解: (x2+mx+n)(x2-3x+4) =x4-3x3+4x2+mx3-3mx2+4mx+nx2-3nx+4n =x4+(m-3)x3+(4-3m+n)x2+(4m-3n)x+4n ∵多项式(x2+mx+n)(x2-3x+4)展开后不含x3和x2项, ∴m-3=0,4-3m+n=0 ∴m=3,n=5.
