Question

How do you factor the expression `x^8 - 1`?

Answer 1

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 `k=sqrt(2)` then:

`(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.