Answer:
(x-4) and (x+2) are terms that can be cancelled
Step-by-step explanation:
we are given
[tex]\frac{2x-8}{x^2-4}\times \frac{x^2+6x+8}{x-4}[/tex]
Firstly, we will factor terms
and then we can cancel common terms
[tex]2x-8=2(x-4)[/tex]
[tex]x^2-4=(x-2)(x+2)[/tex]
[tex]x^2+6x+8=(x+4)(x+2)[/tex]
now, we can replace it
[tex]\frac{2x-8}{x^2-4}\times \frac{x^2+6x+8}{x-4}=\frac{2(x-4)}{(x-2)(x+2)}\times \frac{(x+2)(x+4)}{x-4}[/tex]
now, we can see that
(x-4) and (x+2) are common
so, they will get cancelled
[tex]\frac{2x-8}{x^2-4}\times \frac{x^2+6x+8}{x-4}=\frac{2}{(x-2)}\times \frac{(x+4)}{1}[/tex]
so, we can see that
(x-4) and (x+2) are terms that can be cancelled
