\(0<D\)のとき

\(0<a_{2}^{\;2}-4a_{4}a_{0}\)なので、
\begin{align*} a_{4}x^{4}+a_{2}x^{2}+a_{0} & =a_{4}\left(x^{2}+\frac{a_{2}}{2a_{4}}\right)^{2}-\frac{a_{2}^{2}-4a_{4}a_{0}}{4a_{4}}\\ & =\frac{1}{4a_{4}}\left\{ 4a_{4}^{\;2}\left(x^{2}+\frac{a_{2}}{2a_{4}}\right)^{2}-\left(a_{2}^{\;2}-4a_{4}a_{0}\right)\right\} \\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}+a_{2}\right)^{2}-\left(\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)^{2}\right\} \\ & =\frac{1}{4a_{4}}\left(2a_{4}x^{2}+a_{2}+\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)\left(2a_{4}x^{2}+a_{2}-\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right) \end{align*}

\(0<a_{0}a_{4}\)のとき、

\(D<0\)ならば\(0<a_{0}a_{4}\)が成り立つ。
\begin{align*} a_{4}x^{4}+a_{2}x^{2}+a_{0} & =\frac{1}{4a_{4}}\left(4a_{4}^{\;2}x^{4}+4a_{4}a_{2}x^{2}+4a_{4}a_{0}\right)\\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}\pm2\sqrt{a_{4}a_{0}}\right)^{2}\mp8a_{4}\sqrt{a_{4}a_{0}}x^{2}+4a_{4}a_{2}x^{2}\right\} \\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}\pm2\sqrt{a_{4}a_{0}}\right)^{2}+4a_{4}\left(a_{2}\mp2\sqrt{a_{4}a_{0}}\right)x^{2}\right\} \end{align*} 実数因数分解ができるのは第2項の符号が負のときのみであるので、
\begin{align*} 0 & >\sgn\left(4a_{4}\left(a_{2}\mp2\sqrt{a_{4}a_{0}}\right)x^{2}\right)\\ & =\sgn\left(a_{4}\left(a_{2}\mp2\sqrt{a_{4}a_{0}}\right)\right) \end{align*} 復号によって場合分けをする。

\(a_{4}\left(a_{2}-2\sqrt{a_{4}a_{0}}\right)<0\)のとき、

\begin{align*} a_{4}x^{4}+a_{2}x^{2}+a_{0} & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}+2\sqrt{a_{4}a_{0}}\right)^{2}+4a_{4}\left(a_{2}-2\sqrt{a_{4}a_{0}}\right)x^{2}\right\} \\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}+2\sqrt{a_{4}a_{0}}\right)^{2}-\left(2\sqrt{a_{4}\left(2\sqrt{a_{4}a_{0}}-a_{2}\right)}x\right)^{2}\right\} \\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}+2\sqrt{a_{4}a_{0}}+2\sqrt{a_{4}\left(2\sqrt{a_{4}a_{0}}-a_{2}\right)}x\right)\left(2a_{4}x^{2}+2\sqrt{a_{4}a_{0}}-2\sqrt{2\sqrt{a_{4}a_{0}}-a_{2}}x\right)\right\} \\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}+2\sqrt{a_{4}\left(2\sqrt{a_{4}a_{0}}-a_{2}\right)}x+2\sqrt{a_{4}a_{0}}\right)\left(2a_{4}x^{2}-2\sqrt{2\sqrt{a_{4}a_{0}}-a_{2}}x+2\sqrt{a_{4}a_{0}}\right)\right\} \end{align*}

\(a_{4}\left(a_{2}+2\sqrt{a_{4}a_{0}}\right)<0\)のとき、

\begin{align*} a_{4}x^{4}+a_{2}x^{2}+a_{0} & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}-2\sqrt{a_{4}a_{0}}\right)^{2}+4a_{4}\left(a_{2}+2\sqrt{a_{4}a_{0}}\right)x^{2}\right\} \\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}-2\sqrt{a_{4}a_{0}}\right)^{2}-\left(2\sqrt{-a_{4}\left(2\sqrt{a_{4}a_{0}}+a_{2}\right)}x\right)^{2}\right\} \\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}-2\sqrt{a_{4}a_{0}}+2\sqrt{-a_{4}\left(2\sqrt{a_{4}a_{0}}+a_{2}\right)}x\right)\left(2a_{4}x^{2}-2\sqrt{a_{4}a_{0}}-2\sqrt{-a_{4}\left(2\sqrt{a_{4}a_{0}}+a_{2}\right)}x\right)\right\} \\ & =\frac{1}{4a_{4}}\left\{ \left(2a_{4}x^{2}+2\sqrt{-a_{4}\left(2\sqrt{a_{4}a_{0}}+a_{2}\right)}x-2\sqrt{a_{4}a_{0}}\right)\left(2a_{4}x^{2}-2\sqrt{-a_{4}\left(2\sqrt{a_{4}a_{0}}+a_{2}\right)}x-2\sqrt{a_{4}a_{0}}\right)\right\} \end{align*}

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全ての場合分けに当てはまらないためには、以下が成り立つ必要があるが、
\begin{align*} & a_{2}^{\;2}-4a_{4}a_{0}<0\;\land\;\left(0>a_{0}a_{4}\;\lor\;a_{4}\left(a_{2}-2\sqrt{a_{4}a_{0}}\right)>0\right)\;\land\;\left(0>a_{0}a_{4}\;\lor\;a_{4}\left(a_{2}+2\sqrt{a_{4}a_{0}}\right)>0\right)\\ \Leftrightarrow & a_{2}^{\;2}-4a_{4}a_{0}<0\;\land\;a_{4}\left(a_{2}-2\sqrt{a_{4}a_{0}}\right)>0\;\land\;a_{4}\left(a_{2}+2\sqrt{a_{4}a_{0}}\right)>0 \end{align*} これより、
\begin{align*} 0 & <a_{4}\left(a_{2}-2\sqrt{a_{4}a_{0}}\right)a_{4}\left(a_{2}+2\sqrt{a_{4}a_{0}}\right)\\ & =a_{4}^{\;2}\left(a_{2}^{\;2}-4a_{4}a_{0}\right)\\ & <0 \end{align*} となり矛盾が生じるのでどれかには当てはまる。

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全ての場合分けに当てはまることもある。
例えば\(x^{4}-3x^{2}+1\)は全てに当てはまる。
\begin{align*} x^{4}-3x^{2}+1 & =\left(x^{2}-\frac{3-\sqrt{5}}{2}\right)\left(x^{2}-\frac{3+\sqrt{5}}{2}\right)\\ & =\left(x^{2}+\sqrt{5}x+1\right)\left(x^{2}-\sqrt{5}x+1\right)\\ & =\left(x^{2}+x-1\right)\left(x^{2}-x-1\right) \end{align*}