How do you factor completely #p^4 - 1#?

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Answer 1 · 2017-03-21T00:38:06

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

#color(white)(p^4-1) = (p-1)(p+1)(p-i)(p+i)#

The difference of squares identity can be written:

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

So we find:

#p^4-1 = (p^2)^2 - 1^2#

#color(white)(p^4-1) = (p^2-1)(p^2+1)#

#color(white)(p^4-1) = (p^2-1^2)(p^2+1)#

#color(white)(p^4-1) = (p-1)(p+1)(p^2+1)#

That is as far as we can go with real coefficients, so this is "factored completely over the real numbers".

If we allow complex numbers, then we can use the difference of squares identity again to find:

#p^2+1 = p^2-i^2 = (p-i)(p+i)#

So putting it together we have:

#p^4-1 = (p-1)(p+1)(p-i)(p+i)#

This is "factored completely over the complex numbers".