已知（x²+mx+n）（x²-3x+2）中,不含x²和x项,求m,n的值

(x^2+mx+n)(x^2-3x+2)
x^4-3x^3+2x^2+mx^3-3mx^2+2mx+nx^2-3nx+2n
x^4+(m-3)x^3+(2-3m+n)x^2+(2m-3n)x+2n
因为不含x^2和x项
所以
2-3m+n=0
2m-3n=0
解之得,m=6/7,n=4/7