f(x)=x^5+3x^4+8x^3+11x+m=(x^5+2x^4)+(x^4+2x^3)+(6x^3+12x^2)-(12x^2+24x)+(35x+70)+(m-70)=x4(x+2)+x^3(x+2)+6x^2(x+2)-12x(x+2)+35(x+2)+m-70因此,要f(x)能被x+2整除,需要m-70=0=>m=70