Question

If $p\left(x\right)=x^3-5x^2+4x-3$ and $g\left(x\right)=x-2$, show that $p\left(x\right)$ is not a multiple of $g\left(x\right)$.

Answer 1

Given, $P\left(x\right)=x^3-5x^2+4x-3$

$g\left(x\right)=x-2$

Put $g\left(x\right)=0$

$⟹x=2$

Now, $p\left(2\right)$ should be $0$ if $p\left(x\right)$ is a multiple of $g\left(x\right)$

so,

$p\left(2\right)=2^3-5(2{)}^2+4(2)-3$

$p\left(2\right)=8-20+8-3$

$p\left(2\right)=16-23$

$p\left(2\right)=-7$

since remainder is not zero, $p\left(x\right)$ is a notmultiple of $g\left(x\right)$.