展開する前に
- 符号が違う組合せを探そう
- 掛けたら同じ組合せを探そう
- 足したら同じ組合せを探そう
同じパーツが見つかるように展開することがコツです。
次の式を展開しよう。
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (x\colBX{bisque}{$-1$})(x\colBX{palegreen}{$-2$})(x\colBX{bisque}{$+1$})(x\colBX{palegreen}{$+2$})\\
& \colMM{orange}{ \Darr 符号が違う組合せ}\colMM{green}{ \Darr 符号が違う組合せ}\\
&= (x\colBX{bisque}{$-1$})(x\colBX{bisque}{$+1$}) \times(x\colBX{palegreen}{$-2$})(x\colBX{palegreen}{$+2$})\\
\\
&= (x^2-1)(x^2-4)\\
& \colMM{magenta}{ \Darr 左 \cdot 左 \Darr 右 \cdot 右}\\
&= x^4-5x^2+4\\
& \colMM{magenta}{ \Uarr -4x^2-1x^2・・・外 \cdot 外 + 内 \cdot 内}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan【別解】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{orange}{ \Darr 掛けると +2}\colMM{green}{ \Darr 掛けると +2}\\
& (x\colBX{bisque}{$-1$})(x\colBX{bisque}{$-2$}) \times (x\colBX{palegreen}{$+1$})(x\colBX{palegreen}{$+2$})\\
\\
&= (x^2\colBX{violet}{$-3x$}+2)(x^2\colBX{violet}{$+3x$}+2)\\
\\
&= \{(\colBX{lightcyan}{$x^2+2$})-3x\}\{(\colBX{lightcyan}{$x^2+2$})+3x\}\\
&\colMM{deepskyblue}{ \Darr x^2+2=A\ と置き換える}\\
&\colNS{lightgray}{= (\colBX{lightcyan}{$A$}-3x)(\colBX{lightcyan}{$A$}+3x)}\\
\\
&\colNS{lightgray}{= \colBX{bisque}{$A$}^2-9x^2}\\
& \colMM{orange}{ \Darr A=x^2+2\ に戻す}\\
&= (\colBX{bisque}{$x^2+2$})^2-9x^2\\
\\
&= (x^4+4x^2+4)-9x^2\\
\\
&= x^4-5x^2+4
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (x\colBX{bisque}{$-2$})(x\colBX{palegreen}{$-3$})(x\colBX{bisque}{$+2$})(x\colBX{palegreen}{$+3$})\\
& \colMM{orange}{ \Darr 符号が違う組合せ}\colMM{green}{ \Darr 符号が違う組合せ}\\
&= (x\colBX{bisque}{$-2$})(x\colBX{bisque}{$+2$}) \times(x\colBX{palegreen}{$-3$})(x\colBX{palegreen}{$+3$})\\
\\
&= (x^2-4)(x^2-9)\\
& \colMM{magenta}{ \Darr 左 \cdot 左 \Darr 右 \cdot 右}\\
&= x^4-13x^2+36\\
& \colMM{magenta}{ \Uarr -4x^2-9x^2・・・外 \cdot 外 + 内 \cdot 内}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan【別解】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{orange}{ \Darr 掛けると +6}\colMM{green}{ \Darr 掛けると +6}\\
& (x\colBX{bisque}{$-2$})(x\colBX{bisque}{$-3$}) \times (x\colBX{palegreen}{$+2$})(x\colBX{palegreen}{$+3$})\\
& \colMM{magenta}{ \Darr 符号が違うに注目 \Darr}\\
&= (x^2\colBX{violet}{$-5x$}+6)(x^2\colBX{violet}{$+5x$}+6)\\
\\
&= \{(\colBX{lightcyan}{$x^2+6$})-5x\}\{(\colBX{lightcyan}{$x^2+6$})+5x\}\\
&\colMM{deepskyblue}{ \Darr x^2+6=A\ と置き換える}\\
&\colNS{lightgray}{= (\colBX{lightcyan}{$A$}-5x)(\colBX{lightcyan}{$A$}+5x)}\\
\\
&\colNS{lightgray}{= \colBX{bisque}{$A$}^2-25x^2}\\
& \colMM{orange}{ \Darr A=x^2+6\ に戻す}\\
&= (\colBX{bisque}{$x^2+6$})^2-25x^2\\
\\
&= (x^4+12x^2+36)-25x^2\\
\\
&= x^4-13x^2+36
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{orange}{ \Darr 符号が違う組合せ}\\
& (x\colBX{bisque}{$-2$})(x\colBX{bisque}{$+2$}) \times (x^2+4)\\
