【数学I】例題1.1.6：おき換えを用いた展開（One More）★★

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% 例題I1.1.6：おき換えを用いた展開（One More）★★
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次の式を展開せよ．

(1) $(x-y+2)(x-y-4)$

(2) $(3x+4y-z)(3x-4y+z)$

(3) $(a+b-c+d)(a+b+c-d)$

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% 例題I1.1.6の解答（One More）★★
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(1)

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

(2)

\begin{align*}
&(3x+4y-z)(3x-4y+z)\\
=&\{3x+(4y-z)\}\{3x-(4y-z)\}\\
=&(3x)^2-(4y-z)^2\\
=&9x^2-(16y^2-8yz+z^2)\\
=&9x^2-16y^2+8yz-z^2
\end{align*}

(3)

\begin{align*}
&(a+b+c-d)(a+b-c+d)\\
=&\{(a+b)+(c-d)\}\{(a+b)-(c-d)\}\\
=&(a+b)^2-(c-d)^2\\
=&(a^2+2ab+b^2)-(c^2-2cd+d^2)\\
=&a^2+b^2-c^2-d^2+2ab+2cd
\end{align*}

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% 問題I1.1.6：おき換えを用いた展開（One More）★★
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次の式を展開せよ．

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

(2) $(4a-3b+c)(4a+3b-c)$

(3) $(p+q-r+s)(p+q+r-s)$

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% 問題I1.1.6の解答（One More）★★
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(1)

\begin{align*}
&(x+2y-3)(x+2y+5)\\
=&\{(x+2y)-3\}\{(x+2y)+5\}\\
=&(x+2y)^2+2(x+2y)-15\\
=&x^2+4xy+4y^2+2x+4y-15
\end{align*}

(2)

\begin{align*}
&(4a-3b+c)(4a+3b-c)\\
=&\{4a+(-3b+c)\}\{4a-(-3b+c)\}\\
=&(4a)^2-(-3b+c)^2\\
=&16a^2-(9b^2-6bc+c^2)\\
=&16a^2-9b^2+6bc-c^2
\end{align*}

(3)

\begin{align*}
&(p+q-r+s)(p+q+r-s)\\
=&\{(p+q)+(r-s)\}\{(p+q)-(r-s)\}\\
=&(p+q)^2-(r-s)^2\\
=&(p^2+2pq+q^2)-(r^2-2rs+s^2)\\
=&p^2+q^2-r^2-s^2+2pq+2rs
\end{align*}

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% 例題I1.1.6の側注：おき換えを用いた展開（One More）★★
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% 側注1
$x-y=A$とおくと，

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

% 側注2
$-(4y-z)$をくくると，$4y-z$が共通する部分となる．

% 側注3
$4y-z=A$とおくと，

\begin{align*}
&(3x+A)(3x-A)\\
=&9x^2-A^2
\end{align*}

% 側注4
\begin{align*}
&a+b-c+d\\
=&(a+b)-(c-d)
\end{align*}

% 側注5
$a+b=A,c-d=B$とおくと，$(A+B)(A-B)=A^2-B^2$

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% 問題I1.1.6の側注：おき換えを用いた展開（One More）★★
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% 側注1
$x+2y=A$とおくと，

\begin{align*}
&(A-3)(A+5)\\
=&A^2+2A-15
\end{align*}

% 側注2
$-(3b-c)$をくくると，共通する部分が$3b-c$となる．

% 側注3
$3b-c=A$とおくと，

\begin{align*}
&(4a+A)(4a-A)\\
=&16a^2-A^2
\end{align*}

% 側注4
\begin{align*}
&p+q-r+s\\
=&(p+q)-(r-s)
\end{align*}

% 側注5
$p+q=A,r-s=B$とおくと，$(A+B)(A-B)=A^2-B^2$

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% 例題I1.1.6の考え方・補足など：おき換えを用いた展開（One More）★★
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% 考え方
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分配法則を用いて展開してもよいが，計算に手間が掛かる．共通する部分を探し，それを1つの文字でおき換えるか括弧でくくると，適用できる乗法公式がわかる．

(1) $(x-y+2)(x-y-4)=(A+2)(A-4)$

(2) $(3x+4y-z)(3x-4y+z)=\{3x+(4y-z)\}\{3x-(4y-z)\}=(3x+A)(3x-A)$

(3) $(a+b-c+d)(a+b+c-d)=\{(a+b)+(c-d)\}\{(a+b)-(c-d)\}=(A+B)(A-B)$

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共通する部分を探し，括弧でくくる．