32054中３・多項式の因数分解・計算問題・いろいろな因数分解4

(1) $\;\;\begin{eqnarray}\color{red}ac+ad\color{black}+2(\,c+d\,)=\color{red}a(\,c+d\,)\color{black}+2(\,c+d\,)\end{eqnarray}\;\;$

$\quad\quad c+d=A\;\;$ とおきます。

$\;\begin{eqnarray}a(\,\color{red}c+d\color{black}\,)+2(\,\color{red}c+d\color{black}\,)&=&aA+2A\\[6pt]&=&A(\,a+2\,)\\[6pt]&=&(\,\color{red}c+d\color{black}\,)(\,a+2\,)\\[6pt]&=&(\,a+2\,)(\,c+d\,)\end{eqnarray}\;\;$

答$(\,a+2\,)(\,c+d\,)$

(2) $\;\;\begin{eqnarray}\color{red}xy+x\color{blue}-y-1\color{black}=\color{red}x(\,y+1\,)\color{blue}-(\,y+1\,)\end{eqnarray}\;\;$

$\quad\quad y+1=A\;\;$ とおきます。

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

答$(\,x-1\,)(\,y+1\,)$

(3) $\;\;\begin{eqnarray}\color{red}ab-2bc\color{blue}+3ad-6cd\color{black}=\color{red}b(\,a-2c\,)\color{blue}+3d(\,a-2c\,)\end{eqnarray}\;\;$

$\quad\quad a-2c=A\;\;$ とおきます。

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

答$(\,a-2c\,)(\,b+3d\,)$

(4) $\;\;\begin{eqnarray}\color{red}x^{2}-8x+16\color{black}-4y^{2}=\color{red}(\,x-4\,)^{2}\color{black}-4y^{2}\end{eqnarray}\;\;$

$\quad\quad x-4=A\;\;$ とおきます。

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

答$(\,x+2y-4\,)(\,x-2y-4\,)$

(5) $\;\;\begin{eqnarray}\color{red}4a^{2}-12ab+9b^{2}\color{black}-9c^{2}=\color{red}(\,2a-3b\,)^{2}\color{black}-9c^{2}\end{eqnarray}\;\;$

$\quad\quad 2a-3b=A\;\;$ とおきます。

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

答$(\,2a-3b+3c\,)(\,2a-3b-3c\,)$