Question

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=x^3-6x^2+2x-4,g\left(x\right)=1-\frac{3}{2}x$.

Answer 1

$p\left(x\right)=x^3-6x^2+2x-4$

​$g\left(x\right)=1-\frac{3}{2}x=-\frac{3}{2}\left(x-\frac{2}{3}\right)$

By remainder theorem, when p(x) is divided by $\left(1-\frac{3}{2}x\right)$, then the remainder = $p\left(\frac{2}{3}\right)$.

Putting $x=\frac{2}{3}$ in p(x), we get

$p\left(\frac{2}{3}\right)={\left(\frac{2}{3}\right)}^3-6\times {\left(\frac{2}{3}\right)}^2+2\times \left(\frac{2}{3}\right)-4=\frac{8}{27}-\frac{8}{3}+\frac{4}{3}-4$ $=\frac{8-72+36-108}{27}=-\frac{136}{27}$

∴ Remainder = $-\frac{136}{27}$

Thus, the remainder when p(x) is divided by g(x) is $-\frac{136}{27}$.