Question

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=2x^3+3x^2-11x-3,g\left(x\right)=\left(x+\frac{1}{2}\right)$.

Answer 1

$p\left(x\right)=2x^3+3x^2-11x-3$

​$g\left(x\right)=\left(x+\frac{1}{2}\right)=\left[x-\left(-\frac{1}{2}\right)\right]$

By remainder theorem, when p(x) is divided by $\left(x+\frac{1}{2}\right)$, then the remainder = $p\left(-\frac{1}{2}\right)$.

Putting $x=-\frac{1}{2}$ in p(x), we get

$p\left(-\frac{1}{2}\right)=2\times {\left(-\frac{1}{2}\right)}^3+3\times {\left(-\frac{1}{2}\right)}^2-11\times \left(-\frac{1}{2}\right)-3=-\frac{1}{4}+\frac{3}{4}+\frac{11}{2}-3$ $=\frac{-1+3+22-12}{4}=\frac{12}{4}=3$

∴ Remainder = 3

Thus, the remainder when p(x) is divided by g(x) is 3.