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)#