Factor and solve a cubic

Let $f(x)=x^3-7x^2+4x+12$. Then $f(-1)=-1-7-4+12=0$, so $(x+1)$ is a factor.

The quotient is $x^2-8x+12$. In long division, subtract $x^3+x^2$ to leave $-8x^2+4x+12$; subtract $-8x^2-8x$ to leave $12x+12$; subtract $12x+12$ to leave remainder zero.

$(x+1)(x^2-8x+12)=(x+1)(x-2)(x-6)=0$. Hence $x=-1$, $2$ or $6$.