32053中３・多項式の因数分解・計算問題・いろいろな因数分解3

(1) $\quad x+y=A\;\;$ とおきます。

$\;\;\begin{eqnarray}(\,\color{red} x+y \color{black}\,)^{2}-3y(\,\color{red} x+y \color{black}\,)&=&A^{2}-3yA\\[6pt]&=&A(\,A-3y\,)\\[6pt]&=&(\,\color{red} x+y \color{black}\,)\{(\,\color{red} x+y \color{black}\,)-3y\,\}\\[6pt]&=&(\,x+y\,)(\,x-2y\,)\end{eqnarray}\;\;$

答$(\,x+y\,)(\,x-2y\,)$

(2) $\quad x+1=A\;\;$ とおきます。

$\;\;\begin{eqnarray}(\,\color{red} x+1 \color{black}\,)^{2}+6(\,\color{red} x+1 \color{black}\,)+5&=&A^{2}+6A+5\\[6pt]&=&(\,A+1\,)(\,A+5\,)\\[6pt]&=&\{(\,\color{red} x+1 \color{black}\,)+1\,\}\{(\,\color{red} x+1 \color{black}\,)+5\,\}\\[6pt]&=&(\,x+2\,)(\,x+6\,)\end{eqnarray}\;\;$

答$(\,x+2\,)(\,x+6\,)$

(3) $\quad a-5=A\;\;$ とおきます。

$\;\;\begin{eqnarray}(\,\color{red} a-5 \color{black}\,)^{2}-6(\,\color{red} a-5 \color{black}\,)+9&=&A^{2}-6A+9\\[6pt]&=&(\,A-3\,)^{2}\\[6pt]&=&\{(\,\color{red} a-5 \color{black}\,)-3\,\}^{2}\\[6pt]&=&(\,a-8\,)^{2}\end{eqnarray}\;\;$

答$(\,a-8\,)^{2}$

(4) $\quad x+4y=A\;\;$ とおきます。

$\;\;\begin{eqnarray}(\,\color{red} x+4y \color{black}\,)^{2}-8(\,\color{red} x+4y \color{black}\,)+12&=&A^{2}-8A+12\\[6pt]&=&(\,A-2\,)(\,A-6\,)\\[6pt]&=&\{(\,\color{red} x+4y \color{black}\,)-2\,\}\{(\,\color{red} x+4y \color{black}\,)-6\,\}\\[6pt]&=&(\,x+4y-2\,)(\,x+4y-6\,)\end{eqnarray}\;\;$

答$(\,x+4y-2\,)(\,x+4y-6\,)$

(5) $\quad 2a+b=A,\;\; a-b=B\;\;$ とおきます。

$\;\;\begin{eqnarray}(\,\color{red}2a+b\color{black}\,)^{2}-(\,\color{red}a-b\color{black}\,)^{2}&=&A^{2}-B^{2}\\[6pt]&=&(\,A+B\,)(\,A-B\,)\\[6pt]&=&\{(\,\color{red}2a+b\color{black}\,)+(\,\color{red}a-b\color{black}\,)\}\{(\,\color{red}2a+b\color{black}\,)-(\,\color{red}a-b\color{black}\,)\}\\[6pt]&=&(\,2a+b+a-b\,)(\,2a+b-a+b\,)\\[6pt]&=&3a(\,a+2b\,)\end{eqnarray}\;\;$

答$3a(\,a+2b\,)$