How do you factor $16x^4 - 4y^2$ ?

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Answer 1 · 2015-06-26T04:33:35

$(16x^4-y^2)=4(2x^2-y)(2x^2+y)$

$16x^4-y^2$

Factor out the GCF $4$ .

$4(4x^4-y^2)$

$(4x^4-y^2)$ is in the form of the difference of squares:

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

$a=2x^2 and b=y$

$4((2x^2)^2-y)$ =

$4(2x^2-y)(2x^2+y)$

How do you factor 16x^4 - 4y^216x^4 - 4y^2?