88 Chapter 4: Polynomials: Operations
26. x−3
x−3x2−6x+9
28. x+4
x+4 x2+8x−16
30. x+7
x−2x2+5x−9
x2−2x
32. x+5
x−5x2+0x−25
x2−5x
5x−25
x−3x4+0x3+0x2+0x−81
x4−3x3
27x−81
27x−81
−6x2+9x
2x−3
38. x3+8
x3−3x6+5x3−24
The answer is x3+8.
40. y2−5
y+3 y3+3y2−5y−15
42. t2−2t+3
t+1 t3−t2+t−1
The answer is t2−2t+3+ −4
t+1.
44. 10x3+14x2+21x+34
68x+3
68x−102
46. 3y2−7
5y−9 15y3−27y2−35y+60
48. x+1
2x=5
−2r−7≥8−r
−r≥15
n= 228 and n+ 1 = 228 + 1 = 229; the page numbers are
228 and 229.
56. y3−ay2+a2y−a3
y+a y4+0y3+0y2+0y+a2
58. 5y+2
60. 5x5+5x4−8x2
−8x+2
x5+x4y
−x4y
−x4y−x3y2
The answer is x4−x3y+x2y2−xy3+y4.
64. 2x+(3c+2)
x−1 2x2+3cx −8
Chapter 4 Vocabulary Reinforcement
3. An expression of the type axn, where ais a real-number
constant and nis a nonnegative integer, is a monomial.
Chapter 4 Concept Reinforcement
Chapter 4 Study Guide
=(x−4)3(y2)3
27x12z9
4. 763,000
7.63,000.
6. 3.6×103
8. (3x4−5x2−4)+(x3+3x2+6)
9. (x4−3x2+ 2)(x2−3)
10. We use FOIL.
(y+ 4)(2y+3) = y·2y+y·3+4·2y+4·3
13. (a3b2−5a2b+2ab)−(3a3b2−ab2+4ab)
=a3b2−5a2b+2ab −3a3b2+ab2−4ab
5
15. x−9
Chapter 4 Review Exercises
2. y7·y3·y=y7+3+1 =y11
5. 45
42=4
5−2=4
3
8. (3t4)2=3
2·(t4)2=9·t4·2=9t8
10. 2x
=y
=y3
Chapter 4 Summary and Review: Review Exercises 91
13. 0.00003. 28
14. 8.3×106
15. (3.8×104)(5.5×10−1)=(3.8·5.5) ×(104·10−1)
17. 46 = 4.6×10
18. x2−3x+6=(−1)2−3(−1)+6=1+3+6=10
zero-degree term) are missing.
25. 3x2−2x+3−5x2−1−x
26. −x+1
2+14x4−7x2−1−4x4
27. (3x4−x3+x−4)+(x5+7x3−3x2−5)+(−5x4+6x2−x)=
(3−5)x4+(−1+7)x3+(1−1)x+(−4−5)+x5+(−3+6)x2=
30. (3x5−4x4+3x2+3)−(2x5−4x4+3x3+4x2−5)
=3x5−4x4+3x2+3−2x5+4x4−3x3−4x2+5
t2+7t+12
34. (7x+1)
2=(7x)2+2·7x·1+1
2
36. (3x2+ 4)(3x2−4) = (3x2)2−42=9x4−16
92 Chapter 4: Polynomials: Operations
39. (3y2−2y)2=(3y2)2−2·3y2·2y+(2y)2=
=49
42. x5y−7xy +9x2−8
43. y+w−2y+8w−5=(1−2)y+(1+8)w−5
46. (6x3y2−4x2y−6x)−(−5x3y2+4x2y+6x2−6)
47. p2+pq +q2
50. 3x2−7x+4
51. Locate −1 on the x-axis. Then move vertically to the
graph and horizontally to the y-axis. It appears that the
y-value that is paired with −1 is 0. Thus, the value of
value of y=10x3−10xis −3.75 when x=0.5.
