Question

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

Answer 1

$p\left(x\right)=2x^3+3x^2-11x-3g\left(x\right)=(x+\frac{1}{2})$.

By remainder theorem, when p(x) is divided by ( 3x+2), then the remainder = $p\left(\frac{-1}{2}\right)$.

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

$p\left(\frac{-1}{2}\right)=2(\frac{-1}{2}{)}^3+3(\frac{-1}{2}{)}^2-11\left(\frac{-1}{2}\right)-3=\frac{-2}{8}+\frac{3}{4}+\frac{11}{2}-3=\frac{-2}{8}+\frac{6}{8}+\frac{44}{8}-\frac{24}{8}=\frac{-2+6+44-24}{8}=\frac{24}{8}=3$

∴ Remainder = $3$

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