32031中3・多項式の因数分解・計算問題・因数分解の公式2-1
計算問題 》因数分解の公式2①
次の式を展開しなさい。
(1) $x^{2}+2x+1$
(2) $x^{2}+8x+16$
(3) $x^{2}+12x+36$
(4) $x^{2}-4x+4$
(5) $x^{2}-16x+64$
(6) $x^{2}-20x+100$
解答・解説
(1) $\quad\;\,\begin{eqnarray}x^{2}+2x+1\end{eqnarray}$
$\;\;\begin{eqnarray}&=&x^{2}+2 \times \color{red}1\color{black} \times x+\color{red}1\color{black}^{2}\\[6pt]&=&(\,x+1\,)^{2}\end{eqnarray}\;\;$
答$(\,x+1\,)^{2}$
(2) $\quad\;\,\begin{eqnarray}x^{2}+8x+16\end{eqnarray}$
$\;\;\begin{eqnarray}&=&x^{2}+2 \times \color{red}4\color{black} \times x+\color{red}4\color{black}^{2}\\[6pt]&=&(\,x+4\,)^{2}\end{eqnarray}\;\;$
答$(\,x+4\,)^{2}$
(3) $\quad\;\,\begin{eqnarray}x^{2}+12x+36\end{eqnarray}$
$\;\;\begin{eqnarray}&=&x^{2}+2 \times \color{red}6\color{black} \times x+\color{red}6\color{black}^{2}\\[6pt]&=&(\,x+6\,)^{2}\end{eqnarray}\;\;$
答$(\,x+6\,)^{2}$
(4) $\quad\;\,\begin{eqnarray}x^{2}-4x+4\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}-2 \times \color{red}2\color{black} \times x+\color{red}2\color{black}^{2}\\[6pt]&=&(\,x-2\,)^{2}\end{eqnarray}\;\;$
答$(\,x-2\,)^{2}$
(5) $\quad\;\,\begin{eqnarray}x^{2}-16x+64\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}-2 \times \color{red}8\color{black} \times x+\color{red}8\color{black}^{2}\\[6pt]&=&(\,x-8\,)^{2}\end{eqnarray}\;\;$
答$(\,x-8\,)^{2}$
(6) $\quad\;\,\begin{eqnarray}x^{2}-20x+100\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}-2 \times \color{red}10\color{black} \times x+\color{red}10\color{black}^{2}\\[6pt]&=&(\,x-10\,)^{2}\end{eqnarray}\;\;$
答$(\,x-10\,)^{2}$