Question

Find the value of $k$ if $(2x-3)$ is a factor of $2x^3-9x^2+x+k.$

Answer 1

Given: $(2x-3)$ is a factor of $f\left(x\right)=2x^3-9x^2+x+k$

$(2x-3)=0$ when $x=\frac{3}{2}$

So by factor theorem $f\left(\frac{3}{2}\right)=0$

$\Rightarrow 2\times (\frac{3}{2}{)}^3-9(\frac{3}{2}{)}^2+\frac{3}{2}+k=0$

$\Rightarrow 2\times \left(\frac{27}{8}\right)-9\times \left(\frac{9}{4}\right)+\frac{3}{2}+k=0$

$\Rightarrow \frac{27}{4}-\frac{81}{4}+\frac{6}{4}+k=0$

$\Rightarrow \left(\frac{33}{4}\right)-\left(\frac{81}{4}\right)+k=0$

$\Rightarrow k=\frac{48}{4}=12$