& \colMM{green}{ \Darr 符号が違う組合せ}\\
&= (x^2\colBX{palegreen}{$-4$})(x^2\colBX{palegreen}{$+4$})\\
\\
&\colNS{lightgray}{=(x^2)^2-4^2}\\
\\
&= x^4-16
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{orange}{ \Darr 符号が違う組合せ}\\
& (x\colBX{bisque}{$+y$})^2(x\colBX{bisque}{$-y$})^2\\
& \colMM{red}{ \Darr (ab)^2=a^2b^2\ を逆利用}\\
&= \{(x\colBX{bisque}{$+y$})(x\colBX{bisque}{$-y$})\}^2\\
\\
&= (x^2-y^2)^2\\
& \colMM{red}{ \Darr 左^2 \Darr 右^2}\\
&\colNS{lightgray}{=(x^2)^2-2x^2y^2+(y^2)^2}\\
& \colMM{red}{両方かけて-x^2y^2 \Uarr さらに２倍}\\
&= x^4-2x^2y^2+y^4\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (x\colBX{bisque}{$+1$})(x\colBX{palegreen}{$+2$})(x\colBX{palegreen}{$+3$})(x\colBX{bisque}{$+4$})\\
& \colMM{orange}{ \Darr 足したら\ 5}\colMM{green}{ \Darr 足したら\ 5}\\
&= (x\colBX{bisque}{$+1$})(x\colBX{bisque}{$+4$}) \times(x\colBX{palegreen}{$+2$})(x\colBX{palegreen}{$+3$})\\
\\
&= (\colBX{violet}{$x^2+5x$}+4)(\colBX{violet}{$x^2+5x$}+6)\\
& \colMM{magenta}{ \Darr x^2+5x=A\ と置き換える}\\
&\colNS{lightgray}{=(\colBX{violet}{$A$}+4)(\colBX{violet}{$A$}+6)}\\
\\
&\colNS{lightgray}{=\colBX{lightcyan}{$A$}^2+10\colBX{lightcyan}{$A$}+24}\\
& \colMM{deepskyblue}{ \Darr A=x^2+5x\ に戻す}\\
&= (\colBX{lightcyan}{$x^2+5x$})^2+10(\colBX{lightcyan}{$x^2+5x$})+24\\
\\
&= (x^4+10x^3+25x^2)+10x^2+5x+24\\
\\
&= x^4+10x^3+35x^2+5x+24
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (x\colBX{bisque}{$+1$})(x\colBX{palegreen}{$+5$})(x\colBX{bisque}{$-1$})(x\colBX{palegreen}{$-5$})\\
& \colMM{orange}{ \Darr 符号が違う組合せ}\colMM{green}{ \Darr 符号が違う組合せ}\\
&= (x\colBX{bisque}{$+1$})(x\colBX{bisque}{$-1$}) \times(x\colBX{palegreen}{$+5$})(x\colBX{palegreen}{$-5$})\\
\\
&= (x^2-1)(x^2-25)\\
& \colMM{magenta}{ \Darr 左 \cdot 左 \Darr 右 \cdot 右}\\
&= x^4-26x^2+25\\
& \colMM{magenta}{ \Uarr -25x^2-1x^2・・・外 \cdot 外 + 内 \cdot 内}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan【別解】
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{orange}{ \Darr 掛けると +5}\colMM{green}{ \Darr 掛けると +5}\\
& (x\colBX{bisque}{$+1$})(x\colBX{bisque}{$+5$}) \times (x\colBX{palegreen}{$-1$})(x\colBX{palegreen}{$-5$})\\
& \colMM{magenta}{ \Darr 符号が違うに注目 \Darr}\\
&= (x^2\colBX{violet}{$+6x$}+5)(x^2\colBX{violet}{$-6x$}+5)\\
\\
&= \{(\colBX{lightcyan}{$x^2+5$})+6x\}\{(\colBX{lightcyan}{$x^2+5$})-6x\}\\
&\colMM{deepskyblue}{ \Darr x^2+5=A\ と置き換える}\\
&\colNS{lightgray}{= (\colBX{lightcyan}{$A$}+6x)(\colBX{lightcyan}{$A$}-6x)}\\
\\
&\colNS{lightgray}{= \colBX{bisque}{$A$}^2-36x^2}\\
& \colMM{orange}{ \Darr A=x^2+5\ に戻す}\\
&= (\colBX{bisque}{$x^2+5$})^2-36x^2\\
\\
&= (x^4+10x^2+25)-36x^2\\
\\
&= x^4-26x^2+25
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan
