x=-1,x^3-3x^2+4=0
所以x^3-3x^2+4有因式x+1
x^3-3x^2+4
=(x^3+1)-3x^2+3
=(x+1)(x^2-x+1)-3(x+1)(x-1)
=(x+1)(x^2-x+1-3x+3)
=(x+1)(x^2-4x+4)
=(x+1)(x-2)^2
x^3-3x^2+4
=x^3-3x^2+2x-2x+4
=x(x^2-3x+2)-2(x-2)
=x(x-2)(x-1)-2(x-2)
=(x-2)[x(x-1)-2]
=(x-2)[x^2-x-2]
=(x-2)(x-2)(x+1)
=(x-2)^2(x+1)
可以用估根法,估得原式=0的根为x=-1
再用综合除法得原式=(x+1)(x-2)^2
x^3-3x^2+4
=(x^3-2x^2)-(x^2-4)
=x^2(x-2)-(x-2)(x+2)
=(x-2)(x^2-x-2)
=(x-2)(x-2)(x+1)
(x-2)^2(x+1)
x^3-3x^2+4
=x^3-2x^2-x^2+4
=x^2(x-2)-(x+2)(x-2)
=(x-2)(x^2-x-2)
=(x-2)(x+1)(x-2)