Answer: [tex]\frac{x+6}{x-4}[/tex]
Step-by-step explanation:
First you need know some general formuals of factorization such as
[tex]x^3+y^3=(x+y)(x^2-xy+y^2)[/tex] so you have [tex]x^3+64=X^3+4^3=(x+4)(x^2-4x+4^2)[/tex]
And, [tex]x^2-y^2=(x-y)(x+y)[/tex] so you have [tex]2x^2-32=2(x^2-16)=2(x^2-4^2)=2(x+4)(x-4)[/tex]
If you use them, then you can modify your equations such as
[tex]\frac{x^3+64}{2x^2-32}[/tex] ÷ [tex]\frac{x^2-4x+16}{2x+12}[/tex] = [tex]\frac{(x+4)(x^2-4x+4^2)}{2(x+4)(x-4)}[/tex] ÷ [tex]\frac{x^2-4x+16}{2(x+6)}[/tex] = [tex]\frac{(x+4)(x^2-4x+4^2)}{2(x+4)(x-4)} . \frac{2(x+6)}{{x^2-4x+16}} = \frac{1}{x-4}.\frac{x+6}{1}=\frac{x+6}{x-4}[/tex]
