f(x)=2x^3-3x^2-5x+3
=2x^3-x^2-2x^2-5x+3
=(2x-1)x^2-(2x^2+5x-3)
=(2x-1)x^2-(2x-1)(x+3)
=(2x-1)(x^2-x-3)
=0
因此2x-1=0或x^2-x-3=0
解得x1=1,x2=(1+√13)/2,x3=(1-√13)/2