次の4つの式を展開する問題です。 (1) $(5a + 3b)^2$ (2) $(a + 6b)(a - 5b)$ (3) $(x - 4)^3$ (4) $(x + 3)(x^2 - 3x + 9)$代数学展開多項式2025/7/61. 問題の内容次の4つの式を展開する問題です。(1) (5a+3b)2(5a + 3b)^2(5a+3b)2(2) (a+6b)(a−5b)(a + 6b)(a - 5b)(a+6b)(a−5b)(3) (x−4)3(x - 4)^3(x−4)3(4) (x+3)(x2−3x+9)(x + 3)(x^2 - 3x + 9)(x+3)(x2−3x+9)2. 解き方の手順(1) (5a+3b)2(5a + 3b)^2(5a+3b)2 の展開(5a+3b)2=(5a+3b)(5a+3b)(5a + 3b)^2 = (5a + 3b)(5a + 3b)(5a+3b)2=(5a+3b)(5a+3b)=(5a)(5a)+(5a)(3b)+(3b)(5a)+(3b)(3b)= (5a)(5a) + (5a)(3b) + (3b)(5a) + (3b)(3b)=(5a)(5a)+(5a)(3b)+(3b)(5a)+(3b)(3b)=25a2+15ab+15ab+9b2= 25a^2 + 15ab + 15ab + 9b^2=25a2+15ab+15ab+9b2=25a2+30ab+9b2= 25a^2 + 30ab + 9b^2=25a2+30ab+9b2(2) (a+6b)(a−5b)(a + 6b)(a - 5b)(a+6b)(a−5b) の展開(a+6b)(a−5b)=(a)(a)+(a)(−5b)+(6b)(a)+(6b)(−5b)(a + 6b)(a - 5b) = (a)(a) + (a)(-5b) + (6b)(a) + (6b)(-5b)(a+6b)(a−5b)=(a)(a)+(a)(−5b)+(6b)(a)+(6b)(−5b)=a2−5ab+6ab−30b2= a^2 - 5ab + 6ab - 30b^2=a2−5ab+6ab−30b2=a2+ab−30b2= a^2 + ab - 30b^2=a2+ab−30b2(3) (x−4)3(x - 4)^3(x−4)3 の展開(x−4)3=(x−4)(x−4)(x−4)(x - 4)^3 = (x - 4)(x - 4)(x - 4)(x−4)3=(x−4)(x−4)(x−4)まず、 (x−4)(x−4)=x2−4x−4x+16=x2−8x+16(x - 4)(x - 4) = x^2 - 4x - 4x + 16 = x^2 - 8x + 16(x−4)(x−4)=x2−4x−4x+16=x2−8x+16次に、 (x−4)3=(x−4)(x2−8x+16)=x(x2−8x+16)−4(x2−8x+16)(x - 4)^3 = (x - 4)(x^2 - 8x + 16) = x(x^2 - 8x + 16) - 4(x^2 - 8x + 16)(x−4)3=(x−4)(x2−8x+16)=x(x2−8x+16)−4(x2−8x+16)=x3−8x2+16x−4x2+32x−64= x^3 - 8x^2 + 16x - 4x^2 + 32x - 64=x3−8x2+16x−4x2+32x−64=x3−12x2+48x−64= x^3 - 12x^2 + 48x - 64=x3−12x2+48x−64(4) (x+3)(x2−3x+9)(x + 3)(x^2 - 3x + 9)(x+3)(x2−3x+9) の展開(x+3)(x2−3x+9)=x(x2−3x+9)+3(x2−3x+9)(x + 3)(x^2 - 3x + 9) = x(x^2 - 3x + 9) + 3(x^2 - 3x + 9)(x+3)(x2−3x+9)=x(x2−3x+9)+3(x2−3x+9)=x3−3x2+9x+3x2−9x+27= x^3 - 3x^2 + 9x + 3x^2 - 9x + 27=x3−3x2+9x+3x2−9x+27=x3+27= x^3 + 27=x3+273. 最終的な答え(1) 25a2+30ab+9b225a^2 + 30ab + 9b^225a2+30ab+9b2(2) a2+ab−30b2a^2 + ab - 30b^2a2+ab−30b2(3) x3−12x2+48x−64x^3 - 12x^2 + 48x - 64x3−12x2+48x−64(4) x3+27x^3 + 27x3+27