How do you factor x^9+1x9+1?
1 Answer
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(cosθ+isinθ)n=cosnθ+isinnθ
We find that the nine
cos ((k pi)/9) + i sin ((k pi)/9)cos(kπ9)+isin(kπ9)
for
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(kπ9)−isin(kπ9))(x−cos(−kπ9)−isin(−kπ9))
=(x - cos ((k pi)/9) - i sin((k pi)/9))(x - cos ((k pi)/9) + i sin((k pi)/9))=(x−cos(kπ9)−isin(kπ9))(x−cos(kπ9)+isin(kπ9))
=(x - cos ((k pi)/9))^2 - (i sin((k pi)/9))^2=(x−cos(kπ9))2−(isin(kπ9))2
=x^2 - 2 cos((k pi)/9) x + 1=x2−2cos(kπ9)x+1
for
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)x9+1=(x+1)(x2−2cos(π9)x+1)(x2−2cos(3π9)x+1)(x2−2cos(5π9)x+1)(x2−2cos(7π9)x+1)
