How do you factor completely: #16x^4 - 1#?

1 Answer
Alan P.
Apr 25, 2018

#16x^4-1=color(blue)((4x^2+1)(2x+1)(2x-1)#

Explanation:

Remember
#color(white)("XXX")(a^2-b^2)=(a+b)(a-b)#

#(16x^4-1)# is of this form (with #a=(4x^2)# and #b=1#).

So
#color(white)("XXX")(16x^4-1)=(4x^2+1)(4x^2-1)#

...but we note that #(4x^2-1)# is also of this form (with #a=2x# and #b=1#)
So #(4x^2-1)=(2x+1)(2x-1)#

and, completely factored we have
#color(white)("XXX")(16x^4-1)=(4x^2+1)(2x+1)(2x-1)#

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