【数学I】例題1.1.4：乗法公式を用いた展開（One More）★

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% 例題I1.1.4：乗法公式を用いた展開（One More）★
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次の式を展開せよ．

(1) $(x+1)^2$

(2) $(x-2y)^2$

(3) $(3ab+1)(3ab-1)$

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

(5) $(3x+2)(4x+1)$

(6) $(x+y-z)^2$

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% 例題I1.1.4の解答（One More）★
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(1) $(x+1)^2=x^2+2\cdot x\cdot 1+1^2=x^2+2x+1$

(2) $(x-2y)^2=x^2-2\cdot x\cdot 2y+(2y)^2=x^2-4xy+4y^2$

(3) $(3ab+1)(3ab-1)=(3ab)^2-1^2=9a^2b^2-1$

(4) $(a+2b)(a-3b)=a^2+\{2b+(-3b)\}a+2b\cdot(-3b)=a^2-ab-6b^2$

(5) $(3x+2)(4x+1)=3\cdot 4x^2+(3\cdot 1+2\cdot 4)x+2\cdot 1=12x^2+11x+2$

(6)

\begin{align*}
(x+y-z)^2&=\{x+y+(-z)\}^2\\
&=x^2+y^2+(-z)^2+2\cdot x\cdot y+2\cdot y\cdot(-z)+2\cdot(-z)\cdot x\\
&=x^2+y^2+z^2+2xy-2yz-2zx
\end{align*}

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% 例題I1.1.4の別解
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\begin{align*}
(x+y-z)^2&=\{(x+y)-z\}^2\\
&=(x+y)^2-2(x+y)z+z^2\\
&=(x^2+2xy+y^2)-2xz-2yz+z^2\\
&=x^2+y^2+z^2+2xy-2yz-2zx
\end{align*}

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% 問題I1.1.4：乗法公式を用いた展開（One More）★
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次の式を展開せよ．

(1) $(x+3)^2$

(2) $(k-2)^2$

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

(4) $(x-2y)(x-5y)$

(5) $(4a+2b)(3a+b)$

(6) $(2x-y-z)^2$

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% 問題I1.1.4の解答（One More）★
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(1) $(x+3)^2=x^2+2\cdot x\cdot 3+3^2=x^2+6x+9$

(2) $(k-2)^2=k^2-2\cdot k\cdot 2+2^2=k^2-4k+4$

(3) $(x+2y)(x-2y)=x^2-(2y)^2=x^2-4y^2$

(4) $(x-2y)(x-5y)=x^2+\{(-2y)+(-5y)\}x+(-2y)\cdot(-5y)=x^2-7xy+10y^2$

(5) $(4a+2b)(3a+b)=4\cdot 3a^2+(4\cdot b+2b\cdot 3)a+2b\cdot b=12a^2+10ab+2b^2$

(6)

\begin{align*}
(2x-y-z)^2&=\{2x+(-y)+(-z)\}^2\\
&=(2x)^2+(-y)^2+(-z)^2\\
&+2\cdot(2x)\cdot(-y)+2\cdot(-y)\cdot(-z)+2\cdot(-z)\cdot(2x)\\
&=4x^2+y^2+z^2-4xy+2yz-4zx
\end{align*}

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% 問題I1.1.4の別解
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\begin{align*}
(2x-y-z)^2&=\{(2x-y)-z\}^2=(2x-y)^2-2(2x-y)z+z^2\\
&=(4x^2-4xy+y^2)-4xz+2yz+z^2\\
&=4x^2+y^2+z^2-4xy+2yz-4zx
\end{align*}

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% 例題I1.1.4の側注：乗法公式を用いた展開（One More）★
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% 側注1
\begin{align*}
&(a+b)^2\\
=&a^2+2ab+b^2
\end{align*}

% 側注2
\begin{align*}
&(a-b)^2\\
=&a^2-2ab+b^2
\end{align*}

% 側注3
\begin{align*}
&(a+b)(a-b)\\
=&a^2-b^2
\end{align*}

% 側注4
\begin{align*}
&(x+a)(x+b)\\
=&x^2+(a+b)x+ab
\end{align*}

% 側注5
\begin{align*}
(ax+b)(cx&+d)\\
=acx^2+(ad&+bc)x+bd
\end{align*}

% 側注6
\begin{align*}
&(a+b+c)^2\\
=&a^2+b^2+c^2\\
&+2ab+2bc+2ca
\end{align*}

% 側注7
$x+y=A$とおくと，

\[
(A-z)^2=A^2-2Az+z^2
\]

% 側注8
輪環の順に整理するとよい．

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% 問題I1.1.4の側注：乗法公式を用いた展開（One More）★
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% 側注1
\begin{align*}
&(a+b)^2\\
=&a^2+2ab+b^2
\end{align*}

% 側注2
\begin{align*}
&(a-b)^2\\
=&a^2-2ab+b^2
\end{align*}

% 側注3
\begin{align*}
&(a+b)(a-b)\\
=&a^2-b^2
\end{align*}

% 側注4
\begin{align*}
&(x+a)(x+b)\\
=&x^2+(a+b)x+ab
\end{align*}

% 側注5
\begin{align*}
(ax+b)(cx&+d)\\
=acx^2+(ad&+bc)x+bd
\end{align*}

% 側注6
\begin{align*}
&(a+b+c)^2\\
=&a^2+b^2+c^2\\
&+2ab+2bc+2ca
\end{align*}

% 側注7
$2x-y=A$とおくと，

\[
(A-z)^2=A^2-2Az+z^2
\]

% 側注8
輪環の順に整理するとよい．

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% 例題I1.1.4の考え方・補足など：乗法公式を用いた展開（One More）★
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% 考え方
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分配法則を考えると，公式を用いなくても展開することができるが，後に学ぶ因数分解のためにも公式による手法を覚えておくとよい．

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% 乗法公式
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(i) $(a+b)^2=a^2+2ab+b^2$

(ii) $(a-b)^2=a^2-2ab+b^2$

(iii) $(a+b)(a-b)=a^2-b^2$

(iv) $(x+a)(x+b)=x^2+(a+b)x+ab$

(v) $(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$

(vi) $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

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% 注意
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乗法公式(v)を用いる(5)のような展開は，分配法則を用いても計算の手間に大差がない．乗法公式(v)は後に学ぶ因数分解において大切であるが，分配法則でも同程度の手間で計算できるため，展開公式として無理に暗記する必要はない．なお，(6)は輪環の順ではない$x^2+y^2+z^2+2xy-2yz-2xz$なども正答である．