Question

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

Answer 1

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

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

$1-\frac{3}{2}x=0$

$=>\frac{3}{2}x=1$

$=>x=\frac{2}{3}$

By, remainder theorem, the remainder when $p\left(x\right)$ divided by $g\left(x\right)$ is $p\left(\frac{2}{3}\right)$

Thus,

$p\left(\frac{2}{3}\right)=(\frac{2}{3}{)}^3-6(\frac{2}{3}{)}^2+2\left(\frac{2}{3}\right)-4$

$=\frac{8}{27}-\frac{6\times 4}{9}+\frac{4}{3}-4$

$=\frac{8}{27}-\frac{8}{9}+\frac{4}{3}-4$

$=\frac{8-24+36-108}{27}$

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