Exam
Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Match the polynomial with its binomial factors.
1)
5x2+32x +12
1)
A)
(x +6)(5x +2)
B)
(x +2)(5x +6)
C)
(x –6)(5x –2)
D)
(x +6)(4x +2)
2)
x2+ 4x – 12
2)
A)
(x + 6)(x – 2)
B)
(x – 6)(x + 1)
C)
(x – 6)(x – 1)
D)
(x – 6)(x + 2)
3)
7x2+ x – 180
3)
A)
(7x – 5)(x + 36)
B)
(7x + 36)(x – 5)
C)
(7x + 5)(x – 36)
D)
(7x – 36)(x + 5)
4)
5x2– 39x – 54
4)
A)
(5x + 6)(x – 9)
B)
(x + 5)(x – 9)
C)
(6x + 5)(x – 9)
D)
(x – 6)(x – 9)
5)
x2– 2x – 120
5)
A)
(x – 10)(x – 1)
B)
(x – 10)(x + 1)
C)
(x – 10)(x + 12)
D)
(x + 10)(x – 12)
6)
9x2+ 12x + 4
6)
A)
(9x + 2)(x + 2)
B)
(9x + 2)(x + 1)
C)
(3x + 2)(3x + 2)
D)
(3x – 2)(3x – 2)
7)
x2+ 120x + 119
7)
A)
(x + 119)(x + 1)
B)
(x – 12)(x + 10)
C)
(x + 12)(x – 10)
D)
(x + 120)(x – 1)
8)
x2+4x +4
8)
A)
(x –2)(x –2)
B)
(x +4)(x +4)
C)
(x +2)(x +2)
D)
(x +4)(x – 1)
9)
5x2– 24x + 16
9)
A)
(5x + 12)(5x – 4)
B)
(4 – 5x)(5x – 4)
C)
(x + 4)(5x – 4)
D)
(x – 4)(5x – 4)
10)
9x2+ 47x + 10
10)
A)
(9x + 2)(x + 5)
B)
(9x – 5)(x + 5)
C)
(x – 5)(x + 5)
D)
(x + 2)(x + 5)
11)
x2– x – 6
11)
A)
(x + 3)(x – 2)
B)
(x – 1)(x – 6)
C)
(x + 1)(x – 6)
D)
(x + 2)(x – 3)
12)
12x2+ 17x + 6
12)
A)
(3x + 2)(4x + 3)
B)
(12x – 1)(x + 3)
C)
(12x + 2)(x + 3)
D)
(3x – 2)(4x – 3)
13)
x2– 11x +24
13)
A)
(x + 3)(x – 8)
B)
(x + 3)(x + 8)
C)
(x – 3)(x + 8)
D)
(x – 3)(x – 8)
1
14)
2x2+ 23x + 11
14)
A)
(2x – 1)(x + 1)
B)
(2x + 1)(x + 11)
C)
(2x – 1)(x – 11)
D)
(2x + 11)(x + 1)
15)
20x2+ 3x – 9
15)
A)
(20x – 1)(x – 3)
B)
(20x + 3)(x – 3)
C)
(4x – 3)(5x + 3)
D)
(4x + 3)(5x – 3)
16)
x2–7x +10
16)
A)
(x –10)(x – 1)
B)
(x –2)(x +5)
C)
(x –2)(x –5)
D)
(x +2)(x +5)
17)
x2+ 47x –48
17)
A)
(x – 12)(x + 4)
B)
(x + 12)(x – 4)
C)
(x –48)(x + 1)
D)
(x + 48)(x – 1)
18)
x2+10x +21
18)
A)
(x –3)(x –7)
B)
(x +21)(x + 1)
C)
(x +3)(x +7)
D)
(x –3)(x +7)
19)
15x2– 14x – 8
19)
A)
(15x – 4)(x + 1)
B)
(3x – 4)(5x + 2)
C)
(15x – 4)(x + 2)
D)
(3x + 4)(5x – 2)
Solve.
20)
Throw Me for a Loop Bagels determines that it costs them C =70 +1
4x –1
30x2 dollars to make x
dozen bagels in one day. Find the equivalent expression for C by factoring out 1
4.
20)
A)
C =1
4x280–2
15x
B)
C =1
4280– x –2
15x2
C)
C = – 1
4280– x +2
15x2
D)
C =1
4280+ x –2
15x2
Solve the problem.
21)
A rectangle has a length of x + 3 and a width of x – 3, and has an area of 40 square units. Find the
length and width of the rectangle. (A = LW)
21)
A)
width = 1 unit; length = 40 units
B)
width = 5 units; length = 8 units
C)
width = 4 units; length = 10 units
D)
width = 2 units; length = 20 units
22)
The length of a rectangular frame is 7 cm more than the width. The area inside the frame is 78
square cm. Find the width of the frame.
22)
A)
8 cm
B)
20 cm
C)
6 cm
D)
13 cm
Factor, if possible.
23)
125p3– 1
23)
A)
(5p – 1)(25p2+ 5p + 1)
B)
(125p – 1)(p2+ 5p + 1)
C)
(5p + 1)(25p2– 5p + 1)
D)
(5p – 1)(25p2+ 1)
Determine whether the polynomial is a sum of cubes, a difference of cubes, or neither.
24)
x3+ 125
24)
A)
B)
C)
2
Factor completely, if possible.
25)
x2– x – 30
25)
A)
(x + 6)(x – 5)
B)
Prime
C)
(x + 5)(x – 6)
D)
(x + 1)(x – 30)
Solve the equation.
26)
2x(4x + 8) = 0
26)
A)
0, –1
2
B)
0, –2, 2
C)
0, 2
D)
0, –2
27)
4x(x – 8) = (3x + 3)(x – 8)
27)
A)
–3
B)
3
C)
–8, –3
D)
8, 3
Factor completely.
28)
6xy3+ 5xy2– 6xy
28)
A)
y(6xy – 2)(y + 3x)
B)
xy(3y – 2)(2y + 3)
C)
xy(3y + 2)(2y – 3)
D)
y(3xy – 2)(2y + 3x)
Solve the equation.
29)
(x – 9)(x + 2) = 0
29)
A)
9, –9, 2, –2
B)
9, –2
C)
–9, 2
D)
9, 2
Find the greatest common factor.
30)
9, 12, 24
30)
A)
3
B)
1
C)
6
D)
9
Solve.
31)
The formula for the area of a rectangle is A = LW. Let the trinomial, A =x2+11x +30, represent the
area of a rectangle. Find binomials that could represent the length and width of the rectangle.
31)
A)
(x +5)(x –6)
B)
(x +5)(x +6)
C)
(x –5)(x +6)
D)
(x –5)(x –6)
Factor completely, if possible.
32)
y2+14y +48
32)
A)
(y –8)(y –6)
B)
(y –8)(y +6)
C)
(y +8)(y +6)
D)
(y +48)(y + 1)
33)
36x2– 25
33)
A)
Prime
B)
(6x + 5)(6x – 5)
C)
(6x + 5)2
D)
(6x – 5)2
3
Solve.
34)
A lamp shade has the geometric form as shown. The total surface area S of a lamp shade with a
slant height s and top and bottom base radii of r2 and r1 is given by S =r12+r22+r1s +r2s.
Find an equivalent expression for the total surface area by factoring out a common factor.
34)
A)
S =(r1+ r2)(r1+ r2+ sr1r2)
B)
S =(r1+ r2)(r1+ r2+ s)
C)
S =r1(r1+r22+ sr1+ sr2)
D)
S =(r12+r22+ sr1+ sr2)
Factor out the greatest common factor.
35)
28m8– 16m4+ 8m2
35)
A)
4m2(7m6– 4m2+ 2)
B)
No common factor
C)
m2(28m6– 16m2+ 8)
D)
4(7m8– 4m4+ 2m2)
Determine whether the polynomial is a sum of cubes, a difference of cubes, or neither.
36)
64x3+0.04y3
36)
A)
B)
C)
Factor completely, if possible.
