一元二次方程 因式法

1) 21 = 2x2 + 19x
2x2 + 19x﹣21 = 0
(2x + 21)(x﹣1) = 0
x = -21 / 2或x = 1

2)x(x + 5) = 24
x2 + 5x﹣24 = 0
(x﹣3)(x + 8) = 0
x = 3或x = -8

3)(x﹣3)(x + 7) = -9
x2 + 7x﹣3x﹣21 = -9
x2 + 4x﹣12 = 0
(x﹣2)(x + 6) = 0
x = 2或x = -6

4)(x﹣2)(2x + 4) = 5(x﹣1)
(x﹣2) [2(x + 2)] = 5x﹣5
2(x﹣2)(x + 2) = 5x﹣5
2(x2﹣4) = 5x﹣5
2x2﹣5x﹣3 = 0
(2x + 1)(x﹣3) = 0
x = -1 / 2或x = 3

5)x2 = 3x
x2﹣3x = 0
x(x﹣3) = 0
x = 0或x = 3

6)2x(x + 1) = 3(x + 1)
2x(x + 1)﹣3(x + 1) = 0
(x + 1)(2x﹣3) = 0
x = -1或x = 3 / 2

7)x(2x + 7) = 3(2x + 7)
x(2x + 7)﹣3(2x + 7) = 0
(2x + 7)(x﹣3) = 0
x = -7 / 2或x = 3

8)3(x﹣2)2 = x(x﹣2)
3(x﹣2)2﹣x(x﹣2) = 0
(x﹣2)[3(x﹣2)﹣x] = 0
(x﹣2)(3x﹣6﹣x) = 0
(x﹣2)(2x﹣6) = 0
x = 2或x = 3

9)(x﹣1)(2x + 3) = (x﹣1)(x﹣4)
(x﹣1)(2x + 3)﹣(x﹣1)(x﹣4) = 0
(x﹣1)[(2x + 3)﹣(x﹣4)] = 0
(x﹣1)(2x + 3﹣x + 4) = 0
(x﹣1)(x + 7) = 0
x = 1或x = -7