55. The shaded area is the area of a square with side 20 minus
3
3 x9
Chapter 4 Test 93
58. Let Prepresent the other polynomial. Then we have
(x−1)P=x5−1, or P=x5−1
x−1. We divide to find P.
x−1
x−1
0
Recall that the formula for the area of a rectangle is
Chapter 4 Discussion and Writing Exercises
3. Label the figure as shown.
4. Emma did not divide each term of the polynomial by the
divisor. The first term was divided by 3x, but the second
Chapter 4 Test
10. ab
c3
=(ab)3
c3=a3b3
c3
94 Chapter 4: Polynomials: Operations
16. 1
y8=y−8
19. 5.6×106
3.2×10−11 =5.6
3.2×106
10−11 =1.75×106−(−11) =1.75×1017
25. 7−xis a binomial because it has just 2 terms.
28. 3−x2+2x3+5x2−6x−2x+x5
30. x4+2
3x+5
+4x4+5x2+1
3x
=−x5+0.7x3−0.8x2−21
33. −3x2(4x2−3x−5)=−3x2·4x2−3x2(−3x)−3x2(−5) =
42. (8a2b2−ab +b3)−(−6ab2−7ab −ab3+5b3)
9x10 −16y10
44. (12x4+9x3−15x2)÷(3x2)
Cumulative Review Chapters 1 – 4 95
45. 2x2−4x−2
3x+2 6x3−8x2−14x+13
3x+2.
46. Locate −1 on the x-axis. Then move vertically to the
y-value that is paired with −1 is 3. Thus, the value of
47. When we regard the figure as one large rectangle with
48. Two sides have dimensions aby 5, two other sides have di-
50. (x−5)(x+5)=(x+6)
2
x2−25 = x2+12x+36
Cumulative Review Chapters 1 – 4
9. 3
11. (3.2×10−10)÷(8 ×10−6)= 3.2×10−10
8×10−6=0.4×10−4=
15. 2−[32 ÷(4+2
2)]=2−[32 ÷(4 + 4)]
96 Chapter 4: Polynomials: Operations
16. (x4+3x3−x+7)+(2x5−3x4+x−5)
17. (x2+2xy)+(y2−xy)+(2x2−3y2)
=(1+2)x2+(2−1)xy +(1−3)y2
3x2−3
4x
23. 3y2+5y+6
29. (2x4−3)(2x2+3)=4x6+6x4−6x2−9
32. (18x3+6x2−9x)÷3x=18x3+6x2−9x
33. 3x2−2x−7
x+3 3x3+7x2−13x−21
35. 2
7x=−6
36. 5x−9=36
38. 5.4−1.9x=0.8x
39. x−7
8=3
4
40. 2(2 −3x) = 3(5x+7)
−17 = 21x
41. 1
4x−2
3=3
4+1
3x
−2
3=3
4+1
12x
42. y+5−3y=5y−9
43. 1
4x−7<5−1
2x
4
3·3
4x< 4
3·12
44. 2(x+2) ≥5(2x+3)
−8x
−8≤11
−8Reversing the inequality symbol
45. A=Qx +P
The selling price
is $6.30.
x
47. The area of the surface of the pool is πr2ft2. The area of
48. Familiarize. The page numbers are consecutive integers.
2n=37−1
49. Familiarize. Let l= the length of the room, in feet. Then
50. Familiarize. Let x= the measure of the first angle. Then
Solve.
x=180
18
x=10
51. y2·y−6·y8=y2+(−6)+8 =y4
54. x3x−4
x−5x=x3+(−4)
x−5+1 =x−1
x−4=x−1−(−4) =x−1+4 =x3
Cumulative Review Chapters 1 – 4 99
We let x= 4. Then
56. 32=9
of the picture is (x−2) in.·(x−2) in., or (x2−4x+4) in2.
We subtract to find the area of the frame.
59. (x−3)2+(2x+1)
2=x2−6x+9+4x2+4x+1=
61. First we find (2x2+x−6) ÷(2x−3).
2x2−10x
x−5
63. First we find (x3−4x2−17x+ 60) ÷(x−5).
−12x+60