Question

Find the remainder when $x^3-6x^2+2x-4$ is dividend by $3x-1$.

Answer 1

Let $f\left(x\right)=x^3-6x^2+2x-4$.

The polynomial $f\left(x\right)$ is divided by $3x-1$.

Then $3x-1=0⟹3x=1⟹x=\frac{1}{3}.$

By remainder theorem, the remainder is given by $f\left(\frac{1}{3}\right)$.
Hence, remainder of the given polynomial $f\left(x\right)$ is :

$f\left(\frac{1}{3}\right)$ $=(\frac{1}{3}{)}^3-6(\frac{1}{3}{)}^2+2\left(\frac{1}{3}\right)-4$

$=(\frac{1}{3}{)}^3-6(\frac{1}{3}{)}^2+2\left(\frac{1}{3}\right)-4$

$=\left(\frac{1}{27}\right)-6\left(\frac{1}{9}\right)+2\left(\frac{1}{3}\right)-4$

$=\left(\frac{1}{27}\right)-\left(\frac{6}{9}\right)+\left(\frac{2}{3}\right)-4$

$=\frac{1-(6\times 3)+(2\times 9)-(4\times 27)}{27}$

$=\frac{1-18+18-108}{27}$

$=\frac{-107}{27}$.

That is, the remainder obtained is $\frac{-107}{27}$.