Answer
$\frac{x-4}{3(x+2)}$.
Work Step by Step
First we factor all numerators and denominators of the given expression.
Factor $x^2-6x+8$.
Rewrite the middle term $-6x$ as $-4x-2x$.
$=x^2-4x-2x+8$
Group the terms.
$=(x^2-4x)+(-2x+8)$
Factor each group.
$=x(x-4)-2(x-4)$
Factor out $(x-4)$.
$=(x-4)(x-2)$
Factor $3x+9$.
$=3(x+3)$
Factor $x^2-4$.
$=x^2-2^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x+2)(x-2)$
Substitute all factors into the given expression.
$=\frac{(x-4)(x-2)}{3(x+3)}\div \frac{(x+2)(x-2)}{x+3}$
Replace the division by multiplication by the reciprocal.
$=\frac{(x-4)(x-2)}{3(x+3)}\times \frac{x+3}{(x+2)(x-2)}$
Cancel the common terms $x+3$ and $x-2$.
$=\frac{x-4}{3(x+2)}$.
