Question

When multiplying 2x-8/x^2-4 times x^2+6x+8/x-4, what are the term(s) that can be cancelled

Answer 1

Answer:

(x-4) and (x+2) are terms that can be cancelled

Step-by-step explanation:

we are given

$\frac{2x-8}{x^2-4}\times \frac{x^2+6x+8}{x-4}$

Firstly, we will factor terms

and then we can cancel common terms

$2x-8=2(x-4)$

$x^2-4=(x-2)(x+2)$

$x^2+6x+8=(x+4)(x+2)$

now, we can replace it

$\frac{2x-8}{x^2-4}\times \frac{x^2+6x+8}{x-4}=\frac{2(x-4)}{(x-2)(x+2)}\times \frac{(x+2)(x+4)}{x-4}$

now, we can see that

(x-4) and  (x+2) are common

so, they will get cancelled

$\frac{2x-8}{x^2-4}\times \frac{x^2+6x+8}{x-4}=\frac{2}{(x-2)}\times \frac{(x+4)}{1}$

so, we can see that

(x-4) and (x+2) are terms that can be cancelled