设:(x^4+ax^2-bx+2)/(x^2+3x+2)=cx^2+dx+z用x^2+3x+2乘以cx^2+dx+z展开,对应项系数相等:cx^4+(d+3c)x^3+(z+3d+2c)x^2+(3z+2d)x+2z=x^4+ax^2-bx+2c=1d+3c=0z+3d+2c=a3z+2d=-b2z=2解方程组得到答案:c=1,d=-3,z=1,b=3,a=-6希望我的回答对你有帮助,采纳吧O(∩_∩)O!