（x^2+px+q)(x^2-3x+2)
=x^4-3x^3+2x^2+px^3-3px^2+2px+qx^2-3qx+2q
=x^4+(p-3)x^3+(2-3p+q)x^2+(2p-3q)x+2q
因为不含x^3和x^2项,因此x^3和x^2项系数为0
p-3=0,p=3
2-3p+q=0
2-3*3+q=0
q=7
答：p=3,q=7