若不论x取何值,多项式x³-2x²-4x-1与(x-1)(x²+mx+n)都相等,求m,n
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![若不论x取何值,多项式x³-2x²-4x-1与(x-1)(x²+mx+n)都相等,求m,n](/uploads/image/z/2096145-9-5.jpg?t=%E8%8B%A5%E4%B8%8D%E8%AE%BAx%E5%8F%96%E4%BD%95%E5%80%BC%2C%E5%A4%9A%E9%A1%B9%E5%BC%8Fx%26%23179%3B-2x%26%23178%3B-4x-1%E4%B8%8E%28x-1%29%28x%26%23178%3B%2Bmx%2Bn%29%E9%83%BD%E7%9B%B8%E7%AD%89%2C%E6%B1%82m%2Cn)
若不论x取何值,多项式x³-2x²-4x-1与(x-1)(x²+mx+n)都相等,求m,n
若不论x取何值,多项式x³-2x²-4x-1与(x-1)(x²+mx+n)都相等,求m,n
若不论x取何值,多项式x³-2x²-4x-1与(x-1)(x²+mx+n)都相等,求m,n
(x-1)(x²+mx+n)
=x³+mx²+nx-x²-mx-n
=x³+(m-1)x²+(n-m)x-n
得
m-1=-2
n-m=-4
n=1
方程组无解,所以不存在这样的mn值
原则上应该用待定系数法,但好像题有问题,做不出来