31021中3・多項式の展開・計算問題・展開の公式1-1
計算問題 》展開の公式1①
次の式を展開しなさい。
(1) $(\,x+2\,)(\,x+3\,)$
(2) $(\,x+3\,)(\,x-4\,)$
(3) $(\,x-5\,)(\,x+8\,)$
(4) $(\,x-4\,)(\,x-7\,)$
(5) $(\,x+0.6\,)(\,x-0.9\,)$
(6) $\Bigl(\,x+\dfrac{1}{\;2\;}\,\Bigr)\Bigl(\,x-\dfrac{2}{\;3\;}\,\Bigr)$
解答・解説
(1) $\quad\;\,\begin{eqnarray}(\,x+2\,)(\,x+3\,)\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+(\,2+3\,)x+(\,2\times 3\,)\\[6pt]&=&x^{2}+5x+6\end{eqnarray}\;\;$
答$x^{2}+5x+6$
(2) $\quad\;\,\begin{eqnarray}(\,x+3\,)(\,x-4\,)\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+\{\,3+(\,-4\,)\}x+\{\,3\times(-4)\}\\[6pt]&=&x^{2}-x-12\end{eqnarray}\;\;$
答$x^{2}-x-12$
(3) $\quad\;\,\begin{eqnarray}(\,x-5\,)(\,x+8\,)\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+\{\,(-5)+8\,\}x+\{(\,-5\,) \times 8\,\}\\[6pt]&=&x^{2}+3x-40\end{eqnarray}\;\;$
答$x^{2}+3x-40$
(4) $\quad\;\,\begin{eqnarray}(\,x-4\,)(\,x-7\,)\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+\{\,(-4)+(-7)\,\}x+\{(\,-4\,) \times (-7)\}\\[6pt]&=&x^{2}-11x+28\end{eqnarray}\;\;$
答$x^{2}-11x+28$
(5) $\quad\;\,\begin{eqnarray}(\,x+0.6\,)(\,x-0.9\,)\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+\{\,0.6+(-0.9)\,\}x+\{\,0.6\times (\,-0.9\,)\}\\[6pt]&=&x^{2}-0.3x-0.54\end{eqnarray}\;\;$
答$x^{2}-0.3x-0.54$
(6) $\quad\;\,\begin{eqnarray}\Bigl(\,x+\frac{1}{\;2\;}\,\Bigr)\Bigl(\,x-\frac{2}{\;3\;}\,\Bigr)\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}+\Bigl\{\,\frac{1}{\;2\;}+\Bigl(\,-\frac{2}{\;3\;}\,\Bigr)\Bigr\}x+\Bigl\{\,\frac{1}{\;2\;}\times \Bigl(-\frac{2}{\;3\;}\Bigr)\Bigr\}\\[6pt]&=&x^{2}-\frac{1}{\;6\;}x-\frac{1}{\;3\;}\end{eqnarray}\;\;$
答$x^{2}-\dfrac{1}{\;6\;}x-\dfrac{1}{\;3\;}$