## Algebra 1: Common Core (15th Edition)

$\frac{x^{2}+5x+6}{x^{2}-x-12}$
Given expression : $\frac{(x+3)(x+2)}{(x-4)(x+3)}$ Note that $x=-3$ and $x=4$ are excluded from the solution, as they would cause the denominator to be 0. = $\frac{x(x+2)+3(x+2)}{x(x+3)-4(x+3)}$ =$\frac{x^{2}+2x+3x+6}{x^{2}+3x-4x-12}$ This becomes : $\frac{x^{2}+5x+6}{x^{2}-x-12}$