3x^4-2x^3-9x^2+12x-4=0怎么因式分解的？？？

3x＾4-2x＾3-9x＾2+12x-4=0
2x＾4-2x＾3+x＾4-9x＾2+12x-4=0
2x＾3(x-1)+x＾4-x＾2-(8x＾2-12x+4)=0
2x＾3(x-1)+x＾2(x-1)(x+1)-(8x-4)(x-1)=0
同时约掉x-1,
x=1是方程的一个根
2x＾3+x＾2(x+1)-8x+4=0
3x＾3+x＾2-8x+4=0
3x＾3-3x+x＾2-5x+4=0
3x(x＾2-1)+(x-4)＊(x-1)=0
同上约掉x-1
3x＾2+3x+x-4=0
3x＾2+4x-4=0
(3x-2)(x+2)=0
所以x=2/3.x=-2.x=1

3x^4-2x^3-(9x^2-12x+4)=0
x^3(3x-2)-(3x-2)^2=0
(3x-2)(x^3-3x+2)=0
(3x-2)(x^3-1-3x+3)=0
(3x-2)[(x-1)(x^2+x+1)-3(x-1)]=0
(3x-2)(x-1)(x^2+x+1-3)=0
(3x-2)(x-1)(x^2+x-2)=0
(3x-2)(x-1)^2(x+2)=0
x=2/3,x=1,x=-2

3x^4-2x^3-9x^2+12x-4
=(3x^4-3x^3)+(x^3-x^2)-(8x^2-8x)+(4x-4)
=(3x^3+x^2-8x+4)(x-1)
=[(3x^3-3x^2)+(4x^2-4x)-(4x-4)](x-1)
=(3x^2+4x-4)(x-1)^2
=(3x-2)(x+2)(x-1)^2
=0
x1=1
x2=-2
x3=2/3

3x^4-2x^3-9x^2+12x-4=0
=3x^4-3x^3+x^3-x^2-8x^2+12x-4
=3x^3(x-1)+x^2(x-1)-4(2x^2-3x+1)
=3x^3(x-1)+x^2(x-1)-4(2x-1)(x-1)
=(x-1)(3x^3+x^2-