4次方程式の標準形

\[ X=x+\frac{a_{3}}{4a_{4}} \]

とおくと、

\begin{align*} a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0} & =a_{4}\left\{ \left(X-\frac{a_{3}}{4a_{4}}\right)^{4}+\frac{a_{3}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)^{3}+\frac{a_{2}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)^{2}+\frac{a_{1}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)+\frac{a_{0}}{a_{4}}\right\} \\ & =a_{4}\left\{ X^{4}+\left(-\frac{3a_{3}^{\;2}}{8a_{4}^{\;2}}+\frac{a_{2}}{a_{4}}\right)X^{2}+\left(\frac{a_{3}^{\;3}}{8a_{4}^{\;3}}-\frac{a_{3}a_{2}}{2a_{4}^{\;2}}+\frac{a_{1}}{a_{4}}\right)X-\frac{3a_{3}^{\;4}}{256a_{4}^{\;4}}+\frac{a_{3}^{\;2}a_{2}}{16a_{4}^{\;3}}-\frac{a_{3}a_{1}}{4a_{4}^{\;2}}+\frac{a_{0}}{a_{4}}\right\} \end{align*}

となるので、

\[ \begin{cases} p=-\frac{3a_{3}^{\;2}}{8a_{4}^{\;2}}+\frac{a_{2}}{a_{4}}\\ q=\frac{a_{3}^{\;3}}{8a_{4}^{\;3}}-\frac{a_{3}a_{2}}{2a_{4}^{\;2}}+\frac{a_{1}}{a_{4}}\\ r=-\frac{3a_{3}^{\;4}}{256a_{4}^{\;4}}+\frac{a_{3}^{\;2}a_{2}}{16a_{4}^{\;3}}-\frac{a_{3}a_{1}}{4a_{4}^{\;2}}+\frac{a_{0}}{a_{4}} \end{cases} \]

とおくと、

\begin{align*} 0 & =a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\\ & =a_{4}\left\{ X^{4}+pX^{2}+qX+r\right\} \end{align*}

両辺を\(a_{4}\ne0\)で割って、

\[ X^{4}+pX^{2}+qX+r=0 \]

となる。