Hey! My teacher and I are completely stuck on this Polynomials question. We can complete part i), but part ii) & iiI) is a mystery.
If w=x+x^(-1)
i) 
ii) =\left [ x^{2}+\frac{1}{2}x(1+\sqrt{5})+1 \right ]\left [ x^{2}+\frac{1}{2}x(1-\sqrt{5})+1 \right ])
iii) Show that the roots of 
are the four complex roots of 
and deduce that  : cos36^{\circ}=\frac{1}{4}(\sqrt{5}+1))
\begin{align*} x^4+x^3+x^2+x+1&= x^2(x^2+x+1+x^{-1}+x^{-2})\\ &= x^2 (\omega^2-2 + \omega + 1)\\ &= x^2(\omega^2 + \omega - 1)\end{align*}
\[ \text{The roots of }\omega^2+\omega - 1 = 0\\ \text{are }\omega = \frac{-1\pm \sqrt{5}}{2}\text{ by quadratic formula}\\ \text{so we have}\\ \begin{align*} x^2(\omega^2+\omega - 1) &= x^2\left(\omega - \frac{-1-\sqrt{5}}{2} \right)\left( \omega - \frac{-1+\sqrt{5}}{2} \right)\\ &= x^2 \left(x + \frac12 (1+\sqrt5) + x^{-1} \right)\left( x+ \frac12 (1-\sqrt5) + x^{-1}\right)\\ &= \left(x^2+ \frac12 x (1+\sqrt5) + 1 \right) \left(x^2 + \frac12x (1-\sqrt5)+1 \right)\end{align*} \]
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Sketch solution:
\begin{align*} x^4+x^3+x^2+x+1&=0\\ \implies \frac{x^5-1}{x-1}&=0 \tag{geometric series}\\ \implies x^5-1 &= 0\\ \implies x^5 &= 1\end{align*}
Note that \(36^\circ = \frac\pi5\) and \(72^\circ = \frac{2\pi}{5} \).
\[ \text{Using De Moivre's theorem and the usual approach}\\ \text{the roots of }x^5 = 1\text{ are}\\ x = 1, \, x = \operatorname{cis} \left( \pm \frac{2\pi}{5}\right), \, x = \operatorname{cis} \left( \pm\frac{4\pi}{5} \right) \]
\[ \text{But of course, since }x=1\text{ is not a root of }x^4+x^3+x^2+x+1=0\\ \text{and we know that the quartic equation has four solutions,}\\ \text{it follows that the remaining four roots, i.e. }\operatorname{cis} \left( \pm \frac{2\pi}5 \right)\text{ and }\operatorname{cis}\left( \pm \frac{4\pi}{5} \right)\\ \text{are the solutions of }x^4+x^3+x^2+x+1=0.\]
What requires a bit more work now is figuring whether \( \operatorname{cis} \left( \pm \frac{2\pi}5 \right) \) are the roots of \(x^2 + \frac12 x(1+\sqrt5) +1=0\), or the roots of \(x^2 + \frac12x(1-\sqrt5) - 1 = 0\). Note that because all coefficients are real even in the quadratics, non-real roots come in conjugate pairs.
\[ \text{The quadratic whose roots are }\alpha = \operatorname{cis} \left( -\frac{2\pi}{5} \right) \text{ and }\beta = \operatorname{cis} \left( \frac{2\pi}{5} \right)\\ \text{has the sum of roots }\alpha + \beta = 2\cos \frac{2\pi}{5}.\\ \text{Since }\frac{2\pi}{5}\text{ is a first quadrant angle, }\cos \frac{2\pi}{5} > 0\\ \text{so it follows that }\alpha + \beta > 0. \]
\[ \text{The quadratic }x^2+ \frac12 x(1-\sqrt5) + 1 = 0\\ \text{has sum of roots }\alpha_1+\beta_1 = \frac{\sqrt5 - 1}{2}\\ \text{which is positive}. \]
\[ \text{However the quadratic }x^2+ \frac12 x(1+\sqrt5) + 1 = 0\\ \text{has sum of roots }\alpha_2+\beta_2 = -\frac{\sqrt5 + 1}{2}\\ \text{which is negative.}\]
\[ \text{Hence }x=\operatorname{cis} \left( \pm \frac{2\pi}{5} \right)\text{ must be the two roots of}\\ x^2 + \frac12 x (1-\sqrt5) + 1 = 0\\ \text{and further by the sum of roots,}\\ 2\cos \frac{2\pi}{5} = \frac{\sqrt5 - 1}{2} \implies \boxed{\cos \frac{2\pi}{5} = \frac{\sqrt5 - 1}{4}}\\ \text{as required.}\]
\[ \text{Similarly }2\cos \frac{4\pi}{5} = -\frac{ \sqrt5 + 1}{2}\\ \text{However since }\cos (\pi - x) = -\cos x\text{, we then have}\\ -2 \cos \frac\pi5 = -\frac{\sqrt5+1}{2} \implies \boxed{\cos \frac\pi5 = \frac{\sqrt5 + 1}{2}}\\ \text{as required.}\]
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