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

Question

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.

Answer 1

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

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

Answer 2

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.

Answer 3

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