与えられた8つの2次式を因数分解する問題です。代数学因数分解二次式多項式2025/4/301. 問題の内容与えられた8つの2次式を因数分解する問題です。2. 解き方の手順(1) 2x2+5x+32x^2 + 5x + 32x2+5x+32x2+2x+3x+3=2x(x+1)+3(x+1)=(2x+3)(x+1)2x^2 + 2x + 3x + 3 = 2x(x+1) + 3(x+1) = (2x+3)(x+1)2x2+2x+3x+3=2x(x+1)+3(x+1)=(2x+3)(x+1)(2) 3x2+7x−63x^2 + 7x - 63x2+7x−63x2+9x−2x−6=3x(x+3)−2(x+3)=(3x−2)(x+3)3x^2 + 9x - 2x - 6 = 3x(x+3) - 2(x+3) = (3x-2)(x+3)3x2+9x−2x−6=3x(x+3)−2(x+3)=(3x−2)(x+3)(3) 4x2+x−54x^2 + x - 54x2+x−54x2+5x−4x−5=x(4x+5)−1(4x+5)=(x−1)(4x+5)4x^2 + 5x - 4x - 5 = x(4x+5) - 1(4x+5) = (x-1)(4x+5)4x2+5x−4x−5=x(4x+5)−1(4x+5)=(x−1)(4x+5)(4) 6x2−11x+46x^2 - 11x + 46x2−11x+46x2−8x−3x+4=2x(3x−4)−1(3x−4)=(2x−1)(3x−4)6x^2 - 8x - 3x + 4 = 2x(3x-4) - 1(3x-4) = (2x-1)(3x-4)6x2−8x−3x+4=2x(3x−4)−1(3x−4)=(2x−1)(3x−4)(5) 12x2+7xy+y212x^2 + 7xy + y^212x2+7xy+y212x2+3xy+4xy+y2=3x(4x+y)+y(4x+y)=(3x+y)(4x+y)12x^2 + 3xy + 4xy + y^2 = 3x(4x+y) + y(4x+y) = (3x+y)(4x+y)12x2+3xy+4xy+y2=3x(4x+y)+y(4x+y)=(3x+y)(4x+y)(6) 2x2+xy−3y22x^2 + xy - 3y^22x2+xy−3y22x2+3xy−2xy−3y2=x(2x+3y)−y(2x+3y)=(x−y)(2x+3y)2x^2 + 3xy - 2xy - 3y^2 = x(2x+3y) - y(2x+3y) = (x-y)(2x+3y)2x2+3xy−2xy−3y2=x(2x+3y)−y(2x+3y)=(x−y)(2x+3y)(7) 6x2+7xy+2y26x^2 + 7xy + 2y^26x2+7xy+2y26x2+3xy+4xy+2y2=3x(2x+y)+2y(2x+y)=(3x+2y)(2x+y)6x^2 + 3xy + 4xy + 2y^2 = 3x(2x+y) + 2y(2x+y) = (3x+2y)(2x+y)6x2+3xy+4xy+2y2=3x(2x+y)+2y(2x+y)=(3x+2y)(2x+y)(8) 8x2−2xy−15y28x^2 - 2xy - 15y^28x2−2xy−15y28x2−12xy+10xy−15y2=4x(2x−3y)+5y(2x−3y)=(4x+5y)(2x−3y)8x^2 - 12xy + 10xy - 15y^2 = 4x(2x-3y) + 5y(2x-3y) = (4x+5y)(2x-3y)8x2−12xy+10xy−15y2=4x(2x−3y)+5y(2x−3y)=(4x+5y)(2x−3y)3. 最終的な答え(1) (2x+3)(x+1)(2x+3)(x+1)(2x+3)(x+1)(2) (3x−2)(x+3)(3x-2)(x+3)(3x−2)(x+3)(3) (x−1)(4x+5)(x-1)(4x+5)(x−1)(4x+5)(4) (2x−1)(3x−4)(2x-1)(3x-4)(2x−1)(3x−4)(5) (3x+y)(4x+y)(3x+y)(4x+y)(3x+y)(4x+y)(6) (x−y)(2x+3y)(x-y)(2x+3y)(x−y)(2x+3y)(7) (3x+2y)(2x+y)(3x+2y)(2x+y)(3x+2y)(2x+y)(8) (4x+5y)(2x−3y)(4x+5y)(2x-3y)(4x+5y)(2x−3y)