Question

If the polynomial $5x^3+Mx+N$ is divisible by $x^2+x+1$, then $|M+N|=$

Answer 1

Let $f\left(x\right)=5x^3+Mx+N$
Also, $x^2+x+1=(x-\omega )(x-{\omega }^2)$
$f\left(x\right)$ is divisible by $x^2+x+1$.
Hence using factor theorem, $f\left(\omega \right)=5+M\omega +N=0[\because {\omega }^3=1]$
and
$f\left({\omega }^2\right)=5+M{\omega }^2+N=0[\because {\omega }^6=1]$
$\Rightarrow M=0,N=-5$
$\therefore |M+N|=5$