## Algebra 2 (1st Edition)

$f(x)=x^3+3x^2-6x-8$
Using the Factor Theorem, if $x=a$ is a factor, then $(x-a)$ is a factor of $f(x)$. Hence here $f(x)=(x+4)(x+1)(x-2)=(x^2+5x+4)(x-2)=x^3+5x^2+4x-2x^2-10x-8=x^3+3x^2-6x-8$