Question

Find the remainder when $9x^3-3x^2+x-5$ is divided by $x-\frac{2}{3}$.

• A. $3$
• B. $-3$
• C. $2$
• D. $-2$

Answer 1

The correct option is B $-3$

Given, polynomial $9x^2-3x^2+x-5$ divided by $x-\frac{2}{3}$.

Then divided by $x-\frac{2}{3}$ so $x-\frac{2}{3}=0$ or $x=\frac{2}{3}$.

Replace $x$ by $\frac{2}{3}$, we get,

$p\left(x\right)=9x^3-3x^2+x-5$

$p\left(\frac{2}{3}\right)=9{\left(\frac{2}{3}\right)}^3-3{\left(\frac{2}{3}\right)}^2+\left(\frac{2}{3}\right)-5$

$⟹$ $p\left(\frac{2}{3}\right)=9\times \frac{8}{27}-3\times \frac{4}{9}+\frac{2}{3}-5$

$⟹$ $p\left(\frac{2}{3}\right)=\frac{72}{27}-\frac{12}{9}+\frac{2}{3}-5$

$⟹$ $p\left(\frac{2}{3}\right)=\frac{72-36+18-135}{27}=\frac{-81}{27}=-3$.

Therefore, the required remainder is $-3$.