# Math Help - fifth root of unity

1. ## fifth root of unity

apologies if this is in the wrong place, also this is probably the first of many posts on the roots of unity.

Question:
show x^5-1 = (x-1)(x^2 + ((1+sqrt5)/2)x + 1)(x^2 + ((1-sqrt5)/2)x + 1)

My soln so far:
x^5 -1 = (x-1)(x^4 +x^3+x^2+x+1)

z5 = (x-z5)(x-z5^2)(x-z5^3)(x-z5^4)(x-z5^5)
(zeta 5)

z5= e^ ((2 pi i)/5) = cos(2pi/5)+isin(2pi/5)

no idea what to do.

2. The fifth roots of unity are simply $e^{\frac{2k\pi i}{5}$ for $k=0,1,2,3,4$.

In general, the nth roots of unity are $e^{\frac{2k\pi i}{n}$ for $k=0,1,...,n-1$.

3. Originally Posted by gpenguin
apologies if this is in the wrong place, also this is probably the first of many posts on the roots of unity.

Question:
show x^5-1 = (x-1)(x^2 + ((1+sqrt5)/2)x + 1)(x^2 + ((1-sqrt5)/2)x + 1)

My soln so far:
x^5 -1 = (x-1)(x^4 +x^3+x^2+x+1)
That's a good start. If you now want a purely algebraic solution, notice that $(x^2+\frac12x+1)^2 = x^4+x^3+\frac94x^2 +x+1 = (x^4+x^3+x^2 +x+1) + \frac54x^2$ and therefore $x^4+x^3+x^2 +x+1 = (x^2+\frac12x+1)^2 - \bigl(\frac{\sqrt5}2x\bigr)^2$. Now use the difference of two squares to factorise that as $(x^2+\frac12x+1+\frac{\sqrt5}2x) (x^2+\frac12x+1-\frac{\sqrt5}2x)$.