Question

Using remainder theorem, find the value of $k$ if on dividing $2x^3+3x^2-kx+5$ by $x-2$, leaves a remainder $7$

Answer 1

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

By remainder theorem, when $f\left(x\right)$ is divided by $(x-2)$

$x-2=0,\Rightarrow x=2$

the remainder $=f\left(2\right)$

$=2(2{)}^3+3(2{)}^2-k\left(2\right)+5$

$=16+12-2k+5$

$=33-2k$

According to question

$33-2k=7$

$\Rightarrow 2k=26$

$\Rightarrow k=13$