设3x^3-kx^2+4
=（3x-1)(x²+ax+b)+3
=3x³+3ax²+3bx-x²-ax-b+3
=3x³+(3a-1)x²+(3b-a)x+(3-b)
∴﹛3a-1=-k
3b-a=0
3-b=4
解得b=-1
a=-3
k=10