【数学I】例題1.1.4：乗法公式を用いた展開（One More）★

% 例題I1.1.4：乗法公式を用いた展開 （One More）★
次の式を展開せよ. (1)$(x+1)^2$(2)$(x-2y)^2$(3)$(3ab+1)(3ab-1)$(4)$(a+2b)(a-3b)$(5)$(3x+2)(4x+1)$(6)$(x+y-z)^2$

% 解答（例題I1.1.4）
(1)$(x+1)^2=x^2+2 \cdot x \cdot 1+1^2=x^2+2x+1$(2)$(x-2y)^2=x^2-2 \cdot x \cdot 2y+(2y)^2=x^2-4xy+4y^2$(3)$(3ab+1)(3ab-1)=(3ab)^2-1^2=9a^2b^2-1$(4)$(a+2b)(a-3b)=a^2+\{2b+(-3b)\}a+2b \cdot (-3b)=a^2-ab-6b^2$(5)$(3x+2)(4x+1)=3 \cdot 4x^2+(3 \cdot 1+2 \cdot 4)x+2 \cdot 1 =12x^2+11x+2$(6)$\begin{aligned} (x+y-z)^2&=\{x+y+(-z)\}^2\\ &=x^2+y^2+(-z)^2+2 \cdot x \cdot y+2 \cdot y \cdot (-z)+2 \cdot (-z) \cdot x\\ &=x^2+y^2+z^2+2xy-2yz-2zx \end{aligned}$別解：$\begin{aligned} (x+y-z)^2 &=\{(x+y)-z\}^2 \\ &=(x+y)^2-2(x+y)z+z^2 \\ &=\left(x^2+2xy+y^2\right)-2xz-2yz+z^2 \\ &=x^2+y^2+z^2+2xy-2yz-2zx \end{aligned}$

% 問題I1.1.4
次の式を展開せよ. (1)$(x+3)^2$(2)$(k-2)^2$(3)$(x+2y)(x-2y)$(4)$(x-2y)(x-5y)$(5)$(4a+2b)(3a+b)$(6)$(2x-y-z)^2$

% 解答I1.1.4
(1)$(x+3)^2=x^2+2 \cdot x \cdot 3+3^2=x^2+6x+9}$(2)$(k-2)^2=k^2-2 \cdot k \cdot 2+2^2=k^2-4k+4$(3)$(x+2y)(x-2y)=x^2-(2y)^2=x^2-4y^2$(4)$(x-2y)(x-5y)=x^2+\{(-2y)+(-5y)\}x+(-2y) \cdot  (-5y)=x^2-7xy+10y^2$(5)$(4a+2b)(3a+b)=4 \cdot 3a^2+(4 \cdot b+2b \cdot 3)a+2b \cdot b=12a^2+10ab+2b^2$(6)$\begin{aligned} (2x-y-z)^2&=\{2x+(-y)+(-z)\}^2\\ &=(2x)^2+(-y)^2+(-z)^2\\ &+2 \cdot (2x) \cdot (-y)+2 \cdot (-y) \cdot (-z)+2 \cdot (-z) \cdot (2x)\\ &=4x^2+y^2+z^2-4xy+2yz-4zx \end{aligned}$別解：$\begin{aligned} (2x-y-z)^2 &=\{(2x-y)-z\}^2=(2x-y)^2-2(2x-y)z+z^2\\ &=\left(4x^2-4xy+y^2\right)-4xz+2yz+z^2\\ &=4x^2+y^2+z^2-4xy+2yz-4zx \end{aligned}$