(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