Factor the polynomial $z^5 + 32$ in real factors The question that I have trouble solving is the following:
Factor the polynomial $z^5 + 32$ in real factors. The answer should not use trigonometric functions. (Hint: you are allowed to use the fact that: $cos(\pi/5) = \frac{1+\sqrt{5}}{4}$ and $cos(3\pi/5) = \frac{1-\sqrt{5}}{4}$.
In the previous question you were supposed to find the complex roots to the equation $z^5 = -32$ in polar form, resulting in the roots:
$$z_1 = 2e^{i\frac{\pi}{5}}$$$$z_2 =  2e^{i\frac{3\pi}{5}}$$ $$z_3 = 2e^{i\pi} = -2$$ $$z_4 = 2e^{i\frac{7\pi}{5}}$$$$ z_5 = 2e^{i\frac{9\pi}{5}} $$
My attempt at a solution:
First we know that there were only one real root to the equation $z^5 = -32$, namely $-2$.
So if we write the polynomial as the equation: $z^5 + 32 = 0$ we know that $(z+2)$ must be a factor and we have four potential factors left to find. Since we know that the polynomial $z^5 +32$ only has real coefficients, the non real roots must come in pairs. 
We should then find two roots that are composed of conjugates of the roots with imaginary components to produce the other two real roots... But I am stuck here.
 A: Set $t=2z$, for the moment. Then
\begin{align}
z^5+32=32(t^5+1)
&=32(t+1)(t^4-t^3+t^2-t+1)\\
&=32(t+1)t^2\left(t^2+\frac{1}{t^2}-\left(t+\frac{1}{t}\right)+1\right)\\
&=32(t+1)t^2\left(\left(t+\frac{1}{t}\right)^{\!2}-\left(t+\frac{1}{t}\right)-1\right)
\end{align}
Consider $u^2-u-1=(u-\alpha)(u-\beta)$, where
$$
\alpha=\frac{1+\sqrt{5}}{2},\qquad\beta=\frac{1-\sqrt{5}}{2}
$$
and so
$$
z^5+32=32(t+1)t^2\left(t+\frac{1}{t}-\alpha\right)\left(t+\frac{1}{t}-\beta\right)
$$
which is to say
$$
z^5+32=32(t+1)(t^2-\alpha t+1)(t^2-\beta t+1)=
(z+2)(z^2-2\alpha z+4)(z^2-2\beta z+4)
$$
A: You found the roots so $z^5+32=(z+2)(z-z_1)(z-z_2)(z-z_4)(z-z_5)$.
Notice that $cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$ and that $z_1z_5\in\mathbb{R}$ and $z_2z_4\in\mathbb{R}$
A: HINT:
$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$
$$a^4-a^3b+a^2b^2-ab^3+b^4=(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2$$
If $ab\ne0,$
$$(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2=a^2b^2\left[\left(\frac{a^2+b^2}{ab}\right)^2-\frac{a^2+b^2}{ab}-1\right]$$
So, we can express $a^2+b^2$ in terms of $ab$
A: $z^2 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ where $\overline {\alpha}, \overline {\beta}$ are complex conjugate of $\alpha, \beta$ respectively. Also $(z - \alpha)(z - \overline {\alpha})$ and $(z - \beta)(z- \overline {\beta})$ are real polynomials (why?). Now choose $\alpha = z_1, \beta = z_2.$ Then $\overline {\alpha} = z_5, \overline {\beta} = z_4.$
