Answer
$\frac{3x-24}{x-3} $.
Work Step by Step
First factor all numerators and denominators.
Factor $x^2-6x-16$.
Rewrite the middle term $-6x$ as $-8x+2x$.
$\Rightarrow x^2-8x+2x-16$
Group the terms.
$\Rightarrow (x^2-8x)+(2x-16)$
Factor each group.
$\Rightarrow x(x-8)+2(x-8)$
Factor out $(x-8)$.
$\Rightarrow (x-8)(x+2)$
Factor $x^3+3x^2+2x$.
Factor out $x$.
$\Rightarrow x(x^2+3x+2)$
Rewrite the middle term $3x$ as $2x+1x$.
$\Rightarrow x(x^2+2x+1x+2)$
Group the terms.
$\Rightarrow x[(x^2+2x)+(1x+2)]$
Factor each group.
$\Rightarrow x[x(x+2)+1(x+2)]$
Factor out $(x+2)$.
$\Rightarrow x(x+2)(x+1)$
Factor $x^2-3x-4$.
Rewrite the middle term $-3x$ as $-4x+1x$.
$\Rightarrow x^2-4x+1x-4$
Group the terms.
$\Rightarrow (x^2-4x)+(1x-4)$
Factor each group.
$\Rightarrow x(x-4)+1(x-4)$
Factor out $(x-4)$.
$\Rightarrow (x-4)(x+1)$
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 $(x-4)$.
$\Rightarrow (x-4)(x-3)$
Back substitute all factors into the given fraction.
$\Rightarrow \frac{(x-8)(x+2)}{x(x+2)(x+1)}\cdot(x-4)(x+1)\div \frac{(x-4)(x-3)}{3x}$
Invert the divisor and multiply.
$\Rightarrow \frac{x^2-6x-16}{x^3+3x^2+2x}\cdot(x^2-3x-4)\div \frac{x^2-7x+12}{3x}=\frac{(x-8)(x+2)}{x(x+2)(x+1)}\cdot(x-4)(x+1)\cdot \frac{3x}{(x-4)(x-3)}$
Cancel common factors.
$\Rightarrow \frac{3(x-8)}{x-3} $
Use the distributive property.
$\Rightarrow \frac{3x-24}{x-3} $.
