Answer
$\frac{x-4}{4\left( x+2 \right)}$
Work Step by Step
$\frac{{{x}^{2}}-6x+8}{4x+12}\cdot \frac{x+3}{{{x}^{2}}-4}$
Simplify the expression as,
$\begin{align}
& \frac{{{x}^{2}}-6x+8}{4x+12}\cdot \frac{x+3}{{{x}^{2}}-{{2}^{2}}}=\frac{\left( x-2 \right)\left( x-4 \right)}{4\left( x+3 \right)}\cdot \frac{x+3}{\left( x+2 \right)\left( x-2 \right)} \\
& =\frac{\left( x-4 \right)}{4}\cdot \frac{1}{\left( x+2 \right)} \\
& =\frac{x-4}{4\left( x+2 \right)}
\end{align}$
Thus,
$\frac{{{x}^{2}}-6x+8}{4x+12}\cdot \frac{x+3}{{{x}^{2}}-4}=\frac{x-4}{4\left( x+2 \right)}$
Thus, $\frac{{{x}^{2}}-6x+8}{4x+12}\cdot \frac{x+3}{{{x}^{2}}-4}$ reduces to $\frac{x-4}{4\left( x+2 \right)}$.
