Answer
$-\frac{x+3}{x+4}$ and $x\ne4$.
Work Step by Step
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence here: $\frac{x^2-x-12}{16-x^2}=\frac{x^2+3x-4x-12}{-(x^2-4^2)}=\frac{x(x+3)-4(x+3)}{-(x+4)(x-4)}=\frac{(x+3)(x-4)}{-(x+4)(x-4)}=-\frac{x+3}{x+4}$ and $x\ne4$.
