Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. $-2$

Answer 1

The correct option is C $0$

$g\left(x\right)=0$ implies $x=\frac{-2}{3}$
Thus, remainder $=f\left(\frac{-2}{3}\right)$
$\therefore$ Remainder $=3\times (\frac{-2}{3}{)}^4+2\times (\frac{-2}{3}{)}^3-\frac{1}{3}\times (\frac{-2}{3}{)}^2-\frac{1}{9}\times (\frac{-2}{3}{)}^1+\frac{2}{27}$
$\therefore$ Remainder $=\frac{16}{27}-\frac{16}{27}-\frac{4}{27}+\frac{2}{27}+\frac{2}{27}=0$