试确定a、b,使x4次方+ax²-bx+2能被x²+3x+2整除.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 07:15:24
![试确定a、b,使x4次方+ax²-bx+2能被x²+3x+2整除.](/uploads/image/z/695787-51-7.jpg?t=%E8%AF%95%E7%A1%AE%E5%AE%9Aa%E3%80%81b%2C%E4%BD%BFx4%E6%AC%A1%E6%96%B9%2Bax%26%23178%3B-bx%2B2%E8%83%BD%E8%A2%ABx%26%23178%3B%2B3x%2B2%E6%95%B4%E9%99%A4.)
试确定a、b,使x4次方+ax²-bx+2能被x²+3x+2整除.
试确定a、b,使x4次方+ax²-bx+2能被x²+3x+2整除.
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+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