# How to factor $2x^2 - 8y^2$

How to factor $2x^2 - 8y^2$ ?

So far I got it down to $$2(x^2 - 4y^2),$$ but it's not the answer; I don't think it's factored enough.

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It can also be written as $2(x + 2y)(x - 2y)$, man. –  Ian Mateus Nov 15 '11 at 0:31
The fact that $a^2-b^2=(a-b)(a+b)$ is the most important of the factoring ideas. –  André Nicolas Nov 15 '11 at 0:40

From $2(x^2-4y^2)$, use the difference of squares to write this as

$2(x-2y)(x+2y)$

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In addition to the other answers and comments, a basic observation is that $4y^2$ is the square of $2y$. The problem is about "recognizing" $x^2 - 4y^2$ as a difference of two squares by seeing that the coefficient $4$ is a square and can be absorbed through a doubling of the thing ($y$) being squared.
Notice that $2x^2 - 8y^2 = 0$ when $x = 2y$.
That tells you that $x-2y$ is a factor of $2x^2 - 8y^2$. Doing the long division gives you $2x+4y$ so the complete factorization is $2(x-2y)(x+2y)$
If you also note that $2x^2 - 8y^2 = 0$ when $x = -2y$, you can avoid the division entirely. :-) $2x^2 = 8y^2$ so $x^2 = 4y^2$ so $x = \pm 2y$ so $(x+2y)$ and $(x-2y)$ are factors and comparing coefficients we need to tack on a factor of $2$. –  ShreevatsaR Nov 15 '11 at 5:08