Problem source

\begin{questionparts}
 \item By setting $y=x+x^{-1},$ find the solutions
of 
\[
x^{4}+10x^{3}+26x^{2}+10x+1=0.
\]
\item Solve 
\[
x^{4}+x^{3}-10x^{2}-4x+16=0.
\]		
\end{questionparts}

Solution source

\begin{enumerate}[(i)]
    \item 
    \begin{align*}
        && x^{4}+10x^{3}+26x^{2}+10x+1  &= 0 \\
        \Leftrightarrow && x^2 + 10x + 26 + 10x^{-1} + x^{-2} &= 0 \\
        \Leftrightarrow && (x^2 + x^{-2} + 2) + 10(x+x^{-1}) + 24  &= 0 \\
        \Leftrightarrow && y^2 + 10y + 24  &= 0 \tag{$y = x + x^{-1}$} \\
        \Leftrightarrow && (y+6)(y+4)  &= 0 \\
        \Leftrightarrow && \begin{cases}
            x+x^{-1} = -4 \\
            x+x^{-1} = -6 \\
        \end{cases} \\
        \Leftrightarrow && \begin{cases}
            x^2+4x+1 = 0 \\
            x^2+6x+1 = 0 \\
        \end{cases} \\
        \Leftrightarrow && \boxed{\begin{cases}
            x = -2 \pm \sqrt{3} \\
            x = -3 \pm 2\sqrt{2} \\
        \end{cases}} \\
    \end{align*}

    \item \begin{align*}
        && x^{4}+x^{3}-10x^{2}-4x+16=0 &= 0 \\
        \Leftrightarrow && x^2 + x - 10 - 4x^{-1} + 4x^{-2} &= 0 \\
        \Leftrightarrow && (x^2+4x^{-2} - 4) + (x - 4x^{-1})  - 6 &= 0 \\
        \Leftrightarrow && (x^2+4x^{-2} - 4) + (x - 4x^{-1})  - 6 &= 0 \\
        \Leftrightarrow && z^2 + z  - 6 &= 0 \tag{$z = x -2x^{-1}$} \\
        \Leftrightarrow && (z+3)(z-2) &= 0 \\
        \Leftrightarrow && \begin{cases}
            x-2x^{-1} = -3 \\
            x-2x^{-1} = 2 \\
        \end{cases} \\
        \Leftrightarrow && \begin{cases}
            x^2+3x-2 = 0 \\
            x^2-2x-2 = 0 \\
        \end{cases} \\
        \Leftrightarrow && \boxed{\begin{cases}
            x = \frac{-3 \pm \sqrt{17}}{2} \\
            x = 1 \pm \sqrt{3} \\
        \end{cases}} \\
    \end{align*}
\end{enumerate}