Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

$p(x)=2 x^{3}+3 x^{2}-11 x-3$

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

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

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

$\therefore$ Remainder $=3$

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