【数学I】例題1.1.12：因数分解の工夫（次数が同じ場合）（One More）★★

% 例題I1.1.12：因数分解の工夫(次数が同じ場合) （One More）★★
次の式を因数分解せよ. (1)$x^2+3xy+2y^2-x-3y-2$(2)$2 x^2+5 x y+3 y^2-3 x-5 y-2$

% 解答（例題I1.1.12）
(1)$\begin{aligned} & x^2+3 x y+2 y^2-x-3 y-2 \\=& x^2+(3 y-1)x+\left(2 y^2-3 y-2\right)\\=& x^2+(3 y-1)x+(y-2)(2 y+1)\\=& \{x+(y-2)\}\{x+(2 y+1)\} \\=&(x+y-2)(x+2 y+1)\end{aligned}$(2)$\begin{aligned} & 2 x^2+5 x y+3 y^2-3 x-5 y-2 \\=& 2 x^2+(5 y-3)x+\left(3 y^2-5 y-2\right)\\=& 2 x^2+(5 y-3)x+(y-2)(3 y+1)\\=& \{x+(y-2)\}\{2 x+(3 y+1)\}\\=&(x+y-2)(2 x+3 y+1)\end{aligned}$

% 問題I1.1.12
次の式を因数分解せよ. (1)$x^2+4xy+3y^2-2x-8y-3$(2)$3x^2+11xy+10y^2-x-3y-4$

% 解答I1.1.12
(1)$\begin{aligned} & x^2+4 x y+3 y^2-2x-8y-3 \\=& x^2+(4 y-2)x+\left(3 y^2-8y-3\right)\\=& x^2+(4 y-2)x+(y-3)(3 y+1)\\=& \{x+(y-3)\}\{x+(3 y+1)\} \\=&(x+y-3)(x+3 y+1)\end{aligned}$(2)$\begin{aligned} & 3x^2+11xy+10y^2-x-3y-4 \\ =& 3x^2+(11y-1)x+(10y^2-3y-4)\\ =& 3x^2+(11y-1)x+(2y+1)(5y-4)\\ =& \{x+(2y+1)\}\{3x+(5y-4)\} \\ =&(x+2y+1)(3x+5y-4) \end{aligned}$