How do you factor the expression #x^8 - 1#?
1 Answer
Dec 20, 2015
Use some identities to find:
#x^8-1 = (x-1)(x+1)(x^2+1)(x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1)#
Explanation:
We shall use some identities to help:
The difference of squares identity:
#a^2-b^2 = (a-b)(a+b)#
Consider also:
#(a^2+kab+b^2)(a^2-kab+b^2)=a^4+(2-k^2)a^2b^2+b^4#
In particular if
#(a^2+sqrt(2)ab+b^2)(a^2-sqrt(2)ab+b^2) = a^4+b^4#
So:
#x^8-1#
#=(x^4)^2-1^2#
#=(x^4-1)(x^4+1)#
#=((x^2)^2-1^2)(x^4+1)#
#=(x^2-1)(x^2+1)(x^4+1)#
#=(x^2-1^2)(x^2+1)(x^4+1)#
#=(x-1)(x+1)(x^2+1)(x^4+1)#
#=(x-1)(x+1)(x^2+1)(x^4+1^4)#
#=(x-1)(x+1)(x^2+1)(x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1)#
This is as far as we can go with Real coefficients. The remaining quadratic factors all have Complex zeros.
