2次式の実数の範囲で因数分解
2次式の実数の範囲で因数分解
全て実数の範囲で考える。
全て実数の範囲で考える。
(1)
\[ a^{2}-b^{2}=\left(a+b\right)\left(a-b\right) \](2)
\[ a^{2}\pm2ab+b^{2}=\left(a\pm b\right)^{2} \](3)
\[ a^{2}+b^{2}+c^{2}+2\left(ab+bc+ca\right)=\left(a+b+c\right)^{2} \](4)
\[ \sum_{j=1}^{n}\sum_{k=1}^{n}\left(a_{j}a_{k}\right)=\left(\sum_{k=1}^{n}a_{k}\right)^{2} \](1)
\begin{align*} a^{2}-b^{2} & =a^{2}-ab+ab-b^{2}\\ & =a\left(a-b\right)+b\left(a-b\right)\\ & =\left(a+b\right)\left(a-b\right) \end{align*}(2)
\begin{align*} a^{2}\pm2ab+b^{2} & =a^{2}+b^{2}\pm2ab\\ & =\left(a\pm b\right)^{2}\mp2ab\pm2ab\\ & =\left(a\pm b\right)^{2} \end{align*}(3)
\begin{align*} a^{2}+b^{2}+c^{2}+2\left(ab+bc+ca\right) & =a^{2}+2\left(b+c\right)a+b^{2}+c^{2}+2bc\\ & =a^{2}+2\left(b+c\right)a+\left(b+c\right)^{2}\\ & =\left(a+b+c\right)^{2} \end{align*}(3)-2
(4)で\(n=3\)のときである。(4)
\begin{align*} \sum_{j=1}^{n}\sum_{k=1}^{n}\left(a_{j}a_{k}\right) & =\left(\sum_{j=1}^{n}a_{j}\right)\left(\sum_{k=1}^{n}a_{k}\right)\\ & =\left(\sum_{k=1}^{n}a_{k}\right)^{2} \end{align*}ページ情報
| タイトル | 2次式の実数の範囲で因数分解 |
| URL | https://www.nomuramath.com/getq88d1/ |
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複二次式の定義と因数分解
\[
a_{4}x^{4}+a_{2}x^{2}+a_{0}=\frac{1}{4a_{4}}\left(2a_{4}x^{2}+a_{2}+\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)\left(2a_{4}x^{2}+a_{2}-\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)
\]
n乗根の因数分解
\[
z^{n}-1=\prod_{k=1}^{n}\left(z-e^{\frac{2\pi}{n}ki}\right)
\]
オイラーの4平方恒等式
\[
\left(a_{0}^{\;2}+a_{1}^{\;2}+a_{2}^{\;2}+a_{3}^{\;2}\right)\left(b_{0}^{\;2}+b_{1}^{\;2}+b_{2}^{\;2}+b_{3}^{\;2}\right)=\left(a_{0}b_{0}-a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}\right)^{2}+\left(a_{0}b_{1}+a_{1}b_{0}+a_{2}b_{3}-a_{3}b_{2}\right)^{2}+\left(a_{0}b_{2}-a_{1}b_{3}+a_{2}b_{0}+a_{3}b_{1}\right)^{2}+\left(a_{0}b_{3}+a_{1}b_{2}-a_{2}b_{1}+a_{3}b_{0}\right)^{2}
\]
差積の定義と性質
\[
\Delta\left(x_{1},\cdots,x_{n}\right):=\prod_{1\leq i<j\leq n}\left(x_{i}-x_{j}\right)
\]