We need to factor both numerator and denominator:

`(2x^2-8)/(4x+8)=(2(x^2-4))/(4(x+2))`

At this point we can only cancel the [leading] 2 and 4:

`(2(x^2-4))/(4(x+2))=(x^2-4)/(2(x+2))`

Now, we recognize that the numerator is the difference of 2 squares:

`(x^2-4)/(2(x+2))=((x+2)(x-2))/(2(x+2)`

Now, since the terms in brackets are connected by **multiplication***,* we can cancel the (*x* + 2) from top
and bottom:

`((x+2)(x-2))/(2(x+2))=(x-2)/2`

Get the Daily Math Tweet!

IntMath on Twitter