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

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

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Answer 1 · 2016-05-12T19:46:35

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

Explanation:

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

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

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

$16 = 4^{2}$ $16=4^2$
$y^{4} = {(y^{2})}^{2}$ $y^4=(y^2)^2$

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

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

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

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

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

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