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x^4+1=(x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1)...etc

Posted by 200730690105 at 2009-02-07 04:11:16 on Problem 2126 and last updated at 2009-02-07 04:12:57
如果在x^4+1=(x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1)算是在"实数上的分解",而再进一步分解就就是"复数的范围了",但只要分解后的多项式的系数不出现复数就行了(因此以不分解就够了),如果一定要分解x^2+1,得出的多项式系数必定有复数

高等代数好像有个定理,次数为奇数的多项式就必有实根,为偶数就一定有偶数对共轭复根,而每一对共轭复根都能凑成一个二次实数系数的多项式,如x^4+1就有两对共轭复根,分别可以凑成x^2+sqrt(2)x+1和x^2-sqrt(2)x+1

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