How do you factor $16x^4 - 64x^2$ ?

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Answer 1 · 2015-07-23T23:46:04

$16x^4-64x^2 = 16x^2(x^2-2^2) = 16x^2(x-2)(x+2)$

using the difference of squares identity:

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

First separate out the common factor $16x^2$ to get:

$16x^4-64x^2 = 16x^2(x^2-4)$

Next note that $(x^2-4) = (x^2-2^2)$ is a difference of squares, so we can factor it as $(x-2)(x+2)$

Putting it all together:

$16x^4-64x^2 = 16x^2(x^2-2^2) = 16x^2(x-2)(x+2)$

How do you factor 16x^4 - 64x^216x^4 - 64x^2?