(x^4-2x^3+x^2)-4x^2+4x-12=0
x^2(x-1)^2-(4x^2-8x+4)-4x-8=0
x^2(x-1)^2-4(x-1)^2-4(x+2)=0
(x^2-4)(x-1)^2 -4(x+2)=0
(x+2)[(x-2)(x-1)^2-4]=0
(x+2)(x^3-4x^2+5x-6)=0
(x+2)[(x^3-3x^2)-(x^2-5x+6)]=0
(x+2)[x^2(x-3)-(x-2)(x-3)]=0
(x+2)(x-3)(x^2-x+2)=0
因为x^2-x+2恒正,实根只有两个:
x=-2 或 x=3