Online Test — Factorization
10 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 10
medium
According to the Factor Theorem, (x − a) is a factor of p(x) if:
p(a)=0
p(0)=a
p(1)=a
a=0
Explanation: Factor Theorem: p(a)=0 ⇔ (x−a) is factor
Question 2 of 10
medium
Find the remainder when \(x^{3} - 2x^{2} + 4x - 5\) is divided by (x − 1).
-2
-1
0
1
Explanation: p(1)=1−2+4−5=−2
Question 3 of 10
medium
Which of the following is a factor of \(x^{3} - 4x^{2} + x + 6\)?
(x−1)
(x−2)
(x−3)
(x+2)
Explanation: Test $x=2$: $p(2)=8-16+2+6=0$, so $(x-2)$ is a factor. The other options give non-zero values, e.g. $p(-2)=-8-16-2+6=-20 \neq 0$.
Question 4 of 10
medium
For what value of k is (x − 1) a factor of \(x^{3} - 2x^{2} + kx + 4\)?
-3
-2
2
3
Explanation: p(1)=1−2+k+4=3+k=0→k=−3
Question 5 of 10
medium
The remainder when \(2x^{3} - 3x^{2} + 4x - 1\) is divided by (2x − 1) is:
0
1/4
-1/4
1/2
Explanation: Set $2x-1=0$, so $x=\frac{1}{2}$. Then $p\left(\frac{1}{2}\right)=2 \cdot \frac{1}{8}-3 \cdot \frac{1}{4}+4 \cdot \frac{1}{2}-1=\frac{1}{4}-\frac{3}{4}+2-1=\frac{1}{2}$.
Question 6 of 10
medium
Factorize: \(x^{3} - 6x^{2} + 11x - 6\)
(x−1)(x−2)(x+3)
(x−1)(x−2)(x−3)
(x+1)(x−2)(x−3)
(x+1)(x+2)(x+3)
Explanation: Standard factorization: roots 1,2,3
Question 7 of 10
medium
If p(x) is divided by (x + 3), the remainder is p(___):
3
-3
0
1
Explanation: For (x + a), remainder = p(−a), so for (x+3), remainder = p(−3)
Question 8 of 10
medium
One factor of \(2x^{3} - 3x^{2} - 3x + 2\) is:
(x+1)
(x−2)
(x−1)
(x+2)
Explanation: Test $x=2$: $p(2)=16-12-6+2=0$, so $(x-2)$ is a factor. The other candidates are non-zero, e.g. $p(-2)=-16-12+6+2=-20 \neq 0$.
Question 9 of 10
medium
The remainder when \(x^{4} + x^{3} - 2x^{2} + x + 1\) is divided by (x − 1) is:
0
1
2
3
Explanation: p(1)=1+1−2+1+1=2
Question 10 of 10
medium
Factorize: \(x^{3} - 23x^{2} + 142x - 120\). One factor is (x − 1). The other factors are:
(x−10)(x−12)
(x−8)(x−15)
(x−9)(x−14)
(x−11)(x−13)
Explanation: Divide by (x−1): \(x^{2}-22x+120\) = (x−10)(x−12)