How do you factor #16x^2-1#?

3 Answers
t0hierry
Jun 1, 2016

#(4x -1)(4x +1)#

Explanation:

Both 16 and 1 are squares. This suggests using the formula for the difference of squares.

Shwetank Mauria
Jun 1, 2016

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

Explanation:

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

As such it can be factorized as #(a+b)(a-b)# as such

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

Burglar
Jun 1, 2016

You factor it as a difference of two squares.

Explanation:

Every time you have a difference in the format of

#A^2-B^2# you can factor as #(A-B)(A+B)#.
Then you have to identify if your quantity is the difference of two squared quantities.
It is. In fact we can write it as:

#16x^2-1=(4x)^2-1^2#

Then we can apply our rule

#(4x)^2-1^2=(4x-1)(4x+1)#
and these are your factors.

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