Question

How do you factor `x^9+1`?

Answer 1

`x^9+1 = (x+1)(x^2 - 2 cos((pi)/9) x + 1)(x^2 - 2 cos((3pi)/9) x + 1)(x^2 - 2 cos((5pi)/9) x + 1)(x^2 - 2 cos((7pi)/9) x + 1)`

Explanation:

The easiest way to do this is probably using Complex arithmetic and de Moivre's formula:

`(cos theta + i sin theta)^n = cos n theta + i sin n theta`

We find that the nine `9`th roots of `-1` are:

`cos ((k pi)/9) + i sin ((k pi)/9)`

for `k = +-1, +-3, +-5, +-7, 9`

In order to find factors with Real coefficients, we can pair up the Complex conjugate pairs like this:

`(x - cos ((k pi)/9) - i sin((k pi)/9))(x - cos ((-k pi)/9) - i sin((-k pi)/9))`

`=(x - cos ((k pi)/9) - i sin((k pi)/9))(x - cos ((k pi)/9) + i sin((k pi)/9))`

`=(x - cos ((k pi)/9))^2 - (i sin((k pi)/9))^2`

`=x^2 - 2 cos((k pi)/9) x + 1`

for `k = 1, 3, 5, 7`

Hence:

`x^9+1 = (x+1)(x^2 - 2 cos((pi)/9) x + 1)(x^2 - 2 cos((3pi)/9) x + 1)(x^2 - 2 cos((5pi)/9) x + 1)(x^2 - 2 cos((7pi)/9) x + 1)`