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[函数] 求满足求满足$xp(x-1)=(x-10)p(x)$的多项式$p(x)$

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abababa posted 2018-6-24 19:32 |Read mode
如题,求满足求满足$xp(x-1)=(x-10)p(x)$的多项式$p(x)$

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realnumber posted 2018-6-24 23:18
Last edited by realnumber 2018-6-24 23:28先考虑这样情景,$y=p(x)=a(x-x_1)(x-x_2)\cdots(x-x_n),x_1\le x_2\le \cdots\le x_n$
比较两边零点,依次有$x_1=0,x_2=1,x_3=2,\cdots ,x_{10}=9$
如此$p(x)=ax(x-1)(x-2)\cdots(x-9)$,a为某常数.
有虚根情景应该没有的,虚根就按实部大小排一下.
下面2题也是考虑零点
forum.php?mod=viewthread&tid=5321&extra=page=5
forum.php?mod=viewthread&tid=450

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original poster abababa posted 2018-6-25 17:24
回复 2# realnumber

谢谢,原来是这么算的,需要依次求出所有的零点才行。
最后那个觉得应该是乘以一个多项式:$p(x)=(x-x_1)\cdots(x-x_n)q(x)$,再证明$q(x)$是常数,不过这个就简单了。

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