已知多项式（x^2+px+q)(x^2-3x+2)的乘积中不含x^2和x^3，求q和p

（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

（x^2+px+q)(x^2-3x+2)
=x^4+(p-3)x^3+(2-3p+q)x^2+(2p-3q)x+2q
因为乘积中不含x^2和x^3
所以p-3=0
2-3p+q=0
解得p=3 q=7

(x*2+px+q)＊(x*2-3x+2) =x*4+(p-3)x*3+(q-1)x*2+(2p-3q)x+2q 再让p-3=0 ;q-1=0 算出p=3 q=1