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

1 Answer
Aug 9, 2016

Answer:

#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)#