Show that $(x+2)$ is a factor of $f(x)=x^{3}+4 x^{2}+x-6$
Given: $f(x)=x^{3}+4 x^{2}+x-6$
Now, $f(-2)=(-2)^{3}+4(-2)^{2}+(-2)-6$
$=-8+16-2-6$
$=0$
$\therefore(x+2)$ is a factor of $f(x)=x^{3}+4 x^{2}+x-6$.
Leave a comment
All Study Material