37)
s2+6s +8
37)
A)
(s +8)(s – 1)
B)
(s –4)(s –2)
C)
(s +4)(s +2)
D)
(s +8)(s +6)
Solve the equation.
38)
x2+9=6x
38)
A)
0, –3
B)
–3
C)
–3, 3
D)
3
Factor out the greatest common factor.
39)
xz–xy
39)
A)
x(z + y)
B)
x
C)
x(z – y)
D)
xz(–xy)
Factor, if possible.
40)
4y3+4
40)
A)
cannot be factored
B)
4(y + 1)(y2– y + 1)
C)
4(y – 1)(y2+ y + 1)
D)
4(y + 1)(y2+ y + 1)
4
Solve.
41)
A large crater on the moon is approximately circular. The depth (in feet) of the crater below any
point on a diameter can be modeled by the expression 86
450,000(x2– 400x – 320,000), where x is the
distance to the point on the diameter 1000 feet in from the edge. Write this expression in factored
form.
41)
A)
86
450,000(x – 400)(x + 800)
B)
860
450,000(x – 40)(x + 100)
C)
860
450,000(x + 40)(x – 80)
D)
86
450,000(x + 400)(x – 800)
Factor completely, if possible.
42)
x2– 16x + 256
42)
A)
(x – 16)2
B)
Prime
C)
(x + 16)2
D)
(x + 16)(x – 16)
Factor.
43)
6x2– 17xy + 12y2
43)
A)
(3x + 4y)(2x + 3y)
B)
(x – 4y)(6x – 3y)
C)
(3x – 4y)(2x – 3y)
D)
(6x – 4y)(x – 3y)
Solve the problem.
44)
If an object is propelled upward from a height of 16 feet at an initial velocity of 64 feet per second,
then its height after t seconds is given by the equation h = –16t2+ 64t + 16, where h is in feet. After
how many seconds will the object reach a height of 80 feet?
44)
A)
8 sec
B)
1 sec
C)
4 sec
D)
2 sec
45)
A triangle has an area of 3x2+ 19x – 84. Use factoring to rewrite this expression.
45)
A)
(3x + 28)(x + 3)
B)
(3x – 28)(x – 3)
C)
(3x + 28)(x – 3)
D)
(3x – 28)(x + 3)
Find the missing factor.
46)
x2+ 14x + 45 = (x + 5)( )
46)
A)
x + 40
B)
x – 19
C)
x2+ 9
D)
x + 9
Solve the equation.
47)
x2–16 = 0
47)
A)
0, 4
B)
4
C)
–4
D)
–4, 4
Factor out the greatest common factor.
48)
3x2y5+18x2y4
48)
A)
x2y4(3y +6)
B)
3x4y2(y +6)
C)
y +6
D)
3x2y4(y +6)
Factor completely.
49)
x3y + 8x2y2+ 15xy3
49)
A)
x(xy + 3y2)(x + 5y)
B)
y(x + 3y)(xy + 5y2)
C)
xy(x2+ 8x + 15y2)
D)
xy(x + 3y)(x + 5y)
5
Factor out the greatest common factor.
50)
8z –40
50)
A)
8(z –5)
B)
8(z +5)
C)
5(z –8)
D)
8(z –40)
Find the missing factor.
51)
x2+3x –70 = (x –7)( )
51)
A)
(x +10)
B)
(x +7)
C)
(x –10)
D)
(x –7)
Factor completely.
52)
3x2– 21x + 36
52)
A)
(3x – 9)(x – 4)
B)
3(x – 12)(x + 1)
C)
3(x – 3)(x – 4)
D)
Prime
Factor completely, if possible.
53)
64k2– 25m2
53)
A)
Prime
B)
(8k + 5m)(8k – 5m)
C)
(8k + 5m)2
D)
(8k – 5m)2
54)
x2+ 16x + 64
54)
A)
(x + 8)2
B)
(x – 8)2
C)
Prime
D)
(x + 8)(x – 8)
Factor, if possible.
55)
x6– 27x3
55)
A)
cannot be factored
B)
x3(x – 3)(x2– 3x + 9)
C)
x3(x + 3)(x2– 3x + 9)
D)
x3(x – 3)(x2+ 3x + 9)
Factor completely.
56)
20z2– 3zy – 9y2
56)
A)
(20z – 3y)(z + 3y)
B)
(4z – 3y)(5z + 3y)
C)
(4z + 3y)(5z – 3y)
D)
(20z + y)(z – 9y)
Find the missing factor.
57)
42x2+ 5xy – 25y2= (6x + 5y)( )
57)
A)
7x – 5y
B)
7x + 5y
C)
x – 25xy
D)
5x – 7y
Solve the equation.
58)
(x +9)(x +8) = 0
58)
A)
9, –9, 8, –8
B)
–9, –8
C)
9, –8
D)
9, 8
Factor completely.
59)
2a3+8a2–10a
59)
A)
2a(a +5)(a +1)
B)
2a(a +5)(a –1)
C)
2a(a –5)(a –1)
D)
2a(a –5)(a +1)
Factor out the greatest common factor.
60)
m4n4–m3n6
60)
A)
m3n4(m –n2)
B)
mn(16 –18)
C)
m4n6(m –n2)
D)
m4n4(1 – mn2)
6
Find the missing factor.
61)
x2+ 2x – 48 = (x – 6)( )
61)
A)
x2+ 6
B)
x + 8
C)
x – 8
D)
6– x
Solve the problem.
62)
If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per
second, then its height h after t seconds is given by the equation h = –16t2+112t +128. After how
many seconds does the object hit the ground?
62)
A)
8 sec
B)
7 sec
C)
11 sec
D)
4 sec
Solve for the indicated variable.
63)
K=ax –3x; x
63)
A)
x =K
a+3
B)
x =K–a+3
C)
x =K
a–3
D)
x =K(a –3)
Factor completely, if possible.
64)
1
4–p2
64)
A)
1
4(1 + p)(1 – p)
B)
1
2+ p 1
2– p
C)
1
2– p 1
2– p
D)
1
4+ p 1
4– p
Classify the polynomial as a perfect square trinomial, a difference of squares, or neither.
65)
25t2+16q2
65)
A)
B)
C)
Factor completely, if possible.
66)
x2y2–81
66)
A)
(xy –9)(xy –9)
B)
(x2y2+9)(x2y2–9)
C)
(xy +9)(xy –9)
D)
(xy +81)(xy –81)
67)
x2– 3x – 70
67)
A)
(x + 7)(x – 10)
B)
Prime
C)
(x – 7)(x + 1)
D)
(x – 7)(x + 10)
Factor, if possible.
68)
x3– 1000
68)
A)
(x – 10)(x2+ 10x + 100)
B)
(x – 10)(x2+ 100)
C)
(x + 1000)(x2– 1)
D)
(x + 10)(x2– 10x + 100)
Solve the equation.
69)
15m2– 4m = 0
69)
A)
4
15, –4
15
B)
4
15, 0
C)
0
D)
–4
15, 0
Find the greatest common factor.
70)
15m5, 135m7
70)
A)
135m5
B)
15m2
C)
2025m2
D)
15m5
7
Factor completely, if possible.
71)
4x2+ 12x + 9
71)
A)
(2x – 3)(2x – 3)
B)
Prime
C)
(4x + 3)(x + 3)
D)
(2x + 3)(2x + 3)
Solve the equation.
72)
(x –5)(5x + 20) = 0
72)
A)
–4, 5
B)
–1
5, 1
4
C)
–1
4, 1
5
D)
–5, 4
Solve.
73)
On a certain planet, the height (in feet) of a rock dropped from a height 121 ft above the ground is
given by the expression 121–25t2, where t is time in seconds. Rewrite this expression in factored
form.
73)
A)
(121+25t)(121 –25t)
B)
(11 +5t)(11 –5t)
C)
(11 –5t)(11 –5t)
D)
(11 +5t)(11 +5t)
Solve the equation.
74)
4x2– 24x +36 = 0
74)
A)
–3, 3
B)
0, 3
C)
–3
D)
3
Factor.
