## Algebra 2 (1st Edition)

The solutions are $-3$ and $4$.
$3x^{2}-3x-36=0\qquad$ ...factor out $3$ from left side. $3(x^{2}-x-12)=0\qquad$ ...divide entire expression with $3$. $x^{2}-x-12=0\qquad$ ...rewrite the expression as $(kx+m)(lx+n)$ where $k$ and $l$ are two numbers such that their sum is $-1$ and product is $12$. $\qquad$ ...those numbers are $-4$ and $3$. $\qquad$ ...$-4+3=-1,\ -4\cdot 3=-12$ $\qquad$ ...therefore, $x^{2}-x-12=(x-4)(x+3)$ $\qquad$ ...apply the zero product property and solve for $x$. $x-4=0$ or $x+3=0$ $x=4$ or $x=-3$ The solutions are $-3$ and $4$.