x^{2}-x-12 . 3+x
x^{2}-9 4-x
ans (x-4)(3+X)
(X-3)(4-X)
THERE IN THE BOOK IS (3+X)
(X-3)
$\displaystyle \displaystyle \begin{align*} \frac{x^2 - x - 12}{x^2 - 9} \cdot \frac{3 + x}{4 - x} &= \frac{(x - 4)(x + 3)}{(x - 3)(x + 3)}\cdot \frac{3 + x}{4 - x} \\ &= \frac{-(4 - x)(x + 3)(3 + x)}{(x - 3)(x + 3)(4 - x)} \\ &= \frac{-(3+x)}{x - 3} \end{align*}$
Both your answer and the book's answer are wrong.