75)
6xy +18y
75)
A)
6y(x +18)
B)
6y(x + 3)
C)
3y(x +6)
D)
6y
Factor completely, if possible.
76)
9z2– 6z – 8
76)
A)
(3z – 2)(3z + 4)
B)
(9z + 2)(z – 4)
C)
Prime
D)
(3z + 2)(3z – 4)
Factor completely.
77)
12x4– 17x3+ 6x2
77)
A)
x2(12x – 1)(x – 6)
B)
x(3x + 2)(4x + 3)
C)
(12x2+ 2)(x2+ 3)
D)
x2(3x – 2)(4x – 3)
Classify the polynomial as a perfect square trinomial, a difference of squares, or neither.
78)
36n2–100m2
78)
A)
B)
C)
Solve the problem.
79)
A triangle has an area of 2x2+ 9x – 45. Use factoring to rewrite this expression.
79)
A)
(2x – 15)(x + 3)
B)
(2x – 15)(x – 3)
C)
(2x + 15)(x + 3)
D)
(2x + 15)(x – 3)
Solve for the indicated variable.
80)
B=nx +dx; x
80)
A)
x=B
nd
B)
x=B
n–d
C)
x=B
n+d
D)
x=B–n–d
8
Factor, if possible.
81)
a3–b3
81)
A)
(a +b)(a2+ab +b2)
B)
(a +b)(a2–ab +b2)
C)
(a –b)(a2–ab +b2)
D)
(a –b)(a2+ab +b2)
Factor completely.
82)
u2– 3uv – 70v2
82)
A)
(u – v)(u + 10v)
B)
(u – 7v)(u + v)
C)
(u – 7v)(u + 10v)
D)
(u + 7v)(u – 10v)
83)
x3+ 6x2– 72x
83)
A)
x(x – 12)(x + 6)
B)
(x2+ 12)(x –6)
C)
x(x + 12)(x – 6)
D)
(x2+6)(x – 12)
Classify each equation as linear or quadratic.
84)
–2= (5x –9)(x +12)
84)
A)
Linear
B)
Quadratic
Solve the equation.
85)
4x2=9x
85)
A)
–9
4, 9
4
B)
4
9, 0
C)
9
4, 0
D)
9
4, 4
9
Find the missing factor.
86)
9x2– 37x – 40 = (x – 5)( )
86)
A)
8x + 9
B)
9x + 8
C)
x – 8
D)
x + 9
Solve.
87)
If an object is propelled upward from a height of 240 feet at an initial velocity of 32 feet per second,
then its height h after t seconds is given by the equation h = –16t2+32t +240. Factor to find an
equivalent equation for h.
87)
A)
h = –16(t –3)(t –5)
B)
h = –16(t –3)(t +5)
C)
h = –16(t +3)(t –5)
D)
h = –16(t +3)(t +5)
Find the greatest common factor.
88)
8, 14, 16, 18
88)
A)
2
B)
1
C)
4
D)
8
Solve.
89)
If the area of a square is represented by the expression 4x2–16x +16, find an expression for the
length of each side.
89)
A)
2x +4
B)
2x –4
C)
(2x –4)2
D)
(2x –4)(2x +4)
Factor out the greatest common factor.
90)
–8x3–16x2+12x
90)
A)
–4(2x3+ 4x2– 3x)
B)
–8x(x2+ 2x – 3)
C)
–4x(2x2– 4x + 3)
D)
–4x(2x2+ 4x – 3)
9
Determine whether the polynomial is a sum of cubes, a difference of cubes, or neither.
91)
729x15 – 512y9
91)
A)
B)
C)
Factor completely.
92)
12x2+ 10xy – 12y2
92)
A)
2(6x2+ 5xy – 6y2)
B)
2(3x – 2y)(2x + 3y)
C)
2(3x – 2y)(2x – 3y)
D)
2(3x + 2y)(2x – 3y)
93)
6x2+ 7xt + 2t2
93)
A)
Prime
B)
(2x + t)(3x + 2t)
C)
(2x – t)(3x – 2t)
D)
(6x + t)(x + 2t)
Solve the equation.
94)
x(2x + 10) = 0
94)
A)
0, –1
5
B)
0, 5
C)
0, –5
D)
0, 1
5
Solve.
95)
(x – 9)(x + 5) = 0
95)
A)
9, 5
B)
9, –5
C)
9, –9, 5, –5
D)
–9, 5
Find the missing factor.
96)
x2+14x +48 = (x +8)( )
96)
A)
(x –8)
B)
(x –6)
C)
(x +6)
D)
(x +8)
Factor completely, if possible.
97)
0.04x2+0.16x +0.16
97)
A)
(0.2x +0.4)2
B)
(0.2x2+0.4)2
C)
(0.2x +0.8)2
D)
(0.04x +0.16)2
Factor, if possible.
98)
p6–q3
98)
A)
cannot be factored
B)
(p2–q)(p4+p2q+q2)
C)
(p2+q)(p4–p2q+q2)
D)
(p –q)(p2+pq +q2)
Factor completely, if possible.
99)
12 – 8x +x2
99)
A)
(x – 6)(x + 2)
B)
(x – 6)(x – 2)
C)
(x + 6)(x – 2)
D)
(x + 6)(x + 2)
Factor out the greatest common factor.
100)
18x5+18x
100)
A)
18(x6+ 1)
B)
18x(x4+ 1)
C)
18x(x4)
D)
18(x5+ x)
Solve the equation.
101)
28 + 3y =y2
101)
A)
4, –7
B)
–4, –7
C)
7, 4
D)
7, –4
10
Factor completely.
102)
x2+ 7xy – 144y2
102)
A)
(x – y)(x + 9y)
B)
(x – 16y)(x + y)
C)
(x + 16y)(x – 9y)
D)
(x – 16y)(x + 9y)
103)
7x2y2– 59xy – 36
103)
A)
(7xy + 9)(xy – 4)
B)
(7x + 4y)(x – 9y)
C)
(7xy + 4)(xy – 9)
D)
(7xy – 9)(xy + 4)
104)
3x3+ 6x2– 24x
104)
A)
(3x2+ 6x)(x – 4)
B)
Prime
C)
3x(x + 2)(x – 4)
D)
3x(x – 2)(x + 4)
Factor.
105)
x2– 4x – 96
105)
A)
(x – 96)(x + 1)
B)
(x – 12)(x + 8)
C)
(x – 96)(x – 1)
D)
(x + 12)(x – 8)
Factor completely.
106)
–x2– 6x + 55
106)
A)
–(x + 11)(x – 5)
B)
–(x – 11)(x – 5)
C)
–(x – 11)(x + 1)
D)
–(x – 11)(x + 5)
107)
–30x2– 25x + 30
107)
A)
–5(3x + 2)(2x – 3)
B)
–5(6x – 1)(x + 6)
C)
–5(3x – 2)(2x + 3)
D)
–5(3x – 3)(2x + 2)
Factor completely, if possible.
108)
4m2+29m +45
108)
A)
(m +9)(4m +5)
B)
(m –5)(4m –9)
C)
(m +5)(3m +9)
D)
(m +5)(4m +9)
Factor completely.
109)
6y2+ 27y – 15
109)
A)
(2y – 1)(3y + 15)
B)
3(2y + 1)(y – 5)
C)
3(2y – 1)(y + 5)
D)
(6y – 3)(y + 5)
Solve the equation.
110)
3x(x – 9) = (2x – 3)(x – 9)
110)
A)
3
B)
–9, 3
C)
–3
D)
9, –3
Factor completely, if possible.
111)
64x2+ 81
111)
A)
(8x + 9)(8x – 9)
B)
(8x + 9)2
C)
(8x – 9)2
D)
Prime
Factor.
112)
9x2– 42xy + 49y2
112)
A)
(3x – 7y)2
B)
(3x + 7y)2
C)
(9x + 1)(x + 49)
D)
(3x – 7y)(3x + 7y)
11
Solve the equation.
113)
y2+ 4y =32
113)
A)
–8, –4
B)
–8, 4
C)
–4, 8
D)
8, 4
Factor, if possible.
