Solve:
\begin{equation}
\label{eq:27.21.23}
x^4+Bx^2+Cx+D=(x^2+ux+\alpha)(x^2-ux+\beta)
\end{equation}
So
\begin{equation}
\label{eq:27.21.24}
x^4+Bx^2+Cx+D=x^4+x^2(\beta-u^2+\alpha)+x(u\beta-u\alpha)+\alpha\beta
\end{equation}
So
\begin{align*}
\begin{cases}
B=\beta-u^2+\alpha\\
C=u\beta-u\alpha\\
D=\alpha\beta\\
\end{cases}
\end{align*}
So $Bu=\beta u-u^3+\alpha u$.So $Bu+u^3=\beta u+\alpha u$.And $C=\beta u-\alpha u$.So $\beta u=\frac{Bu+u^3+C}{2}$.$\alpha u=\frac{Bu+u^3-C}{2}$.So
\begin{equation}\alpha\beta
u^2=Du^2=(\frac{Bu+u^3+C}{2})(\frac{Bu+u^3-C}{2})\end{equation}
So
\begin{equation}
\label{eq:28.2.26}
(u^2)^3+2B(u^2)^2+(B^2-4D)u^2-C^2=0
\end{equation}
According to Cardano 's formula,$u^2$ can be solved.Then $\alpha,\beta$ can be solved.