Exam
Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Use synthetic division to decide whether the given number is a solution of the given equation.
1)
x3+ 6x2– 14x + 16; x = 2 + i
1)
A)
Yes
B)
No
2)
–6x3+ 9x2+ x – 4; x =1
2)
A)
Yes
B)
No
3)
9x3+ x2+ 8x – 9; x =1
3)
A)
Yes
B)
No
4)
x2+ 2x + 5; x =1+ 2i
4)
A)
Yes
B)
No
5)
x2– 6x + 25; x =3– 4i
5)
A)
Yes
B)
No
6)
x4– 3x2– 4; x = –2
6)
A)
Yes
B)
No
1
Use synthetic division to find the quotient.
7)
(–6x3+ 2x2+ 5x – 10) ÷ (x – 2)
7)
A)
–6x2– 10x – 15 +–40
x – 2
B)
–6x2– 10x – 25 +–40
x – 2
C)
–6x2– 10x – 5 +–10
x – 2
D)
–6x2– 10x – 15 +–25
x – 2
Use the remainder theorem to find P(k).
8)
k = –2; P(x) = –x3+ 2x2– 5
8)
A)
–14
B)
11
C)
–11
D)
3
Use synthetic division to find the quotient.
9)
x3+8
3x2– 4x + 1
x –1
3
9)
A)
x2+13
3x +–43
x + 1
B)
x2+ 3x – 6
C)
x2+ 3x – 3
D)
x2+13
3x +14
9+16
27
Use the remainder theorem to find P(k).
10)
k =2; P(x) = –2x5– 2x3– 4x2+ 5
10)
A)
–90
B)
–91
C)
59
D)
69
2
11)
k = –3; P(x) =x2+ 3x – 5
11)
A)
–5
B)
–13
C)
5
D)
–23
Use synthetic division to find the quotient.
12)
(x4+256) ÷ (x –4)
12)
A)
x3–4x2+16x –64 +512
x –4
B)
x3+4x2+16x +64 +512
x –4
C)
x3+4x2+16x +64
D)
x3+4x2+16x +64 +256
x –4
13)
x4+ 8x3+ 14x2+ 13x + 6
x + 6
13)
A)
x3+ 2x2+ 4x
B)
x3+ 2x2+ 4x + 3
C)
x3+ 2x2+ 2x + 1
D)
x3+ 3x2+ 2x – 1
14)
x3– x2+ 6
x + 2
14)
A)
x2– 3x + 6 +–6
x + 2
B)
x2– 2x + 6 +6
x + 2
C)
x2– 3x + 6 +6
x + 2
D)
x2+ 3x + 6 +–6
x + 2
3
Use the remainder theorem to find P(k).
15)
k = – 1
2; P(x) =6x3– 27x2– 13x
15)
A)
2
B)
0
C)
–1
D)
–2
16)
k = –4; P(x) =x3– 2x2+ 4x + 3
16)
A)
–115
B)
16
C)
19
D)
–109
Use synthetic division to decide whether the given number is a solution of the given equation.
17)
3x4– 10x3– 2x + 1; x =1
3
17)
A)
Yes
B)
No
Use the remainder theorem to find P(k).
18)
k = –3; P(x) =7x4+ 8x3+ 6x2– 7x + 66
18)
A)
642
B)
–1128
C)
492
D)
792
19)
k = –3; P(x) =3x3– 6x2– 4x + 7
19)
A)
–116
B)
–140
C)
–46
D)
24
4
Use synthetic division to find the quotient.
20)
x3– 1
x – 1
20)
A)
x2+ x + 1 +1
x – 1
B)
x3+x2+ x + 1 +1
x – 1
C)
x2+ x + 1
D)
x3+x2+ x + 1
Use synthetic division to decide whether the given number is a solution of the given equation.
21)
x3+ 3x2+ 25x + 75; x =5i
21)
A)
Yes
B)
No
Use the remainder theorem to find P(k).
22)
k =4+ i; P(x) =x3+5
22)
A)
52 +47i
B)
52 +48i
C)
57 +48i
D)
57 +47i
Use synthetic division to find the quotient.
23)
(2x3+ x2– 2x + 2) ÷ (x + 2)
23)
A)
2x2– 3x – 4 +– 6
x + 2
B)
2x2– 3x – 4
C)
2x2– 3x + 4
D)
2x2– 3x + 4 +– 6
x + 2
5
24)
5x3– 33x2+ 22x – 24
x – 6
24)
A)
5
6x2+ – 11
2x +11
3
B)
5x – 3
C)
–5x2+ 6x + 4
D)
5x2– 3x + 4
Use synthetic division to decide whether the given number is a solution of the given equation.
25)
3x4+ 3x2– 6; x =2
3
25)
A)
Yes
B)
No
Use synthetic division to find the quotient.
26)
(3x4– 2x3– 10x2+ 15) ÷ (x – 2)
26)
A)
3x3+ 4x2– 2x – 4 +–11
x – 2
B)
3x3+ 4x2– 2x – 4 +7
x – 2
C)
3x3+ 4x2– 2x + 4 +–8
x – 2
D)
3x3+ 4x2– x – 3 +1
x – 2
Use the remainder theorem to find P(k).
27)
k = –3; P(x) =x6+ 2x5+ 3x4+ 4x3– 2x2+ 3x – 5
27)
A)
–346
B)
347
C)
346
D)
1552
28)
k = –5+ 2i; P(x) =x2– 2x – 3
28)
A)
4
B)
31 – 14i
C)
28 – 24i
D)
28 – 14i
6
Use synthetic division to decide whether the given number is a solution of the given equation.
29)
–x4+ 9x2– x + 6; x =1
29)
A)
Yes
B)
No
Use synthetic division to find the quotient.
30)
x5+ 7x4+ 8x3– 8x2+ 12x + 13
x + 5
30)
A)
x3+ 2x2– 2x + 2 +3
x + 5
B)
x4+ 2x3– 2x2+ 2x + 2 +3
x + 5
C)
x4+ 2x3– 2x2+ 2x – 2 +5
x + 5
D)
x4+ 2x3– 2x2+ 2x + 3
7
Answer Key
Testname: C16
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