【数学I】例題1.1.13：因数分解の工夫(おき換え)（One More）★★

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% 例題I1.1.13：因数分解の工夫（おき換え）（One More）★★
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次の式を因数分解せよ．

(1) $(x+4y)^2-7(x+4y)+10$

(2) $(x^2-3x)(x^2-3x-2)-8$

(3) $(x+1)(x+2)(x+3)(x+4)-24$

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% 例題I1.1.13の解答（One More）★★
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(1)

\begin{align*}
&(x+4y)^2-7(x+4y)+10\\
=&\{(x+4y)-2\}\{(x+4y)-5\}\\
=&(x+4y-2)(x+4y-5)
\end{align*}

(2)

\begin{align*}
&(x^2-3x)(x^2-3x-2)-8\\
=&(x^2-3x)\{(x^2-3x)-2\}-8\\
=&(x^2-3x)^2-2(x^2-3x)-8\\
=&\{(x^2-3x)-4\}\{(x^2-3x)+2\}\\
=&(x^2-3x-4)(x^2-3x+2)\\
=&(x+1)(x-4)(x-1)(x-2)
\end{align*}

(3)

\begin{align*}
&(x+1)(x+2)(x+3)(x+4)-24\\
=&(x+1)(x+4)\times(x+2)(x+3)-24\\
=&(x^2+5x+4)(x^2+5x+6)-24\\
=&\{(x^2+5x)+4\}\{(x^2+5x)+6\}-24\\
=&(x^2+5x)^2+10(x^2+5x)\\
=&(x^2+5x)\{(x^2+5x)+10\}\\
=&x(x+5)(x^2+5x+10)
\end{align*}

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% 問題I1.1.13：因数分解の工夫（おき換え）（One More）★★
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次の式を因数分解せよ．

(1) $(x+3y)^2-5(x+3y)+6$

(2) $(x^2-2x)(x^2-2x-3)-4$

(3) $(x-1)(x+1)(x+4)(x+6)+24$

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% 問題I1.1.13の解答（One More）★★
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(1)

\begin{align*}
&(x+3y)^2-5(x+3y)+6\\
=&\{(x+3y)-2\}\{(x+3y)-3\}\\
=&(x+3y-2)(x+3y-3)
\end{align*}

(2)

\begin{align*}
&(x^2-2x)(x^2-2x-3)-4\\
=&(x^2-2x)\{(x^2-2x)-3\}-4\\
=&(x^2-2x)^2-3(x^2-2x)-4\\
=&\{(x^2-2x)-4\}\{(x^2-2x)+1\}\\
=&(x^2-2x-4)(x^2-2x+1)\\
=&(x^2-2x-4)(x-1)^2
\end{align*}

(3)

\begin{align*}
&(x-1)(x+1)(x+4)(x+6)+24\\
=&(x-1)(x+6)\times(x+1)(x+4)+24\\
=&(x^2+5x-6)(x^2+5x+4)+24\\
=&\{(x^2+5x)-6\}\{(x^2+5x)+4\}+24\\
=&(x^2+5x)^2-2(x^2+5x)\\
=&(x^2+5x)\{(x^2+5x)-2\}\\
=&x(x+5)(x^2+5x-2)
\end{align*}

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% 例題I1.1.13の側注：因数分解の工夫（おき換え）（One More）★★
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% 側注1
$x+4y=A$とおくと，
\begin{align*}
&A^2-7A+10\\
=&(A-2)(A-5)
\end{align*}

% 側注2
$x^2-3x=A$とおくと，
\begin{align*}
&A(A-2)-8\\
=&A^2-2A-8\\
=&(A-4)(A+2)
\end{align*}

% 側注3
共通する部分が現れるように，組み合わせを工夫する．

% 側注4
$x^2+5x=A$とおくと，
\begin{align*}
&(A+4)(A+6)-24\\
=&A^2+10A\\
=&A(A+10)
\end{align*}

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% 問題I1.1.13の側注：因数分解の工夫（おき換え）（One More）★★
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% 側注1
$x+3y=A$とおくと，
\begin{align*}
&A^2-5A+6\\
=&(A-2)(A-3)
\end{align*}

% 側注2
$x^2-2x=A$とおくと，
\begin{align*}
&A(A-3)-4\\
=&A^2-3A-4\\
=&(A-4)(A+1)
\end{align*}

% 側注3
共通する部分が現れるように，組み合わせを工夫する．

% 側注4
$x^2+5x=A$とおくと，
\begin{align*}
&(A-6)(A+4)+24\\
=&A^2-2A\\
=&A(A-2)
\end{align*}

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% 例題I1.1.13の考え方・補足など：因数分解の工夫（おき換え）（One More）★★
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% 考え方
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(1)は，展開してから1つの文字に着目して因数分解してもよいが，計算に手間が掛かる．そこで，式の特徴に合わせて，共通する部分を見つけることを考える．(2)はそのままでも共通する部分があるが，(3)はそのままでは共通する部分がないので，組み合わせを工夫して共通する部分を見つける．

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共通する部分を探し，くくり出しや因数分解の公式の適用を考える．

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(2)や(3)において，それぞれ$(x^2-3x-4)(x^2-3x+2),(x^2+5x)(x^2+5x+10)$と答えてしまわないように注意すること．因数分解はできるところまでするように心掛けよう（確認する習慣をつけるとよい）．