Find exact value of $\cos (\frac{2\pi}{5})$ using complex numbers. 
Factorise $z^5-1$ over the real field.
  Show that $\cos \frac{2\pi}{5}$ is a root of the equation $4x^2+2x-1=0$ and hence find its exact value.

I have worked out that $$ z^5-1=(z-1)(z^4+z^3+z^2+z+1)\\=(z-1)(z^2-2z\cos72^o+1)  (z^2+2z\cos 72^o+1) $$
 A: Find the polynomials that cancel $\cos\left(\frac{2\pi}{5}\right)$ :
We know that the solution of $z^5=1$ are given by $\alpha^k=e^{\frac{2ik\pi}{5}}$, $k=0,1,2,3,4$. We can observe that $\alpha^{4}=\alpha^{-1}$ and $\alpha^{3}=\alpha^{-2}$. Moreover,
$$(*):\quad 0=1+\alpha+\alpha^2+\alpha^3+\alpha^4=1+\alpha+\alpha^{-1}+\alpha^2+\alpha^{-2}$$
and $\alpha+\alpha^{-1}=2\cos\left(\frac{2\pi}{5}\right)$.
Let set $X=\alpha+\alpha^{-1}$, then
$$X^2=(\alpha+\alpha^{-1})^2=\alpha^2+\alpha^{-2}+2$$
and so 
$$\alpha^2+\alpha^{-2}=X^2-2$$
if we replace in $(*)$ we obtain
$$0=X+X^2-2+1=X^2+X-1.$$
Now set $x=\frac{X}{2}=\cos\left(\frac{2\pi}{5}\right)$, therefore, $X=2x$ and so
$$0=(2x)^2+2x-1=4x^4+2x-1$$
and thus, $\cos\left(\frac{2\pi}{5}\right)$ is solution of
$$4x^2+2x-1=0$$
Find the value of $\cos\left(\frac{2\pi}{5}\right)$ :
$$4x^2+2x-1=0\iff x^2+\frac{1}{2}x-\frac{1}{4}=0\iff \left(x+\frac{1}{4}\right)^2-\frac{1}{16}-\frac{1}{4}=0\iff\left(x+\frac{1}{4}\right)^2=\frac{5}{16}\iff x+\frac{1}{4}=\pm\frac{\sqrt 5}{4}\iff x=\frac{-1\pm\sqrt 5}{4}$$
But $\cos\left(\frac{2\pi}{5}\right)>0$ because $\frac{2\pi}{5}<\frac{\pi}{2}$, then,
$$\cos\left(\frac{2\pi}{5}\right)=\frac{-1+\sqrt 5}{4}$$
