Question

How do you divide `(x^2-9)/(4x+12)div(x-3)/6`?

Answer 1

`((x - 3)(x + 3))/(2(x + 6)) . (6)/(x - 3)` = `(6(x + 3))/(2(x + 6))`

Answer 2

First, notice that `x^2-9 = x^2-3^2 = (x-3)(x+3)`
This comes from a standard identity: `(a^2 - b^2) = (a-b)(a+b)`

So `(x^2 - 9)/(x-3) = ((x-3)(x+3))/(x-3) = (x+3)`

Hence

`(x^2-9)/(4x+12)-:(x-3)/6`

`=(x+3)/(4x+12)-:1/6 =(6(x+3))/(4(x+3)) = 6/4 = 3/2`