Answer:
[tex]\frac{(x+2)(x+3)^2(x-3)}{(x-4)^2(x+4)}[/tex]
Step-by-step explanation:
i'm going to assume what you have looks like
[tex]\frac{x^2-9}{\frac{x^2-16}{\frac{x^2+5x+6}{x-4}}}[/tex]
Let's start by factoring everything
x²-9= (x+3)(x-3)
x²-16= (x+4)(x-4)
x²+5x+6=(x+3)(x+2)
x-4= x-4
which means what we have looks like
[tex]\frac{(x+3)(x-3)}{\frac{(x+4)(x-4)}{\frac{(x+3)(x+2)}{x-4}}}=\frac{(x+3)(x-3)}{(x+4)(x-4)}\div\frac{x-4}{(x+3)(x+2)}=\frac{(x+3)(x-3)}{(x+4)(x-4)}*\frac{(x+3)(x+2)}{x-4}=\frac{(x+3)^2(x-3)(x+2)}{(x+4)(x-4)^2}[/tex]
And I think that's as simplified as it can get
