试确定a、b,使x4次方+ax²-bx+2能被x²+3x+2整除.

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=...