Answer:
[tex]\large\boxed{\dfrac{x-4}{x+2}}[/tex]
Step-by-step explanation:
[tex]\dfrac{x^2-16}{x^2+6x+8}=\dfrac{x^2-4^2}{x^2+4x+2x+8}\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{(x-4)(x+4)}{x(x+4)+2(x+4)}=\dfrac{(x-4)(x+4)}{(x+4)(x+2)}\qquad\text{cancel}\ (x+4)\\\\=\dfrac{x-4}{x+2}[/tex]
