Interesting Maths Q4 (Easy)

http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-136.jpg

圖片參考：http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-136.jpg


2011-11-06 12:25:43 補充：
不明白你所指的是那部份?

2011-11-06 19:21:45 補充：
沒甚麼問題啊!而且根據一般因式分解,也不包含複數的.

2011-11-06 21:01:45 補充：
我不明白你的意思.我看不出有甚麼問題,也沒有不工整的問題,你不是要把所有的(x - c)的因式找出來嗎?現在十二個都找出來了.還有另一方法:
http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-142.jpg

2011-11-06 21:45:14 補充：
Can you show me reference to the definition of Cfactorization?

第二行的最後一個括號:

x^4-x^2+1

2011-11-06 12:35:15 補充：
你為何不把它寫作a^2+b^2的模式

(for further factorizing)

2011-11-06 12:45:30 補充：
還有，根據因式分解的定義，不是應該不包括方開根嗎？

2011-11-06 12:47:43 補充：
*開方根

______________

2011-11-06 20:29:31 補充：
Cfactorize 包含複數，但不包含開方根啊！不是嗎？

2011-11-06 21:37:55 補充：
I think no need to change x^2-x+1 into (x-1/2+√3/2i)(x-1/2-√3/2i)

since Cfactorization has complex no. but no need to have square root.

Isn't it?

C i think is complex~_~"

x^12-1
=(x^6+1)(x^6-1)
=[(x^2+1)(x^4+x^2+1)][(x^3+1)(x^3-1)]
=(x^2+1)(x^4+x^2+1)[(x+1)(x^2-x+1)][(x-1)(x^2+x+1)]
=(x+1)(x-1)(x^2+1)(x^2+x+1)(x^2-x+1)(x^4+x^2+1)

2011-11-04 22:07:47 補充：
Sorry should be
x^12-1
=(x^6+1)(x^6-1)
=[(x^2+1)(x^4-x^2+1)][(x^3+1)(x^3-1)]
=(x^2+1)(x^4-x^2+1)[(x+1)(x^2-x+1)][(x-1)(x^2+x+1)]
=(x+1)(x-1)(x^2+1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1)