How do you factor 1-x^8 ?

2 Answers
Nov 15, 2016

(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

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.