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

$\frac{3m+4}{m-7}$ In the original equation, the denominator is zero when when $m=-4$ or $m=7$
Given Expression : $\frac{-1(3m^{2}-16m-16)}{m^{2}-3m-28}$ = $\frac{-1(-3m-4)(m+4)}{(m-7)(m+4)}$ (After factoring the quadratic expressions in the numerator and the denominator) Thus, this becomes : $\frac{-1(-3m-4)}{m-7}$ (The common factor $(m+4)$ gets cancelled off) Answer : $\frac{3m+4}{m-7}$