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

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Answer 1 · 2016-05-12T19:46:35

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

First, notice that both terms have a common factor of $x^2$ , so we can factor it out:

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

Focusing on just the $(16-y^4)$ , notice that both of these are squared terms:

This will be useful since $(16-y^4)$ is a difference of squares, which can be factored as: $a^2-b^2=(a+b)(a-b)$ . Thus, we see that:

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

Note that $(4-y^2)$ is also a difference of squares, since $y^2$ is obviously squared and $4=2^2$ . We see that:

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

How do you factor 16x^2-x^2y^416x^2-x^2y^4?