次の式を因数分解しよう。
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【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{red}{x:次数3\ \searrow} & \colMM{red}{ \darr\ y:次数1\ \darr}\\ & x^3-x^2+xy-2x+y\\ & \colMM{red}{ \Darr 次数が低い\ y\ (の個数)で整理!}\\ &= (xy+\textcolor{lightgray}{1}y)+(x^3-x^2-2x)\\ & \colMM{orange}{ \Darr ①共通}\colMM{green}{ \Darr①共通}\\ &= (x+1)y + x(x^2-x-2)\\ & \colMM{green}{ \Darr②たすきがけ}\\ &= \colBX{violet}{$(x+1)$}\,y +x\,\colBX{violet}{$(x+1)$}\,(x-2)\\ & \colMM{magenta}{ \Darr①共通}\\ &= \colBX{violet}{$(x+1)$}\,\left\{ y+x(x-2) \right\}\\ \\ &= (x+1)(x^2-2x+y) \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
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【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{red}{x:次数3\ \searrow} & \colMM{red}{ \darr\ y:次数1\ \darr\ y:次数1\ \Darr}\\ & x^3+x^2y+x^2+xy-2x-2y\\ & \colMM{red}{ \Darr 次数が低い\ y\ (の個数)で整理!}\\ &= (x^2y+xy-2y)+(x^3+x^2-2x)\\ & \colMM{orange}{ \Darr ①共通}\colMM{green}{ \Darr①共通}\\ &= \colBX{violet}{$(x^2+x-2)$}\,y + x\,\colBX{violet}{$(x^2+x-2)$}\\ & \colMM{magenta}{ \Darr①共通}\\ &= \colBX{violet}{$(x^2+x-2)$}\,(y+x)\\ & \colMM{green}{ \Darr②たすきがけ}\\ &= (x-1)(x+2)(x+y) \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
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【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{red}{a:次数3\ \searrow} & \colMM{red}{ c:次数2\darr \ \darr\ b:次数1}\\ & a^3+ab-ac^2+bc\\ & \colMM{red}{ \Darr 次数が低い\ b\ (の個数)で整理!}\\ &= (ab+bc)+(a^3-ac^2)\\ & \colMM{orange}{ \Darr ①共通}\colMM{green}{ \Darr①共通}\\ &= (a+c)b + a(a^2-c^2)\\ & \colMM{green}{ \Darr②公式}\\ &= \colBX{violet}{$(a+c)$}\,b + a\,\colBX{violet}{$(a+c)$}(a-c)\\ & \colMM{magenta}{ \Darr ①共通}\\ &= \colBX{violet}{$(a+c)$}\,\left\{b+a(a-c)\right\}\\ \\ &= (a+c)(a^2+b-ac) \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan