Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
Question:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

$p(x)=x^{3}-6 x^{2}+2 x-4, g(x)=1-\frac{3}{2} x$

Solution:

$p(x)=x^{3}-6 x^{2}+2 x-4$

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

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

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

$\therefore$ Remainder $=-\frac{136}{27}$

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