Question

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

Answer 1

$p\left(x\right)=x^3-6x^2+2x-4g\left(x\right)=1-\frac{3}{2}x1-\frac{3}{2}x=0\frac{3x}{2}=1$

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

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

$p\left(\frac{2}{3}\right)=(\frac{2}{3}{)}^3-6(\frac{2}{3}{)}^2+2\left(\frac{2}{3}\right)-4=\frac{8}{27}-\frac{24}{9}+\frac{4}{3}-4=\frac{8}{27}-\frac{72}{27}+\frac{36}{27}-\frac{108}{27}=\frac{8-72+36-108}{27}=\frac{-136}{27}$

∴ Remainder = $\frac{-136}{27}$

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