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

Question

2090 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-24T22:51:47

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

Answer 2 · 2015-05-24T22:55:29

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

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

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

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