∵有因式x-2
∴x-2=0时x^3-6x^2+11x+2m=0
即x=2时x^3-6x^2+11x+2m=0
∴2³-24+22+2m=0
m=-3
∴x³-6x²+11x-6
=x³-2x²-4x²+8x+3x-6
=x²(x-2)-4x(x-2)+3(x-2)
=(x-2)(x²-4x+3)
=(x-2)(x-3)(x-1)
(x^3-6x^2+11x+2m)/(x-2)=x^2-4x+3
所以，原=(x-2)(x^2-4x+3)=(x-2)(x-1)(x-3)
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