【数学I】例題1.1.3：多項式の乗法（One More）★

% 例題I1.1.3：多項式の乗法 （One More）★
次の計算をせよ. (1)$4x^3y^2\times(-2xy^3)^2$(2)$2ab^2c(3a^3+2b+c^2)$(3)$(x+2)(2x^2-x+5)$(4)$(2x^3-4x^2+x)(2-x+x^2)$

% 解答（例題I1.1.3）
(1)$\begin{aligned} 4x^3y^2\times(-2xy^3)^2&=4x^3y^2\times(-2)^2x^2(y^3)^2 =4x^3y^2\times 4x^2y^6\\ &=4 \cdot 4x^{3+2}y^{2+6} =16x^5y^8 \end{aligned}$(2)$\begin{aligned} 2ab^2c(3a^3+2b+c^2)&=2ab^2c \cdot 3a^3+2ab^2c \cdot 2b+2ab^2c \cdot c^2\\ &=6a^4b^2c+4ab^3c+2ab^2c^3 \end{aligned}$(3)$\begin{aligned} (x+2)(2x^2-x+5)&=x(2x^2-x+5)+2(2x^2-x+5)\\ &=2x^3-x^2+5x+4x^2-2x+10\\ &=2x^3+3x^2+3x+10 \end{aligned}$(4)$\begin{aligned} &(2x^3-4x^2+x)(2-x+x^2)\\ =&(2x^3-4x^2+x)(x^2-x+2)\\ =& 2x^3(x^2-x+2)-4x^2(x^2-x+2)+x(x^2-x+2)\\ =& 2x^5-2x^4+4x^3-4x^4+4x^3-8x^2+x^3-x^2+2x\\ =&2x^5-6x^4+9x^3-9x^2+2x \end{aligned}$

% 問題I1.1.3
次の計算をせよ. (1)$3x^2y \times(-4xy^2)^2$(2)$5abc^2(2a^2-3b+4c)$(3)$(x-3)(x^2+4x-7)$(4)$(x^3-2x+5)(3x^2-x+4)$

% 解答I1.1.3
(1)$\begin{aligned} 3x^2y \times(-4xy^2)^2 &=3x^2y \times(-4)^2x^2(y^2)^2 =3x^2y \times 16x^2y^4\\ &=3 \cdot 16x^{2+2}y^{1+4} =48x^4y^5 \end{aligned}$(2)$\begin{aligned} 5abc^2(2a^2-3b+4c)&=5abc^2 \cdot 2a^2+5abc^2 \cdot (-3b)+5abc^2 \cdot 4c\\ &=10a^3bc^2-15ab^2c^2+20abc^3 \end{aligned}$(3)$\begin{aligned} (x-3)(x^2+4x-7)&=x(x^2+4x-7)-3(x^2+4x-7)\\ &=x^3+4x^2-7x-3x^2-12x+21\\ &=x^3+x^2-19x+21 \end{aligned}$(4)$\begin{aligned} &(x^3-2x+5)(3x^2-x+4)\\ =& x^3(3x^2-x+4)-2x(3x^2-x+4)+5(3x^2-x+4)\\ =& 3x^5-x^4+4x^3-6x^3+2x^2-8x+15x^2-5x+20\\ =& 3x^5-x^4-2x^3+17x^2-13x+20 \end{aligned}$