【数学I】例題1.1.8：因数分解の基本（One More）★

% 例題I1.1.8：因数分解の基本 （One More）★
次の式を因数分解せよ. (1)$3x^3y+9x^2y+15xy^3$(2)$xy-y-x+1$(3)$x^2+10x+25$(4)$9a^3-6a^2+a$(5)$16x^2-(x+1)^2$(6)$a^2+5a+6$

% 解答（例題I1.1.8）
(1)$3x^3y+9x^2y+15xy^3=3xy(x^2+3x+5y^2)$(2)$xy-y-x+1=(x-1)y-(x-1)=(x-1)(y-1)$(3)$x^2+10x+25=x^2+2 \cdot x \cdot 5+5^2=(x+5)^2$(4)$9a^3-6a^2+a=a(9a^2-6a+1)=a\{(3a)^2-2 \cdot 3a \cdot 1+1^2\}=a(3a-1)^2$(5)$\begin{aligned} 16x^2-(x+1)^2&=\{4x+(x+1)\}\{4x-(x+1)\}\\ &=(4x+x+1)(4x-x-1)\\ &=(5x+1)(3x-1) \end{aligned}$(6)$a^2+5a+6=a^2+(2+3)a+2 \cdot 3=(a+2)(a+3)$

% 問題I1.1.8
次の式を因数分解せよ. (1)$4x^3y^2+8x^2y^2+12xy^3$(2)$ab-b-a+1$(3)$x^2-14x+49$(4)$25m^3-20m^2+4m$(5)$9y^2-(y-2)^2$(6)$x^2+7x+10$

% 解答I1.1.8
(1)$4x^3y^2+8x^2y^2+12xy^3=4xy^2(x^2+2x+3y)$(2)$ab-b-a+1=(a-1)b-(a-1)=(a-1)(b-1)$(3)$x^2-14x+49=x^2-2 \cdot x \cdot 7+7^2=(x-7)^2$(4)$\begin{aligned} 25m^3-20m^2+4m&=m(25m^2-20m+4)\\ &=m\{(5m)^2-2 \cdot 5m \cdot 2+2^2\}\\ &m(5m-2)^2 \end{aligned}$(5)$\begin{aligned} 9y^2-(y-2)^2&=\{3y+(y-2)\}\{3y-(y-2)\}\\ &=(3y+y-2)(3y-y+2)\\ &=(4y-2)(2y+2)\\ &=4(2y-1)(y+1) \end{aligned}$(6)$x^2+7x+10=x^2+(2+5)x+5 \cdot 2=(x+2)(x+5)$