用费拉里法解方程的思路，是将一个一元四次方程，拆分成两个一元二次方程 也就是
$$ \large \begin{align*} & x^4+bx^3+cx^2+dx+e\\[4mm] = & (x^2+s_{1}x+t_{1})(x^2+s_{2}x+t_{2})\\[4mm] = & 0 \end{align*} $$解这2个方程$ x^2+s_{1}x+t_{1}=0,x^2+s_{2}x+t_{2}=0 $，即可得到一元四次方程的4个根 令
$$ \large \left\{\begin{array}{l} p= bd-4e-\dfrac{c^2}{3}\\[4mm] q=e(4c-b^2)-d^2+\dfrac{c}{3}(bd-4e)-\dfrac{2}{27}c^3\\[4mm] \Delta=\dfrac{q^2}{4} +\dfrac{p^3}{27} \end{array}\right. $$一、$ \Delta \geqslant 0 $时
$$ \large \left\{\begin{array}{l} y=\sqrt[3]{-\dfrac{q}{2}+\sqrt{\Delta}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\Delta}}+\dfrac{c}{3}\\[4mm] s_{1}=\dfrac{1}{2}\left( b+\sqrt{4y+b^2-4c} \right)\\[4mm] t_{1}=\dfrac{y}{2}+\dfrac{0.5by-d}{\sqrt{4y+b^2-4c} }\\[4mm] s_{2}=\dfrac{1}{2}\left( b-\sqrt{4y+b^2-4c} \right)\\[4mm] t_{2}=\dfrac{y}{2}-\dfrac{0.5by-d}{\sqrt{4y+b^2-4c} } \end{array}\right. $$至此，已将方程分解成了2个一元二次方程
$$ \large \left\{\begin{array}{l} x^2+s_{1}x+t_{1}=0\\[4mm] x^2+s_{2}x+t_{2}=0 \end{array}\right. $$解这两个一元二次方程得到的4个根，即为一元四次方程的根 4个根分别为
$$ \large \left\{\begin{array}{l} x_{1,2}=\dfrac{-s_{1} \pm \sqrt{s_{1}^{2}-4t_{1}}}{2}\\[4mm] x_{3,4}=\dfrac{-s_{2} \pm \sqrt{s_{2}^{2}-4t_{2}}}{2} \end{array}\right. $$二、$ \Delta<0 $ 时 $ \Delta=\dfrac{q^2}{4} +\dfrac{p^3}{27}<0 $，且$ p,q $都是实数，可得出$ p<0 $
$$ \large \left\{\begin{array}{l} T=\dfrac{q}{2p}\sqrt{-\dfrac{27}{p}}\\[4mm] \theta=\arccos T\\[4mm] y=2\sqrt{-\dfrac{p}{3}}\cos\left(\dfrac{\theta}{3}\right)+\dfrac{c}{3}\\[4mm] s_{1}=\dfrac{1}{2}\left( b+\sqrt{4y+b^2-4c} \right)\\[4mm] t_{1}=\dfrac{y}{2}+\dfrac{0.5by-d}{\sqrt{4y+b^2-4c} }\\[4mm] s_{2}=\dfrac{1}{2}\left( b-\sqrt{4y+b^2-4c} \right)\\[4mm] t_{2}=\dfrac{y}{2}-\dfrac{0.5by-d}{\sqrt{4y+b^2-4c} } \end{array}\right. $$方程的4个根分别为
$$ \large \left\{\begin{array}{l} x_{1,2}=\dfrac{-s_{1} \pm \sqrt{s_{1}^{2}-4t_{1}}}{2}\\[4mm] x_{3,4}=\dfrac{-s_{2} \pm \sqrt{s_{2}^{2}-4t_{2}}}{2} \end{array}\right. $$