Question

Polynomial $P\left(x\right)$ contains only terms of odd degree. When $P\left(x\right)$ is divided by $(x-3)$, the remainder is $6$. If $P\left(x\right)$ is divided by $(x^2-9)$, then the remainder is $g\left(x\right)$. Then the value of $g\left(2\right)$ is

Answer 1

$P\left(x\right)$ contains only terms of odd degree.
So, $P\left(x\right)$ is an odd function.
$\Rightarrow P(-3)=-P\left(3\right)=-6$

Let $P\left(x\right)=Q\left(x\right)(x^2-9)+ax+b$
where $Q\left(x\right)$ is the quotient and $ax+b=g\left(x\right)$ is the remainder.

Now, $P\left(3\right)=3a+b=6\cdots \left(1\right)$
$P(-3)=-3a+b=-6\cdots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$a=2,b=0$
$\therefore g\left(x\right)=2x$
$\Rightarrow g\left(2\right)=4$