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【数学I】例題1.1.7:掛ける順序や組み合わせを工夫した展開(One More)★★

【数学I】例題1.1.7:掛ける順序や組み合わせを工夫した展開(One More)
【数学I】例題1.1.7:掛ける順序や組み合わせを工夫した展開の例題ページ
問題の解答

【数学I】問題1.1.7:掛ける順序や組み合わせを工夫した展開の解答
検索用コード(LaTeX)
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% 例題I1.1.7:掛ける順序や組み合わせを工夫した展開(One More)★★
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次の式を展開せよ.

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

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

(3) $(p-q)^2(p+q)^2(p^2+q^2)^2$

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% 例題I1.1.7の解答(One More)★★
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(1)

\begin{align*}
&(x-1)(x-4)(x+3)(x+6)\\
=&(x-1)(x+3)\times(x-4)(x+6)\\
=&(x^2+2x-3)(x^2+2x-24)\\
=&\{(x^2+2x)-3\}\{(x^2+2x)-24\}\\
=&(x^2+2x)^2-27(x^2+2x)+72\\
=&x^4+4x^3+4x^2-27x^2-54x+72\\
=&x^4+4x^3-23x^2-54x+72
\end{align*}

(2)

\begin{align*}
&(x-2)(x+2)(x^2+4)(x^4+4)\\
=&(x^2-4)(x^2+4)(x^4+4)\\
=&(x^4-16)(x^4+4)\\
=&x^8+4x^4-16x^4-64\\
=&x^8-12x^4-64
\end{align*}

(3)

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

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% 問題I1.1.7:掛ける順序や組み合わせを工夫した展開(One More)★★
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次の式を展開せよ.

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

(2) $(x-3)(x+3)(x^2+9)(x^4+9)$

(3) $(m-n)^2(m^2+mn+n^2)^2$

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% 問題I1.1.7の解答(One More)★★
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(1)

\begin{align*}
&(x-4)(x-5)(x+2)(x+3)\\
=&(x-4)(x+2)\times(x-5)(x+3)\\
=&(x^2-2x-8)(x^2-2x-15)\\
=&\{(x^2-2x)-8\}\{(x^2-2x)-15\}\\
=&(x^2-2x)^2-23(x^2-2x)+120\\
=&x^4-4x^3+4x^2-23x^2+46x+120\\
=&x^4-4x^3-19x^2+46x+120
\end{align*}

(2)

\begin{align*}
&(x-3)(x+3)(x^2+9)(x^4+9)\\
=&(x^2-9)(x^2+9)(x^4+9)\\
=&(x^4-81)(x^4+9)\\
=&x^8+9x^4-81x^4-729\\
=&x^8-72x^4-729
\end{align*}

(3)

\begin{align*}
&(m-n)^2(m^2+mn+n^2)^2\\
=&\{(m-n)(m^2+mn+n^2)\}^2\\
=&(m^3-n^3)^2\\
=&m^6-2m^3n^3+n^6
\end{align*}

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