Question

If (x + 2) is a factor of p(x) = ax³ + bx² + x − 6 and p(x) when divided by (x − 2) leaves a remainder 4, prove that a = 0 and b = 2.

Answer 1

Let:
$p\left(x\right)=a{x}^3+b{x}^2+x-6$
Now,
$x+2=0\Rightarrow x=-2$

$$\begin{gathered} \left(x+2\right)\text{is a factor}p\left(x\right). \\ \Rightarrow p\left(-2\right)=0 \\ \Rightarrow a\times {\left(-2\right)}^3+b\times {\left(-2\right)}^2+\left(-2\right)-6=0 \\ \Rightarrow -8a+4b-8=0 \\ \Rightarrow 8a-4b=-8...\left(\mathrm{i}\right) \end{gathered}$$

Also,
$x-2=0\Rightarrow x=2$
When p(x) is divided by (x $-$ 2), the remainder is 4.
p(2) = 4

$$\begin{gathered} \mathrm{Here}, \\ p\left(2\right)=a\times 2^3+b\times 2^2+2-6 \\ =8a+4b-4 \end{gathered}$$

Thus, we have:
$$\begin{gathered} 8a+4b-4=4 \\ 8a+4b=8...\left(\mathrm{ii}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ 16a=0 \\ \Rightarrow a=0 \end{gathered}$$

By substituting the value of a, we get the value of b, i.e., 2.
∴ a = 0 and b = 2