114)
27x3–x6
114)
A)
x3(3 – x)(9+3x +x2)
B)
x3(3 – x)(9–3x +x2)
C)
x3(3 + x)(9–3x +x2)
D)
cannot be factored
Factor completely.
115)
3x6+ 21x5+ 36x4
115)
A)
x4(x + 4)(3x + 9)
B)
x4(3x + 12)(x + 3)
C)
3x4(x + 4)(x + 3)
D)
34(x2+ 7x + 12)
Classify each equation as linear or quadratic.
116)
6= –10x2+10x
116)
A)
Quadratic
B)
Linear
Factor.
117)
8–4x –2t +xt
117)
A)
(2 +x)(4–t)
B)
(2 –x)(4+t)
C)
(2 –x)(4–t)
D)
(2 +x)(4+t)
118)
t2b2– 1
118)
A)
(t –b)(t+b)
B)
(tb + 1)(tb – 1)
C)
(tb + 1)2
D)
(tb – 1)2
Solve the equation.
119)
x2– 8x + 16 = 0
119)
A)
4
B)
–4, 4
C)
0, 4
D)
–4
Factor by grouping.
120)
8–4t –2x +tx
120)
A)
(2 –t)(4+x)
B)
(2 +t)(4–x)
C)
(2 –t)(4–x)
D)
(2 +t)(4+x)
Factor completely.
121)
16x3+ 24x2+ 9x
121)
A)
x(16x + 1)(x + 9)
B)
x(4x + 3)(4x + 3)
C)
x(4x – 3)(4x – 3)
D)
(16x2+ 3)(x + 3)
Solve the equation.
122)
(3x – 8)(x + 7) =27x – 71
122)
A)
8
3, 71
27, –7
B)
71
26, 71
24
C)
71
24, –7
D)
5
3, 3
Find the greatest common factor.
123)
42, 24
123)
A)
66
B)
6
C)
3
D)
7
12
Factor completely, if possible.
124)
r2– 8r + 16
124)
A)
(r – 8)(r + 8)
B)
(r – 4)(r + 4)
C)
(r – 4)2
D)
(r + 4)2
Determine whether the polynomial is a sum of cubes, a difference of cubes, or neither.
125)
5x6– 512y3
125)
A)
B)
C)
Find the missing factor.
126)
5x2– 43x – 18 = (5x + 2)( )
126)
A)
x + 9
B)
5x – 9
C)
5x + 16
D)
x – 9
Factor completely, if possible.
127)
49x2+ 126xy + 81y2
127)
A)
(7x + 9y)2
B)
(7x – 9y)2
C)
(49x + 1)(x + 81)
D)
(7x + 9y)(7x – 9y)
Factor completely.
128)
9z3+ 6z2– 8z
128)
A)
(z2– 2)(9z + 4)
B)
z(9z + 1)(z – 8)
C)
(3z – 4)(3z + 2)
D)
z(3z + 4)(3z – 2)
Find the missing factor.
129)
5x2+ 36x + 36 = (5x + 6)( )
129)
A)
5x + 6
B)
5x + 30
C)
x + 6
D)
x – 6
Solve the equation.
130)
6x(x +6) = (2x +4)(x +6)
130)
A)
–6, 1
B)
–6, 2
C)
6, 2
D)
6, 1
Factor completely, if possible.
131)
36y4– 25
131)
A)
Prime
B)
(6y2– 5)2
C)
(6y2+ 5)2
D)
(6y2+ 5)(6y2– 5)
Solve the problem.
132)
The product of two consecutive integers is 29 more than their sum. Find the integers.
132)
A)
5, 6 or –5, –4
B)
6, 7
C)
–5, –4
D)
6, 7 or –5, –4
13
Solve.
133)
Acme Computer Products finds that the marginal revenue R in producing x computer networking
connectors is given by the equation R =50 –1
50x2 dollars/connector. Find an equivalent expression
for R by factoring out 1
50x.
133)
A)
R =1
50x(2500 – x)
B)
R =1
50x1
x– x
C)
R =1
50x2500
x+ x
D)
R =1
50x2500
x– x
Solve the equation.
134)
(x –7)(x –6) = 0
134)
A)
7, –7, 6, –6
B)
–7, –6
C)
7, –6
D)
7, 6
Factor.
135)
49x2y +84x3+ 7xy2
135)
A)
7x(3x + y)(4x + y)
B)
(21x + 7y)(4x + y)
C)
7(3 + xy)(4 + xy)
D)
7x(3x – y)(4x – y)
Solve the equation.
136)
(y – 7)(5y + 13) = 0
136)
A)
–7, 5
13
B)
–5
13, 7
C)
–7, 13
5
D)
–13
5, 7
Find the missing factor.
137)
3x2+ 13x – 56 = (3x – 8)( )
137)
A)
8– 3x
B)
3x – 64
C)
x + 7
D)
x – 7
Factor completely, if possible.
138)
49m2–4
25
138)
A)
7m +2
57m –2
5
B)
Prime
C)
7m +2
52
D)
7m –2
52
Solve the equation.
139)
4d2+ 20d + 24 = 0
139)
A)
1
3, 1
2
B)
–3, –2
C)
3, 2
D)
–1
3, –2
Factor completely.
140)
9x2– 27xy – 36y2
140)
A)
9(x + y)(x – 4y)
B)
9(x – y)(x + 4y)
C)
Prime
D)
(9x – 9y)(x + 4y)
14
Solve.
141)
A square garden has sides of length 2. The garden is enlarged so that its sides have length 9n. Find
an expression, in factored form, for the area added to the original garden.
141)
A)
(n +2)(n –2)
B)
(9n –2)2
C)
(9n +2)(9n –2)
D)
(81n +2)(81n –2)
Solve.
142)
Kerri’s teacher asks her to solve the following problem: “Factor the following trinomial, if possible.
x2+9x +14“ What factorization could Kerri use to represent her solution?
142)
A)
(x +7)(x +2)
B)
(x –7)(x –2)
C)
(x +7)(x –2)
D)
(x –7)(x +2)
Factor completely, if possible.
143)
x2+ 9x – 22
143)
A)
(x – 11)(x + 2)
B)
(x – 11)(x + 1)
C)
(x + 11)(x – 2)
D)
Prime
Factor.
144)
121– 144m2
144)
A)
(11 + 12m)(11 – 12m)
B)
(11 + 12m)2
C)
(11m – 12)(11m + 12)
D)
(11 – 12m)2
Solve the equation.
145)
4x2+ 16x +16 = 0
145)
A)
0, –2
B)
–2
C)
2
D)
–2, 2
Factor out the greatest common factor.
146)
15wx –20wy –15wz
146)
A)
5w(3x –4y –3z)
B)
5w(3x –20wy –15wz)
C)
5(3wx –4wy –3wz)
D)
15w(x –20y –15z)
Factor completely.
147)
u2– 2uv – 15v2
147)
A)
(u + 3v)(u – 5v)
B)
(u – v)(u + 5v)
C)
(u – 3v)(u + v)
D)
(u – 3v)(u + 5v)
Factor completely, if possible.
148)
u2– 5u +6
148)
A)
(u + 3)(u + 2)
B)
(u + 3)(u – 2)
C)
(u – 3)(u + 2)
D)
(u – 3)(u – 2)
Factor completely.
149)
x2+ 2xy – 24y2
149)
A)
(x – 6y)(x + 4y)
B)
(x – y)(x + 4y)
C)
(x – 6y)(x + y)
D)
(x + 6y)(x – 4y)
Factor out the greatest common factor.
150)
5x2+40x
150)
A)
5x2(x +8)
B)
5x(x +40)
C)
x(5x +40)
D)
5x(x +8)
15
Solve the equation.
151)
x2=64
151)
A)
8
B)
–8, 8
C)
–8
D)
0, –8
Factor out the greatest common factor.
152)
5t2–25t –25
152)
A)
5(t2–5t –5)
B)
5(t2–25t –25)
C)
5(t2– 20t – 20)
D)
5t(t2–5t –5)
Classify each equation as linear or quadratic.
