与えられた7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2025/7/311. 問題の内容与えられた7つの2次式を因数分解する問題です。2. 解き方の手順(1) 4x2−7x+34x^2 - 7x + 34x2−7x+34x2−7x+3=(4x−3)(x−1)4x^2 - 7x + 3 = (4x - 3)(x - 1)4x2−7x+3=(4x−3)(x−1)(2) 4x2+11x−34x^2 + 11x - 34x2+11x−34x2+11x−3=(4x−1)(x+3)4x^2 + 11x - 3 = (4x - 1)(x + 3)4x2+11x−3=(4x−1)(x+3)(3) 4x2−11x−34x^2 - 11x - 34x2−11x−34x2−11x−3=(4x+1)(x−3)4x^2 - 11x - 3 = (4x + 1)(x - 3)4x2−11x−3=(4x+1)(x−3)(4) 4x2+8x+34x^2 + 8x + 34x2+8x+34x2+8x+3=(2x+1)(2x+3)4x^2 + 8x + 3 = (2x + 1)(2x + 3)4x2+8x+3=(2x+1)(2x+3)(5) 4x2−8x+34x^2 - 8x + 34x2−8x+34x2−8x+3=(2x−1)(2x−3)4x^2 - 8x + 3 = (2x - 1)(2x - 3)4x2−8x+3=(2x−1)(2x−3)(6) 4x2+4x−34x^2 + 4x - 34x2+4x−34x2+4x−3=(2x−1)(2x+3)4x^2 + 4x - 3 = (2x - 1)(2x + 3)4x2+4x−3=(2x−1)(2x+3)(7) 4x2−4x−34x^2 - 4x - 34x2−4x−34x2−4x−3=(2x+1)(2x−3)4x^2 - 4x - 3 = (2x + 1)(2x - 3)4x2−4x−3=(2x+1)(2x−3)3. 最終的な答え(1) (4x−3)(x−1)(4x - 3)(x - 1)(4x−3)(x−1)(2) (4x−1)(x+3)(4x - 1)(x + 3)(4x−1)(x+3)(3) (4x+1)(x−3)(4x + 1)(x - 3)(4x+1)(x−3)(4) (2x+1)(2x+3)(2x + 1)(2x + 3)(2x+1)(2x+3)(5) (2x−1)(2x−3)(2x - 1)(2x - 3)(2x−1)(2x−3)(6) (2x−1)(2x+3)(2x - 1)(2x + 3)(2x−1)(2x+3)(7) (2x+1)(2x−3)(2x + 1)(2x - 3)(2x+1)(2x−3)