【数学I】例題1.1.11：因数分解の工夫（次数の低い文字に着目）（One More）★★

% 例題I1.1.11：因数分解の工夫（次数の低い文字に着目） （One More）★★
次の式を因数分解せよ. (1)$4a^2+2ab+b-1$(2)$x^3+x^2 y+2 x y+y^2-1$(3)$a^2+2ab+ac+b^2+bc$

% 解答（例題I1.1.11）
(1)$\begin{aligned} 4a^2+2ab+b-1 &=(2a+1)b+4a^2-1 \\ &=(2a+1)b+(2a+1)(2a-1)\\ &=(2a+1)(2a+b-1) \end{aligned}$(2)$\begin{aligned} & x^3+x^2 y+2 x y+y^2-1 \\ =& y^2+\left(x^2+2 x\right)y+\left(x^3-1\right)\\ =& y^2+\left(x^2+2 x\right)y+(x-1)\left(x^2+x+1\right)\\ =& \left\{y+\left(x^2+x+1\right)\right\}\{y+(x-1)\} \\ =& \left(x^2+x+y+1\right)(x+y-1)\end{aligned}$(3)$\begin{aligned} & a^2+2ab+ac+b^2+bc \\=&(a+b)c+\left(a^2+2ab+b^2\right)\\=&(a+b)c+(a+b)^2 \\=&(a+b)\{c+(a+b)\} \\=&(a+b)(a+b+c)\end{aligned}$

% 問題I1.1.11
次の式を因数分解せよ. (1)$9x^2+3xy+y-1$(2)$x^3+x^2y+3xy+y^2+2y-8$(3)$xy+xz-y^2-z^2-2yz$

% 解答I1.1.11
(1)$\begin{aligned} 9x^2+3xy+y-1 &=(3x+1)y+9x^2-1 \\ &=(3x+1)y+(3x+1)(3x-1)\\ &=(3x+1)(3x+y-1) \end{aligned}$(2)$\begin{aligned} & x^3+x^2y+3xy+y^2+2y-8 \\ =& y^2+(x^2+3x+2)y+(x^3-8)\\ =& y^2+(x^2+3x+2)y+(x-2)(x^2+2x+4)\\ =& \left\{y+(x^2+2x+4)\right\}\{y+(x-2)\} \\ =& \left(x^2+2x+y+4\right)(x+y-2) \end{aligned}$(3)$\begin{aligned} & xy+xz-y^2-z^2-2yz \\=&(y+z)x-(y^2+2yz+z^2)\\=&(y+z)x-(y+z)^2 \\=&(y+z)\{x-(y+z)\} \\=&(y+z)(x-y-z)\end{aligned}$