153)
6x2= (6x –7)(x –4)
153)
A)
Quadratic
B)
Linear
Solve for the indicated variable.
154)
4nx + x =A; x
154)
A)
x =A–4n + 1
B)
x =A
4n – 1
C)
x =A
4n + 1
D)
x =A–4n –1
Find the greatest common factor.
155)
64a10b3, 88a6b10
155)
A)
8a6b3
B)
4a4b7
C)
704a10b10
D)
8a10b10
Solve.
156)
Sand is poured carefully into a pile that forms a right circular cone as shown. The total surface area
of the cone with slant height s and base radius r is given by the polynomial S =r2+rs. Find an
equivalent expression for S by factoring out a common factor.
156)
A)
S =rs(r + 1)
B)
S =r(r + s)
C)
S = r(r2+s)
D)
S =r(r + s)
Factor completely.
157)
9x2+ 18xt + 8t2
157)
A)
(3x + 2t)(3x + 4t)
B)
Prime
C)
(3x – 2t)(3x – 4t)
D)
(9x + 2t)(x + 4t)
Factor.
158)
32x2+ 80x + 50
158)
A)
2(x + 5)(x – 5)
B)
2(16x2+ 40x +25)
C)
2(4x + 5)2
D)
2(4x – 5)2
Factor completely.
159)
3x2– 3x – 18
159)
A)
(3x + 6)(x – 3)
B)
3(x + 2)(x – 3)
C)
Prime
D)
3(x – 2)(x + 3)
16
Solve.
160)
A bullet is fired straight up into the air with a muzzle velocity of 282 meters per second, and after t
seconds its height above the ground is h =1128 +282t – 4.6t2 meters. Find an equivalent expression
for the height of the bullet by factoring out a common factor.
160)
A)
h = 2(564 +141t – 2.3t2)
B)
h =4(282+70.5t – 2.3t2)
C)
h = 2(564 +141t + 2.3t2)
D)
h =1128(1 +70.5t –1.2t2)
Factor, if possible.
161)
125x3+ 1
161)
A)
(5x – 1)(25x2+ 5x + 1)
B)
(125x + 1)(x2– 5x + 1)
C)
(5x + 1)(25x2– 5x + 1)
D)
(5x + 1)(25x2+ 1)
Solve the equation.
162)
(2y + 25)(2y + 29) = 0
162)
A)
–25
2, –29
2
B)
23, 27
C)
25
2, 29
2
D)
–2
23, –2
29
Factor completely.
163)
12x2– 17xy + 6y2
163)
A)
(12x + 3y)(x + 2y)
B)
(4x – 3y)(3x – 2y)
C)
(4x + 3y)(3x – 2y)
D)
(4x + 3y)(3x + 2y)
Solve the problem.
164)
A parallelogram has an area of 2x2+ 7x – 39. Use factoring to rewrite this expression.
164)
A)
(2x – 13)(x – 3)
B)
(2x + 13)(x + 3)
C)
(2x + 13)(x – 3)
D)
(2x – 13)(x + 3)
Factor by grouping.
165)
24x2+ 20xy + 18xy + 15y2
165)
A)
(4x – 3y)(6x + 5y)
B)
(24x + 3y)(x + 5y)
C)
(4x + 3y)(6x + 5y)
D)
(4x + 3)(6x + 5)
Factor completely.
166)
x2– 3xy – 28y2
166)
A)
(x + 7y)(x – 4y)
B)
(x – 7y)(x + 4y)
C)
(x – y)(x + 4y)
D)
(x – 7y)(x + y)
Factor out the greatest common factor.
167)
t6–20t4
167)
A)
t4(t2+20)
B)
t(t5–20t3)
C)
t2(t4–20)
D)
t4(t2–20)
Factor, if possible.
168)
5p3–320
168)
A)
5(p +4)(p2–4p +16)
B)
5(p –4)(p2–4p +16)
C)
5(p –20)(p2+20p +80)
D)
5(p –4)(p2+4p +16)
17
Solve.
169)
A science student watching a group of ants moving along a straight line determines that their
position x along the line as a function of time t is given by x = t4 +3t3 –13t2 +39t cm. Give an
equivalent expression for x by factoring out a common factor.
169)
A)
x = t(t3+3t2+ 26t)
B)
x = t(t2+3t – 26)
C)
x = t(t2+3t + 26)
D)
x = t(t3+3t2–13t + 39)
Find the missing factor.
170)
x2+ 4x – 32 = (x + 8)( )
170)
A)
x – 4
B)
4– x
C)
x + 4
D)
–x – 4
Solve the equation.
171)
x2–2x = 0
171)
A)
2, 0
B)
4, 0
C)
–2, 0
D)
–2, 1
Find the greatest common factor.
172)
2x(3x + 4), (3x + 4)
172)
A)
3x
B)
3x + 4
C)
2x + 1
D)
2x
Factor completely, if possible.
173)
3x2+ 16x – 12
173)
A)
(x + 6)(3x – 2)
B)
(2 – 3x)(3x – 2)
C)
(x – 6)(3x – 2)
D)
(3x – 14)(3x – 2)
Factor by grouping.
174)
4m(4– m) + 7n(4– m)
174)
A)
(4m – 7n)(4– m)
B)
No common factor (except 1)
C)
m(4+ 7n)(4– 1)
D)
(4m + 7n)(4– m)
Factor completely, if possible.
175)
x2+ 6xy + 9y2
175)
A)
(x – 3y)2
B)
(x + 3y)2
C)
(x + 3y)(x – 3y)
D)
Prime
Solve.
176)
Sam’s teacher asks him to solve the following problem: “Factor the following trinomial, if possible:
x2+13x –21“. What factorization could Sam use to represent his solution?
176)
A)
(x –3)(x +7)
B)
(x –3)(x –7)
C)
Prime
D)
(x +3)(x –7)
Solve the equation.
177)
6r2=3r
177)
A)
0, 2
B)
0
C)
1
2
D)
0, 1
2
Solve.
178)
The growth of revenue R over a 12–year time period for Acme Computer Products is given by
R =0.8t +32t2thousand dollars, where t is the time in years. Find an equivalent expression for R by
factoring out a common factor.
178)
A)
R =4t(0.1 +8t)
B)
R = –8t(0.1 –4t)
C)
R =32(0.1t + t2)
D)
R =8t(0.1 +4t)
18
Determine whether the polynomial is a sum of cubes, a difference of cubes, or neither.
179)
x6+y12z18
179)
A)
B)
C)
Factor completely.
180)
x5– 7x4+ 12x3
180)
A)
x3(x – 4)(x + 3)
B)
x–4(x2– 7x + 12)
C)
x3(x + 4)(x – 3)
D)
x3(x – 4)(x – 3)
181)
p2–15pq +54q2
181)
A)
(pq –6)(pq –9)
B)
(p +6q)(p +9q)
C)
(p –6q)(p –9q)
D)
(p –54q)(p – q)
Provide an appropriate response.
182)
In the right triangle shown, find the length of the missing side.
6 m 10 m
182)
A)
8 m
B)
7 m
C)
10 m
D)
9 m
Solve.
183)
A furniture store sells two sizes of cube–shaped stackable storage cabinets. The interior storage
space of the smaller cube is x in. by x in. by x in. The interior storage space of the larger cube is y in.
by y in. by in. Write an expression, in factored form, that shows the amount that the storage space
of the larger cube exceeds the storage space of the smaller cube.
183)
A)
(y – x)(y2– yx +x2)
B)
(y – x)(y2+ yx +x2)
C)
(y + x)(y2– yx +x2)
D)
(x – y)(x2+ xy +y2)
Solve the problem.
184)
If an object is propelled upward from ground level with an initial velocity of 43.3 feet per second,
its height h in feet t seconds later is given by the equation h = –16t2+43.3t. After how many
seconds does the object hit the ground? Round to the nearest tenths.
184)
A)
2.7 sec
B)
1.4 sec
C)
0.4 sec
D)
5.4 sec
Factor completely, if possible.
