Problem source

Show that setting $z - z^{-1}=w$ in the quartic equation
\[
z^4 +5z^3 +4z^2 -5z +1=0
\]
results in the quadratic equation $w^2+5w+6=0$. Hence  
solve the above quartic equation.
Solve similarly the equation
\[
2z^8 -3z^7-12z^6 +12z^5 +22z^4-12z^3 -12 z^2 +3z +2=0 \;.
\]

Solution source

\begin{align*}
&& 0 &= z^4 +5z^3 +4z^2 -5z +1 \\
&&0 &= z^2 + z^{-2} + 5(z-z^{-1}) + 4 \\
&&&= (z-z^{-1})^2+2+5(z-z^{-1})+4 \\
&&&= w^2 + 5w + 6 \\
&&&= (w+3)(w+2) \\
\Rightarrow && 0 &= z-z^{-1}+3 \\
\Rightarrow && 0 &= z^2+3z-1 \\
\Rightarrow && z &= \frac{-3 \pm \sqrt{3^2+4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \\
\Rightarrow && 0 &= z-z^{-1}+2 \\
\Rightarrow && 0 &= z^2+2z-1 \\
\Rightarrow && z &= \frac{-2 \pm \sqrt{2^2+4}}{2} = - 1 \pm \sqrt{2} \\
\end{align*}

\begin{align*}
&&0 &= 2z^8 -3z^7-12z^6 +12z^5 +22z^4-12z^3 -12 z^2 +3z +2 \\
&& 0 &= 2(z^4+z^{-4}) - 3(z^3-z^{-3})-12(z^2+z^{-2})+12(z-z^{-1})+22 \\
&&&= 2\left ((z-z^{-1})^4+4(z^2+z^{-2})-6\right)-3 \left ((z-z^{-1})^3+3(z-z^{-1}) \right)-12 \left ((z-z^{-1})^2+2 \right)+12(z-z^{-1})+22 \\
&&&= 2(w^4+4(w^2+2)-6)-3w^3-9w-12w^2-24+12w+22 \\
&&&= 2 w^4-3w^3-4w^2+3w+2 \\
\Rightarrow && 0 &= 2(w^2+w^{-2})-3(w-w^{-1})-4 \\
&&&= 2((w-w^{-1})^2+2)-3(w-w^{-1})-4 \\
&&&= 2x^2-3x \\
&&&= x(2x-3) \\
\Rightarrow && 0 &= w -w^{-1}  \\
\Rightarrow && w &= \pm 1 \\
\Rightarrow && \pm 1 &= z-z^{-1} \\
\Rightarrow && 0 &= z^2 \mp z-1 \\
\Rightarrow && z &= \frac{\pm 1 \pm \sqrt{5}}{2} \\
\Rightarrow && \frac32 &= w-w^{-1} \\
\Rightarrow && 0 &= 2w^2-3w -2 \\
&&&= (2w+1)(w-2) \\
\Rightarrow && 2 &= z-z^{-1} \\
\Rightarrow && 0 &= z^2-2z-1 \\
\Rightarrow && z &= 1 \pm \sqrt{2} \\
\Rightarrow && -\frac12 &= z-z^{-1} \\
\Rightarrow && 0 &= 2z^2+z-2 \\
\Rightarrow && z &= \frac{-1 \pm \sqrt{17}}{4} \\
\Rightarrow && z &\in \left \{ \frac{\pm 1 \pm \sqrt{5}}{2}, 1 \pm \sqrt{2}, \frac{-1 \pm \sqrt{17}}{4} \right \}
\end{align*}