超高難度因式分解(要有過程)

P = x^8 + 98x^4 + 1
= x^8 + 98x^4 + 49^2 + 1 – 49^2(類似一元二次方的配方法)
= (x^4 + 49)^2 – 2400
= (x^4 + 49)^2 – (20√6)^2
= (x^4 + 49 + 20√6)(x^4 + 49 – 20√6)
因49 +/– 20√6 = 5^2 +/– 20√6 + (2√6)^2 = (5 +/– 2√6)^2，再用配方法
P = [x^4 + 2(5 + 2√6)x^2 + (5 + 2√6)^2 – 2(5 + 2√6)x^2][ x^4 + 2(5 – 2√6)x^2 + (5 – 2√6)^2 – 2(5 – 2√6)x^2]
因2(5 +/– 2√6) = 10 +/– 4√6 = 4 +/– 4√6 + (√6)^2 = (2 +/– √6)^2
P = [(x^2 + 5 + 2√6)^2 – (2 + √6)^2x^2][(x^2 + 5 – 2√6)^2 – (2 – √6)^2x^2]
= [x^2 + (2 + √6)x + 5 + 2√6][x^2 – (2 + √6)x + 5 + 2√6][x^2 + (2 – √6)x + 5 – 2√6][x^2 – (2 – √6)x + 5 – 2√6]
以上四項一元二次方為含無理數的因式,利用判別式得知無法再分解為不含複數的因式.八項一元一次方因式為x – [(2 +/– √6)/2](+/– 1 +/– i)
紅色兩項相乘得:
[x^2 + (2 + √6)x + 5 + 2√6][x^2 + (2 – √6)x + 5 – 2√6]
= x^4 + (2 + √6)x^3 + (5 + 2√6)x^2 + (2 – √6)x^3 + (2 + √6)(2 – √6)x^2 + (5 + 2√6)(2 – √6)x + (5 – 2√6)x^2 + (2 + √6)(5– 2√6)x + (5 + 2√6)(5 – 2√6)
= x^4 + (2 + √6 + 2 – √6)x^3 + (5 + 2√6 + 4 – 6 + 5 – 2√6)x^2 + (10 – √6 – 12 + 10 + √6 – 12)x + (25 – 24)
= x^4 + 4x^3 + 8x^2 – 4x + 1
餘下兩項相乘得 : x^4 – 4x^3 + 8x^2 + 4x + 1
這兩項一元四次方為不含無理數的因式.

硬湊出來吧：

x^8 + 98x^4 + 1

= (x^4 + 8x^2 + 1)^2 - [4x(x^2-1)]^2

Let's think:
(x^4 - 4x^3 +8x^2 + 4x + 1)(x^4 + 4x^3 + 8x^2 - 4x + 1)
=(x^4+8x^2+1-(4x^3-4x))(x^4+8x^2+1+(4x^3-4x))
=(x^4+8x^2+1)^2-(4x^3-4x)^2
=...
之後推算下去即可,就能知道計算過程,懶的算直接公佈最後一步驟用了平方差公式

2009-12-23 20:49:44 補充：
想起來了,是補項!!

2009-12-25 22:35:38 補充：
樓下算法超威的!!

I offer the Ans.

x^8 + 98x^4 + 1 can be factored into two polynomials.



(x^4 - 4x^3 +8x^2 + 4x + 1)(x^4 + 4x^3 + 8x^2 - 4x + 1).



If you multiply those together you will end up with x^8 + 98x^4 + 1.


but why, i dont know. Let's think.

如果你是國中生，這種題目算資優題了