Answer
$\frac{3m+4}{m-7}$
In the original equation, the denominator is zero when when $m=-4$ or $m=7$
Work Step by Step
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 : $\frac{3m+4}{m-7}$
