MQ9 --- Factorization

您好！我是 lop，高興能解答您的問題。

Let the factorized form of x⁴ + 4 be (x³+ax²+bx+c)(x+d), a, b, c and d are integers ( If we can't factorize x⁴ + 4 to (x³+ax²+bx+c)(x+d), we can try factorize x⁴ + 4 to (x²+ax+b)(x²+cx+d) ).

a + d = 0 --- (1)
b + ad = 0 --- (2)
c + bd = 0 --- (3)
cd = 4 --- (4)

From (1) :

d = -a --- (5)

Sub (5) into (2)

b + a(-a) = 0
b = a²--- (6)

Sub (6) into (3)

c + bd = 0
c + a²(-a) = 0
c = a³--- (7)

Sub (5) and (7) into (4)

-a⁴ = 4
a⁴ = -4
a²= √(-4) (rej.)

∴ x⁴ + 4 can't factorize to a form of (x³+ax²+bx+c)(x+d).

========================================================

Then we try x⁴ + 4 = (x²+ax+b)(x²+cx+d)

a + c = 0 --- (!)
ac + b + d = 0 --- (@)
bc + ad = 0 --- (#)
bd = 4 --- ($)

From (!) :

c = -a --- (%)

Sub (%) into (@) and (#)

-a²+ b + d = 0 --- (^)
b(-a) + ad = 0
a(d-b) = 0

If a = 0, then c = 0 , then

b + d = 0
b = -d

bd = 4
-d²= 4
d²= -4 (rej.)

∴ a ≠ 0.

a(d-b) = 0
d - b = 0
b = d --- (&)

Sub (&) into ($)

b²= 4
b = -2 or 2

If b = -2, d = -2

-a²+ b + d = 0
-a²-2 - 2 = 0
a²= -4 (rej.)

If b = 2, d = 2

-a²+ b + d = 0
-a²+ 2 + 2 = 0
a²= 4
a = -2 or 2
When a = -2, c = 2; When a = 2, c = -2

∴ (x⁴ + 4) = (x²+2x+2)(x²+2x+2)

========================================================

It is easy to see that we can't factorize x²+2x+2 or x²+2x+2 anymore, so the answer is

(x⁴ + 4) = (x²+2x+2)(x²+2x+2)

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2012-05-16 17:51:23 補充：
Corrections,

If b = 2, d = 2

-a²+ b + d = 0
-a²+ 2 + 2 = 0
a²= 4
a = -2 or 2
When a = -2, c = 2; When a = 2, c = -2

∴ (x⁴ + 4) = (x²+2x+2)(x²-2x+2)

It is easy to see that we can't factorize x²+2x+2 or x²-2x+2 anymore, so the answer is

(x⁴ + 4) = (x²+2x+2)(x²-2x+2)

2012-05-16 17:51:43 補充：
Sorry I made a careless mistake

2012-05-16 17:55:53 補充：
Re ☂雨後陽光☀,

I have thought that to answer in your simplest way but I think my way is much better because it can also work in any factorizing questions. Actually, I couldn't think te answer at first !

x^4 + 4
= (x^2)^2 + 2^2
= (x^2 + 2 )^2 - 4x^2
= (x^2 + 2 )^2 - (2x)^2
= (x^2 + 2 + 2x) (x^2 - 2x + 2)

在http://hk.knowledge.yahoo.com/question/question?qid=7012051500458#ooa_hash中
我那句'complex no.'意思是不需用到complex no.（留意 "=_="）
對於引起誤會，我謹此至歉

現鄭重聲明：
此題只是一般因式分解
不需涉及complex no.
若需涉及complex no.
我會注明complex factorization 或 Cfactorization

2012-05-17 16:30:19 補充：
其實我相信大部分人對因式分解的題目都不會想到用complex no.
並不是因complex no. 深
而是主流對因式分解的理解都在Z[x]內
我想沒有必要在問因式分解問題前先說明「規則」
不需簡單複雜化
而涉及complex no. 的因式分解應特別注明

不知大家有何意見？

2012-05-18 07:39:18 補充：
我會仔細考量你的意見。
謝謝！

Comment: You have to specify cleary the integral domain in which you want to do your factorization, or that it's an elementary/junior high school problem. No one knows exactly what field you are working on, unless you tell us.

2012-05-17 03:23:51 補充：
我之所以說上面的 comment，其實是因為你前面另一題因子分解問題：

http://hk.knowledge.yahoo.com/question/question?qid=7012051500458#ooa_hash

在 CRebecca 回了你一個在 Z[x] 的因子分解後，由於他用了 x^3 = 1 的其中一個 complex root 的技巧作證明；你只補充了一句 complex number；於是誤會就來了，CRebecca 以為你是想在 C[x] 作因子分解，就有兩個追加補充。你在 comment 時反而說在意見欄中的分解更簡單！天啊，那根本就是回答者原來的答案～～＠＠

2012-05-17 11:40:01 補充：
我倒是有不同的看法。

我認為，這不是需不需要的問題，其實他的證明是有問題的，錯在他只看一個 complex root，他應該寫 x^2 + x + 1 的兩個複數根，都為原式的根 in C。

當然，證明可以不繞到 complex numbers，但是 complex numbers 也不是什麼難的東西吧，都在高中學過，所以我才說，那你應該說只用國中或小學程度的因子分解方法。


你其實不必這廚鄭重，在網路發問，很難說涓楚，人家可以用什麼方法來解決的，我只是說，你如果認為人家不需要用高中的方法〔他的方法也只是初等的，並沒高深的地方〕，你連這也想更初等，就該說出來了。

2012-05-18 07:18:11 補充：
OK, do whatever you wish. I would not say anything after this post.

If I were you, I'd post as follows:

factorize in Z[x] the polynomial, x^4 + 4.

There would be no ambiguity.

So complicated= ="But the method seems useful since it can be applied to all similar questions

x⁴ + 4

= x⁴+ 4x² + 4 - 4x²

= (x² + 2)² - (2x)²

= (x² - 2x + 2) (x² + 2x + 2)

2012-05-16 17:20:31 補充：
雖然 lop 知識長的方法很繁沉,但這是個重要方法,有一般用途,很值得參考。

而以上的方法雖簡,但缺乏一般性,沒什益處的。