

34. 10z2−21z−10

(z+ )(10z+ ) or (2z+ )(5z+).

The factors can also be written as 10, −1 and 1,

(4) Split the middle term: 7x=3x+4x.

36. x2−10xy +24y2

37. 6z3+3z2+2z+1=3z2(2z+1)+(2z+1)

39. 4y2−7yz −15z2

We will use the FOIL method.

(4y+)(y+ ) or (2y+ )(2y+).

−z,15zand z,−15zand −3z,5zand 3z,−5z.

The factors can also be written as 15z,−zand

40. 3x3+21x2+30x=3x(x2+7x+ 10)

41. x3−3x2−2x+6=x2(x−3) −2(x−3)

43. y2+6y+8

We will use the FOIL method to factor 2y2+11y+ 15.

(3) The constant term and the middle term are both

positive, so we look for pairs of positive factors of

Chapter 5 Mid-Chapter Review 109

(4) Look for combinations of factors from steps (2) and

45. x3−7x2+4x−28 = x2(x−7)+4(x−7)

(y−4)(−y−1)

47. 16x2−16x−60 = 4(4x2−4x−15)

48. 10a2−11ab +3b2

We will use the FOIL method.

We can also write these factors as 3b,−band −3b,

b.

51. 4x2+11xy +6y2

(3) Look for a factorization of 24y2in which the sum

of the factors is 11y. The numbers we want are 3y

52. 6−5z−6z2=−6z2−5z+6=−1(6z2+5z−6)

We will use the FOIL method to factor 6z2+5z−6.

(1) There is no common factor (other than 1 or −1.)

(−2z−3)(3z−2),or

(2z+ 3)(−3z+2)

55. 9x2−6xy −8y2

We will use the ac-method.

110 Chapter 5: Polynomials: Factoring

56. −3+8z+3z2=3z2+8z−3

1.

57. m2−6mn −16n2

58. 2w2−12w+18=2(w2−6w+9)

59. 18t3−18t2+4t=2t(9t2−9t+2)

(4) Split the middle term: −9t=−6t−3t.

60. 5z3+15z2+z+3=5z2(z+3)+(z+3)

61. −14 + 5t+t2,ort2+5t−14

62. 4t2−20t+25

(3) Factor the last term, 25. The middle term is neg-

63. t2+4t−12

64. 12+5z−2z2=−1(2z2−5z−12)

(3) Factor the last term, −12. The possibilities are −1,

(4) Look for combinations of factors from steps (2) and

65. 12+4y−y2=−1(y2−4y−12)

(y−6)(−y−2)

b)(cx +d)=acx3+adx2+bcx +bd.

67. There is a finite number of pairs of numbers with the cor-

68. Since both constants are negative, the middle term will be

negative so (x−17)(x−18) cannot be a factorization of

Exercise Set 5.5 111

Exercise Set 5.5

14. (x+1)

2

26. 2x2−40x+ 200 = 2(x2−20x+ 100) = 2(x−10)2

36. (1 −a5)2

52. Yes

54. (q+ 1)(q−1)

64. (9 + w)(9 −w)

66. (5x+ 2)(5x−2)

82. 4x4−64=4(x4−16)

=4(x2+ 4)(x2−4)

88. x+1

To find the x-intercept, let y= 0 and solve for x.

96.

y

3x  5y  15

98.

100. The shaded region consists of a semicircle with radius x

and a rectangle with length yand width x+x,or2x.The

106. 2(81x2−41)

116. 1.28x2−2 = 2(0.64x2−1) = 2[(0.8x)2−12]=

120. a2n−49b2n=(an)2−(7bn)2=

2= 16.

Exercise Set 5.6

correct.

RC8. 1000x12 =10x4·10x4·10x4, so the cube root of 1000x12

10. p3−27 = (p−3)(p2+3p+9)

22. 3z3−3=3(z3−1) = 3(z−1)(z2+z+1)

30. y3+0.125 = (y+0.5)(y2−0.5y+0.25)

40. a9+64b9=(a3+4b3)(a6−4a3b3+16b6)

Exercise Set 5.7 113

52. The model shows a cube with volume a3from which a

54. 8

27x3+1

64y3=2

3x+1

4y4

9x2−1

6xy +1

16y2

60. y4−8y3−y+8

Exercise Set 5.7

10. x3−5x2−25x+ 125 = x2(x−5) −25(x−5) =

22. 45 −3x−6x2=−3(−15 + x+2x2)=

3(5 −2x)(3 + x).

24. x2+8x+5

30. x6−2x5+7x4=x4(x2−2x+7)

36. xy(x−y)

38. 10p4q4+35p3q3+10p2q2=5p2q2(2p2q2+7pq +2)

46. 2x2−4x+xy −2y=2x(x−2) + y(x−2) =

54. (0.1x2−0.5y2)2,or0.01(x2−5y2)2

114 Chapter 5: Polynomials: Factoring

70. 16p3+54q3= 2(8p3+27q3)

= 2(2p+3q)(4p2−6pq +9q2)

74. 81a4−b4=(9a2+b2)(9a2−b2)=

86. 3(−2) −2+|−4−(−1)|=3(−2) −2+|−3|

88. −1

3>−1

2

92. 1

5x2−x+4

5=1

5(x2−5x+4)= 1

5(x−4)(x−1)

94. x3+x2−(4x+4) = x3+x2−4x−4

Exercise Set 5.8

2. −2, 7

26. x2+7x+6=0

(x+ 6)(x+1)=0

30. x2−9x+14=0

32. x2−3x=0

x(x−3)=0

38. 4x2−9=0

Exercise Set 5.8 115

40. 0=25+x2+10x

x=0 or x=8

7

46. 3x2−7x=20

48. 2y2+12y=−10

(8a+ 9)(8a−9) = 0

x−3=0 or x +2=0

62. We let y= 0 and solve for x.

0=x2+2x−8

0=(x+ 4)(x−2)

64. The solutions of the equation are the first coordinates of

72. 1

116 Chapter 5: Polynomials: Factoring

76. (t−5)2= 2(5 −t)

78. x2−25

36 =0

x+5

9or x=5

9

Exercise Set 5.9

4. Solve: 5w·w= 320

6. Solve: 1

2b(b−3) = 35

8. Solve: 1

2x(x+5)=42

10. Solve: 232−23 = N

14. Solve: N=1

x=14 or x=−16

The integers are 14 and 16 or −16 and −14.

26. Let h= the length of the hypotenuse, in feet. Then

the leg is 10 ft.

30. Let x= the length of the unknown leg and x+ 2 = the

length of the hypotenuse, in cm.