Question

Divide: $f\left(x\right)=9x^3-3x^2+x-5$ by $g\left(x\right)=x-\frac{2}{3}$

Answer 1

Let us denote the given polynomials as

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

$g\left(x\right)==x-\frac{2}{3}$

We have to find the remainder when $f\left(x\right)$ is divided by $g\left(x\right).$.

By the remainder theorem, when $f\left(x\right)$ is divided by $g\left(x\right)$ the remainder is

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

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

$=\frac{8}{3}-\frac{4}{3}+\frac{2}{3}-5$

$=\frac{8-4+2}{3}-5$

$=2-5=-3$

Remainder by actual division

Remainder is $-3$