【数学I】例題1.1.13：因数分解の工夫(おき換え)（One More）★★

% 例題I1.1.13：因数分解の工夫(おき換え) （One More）★★
次の式を因数分解せよ. (1)$(x+4 y)^2-7(x+4 y)+10$(2)$\left(x^2-3 x\right)\left(x^2-3 x-2\right)-8$(3)$(x+1)(x+2)(x+3)(x+4)-24$

% 解答（例題I1.1.13）
(1)$\begin{aligned} &(x+4 y)^2-7(x+4 y)+10 \\ =& \{(x+4 y)-2\}\{(x+4 y)-5\} \\ =&(x+4 y-2)(x+4 y-5) \end{aligned}$(2)$\begin{aligned} & \left(x^2-3 x\right)\left(x^2-3 x-2\right)-8 \\=& \left(x^2-3 x\right)\left\{\left(x^2-3 x\right)-2\right\}-8 \\=& \left(x^2-3 x\right)^2-2\left(x^2-3 x\right)-8 \\=& \left\{\left(x^2-3 x\right)-4\right\}\left\{\left(x^2-3 x\right)+2\right\} \\=& \left(x^2-3 x-4\right)\left(x^2-3 x+2\right)\\=&(x+1)(x-4)(x-1)(x-2)\end{aligned}$(3)$\begin{aligned} &(x+1)(x+2)(x+3)(x+4)-24 \\=&(x+1)(x+4)\times(x+2)(x+3)-24 \\=& \left(x^2+5 x+4\right)\left(x^2+5 x+6\right)-24 \\=& \left\{\left(x^2+5 x\right)+4\right\}\left\{\left(x^2+5 x\right)+6\right\}-24 \\=& \left(x^2+5 x\right)^2+10\left(x^2+5 x\right)\\=& \left(x^2+5 x\right)\left\{\left(x^2+5 x\right)+10\right\} \\=& x(x+5)\left(x^2+5 x+10\right)\end{aligned}$

% 問題I1.1.13
次の式を因数分解せよ. (1)$(x+3y)^2-5(x+3y)+6$(2)$\left(x^2-2x\right)\left(x^2-2x-3\right)-4$(3)$(x-1)(x+1)(x+4)(x+6)+24$

% 解答I1.1.13
(1)$\begin{aligned} &(x+3y)^2-5(x+3y)+6 \\ =& \{(x+3y)-2\}\{(x+3y)-3\} \\ =&(x+3y-2)(x+3y-3) \end{aligned}$(2)$\begin{aligned} & \left(x^2-2x\right)\left(x^2-2x-3\right)-4 \\=& \left(x^2-2x\right)\left\{\left(x^2-2x\right)-3\right\}-4 \\=& \left(x^2-2x\right)^2-3\left(x^2-2x\right)-4 \\=& \left\{\left(x^2-2x\right)-4\right\}\left\{\left(x^2-2x\right)+1\right\} \\=& \left(x^2-2x-4\right)\left(x^2-2x+1\right)\\=& \left(x^2-2x-4\right)(x-1)^2\end{aligned}$(3)$\begin{aligned} &(x-1)(x+1)(x+4)(x+6)+24 \\=&(x-1)(x+6)\times(x+1)(x+4)+24 \\=& \left(x^2+5x-6\right)\left(x^2+5x+4\right)+24 \\=& \left\{\left(x^2+5x\right)-6\right\}\left\{\left(x^2+5x\right)+4\right\}+24 \\=& \left(x^2+5x\right)^2-2\left(x^2+5x\right)\\=& \left(x^2+5x\right)\left\{\left(x^2+5x\right)-2\right\} \\=& x(x+5)\left(x^2+5x-2\right)\end{aligned}$