与えられた6つの式を因数分解する問題です。代数学因数分解二次式2025/8/81. 問題の内容与えられた6つの式を因数分解する問題です。2. 解き方の手順(1) 2x2+11x+52x^2 + 11x + 52x2+11x+5 2x2+11x+5=(2x+1)(x+5)2x^2 + 11x + 5 = (2x + 1)(x + 5)2x2+11x+5=(2x+1)(x+5)(2) 2x2−11x+52x^2 - 11x + 52x2−11x+5 2x2−11x+5=(2x−1)(x−5)2x^2 - 11x + 5 = (2x - 1)(x - 5)2x2−11x+5=(2x−1)(x−5)(3) 2x2−x−62x^2 - x - 62x2−x−6 2x2−x−6=(2x+3)(x−2)2x^2 - x - 6 = (2x + 3)(x - 2)2x2−x−6=(2x+3)(x−2)(4) 2x2+xy−6y22x^2 + xy - 6y^22x2+xy−6y2 2x2+xy−6y2=(2x−3y)(x+2y)2x^2 + xy - 6y^2 = (2x - 3y)(x + 2y)2x2+xy−6y2=(2x−3y)(x+2y)(5) 4t2+8t+34t^2 + 8t + 34t2+8t+3 4t2+8t+3=(2t+1)(2t+3)4t^2 + 8t + 3 = (2t + 1)(2t + 3)4t2+8t+3=(2t+1)(2t+3)(6) 10x2−23x+1210x^2 - 23x + 1210x2−23x+12 10x2−23x+12=(2x−3)(5x−4)10x^2 - 23x + 12 = (2x - 3)(5x - 4)10x2−23x+12=(2x−3)(5x−4)3. 最終的な答え(1) (2x+1)(x+5)(2x + 1)(x + 5)(2x+1)(x+5)(2) (2x−1)(x−5)(2x - 1)(x - 5)(2x−1)(x−5)(3) (2x+3)(x−2)(2x + 3)(x - 2)(2x+3)(x−2)(4) (2x−3y)(x+2y)(2x - 3y)(x + 2y)(2x−3y)(x+2y)(5) (2t+1)(2t+3)(2t + 1)(2t + 3)(2t+1)(2t+3)(6) (2x−3)(5x−4)(2x - 3)(5x - 4)(2x−3)(5x−4)