How do you factor $7x^2y^2-63y^4$ ?

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Answer 1 · 2017-08-23T00:00:54

$7x^2y^2-63y^4 = 7y^2(x-3y)(x+3y)$

The difference of squares identity can be written:

$a^2-b^2 = (a-b)(a+b)$

We will use this with $a=x^2$ and $b=(3y)^2$ , but first separate out the common factor $7y^2$ ...

$7x^2y^2-63y^4 = 7y^2(x^2-9y^2)$

$color(white)(7x^2y^2-63y^4) = 7y^2(x^2-(3y)^2)$

$color(white)(7x^2y^2-63y^4) = 7y^2(x-3y)(x+3y)$

How do you factor 7x^2y^2-63y^47x^2y^2-63y^4 ?