32021中３・多項式の因数分解・計算問題・因数分解の公式1-1
banno計算問題 》因数分解の公式１①
次の式を因数分解しなさい。
(1) $x^{2}+5x+6$
(2) $x^{2}+8x+12$
(3) $x^{2}+12x+32$
(4) $x^{2}-6x+8$
(5) $x^{2}-10x+24$
(6) $x^{2}-16x+48$
解答・解説
(1) $\quad\;\,\begin{eqnarray}x^{2}+5x+6\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+ \color{red}(\,2+3\,)\color{black}x+ \color{red}(\,2\times 3\,)\\[6pt]&=&(\,x+2\,)(\,x+3\,)\end{eqnarray}\;\;$
答$(\,x+2\,)(\,x+3\,)$
(2) $\quad\;\,\begin{eqnarray}x^{2}+8x+12\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+ \color{red}(\,2+6\,)\color{black}x+\color{red}(\,2\times 6\,)\\[6pt]&=&(\,x+2\,)(\,x+6\,)\end{eqnarray}\;\;$
答$(\,x+2\,)(\,x+6\,)$
(3) $\quad\;\,\begin{eqnarray}x^{2}+12x+32\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+ \color{red}(\,4+8\,)\color{black}x+\color{red}(\,4\times 8\,)\\[6pt]&=&(\,x+4\,)(\,x+8\,)\end{eqnarray}\;\;$
答$(\,x+4\,)(\,x+8\,)$
(4) $\quad\;\,\begin{eqnarray}x^{2}-6x+8\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+ \color{red}\{(-2)+(-4)\} \color{black}x+ \color{red}\{(-2) \times (-4)\} \\[6pt]&=&(\,x-2\,)(\,x-4\,)\end{eqnarray}\;\;$
答$(\,x-2\,)(\,x-4\,)$
(5) $\quad\;\,\begin{eqnarray}x^{2}-10x+24\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+ \color{red}\{(-4)+(-6)\} \color{black}x+ \color{red}\{(-4) \times (-6)\} \\[6pt]&=&(\,x-4\,)(\,x-6\,)\end{eqnarray}\;\;$
答$(\,x-4\,)(\,x-6\,)$
(6) $\quad\;\,\begin{eqnarray}x^{2}-16x+48\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+ \color{red}\{(\,-4\,)+(\,-12\,)\} \color{black}x+ \color{red}\{(\,-4\,) \times (\,-12\,)\}\\[6pt]&=&(\,x-4\,)(\,x-12\,)\end{eqnarray}\;\;$
答$(\,x-4\,)(\,x-12\,)$
