Question

If $p\left(x\right)=2x^4-3x^3-5x^2+8x-3$, find the remainder when $p\left(x\right)$ is divided by $(x-\frac{3}{2})$. Also, verify the result by actual division.

Answer 1

$p\left(x\right)=2x^4-3x^3-5x^2+8x-3$

$g\left(x\right)=x-\frac{3}{2}$

$g\left(x\right)=o,x-\frac{3}{2}=0,x=\frac{3}{2}$

By remainder theorem,

When $p\left(x\right)$ is divided by $g\left(x\right)$

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

$=2(\frac{3}{2}{)}^4-3(\frac{3}{2}{)}^3-5(\frac{3}{2}{)}^2+8(\frac{3}{2})-3=-\frac{9}{4}$

Now,

if we divide $2x^4-3x^3-5x^2+8x-3$ by $x-\frac{3}{2}$ by actual division, the remainder is $-\frac{9}{4}$ and the quotient is $2x^3-5x+\frac{1}{2}$

Hemce, verified.