Question

试确定a、b,使x4次方+ax²-bx+2能被x²+3x+2整除.

Answer 1

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+2

c=1

d+3c=0

z+3d+2c=a

3z+2d=-b

2z=2

解方程组

得到答案：c=1,d=-3,z=1,b=3,a=-6