Question

Find the remainder when $f\left(x\right)=3x^4+2x^3-\frac{x^2}{3}-\frac{x}{9}+\frac{2}{27}$ is divided by $g\left(x\right)=x+\frac{2}{3}$.

Answer 1

As we know that from the remainder theorem, the remainder of the division of the polynomial $f\left(x\right)$ by a linear polynomial $(x-a)$ is $f\left(a\right)$
Given polynomial $f\left(x\right)=3x^4+2x^3-\frac{x^2}{3}-\frac{x}{9}+\frac{2}{27}$
and the linear polynomial
$g\left(x\right)=x+\frac{2}{3}=x-\left(\frac{-2}{3}\right)$
so, $a=\frac{-2}{3}$
Thus, $f\left(a\right)=f\left(\frac{-2}{3}\right)$
$=3(\frac{-2}{3}{)}^4+2(\frac{-2}{3}{)}^3-\frac{(\frac{-2}{3}{)}^2}{3}-\frac{\left(\frac{-2}{3}\right)}{9}+\frac{2}{27}$
$=3\left(\frac{16}{81}\right)-2\left(\frac{8}{27}\right)-\frac{1}{3}\left(\frac{4}{9}\right)+\frac{1}{9}\left(\frac{2}{3}\right)+\frac{2}{27}$
$=\frac{16}{27}-\frac{16}{27}-\frac{4}{27}+\frac{2}{27}+\frac{2}{27}$
$\therefore f\left(\frac{-2}{3}\right)=0$
Hence, remainder is $0$.
We can verify the above result of remainder by actual division also

Thus, remainder is $0$