185)
t2+2
7t +1
49
185)
A)
t +1
7t –1
7
B)
t +1
72
C)
Prime
D)
t –1
72
Solve the equation.
186)
24 –n2=2n
186)
A)
–6, –4
B)
4, 6
C)
4, –6
D)
6, –4
187)
r(r –18) = –81
187)
A)
81, 9
B)
–9
C)
9
D)
±9
19
Provide an appropriate response.
188)
An object is dropped from a height of 36 feet. The expression 36 – 16t2 gives the height of the falling
object after t seconds. The time it takes for the object to reach the ground may be found by solving
36 – 16t2= 0. Solve for t.
188)
A)
2
3 sec
B)
4 sec
C)
3 sec
D)
3
2, or 1.5 sec
Solve for the indicated variable.
189)
A=pz +dz; z
189)
A)
z=A
pd
B)
z=A
p+d
C)
z=A–p–d
D)
z=A
p–d
Factor out the greatest common factor.
190)
90x8y9– 63x2y7+ 54x5y2
190)
A)
No common factor
B)
9(10x8y9– 7x2y7+ 6x5y2)
C)
9x2(10x6y9– 7y7+ 6x3y2)
D)
9x2y2(10x6y7– 7y5+ 6x3)
191)
2s6+8s4
191)
A)
s4(2s2+8)
B)
8(s2+ 4s)
C)
2(s6+ 4s4)
D)
2s4(s2+ 4)
Solve the equation.
192)
–5x2=25x
192)
A)
5, 0
B)
–5, 0
C)
–5, 5
D)
5, 5
193)
x2+ 10x – 24 = 0
193)
A)
12, –2
B)
12, 2
C)
–12, 1
D)
–12, 2
Find the greatest common factor.
194)
8s, 12
194)
A)
24s
B)
4s
C)
8
D)
4
Find the missing factor.
195)
56x3– 231x2– 245x =7x(8x + 7)( )
195)
A)
x + 35
B)
x – 5
C)
x + 5
D)
x – 35
Solve the equation.
196)
10b2+ 31b + 12 = –3
196)
A)
–2
5, –3
5
B)
–5
2, –3
5
C)
5
2, 3
5
D)
2
5, 5
3
Factor completely.
197)
30x2– 43xy + 15y2
197)
A)
(5xy – 3)(6xy – 5)
B)
(5x + 3y)(6x – 5y)
C)
(30x – 3y)(x – 5y)
D)
(5x – 3y)(6x – 5y)
20
Factor by grouping.
198)
20r2+45ry –4xr –9xy
198)
A)
(9r +4y)(5r – x)
B)
(4r +9y)(5x – r)
C)
(4r +9y)(5r – x)
D)
(4r +9y)(x –5r)
Factor completely.
199)
x3– x2– 30x
199)
A)
x(x + 6)(x – 5)
B)
Prime
C)
x(x + 5)(x – 6)
D)
(x2+ 1)(x – 30)
Solve the equation.
200)
x2– x =20
200)
A)
4, 5
B)
1, 20
C)
–4, 5
D)
–4, –5
Solve.
201)
A game show awards points to contestants based on how many rounds the contestants make it
through. For the first 15 rounds, contestants get p points per round. After that, they get twice as
many points for each additional round they pass. During one episode of the show, x contestants
made it through 15 rounds, and y contestants made it through 18 rounds. Find an expression in
factored form for the total number of points awarded during that episode.
201)
A)
3p(5yp +7xp) points
B)
3p(5x +7y) points
C)
3(5xp +3yp) points
D)
15p(5x +2y) points
Solve the problem.
202)
A parallelogram has an area of 3x2+ 4x – 95. Use factoring to rewrite this expression.
202)
A)
(3x – 19)(x + 5)
B)
(3x + 19)(x – 5)
C)
(3x + 19)(x + 5)
D)
(3x – 19)(x – 5)
Factor.
203)
x3– 216
203)
A)
(x – 6)3
B)
Prime polynomial
C)
(x – 6)(x2+ 6x + 36)
D)
(x + 6)(x2– 6x + 36)
204)
4x3+ 4x2– 24x
204)
A)
4x(x + 2)(x – 3)
B)
4x(x – 2)(x + 3)
C)
(4x2+ 8)(x – 3)
D)
(4x2+ 8x)(x – 3)
Factor completely, if possible.
205)
4s2– 121t4
205)
A)
(2s + 11t2)2
B)
Prime
C)
(2s + 11t2)(2s – 11t2)
D)
(2s – 11t2)2
Factor completely.
206)
4x3+ 8x2y – 32xy2
206)
A)
(4x2+ 8xy)(x – 4y)
B)
4x(x – 2y)(x + 4y)
C)
Prime
D)
4x(x + 2y)(x – 4y)
Factor completely, if possible.
207)
x2+ 71x + 72
207)
A)
Prime
B)
(x + 72)(x – 1)
C)
(x – 12)(x + 6)
D)
(x + 12)(x – 6)
21
Factor by grouping.
208)
2x(5x + 4) – 5(5x + 4)
208)
A)
(10x + 5)(x – 4)
B)
(2x – 5)(5x + 4)
C)
(2x + 5)(5x – 4)
D)
(10x – 5)(x + 4)
Factor completely, if possible.
209)
z2– 8z + 16
209)
A)
(z – 8)(z + 8)
B)
(z + 4)2
C)
(z – 4)(z + 4)
D)
(z – 4)2
Factor by grouping.
210)
12a3– 18a2b + 10ab2– 15b3
210)
A)
(12a2+ 5b2)(a – 3b)
B)
(6a2+ 5b)(2a – 3b)
C)
(6a2+ 5b2)(2a – 3b)
D)
(6a2– 5b2)(2a + 3b)
Factor completely, if possible.
211)
p2–9p +20
211)
A)
(p –4)(p +5)
B)
(p +4)(p +5)
C)
(p –4)(p –5)
D)
(p –20)(p – 1)
Factor out the greatest common factor.
212)
5x +15
212)
A)
5(x +15)
B)
3(x +5)
C)
5
D)
5(x + 3)
Factor completely.
213)
p4–11p3+30p2
213)
A)
(p2–6)(p2–5)
B)
p2(p +6)(p +5)
C)
p2(p –30)(p – 1)
D)
p2(p –6)(p –5)
Find the missing factor.
214)
49x2– 14x – 48 = (7x – 8)( )
214)
A)
6x – 7
B)
6x + 7
C)
7x – 6
D)
7x + 6
Solve the equation.
215)
(x +7)(x –5) = –11
215)
A)
7, –5
B)
–6, 4
C)
–7, 5
D)
4, 6
Factor completely, if possible.
216)
6z2+ 5z – 6
216)
A)
(3z – 2)(2z + 3)
B)
(6z – 2)(z + 3)
C)
(3z + 2)(2z – 3)
D)
Prime
Solve.
217)
A square piece of cardboard is used to make an open box by cutting 3 in. by 3 in. squares from each
corner and turning up the sides. The volume of the resulting box can be modeled by the polynomial
3x2–36x +108, where x is the original length of each side of the piece of cardboard. Write the
polynomial in factored form.
217)
A)
3(x +6)2
B)
3(x –6)2
C)
3(x –6)(x +6)
D)
3(x –3)2
22
Solve the equation.
218)
x(x –5) =14
218)
A)
2, –7
B)
–2, –7
C)
2, 7
D)
–2, 7
Factor, if possible.
219)
t3+ 125
219)
A)
(t + 5)(t2– 5t + 25)
B)
(t – 125)(t2– 1)
C)
(t – 5)(t2+ 5t + 25)
D)
(t + 5)(t2+ 25)
220)
z4– 8
220)
A)
(z – 2)(z3– 2z2+ 4)
B)
(z – 2)(z3+ 2z2+ 4)
C)
(z2– 2)(z2+ 2)
D)
cannot be factored
Solve the equation.
