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

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Answer 1 · 2015-07-17T20:37:27

Use difference of squares identity twice to find:

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

The difference of squares identity is:

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

So we find:

#x^4 - 1#

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

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

Note that #x^2+1# has no simpler linear factors with real coefficients since #x^2+1 >= 1 > 0# for all #x in RR#