Answer
$\frac{x}{x-4}$.
Work Step by Step
First factor all numerators and denominators.
Factor $x^2-16$.
$\Rightarrow x^2-4^2$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$\Rightarrow (x+4)(x-4)$
Factor $x^2+7x+12$
Rewrite the middle term $7x$ as $4x+3x$
$\Rightarrow x^2+4x+3x+12$
Group the terms.
$\Rightarrow (x^2+4x)+(3x+12)$
Factor each group.
$\Rightarrow x(x+4)+3(x+4)$
Factor out $(x+4)$.
$\Rightarrow (x+4)(x+3)$
Factor $x^2+3x$.
Factor out $x$.
$\Rightarrow x(x+3)$
Back substitute all factors into the fraction.
$\Rightarrow \Rightarrow \frac{x^2}{x^2-16}\cdot\frac{x^2+7x+12}{x^2+3x}=\frac{x^2}{(x+4)(x-4)}\cdot\frac{(x+4)(x+3)}{x(x+3)}$
Cancel the common factors $x$, $x+4$, $x+3$: .
$\Rightarrow \frac{x}{x-4}$.