221)
78n2+ 12n = 0
221)
A)
–2
13, 0
B)
0
C)
2
13, 0
D)
2
13, –2
13
Factor completely, if possible.
222)
10 + 11z – 6z2
222)
A)
(2 + 3z)(5– 2z)
B)
(10 + z)(1 – 6z)
C)
(10 + 3z)(1 – 2z)
D)
(2 – 3z)(5+ 2z)
Factor completely.
223)
6x4+ 21x3– 12x2
223)
A)
(6x2– 3)(x2+ 4)
B)
x2(2x – 1)(3x + 12)
C)
3x2(2x – 1)(x + 4)
D)
3x(2x + 1)(x – 4)
Solve.
224)
(x +6)(x –5) = –10
224)
A)
6, –5
B)
–5, 4
C)
–6, 5
D)
4, 5
Find the greatest common factor.
225)
–r9s6, –rs5
225)
A)
s5
B)
rs5
C)
r9s6
D)
r9s5
Solve the equation.
226)
(x +6)(x –8) = 0
226)
A)
6, –8
B)
6, –6, 8, –8
C)
–6, 8
D)
6, 8
Factor out the greatest common factor.
227)
56x8y9– 24x5y7+ 40x2y5
227)
A)
No common factor
B)
8x2(7x6y9– 3x3y7+ 5y5)
C)
8(7x8y9– 3x5y7+ 5x2y5)
D)
8x2y5(7x6y4– 3x3y2+ 5)
23
Factor completely.
228)
8x2y2– 18xy2+ 9y2
228)
A)
(2x – 3y)(4x – 3y)
B)
y2(x – 3)(8x – 3)
C)
(4x – 3y)(2x – 3y)
D)
y2(4x – 3)(2x – 3)
Factor by grouping.
229)
m2(x – y) + 5(y – x)
229)
A)
(x + 5)(m2– y)
B)
m2 (x – y + 5)
C)
(x – y) (m2– 5)
D)
(x – y) (m2+ 5)
Classify the polynomial as a perfect square trinomial, a difference of squares, or neither.
230)
36n2–81
230)
A)
B)
C)
Solve.
231)
If the area of a square is represented by expression x2–6x +9, find an expression for the length of
each side.
231)
A)
x +3
B)
(x +3)2
C)
(x –3)2
D)
x –3
Factor.
232)
–36 +x2+5x
232)
A)
(x – 9)(x – 1)
B)
(x + 9)(x – 4)
C)
(x – 9)(x + 4)
D)
(x + 9)(x + 1)
Factor by grouping.
233)
20x2– 25x + 8x – 10
233)
A)
(5x – 2)(4x + 5)
B)
(5x + 2)(4x – 5)
C)
(20x + 2)(x – 5)
D)
(20x – 2)(x + 5)
Find the greatest common factor.
234)
104x4, 52x8
234)
A)
26x4
B)
52x4
C)
78x4
D)
104x4
Factor completely, if possible.
235)
x2– x – 48
235)
A)
Prime
B)
(x + 6)(x – 8)
C)
(x – 48)(x + 1)
D)
(x – 6)(x + 8)
Solve the equation.
236)
6x4=24x3
236)
A)
0, 4
B)
–4, 6
C)
4, 6
D)
0, –4
Solve.
237)
The formula for the volume of a rectangular solid is V = LWH. Let the trinomial,
V =7x3+ 7x2–84x, represent the volume of a rectangular solid. Find monomials and/or binomials
that could represent the length, width, and height.
237)
A)
7x(x +4)(x –3)
B)
7x(x –4)(x –3)
C)
7x(x +4)(x +3)
D)
7x(x –4)(x +3)
24
Solve the equation.
238)
2x(2x –8) = –16
238)
A)
1
2
B)
±2
C)
2
D)
–0.5
Classify each equation as linear or quadratic.
239)
5x –9
–6= –5
239)
A)
Linear
B)
Quadratic
Factor completely, if possible.
240)
x4–81
240)
A)
Prime
B)
(x +3)2(x –3)2
C)
(x2+9)(x +3)(x –3)
D)
(x2–9)(x +3)(x –3)
Classify the polynomial as a perfect square trinomial, a difference of squares, or neither.
241)
9t3–100q3
241)
A)
B)
C)
Factor completely, if possible.
242)
6x2– 23x – 35
242)
A)
(x – 7)(x – 5)
B)
(x + 6)(x – 5)
C)
(7x + 6)(x – 5)
D)
(6x + 7)(x – 5)
243)
x2– 1.8x + 0.81
243)
A)
(x – 0.9)2
B)
(x + 0.9)(x – 0.9)
C)
(x + 0.9)2
D)
Prime
Find the missing factor.
244)
6x2– 7xy – 20y2= (2x – 5y)( )
244)
A)
4x – 3y
B)
3+ 4xy
C)
4y – 3x
D)
3x + 4y
Provide an appropriate response.
245)
A rectangular photograph measures 24 inches by 41 inches. The photographer wishes to frame the
photograph with a frame x inches wide, as shown. Write an expression in factored form for the area
of the frame in terms of x.
41 in.
24 in.
245)
A)
2x(2x +65) sq in.
B)
2(2x2+65x +984) sq in.
C)
4x(x –65) sq in.
D)
(2x +24)(2x +41) sq in.
25
Solve the equation.
246)
84n2+ 91n = 0
246)
A)
–13
12, 91
B)
–13
12
C)
–13
12, 0
D)
0
Classify each equation as linear or quadratic.
247)
8x =2x2
247)
A)
Quadratic
B)
Linear
Factor completely, if possible.
248)
y2–25
248)
A)
(y2+5)(y2–5)
B)
(y +5)(y –5)
C)
(y +25)(y –25)
D)
(y –5)(y –5)
Solve the equation.
249)
2x2+12 =10x
249)
A)
2, 3
B)
–3, 2
C)
–2, 3
D)
–3, –2
Factor completely.
250)
80x3+ 84x2– 20x
250)
A)
x(16x + 20)(5x – 1)
B)
4x(4x + 5)(5x – 1)
C)
x(4x + 5)(20x – 4)
D)
4(4x2+ 5)(5x – 1)
Factor, if possible.
251)
27 +n3
251)
A)
(3 + n)(9+n2)
B)
(3 –n)(9+3n +n2)
C)
(3 + n)(9–3n +n2)
D)
(3 + n)(9+3n +n2)
Solve the problem.
252)
A rectangle has a length of x + 2 and a width of x – 2, and has an area of 60 square units. Find the
length and width of the rectangle. (A = LW)
252)
A)
width = 6 units; length = 10 units
B)
width = 3 units; length = 20 units
C)
width = 5 units; length = 12 units
D)
width = 4 units; length = 15 units
Factor completely.
253)
x5–4x4y –21x3y2
253)
A)
x2(x –7y)(x +3y)
B)
x3(x –3y)(x +7y)
C)
x3(x +7y)(x +3y)
D)
x3(x –7y)(x +3y)
Find the missing factor.
254)
2x2+ 7x + 6 = (x + 2)( )
254)
A)
x – 4
B)
2x + 3
C)
2x – 4
D)
x + 3
Factor completely.
255)
6x3– 26x2– 20x
255)
A)
(x2– 5)(6x + 4)
B)
x(3x + 2)(2x – 10)
C)
2(3x – 2)(x + 5)
D)
2x(3x + 2)(x – 5)
26
Factor out the greatest common factor.
256)
12s7t3+4s5t4
256)
A)
s5t3(12s +4t)
B)
4s2t(3s + t)
C)
4s7t4(3s2+ t)
D)
4s5t3(3s2+ t)
Factor completely.
257)
24a2– 20a – 24
257)
A)
(12a + 8)(2a – 3)
B)
4(3a – 2)(2a + 3)
C)
(3a + 2)(8a – 12)
D)
4(3a + 2)(2a – 3)
Factor out the greatest common factor.
258)
6m9– 8m4– 12m2
258)
A)
2(3m9– 4m4– 6m2)
B)
m2(6m7– 8m2– 12)
C)
No common factor
D)
2m2(3m7– 4m2– 6)
Solve.
