How do you factor #1-x^8# ?

2 Answers
Nov 15, 2016

Answer:

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

Explanation:

Remember #(a^2-b^2)=(a-b)(a+b)#
#(1-x^8)=(1-x^4)(1+x^4)#
=#(1-x^2)(1+x^2)(1+x^4)#
=#(1-x)(1+x)(1+x^2)(1+x^4)#

Nov 15, 2016

Answer:

#1-x^8 = (1-x)(1+x)(1+x^2)(1-sqrt(2)x+x^2)(1+sqrt(2)x+x^2)#

Explanation:

The difference of squares identity can be written:

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

Note also that:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

In particular, putting #k = sqrt(2)# we find:

#(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2) = a^4+b^4#

So we find:

#1 - x^8 = 1^2-(x^4)^2#

#color(white)(1-x^8) = (1-x^4)(1+x^4)#

#color(white)(1-x^8) = (1^2-(x^2)^2)(1^4+x^4)#

#color(white)(1-x^8) = (1-x^2)(1+x^2)(1^2-sqrt(2)x+x^2)(1^2+sqrt(2)x+x^2)#

#color(white)(1-x^8) = (1^2-x^2)(1+x^2)(1-sqrt(2)x+x^2)(1+sqrt(2)x+x^2)#

#color(white)(1-x^8) = (1-x)(1+x)(1+x^2)(1-sqrt(2)x+x^2)(1+sqrt(2)x+x^2)#

The remaining quadratic factors have no linear factors with Real coefficients.