Question

Find the remainder when $2x^2-3x+5$ is divided by $2x-3$. Does it exactly divide the polynomial? State reason

Answer 1

Given polynomial $2x^2-3x+5$ is divided by $2x-3$

Then $2x-3=0$ or $x=\frac{3}{2}$

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

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

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

$\Rightarrow p\left(\frac{3}{2}\right)=\frac{18}{4}-\frac{9}{2}+5$

$\Rightarrow p\left(\frac{3}{2}\right)=\frac{18-18+20}{4}$

$\Rightarrow p\left(\frac{3}{2}\right)=5$

Thus remainder is $5$

Then the value of polynomial $2x^2-3x+5$ is not zero by polynomial $2x-3$

So $2x-3$ is not the factor of polynomial $2x^2-3x+5$