259)
For a 30 yard field goal, one possible path the ball could follow is modeled by the expression
–1
21 (x2+ 3x – 54), where x is the ground distance (in yards) of the ball to the goal post. Factor
this expression.
259)
A)
–1
21 (x – 9)(x + 6)
B)
–1
21 (x – 9)(x + 1)
C)
–1
21 (x + 3)(x – 6)
D)
–1
21 (x + 9)(x – 6)
Determine whether the polynomial is a sum of cubes, a difference of cubes, or neither.
260)
x3– 343
260)
A)
B)
C)
Solve the problem.
261)
Two cars leave an intersection. One car travels north; the other east. When the car traveling north
had gone 15 miles, the distance between the cars was 5 miles more than the distance traveled by the
car heading east. How far had the eastbound car traveled?
261)
A)
20 mi
B)
25 mi
C)
30 mi
D)
15 mi
Factor, if possible.
262)
a6– 8
262)
A)
(a2+ 2)(a4– 2a2+ 4)
B)
(a2– 2)(a4+ 2a2+ 4)
C)
cannot be factored
D)
(a2– 2)(a4– 2a2+ 4)
Find the missing factor.
263)
x2–8x +15 = (x –3)( )
263)
A)
(x +3)
B)
(x –5)
C)
(x –3)
D)
(x +5)
27
Solve the equation.
264)
y2=1
49
264)
A)
–1
7, 1
7
B)
–7, 7
C)
1
7
D)
–1
49 , 1
49
Classify the polynomial as a perfect square trinomial, a difference of squares, or neither.
265)
4x2+32x +64
265)
A)
B)
C)
Solve.
266)
4x2–3x =7
266)
A)
7
4, –1
B)
4
7, 0
C)
4
7, –1
D)
7
4, 1
Factor completely.
267)
6x2– 26x – 20
267)
A)
2(3x – 2)(x + 5)
B)
(6x + 4)(x – 5)
C)
(3x + 2)(2x – 10)
D)
2(3x + 2)(x – 5)
Solve the equation.
268)
–2(b + 11) = 0
268)
A)
11, 0
B)
1, –11
C)
–11
D)
–2, –11
Solve the problem.
269)
The length of a rectangle is 6 inches more than its width. If 3 inches are taken from the length and
added to the width, the figure becomes a square with an area of 81 square inches. What are the
dimensions of the original figure?
269)
A)
9 in. by 9 in.
B)
3 in. by 9 in.
C)
6 in. by 9 in.
D)
6 in. by 12 in.
Solve the equation.
270)
64k2– 4 = 0
270)
A)
2, 0
B)
4, –1
4
C)
1
4, –1
4
D)
4, 0
Provide an appropriate response.
271)
The energy it takes to lift an object with a mass of 4 kg from level y1 to level y2 is 4gy2–4gy1,
where g is a constant. Factor this expression.
271)
A)
4(y2–y1)
B)
4
g(y2–y1)
C)
g
4(y2–y1)
D)
4g(y2–y1)
Find the greatest common factor.
272)
15m5, 135m9, 225m6
272)
A)
135m5
B)
2025m4
C)
15m5
D)
15m4
Factor by grouping.
273)
r3+r2+9r +9
273)
A)
(r2+9)(r –9)
B)
(r2+9)(r +9)
C)
(r2+9)(r + 1)
D)
(r2+ 1)(r +9)
28
Provide an appropriate response.
274)
Find the greatest common factor of 15x4 and 21x3.
274)
A)
7x4
B)
105x5
C)
3x3
D)
5x3
Solve the equation.
275)
t2+25 = –10t
275)
A)
–5, 5
B)
0, –5
C)
5
D)
–5
Solve.
276)
In the first round of a game show, a contestant earned 120 points for each of x questions she
answered correctly and lost 90 points for each of y questions she answered incorrectly. She then
gained 120 points for each of z questions she answered correctly in the bonus round. Find in
factored form the amount of points the contestant earned.
276)
A)
30(4x –9y +12z) points
B)
12(x –9y +12z) points
C)
30(4x –3y +4z) points
D)
3(4x –3y +4z) points
Solve the equation.
277)
21 –x2+ 4x = 0
277)
A)
7, –3
B)
7, 3
C)
3, –7
D)
–3, –7
278)
2x2– 14x + 24 = 0
278)
A)
0, 3, 4
B)
–3, –4
C)
2, 3, 4
D)
3, 4
Factor, if possible.
279)
343–x3
279)
A)
(7 + x)(49 –7x +x2)
B)
(7 – x)(49 +x2)
C)
(7 – x)(49 –7x +x2)
D)
(7 – x)(49 +7x +x2)
Classify the polynomial as a perfect square trinomial, a difference of squares, or neither.
280)
25x2–50x +25
280)
A)
B)
C)
Solve the problem.
281)
The product of two consecutive integers is 7 less than 7 times their sum. Find the integers.
281)
A)
0, 1 or 13, 14
B)
13, 14
C)
0, 1 or 14, 15
D)
0, 1
Solve.
282)
A drug testing company approves a new drug and has confirmed by independent research that the
strength of the reaction to the drug is given by the formula S =1
2d2 –1
3d3, where d is the dosage
amount of the drug and is a constant. Find an equivalent expression for S by factoring out a
common factor.
282)
A)
S =d2
31
2– d
B)
S =d2
2(– d)
C)
S =d21
2–1
3d
D)
S =d21
2+1
3d
29
Find the missing factor.
283)
x2– 11x + 28 = (x – 7)( )
283)
A)
x2+ 7
B)
7– x
C)
x – 4
D)
x + 4
Factor completely, if possible.
284)
15y2+ 29y + 12
284)
A)
(15y + 4)(y + 3)
B)
(3y + 4)(5y + 3)
C)
Prime
D)
(3y – 4)(5y – 3)
Factor completely.
285)
4x2+ 12xy + 9y2
285)
A)
(2x – 3y)(2x – 3y)
B)
(4x + 3y)(x + 3y)
C)
(2x + 3y)(2x + 3y)
D)
(4x + y)(x + 9y)
Solve the problem.
286)
A box with length x has a volume of V =4x2+22x +30. Rewrite this volume equation in factored
form.
286)
A)
2(2x –5)(x –3)
B)
2(2x –5)(x +3)
C)
2(2x +5)(x –3)
D)
2(2x +5)(x +3)
Factor completely, if possible.
287)
6x2+ 31x + 28
287)
A)
(6x + 7)(x + 4)
B)
(x – 24)(x + 4)
C)
(x + 7)(x + 4)
D)
(6x – 24)(x + 4)
Solve the equation.
288)
8y2+ 22y + 15 = 0
288)
A)
–5
8, –1
5
B)
5
4, –3
2
C)
5
4, 3
2
D)
–5
4, –3
2
Factor completely, if possible.
289)
36x2– 60x + 25
289)
A)
(6x – 5)(6x + 5)
B)
(6x + 5)2
C)
(6x – 5)2
D)
Prime
Factor, if possible.
290)
4x3+32
290)
A)
4(x +8)(x2–8x +16)
B)
4(x +2)(x2+2x +4)
C)
4(x +2)(x2–2x +4)
D)
4(x –2)(x2+2x +4)
Factor by grouping.
291)
t(9– m) + s(9– m)
291)
A)
No common factor (except 1)
B)
(t – s)(9– m)
C)
t(9– m) + s
D)
(t + s)(9– m)
Factor completely, if possible.
292)
4x2–5x +6
292)
A)
(4x – 1)(x –6)
B)
Prime
C)
(4x –6)(x – 1)
D)
(2x –6)(2x – 1)
30
Solve the equation.
293)
(x – 7)2+x2=(x + 1)2
293)
A)
4, 12
B)
–4, –12
C)
7, 12
D)
7, –1
31
Answer Key
Testname: C6
32
Answer Key
Testname: C6
33
Answer Key
Testname: C6
34
Answer Key
Testname: C6
Answer Key
Testname: C6
36
Answer Key
Testname: C6
37