Question

Find the remainder when $p\left(x\right)=x^3-6x^2+2x-4$ is divided by $g\left(x\right)=3x-1$

Answer 1

Here, $g\left(x\right)=3x-1$. To apply Remainder theorem, $(3x-1)$ should be converted to $(x-a)$ form.
$\therefore 3x-1=\frac{1}{3}(3x-1)=x-\frac{1}{3}\therefore g\left(x\right)=(x-\frac{1}{3})$
By remainder theorem, $r\left(x\right)=p\left(a\right)=p\left(\frac{1}{3}\right)$
$p\left(x\right)=x^3-6x^2+2x-4\Rightarrow p\left(\frac{1}{3}\right)={\left(\frac{1}{3}\right)}^3-6{\left(\frac{1}{3}\right)}^2+2\left(\frac{1}{3}\right)-4$
$=\frac{1}{27}-\frac{6}{9}+\frac{2}{3}-4=\frac{1-18+18-108}{27}=\frac{-107}{27}$
$\therefore$ the remainder $p\left(\frac{1}{3}\right)=-\frac{107}{27}$