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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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    The fact that $a^2-b^2=(a-b)(a+b)$ is the most important of the factoring ideas.2011-11-15

3 Answers 3

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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.

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

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    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 $2$.2011-11-15