(L.Euler 1770,Vollst.Anleitung zur Algebra ,St.Petersburg,Opera Omnia,vol.I).Consider an equation of degree four with symmetric coefficients,e.g.,
\begin{equation}
\label{eq:28.13.35}
x^4+5x^3+8x^2+5x+1=0
\end{equation}.Decompose the polynomial as $(x^2+rx+1)(x^2+sx+1)$ and find the four solutions of \ref{eq:28.13.35}.
Solve:$x^4+5x^3+8x^2+5x+1=x^4+x^3(s+r)+x^2(2+rs)+x(r+s)+1$.So
\begin{align*}
\begin{cases}
s+r=5\\
rs=6\\
\end{cases}
\end{align*}
Without the loss of of generality ,let $s=2,r=3.$So
\begin{equation}
\label{eq:28.13.42}
x^4+5x^3+8x^2+5x+1=(x^2+2x+1)(x^2+3x+1)=0
\end{equation}
So $x_1=-1,x_2=-1,x_3=\frac{-3+\sqrt{5}}{2},x_4=\frac{-3-\sqrt{5}}{2}$.