高次方程式求解

(1) 4x^5 - 9x^3 + 6x^2 - 13x + 6 = 0
(4x^5 - 9x^3) + (6x^2 - 13x + 6) = 0
(2x + 3)(2x - 3)x^3 + (2x - 3)(3x - 2) = 0
(2x - 3)(2x^4 + 3x^3 + 3x - 2) = 0
(2x - 3)[(2x^4 - 2) + (3x^3 + 3x)] = 0
(2x - 3)[2(x^2 + 1)(x^2 - 1) + 3x(x^2 + 1) = 0
(2x - 3)(x^2 + 1)(2x^2 - 2 + 3x) = 0
(2x - 3)(x^2 + 1)(2x - 1)(x + 2) = 0
得 x = 3/2, 1/2, -2

(2) (x^2 + 14x + 24)(x^2 + 11x + 24) = 4x^2
[(x^2 + 24) + 14x][(x^2 + 24) + 11x] - 4x^2 = 0
(x^2 + 24)^2 + 25x(x^2 + 24) + 154x^2 - 4x^2 = 0
(x^2 + 24)^2 + 25x(x^2 + 24) + 150x^2 = 0
[(x^2 + 24) + 10x][(x^2 + 24) + 15x] = 0
(x^2 + 10x + 24)(x^2 + 15x + 24) = 0
(x + 4)(x + 6)(x^2 + 15x + 24) = 0
得 x = -4, -6, (-15 + √129)/2, (-15 - √129)/2

1.
......
(2x - 3)(x + 2)(2x - 1)(x² + 1) = 0
x = 3/2, -2, 1/2, i, -i

(There are 3 real roots and 2 imaginary roots.)

(1)4x^5-9x^3+6x^2-13x+6=0，
(2x-3)(2x^4+3x^3+3x-2)=0，
(2x-3)(2x-1)(x^3+2x^2+x+2)=0，
(2x-3)(2x-1)(x+2)(x^2+1)=0，x=3/2，1/2，-2

2014-01-16 02:05:47 補充：
(2)(x^2+14x+24)(x^2+11x+24)=4x^2，
x^4+25x^3+198x^2+600x+576=0，
(x+4)(x^3+21x^2+114x+144)=0，
(x+4)(x+6)(x^2+15x+24)=0
x=-6，-4，(-15+-(15^2-4*24)^(1/2))/2