Question

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

Answer 1

Explanation:

We shall use some identities to help:

The difference of squares identity:

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

Consider also:

$(a^2+kab+b^2)(a^2-kab+b^2)=a^4+(2-k^2)a^2{b}^2+b^4$

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:

$x^8-1$

$={\left(x^4\right)}^2-1^2$

$\begin{array}{c}=(x^4-1)(x^4+1)\\ =({\left(x^2\right)}^2-1^2)(x^4+1)\\ =(x^2-1)(x^2+1)(x^4+1)\\ =(x^2-1^2)(x^2+1)(x^4+1)\\ =(x-1)(x+1)(x^2+1)(x^4+1)\\ =(x-1)(x+1)(x^2+1)(x^4+1^4)\\ =(x-1)(x+1)(x^2+1)(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)\end{array}$

This is as far as we can go with Real coefficients. The

remaining quadratic factors all have Complex zeros.