Answer:
[tex]\dfrac{x^3-5x^2 +4x}{2x+8}[/tex]
Step-by-step explanation:
The basic idea is to factor each expression and cancel common factors from numerator and denominator.
[tex]\dfrac{x^2-16}{2x+8}\cdot\dfrac{x^3-2x^2 +x}{x^2+3x-4}=\dfrac{(x-4)(x+4)}{2(x+4)}\cdot\dfrac{x(x-1)^2}{(x+4)(x-1)}\\\\=\dfrac{(x-4)x(x-1)}{2(x+4)}\cdot\dfrac{(x+4)(x-1)}{(x+4)(x-1)}\\\\=\boxed{\dfrac{x^3-5x^2 +4x}{2x+8}}[/tex]
