If (x − 2) and (x + 3) are factors of \(x^{3} + ax^{2} + bx - 30\), find a + b.
3
4
5
6
Explanation: Since $(x-2)$ and $(x+3)$ are factors: $p(2)=8+4a+2b-30=0 \Rightarrow 2a+b=11$ and $p(-3)=-27+9a-3b-30=0 \Rightarrow 3a-b=19$. Adding gives $5a=30$, so $a=6$ and $b=-1$. Therefore $a+b=5$.
Question 2 of 6medium
The polynomial \(x^{3} + 3x^{2} - 4x - 12\) has how many distinct real zeros?
Find the remainder when \(2x^{3} - 5x^{2} - 4x + 3\) is divided by (2x − 1).
0
1
2
3
Explanation: The remainder on division by $(2x-1)$ is $p\left(\frac{1}{2}\right)$. Compute $p\left(\frac{1}{2}\right)=2 \cdot \frac{1}{8}-5 \cdot \frac{1}{4}-4 \cdot \frac{1}{2}+3=\frac{1}{4}-\frac{5}{4}-2+3=0$. So the remainder is $0$.
Question 4 of 6medium
Find the value of k such that \(x^{3} + kx^{2} - 2x + 4\) is divisible by (x + 2).
0
1
2
-2
Explanation: For $(x+2)$ to divide the polynomial, $p(-2)=0$. Here $p(-2)=(-2)^{3}+k(-2)^{2}-2(-2)+4=-8+4k+4+4=4k$. Setting $4k=0$ gives $k=0$.
Question 5 of 6medium
Factorize completely: \(x^{4} - 10x^{2} + 9\)
(\(x^{2}+1\))(\(x^{2}-9\))
(\(x^{2}-1\))(\(x^{2}-9\))
(x-1)(x+1)(x-3)(x+3)
Both B and C
Explanation: Let y=\(x^{2}\): \(y^{2}-10y+9\)=(y-1)(y-9)=(\(x^{2}-1\))(\(x^{2}-9\))=(x-1)(x+1)(x-3)(x+3). Both B and C are equivalent forms
Question 6 of 6medium
For what value of k does \(2x^{3} + kx^{2} - 13x + 6\) have (2x − 1) as a factor?
1
11
-5
5
Explanation: For $(2x-1)$ to be a factor, $p\left(\frac{1}{2}\right)=0$. Here $p\left(\frac{1}{2}\right)=2 \cdot \frac{1}{8}+k \cdot \frac{1}{4}-13 \cdot \frac{1}{2}+6=\frac{1}{4}+\frac{k}{4}-\frac{13}{2}+6=\frac{k}{4}-\frac{1}{4}$. Setting this to $0$ gives $k=1$.
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