32051中３・多項式の因数分解・計算問題・いろいろな因数分解1

(1) $\quad\;\,\begin{eqnarray}2x^{2}-10x+12\end{eqnarray}\;\;$

$\;\;\begin{eqnarray}&=&2(\,x^{2}-5x+6\,)\\[6pt]&=&2\{\,x^{2}+(-2-3)x+(-2) \times (-3)\,\}\\[6pt]&=&2(\,x-2\,)(\,x-3\,)\end{eqnarray}\;\;$

答$2(\,x-2\,)(\,x-3\,)$

(2) $\quad\;\,\begin{eqnarray}-x^{2}-8xy-12y^{2}\end{eqnarray}\;\;$

$\;\;\begin{eqnarray}&=&-(\,x^{2}+8xy+12y^{2}\,) \\[6pt]&=&-\{\,x^{2}+(2y+6y)x+(2y) \times (6y)\,\} \\[6pt]&=&-(\,x+2y\,)(\,x+6y\,)\end{eqnarray}\;\;$

答$-(\,x+2y\,)(\,x+6y\,)$

(3) $\quad\;\,\begin{eqnarray}-3x^{2}+24x-48\end{eqnarray}\;\;$

$\;\;\begin{eqnarray}&=& -3(\,x^{2}-8x+16\,) \\[6pt]&=&-3(\,x^{2}-2 \times 4 \times x + 4^{2}\,) \\[6pt]&=&-3(\,x-4\,)^{2}\end{eqnarray}\;\;$

答$-3(\,x-4\,)^{2}$

(4) $\quad\;\,\begin{eqnarray}8x^{2}-40xy+50y^{2}\end{eqnarray}\;\;$

$\;\;\begin{eqnarray}&=&2(\,4x^{2}-20xy+25y^{2}\,) \\[6pt]&=& 2\{\,(2x)^{2}-2 \times 5y \times 2x+ (5y)^{2}\} \\[6pt]&=&2(\,2x-5y\,)^{2}\end{eqnarray}\;\;$

答$2(\,2x-5y\,)^{2}$

(5) $\quad\;\,\begin{eqnarray}50x^{2}-18y^{2}\end{eqnarray}\;\;$

$\;\;\begin{eqnarray}&=&2(\,25x^{2}-9y^{2}\,) \\[6pt]&=&2\{(\,5x\,)^{2}-(\,3y\,)^{2}\} \\[6pt]&=&2(\,5x+3y\,)(\,5x-3y\,)\end{eqnarray}\;\;$

答$2(\,5x+3y\,)(\,5x-3y\,)$