f(x)=1/2x^5+1/2x^4-3x^3-7x^2-11/2x-3/2=(x^5+2x^4-6x^3-14x^2-11x-3)/2=(x+1)(x^4-6x^2-8x-3)/2=(x+1)^2(x^3-x^2-5x-3)/2=(x+1)^3(x^2-2x-3)/2=(x+1)^4(x-3)/2所以f（x）=0的跟为x=-1或x=3