Question

The polynomial $p\left(x\right)=a{x}^3-3x^2+4$ and $g\left(x\right)=2x^3-5x+a$, when divided by $(x-2)$ and $(x-3)$ leaves the remainders $p$ and $q$, respectively. If $p-2q=4$, then find the value of $a$.

Answer 1

Let $f\left(x\right)=a{x}^3-3x^2+4$ and $g\left(x\right)=2x^3-5x+a$

By remainder theorem, $f\left(2\right)=p$ ......$\left(1\right)$

$x-2=0\Rightarrow x=2$

and $g\left(2\right)=q$ .......$\left(2\right)$

$\Rightarrow x-2=0\Rightarrow x=2$

Eqn$\left(1\right)$ gives

$a{2}^3-32^2+4=p$

$\Rightarrow 8a-8=p$ .....$\left(3\right)$

$\left(2\right)$ gives

$22^3-5\times 2+a=q$

$\Rightarrow 6+a=q$ .....$\left(4\right)$

According to the question,

$p-2q=4$

$\Rightarrow (8a-8)-2(6+a)=4$

$\Rightarrow 6a=24$ or $a=4$