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在实数域上，奇次一定有一个根，偶次有共轭虚根,总是可以分解成两个n/2的多项式。

Posted by yangguo981 at 2006-05-16 22:28:26 on Problem 2126
In Reply To:Re:这结论是怎么得出来的，n>2就一定可以被分解？ Posted by:hahaxiao at 2005-09-16 11:35:23

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那x^2+1=0不也可以分成(x+sqrt(i))*(x-sqrt(i)),而且根据题目说只要是coefficients是实数不就可以了,那为什么x^2+1=0不能分解 (0) xrjxrjxrj 2006-10-28 11:25:53
- Re:那x^2+1=0不也可以分成(x+sqrt(i))*(x-sqrt(i)),而且根据题目说只要是coefficients是实数不就可以了,那为什么x^2+1=0不能分解 (13) xrjxrjxrj 2006-10-28 11:43:19
  - Re:那x^2+1=0不也可以分成(x+sqrt(i))*(x-sqrt(i)),而且根据题目说只要是coefficients是实数不就可以了,那为什么x^2+1=0不能分解 (18) chgsh8089 2006-11-24 22:38:24
题目说了系数一定要是实数吧？这样分解符合题意么？ (0) rovingcloud 2007-03-06 01:25:20
- 这纯属我的主观臆断……当我没发过 -,- (0) rovingcloud 2007-03-06 18:36:31
- Re:无理数也是实数,还有题目说 given two polynomials, one may sum them up, multiply them, or even divide one of them by the other. (118) HG 2009-05-25 20:27:28
Re:为什么说“偶次有共轭虚根”？请证明一下…… (0) LyuuSyouHa 2009-05-29 16:42:59

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