答：2x^4-3x^3-x^2-3x+2=0(2x^4-4x^3)+(x^3-2x^2)+(x^2-3x+2)=02(x-2)x^3+(x-2)x^2+(x-1)(x-2)=0(x-2)(2x^3+x^2+x-1)=0所以：x-2=0或者2x^3+x^2+x-1=0解得：x=22x^3+x^2+x-1=0(2x^3-x^2)+2x^2+x-1=0(2x-1)x^2+(2x-1)...