From formula a^2-b^2 = (a-b)(a+b) :
x^2 - 16 = (x-4)(x+4)
In 2x+8 factor out "2", you'll reach 2(x+4). So:
[tex]\frac{8}{2x+8} \cdot \frac{x^2-16}{4}=\frac{8(x-4)(x+4)}{2(x+4) \cdot 4}= \frac{8(x+4)(x-4)}{8(x+4)}= \boxed{x-4}[/tex]
(In finish you delete 8(x+4) from nominator and denominator)
