純數-多項式II

part c: 我相信你能夠證明only if part，即給了α是f(x) = 0的二重根，因此證明α是g(x) = 0的根。

你的問題是證明if part，即由α是g(x) = 0推論至α是f(x) = 0的二重根。

根據part b，(4x + a)f'(x) = 16f(x) + g(x) ... (1)

由於題目給了α是f(x) = 0的根，條件亦給了α是g(x) = 0的根

因此，式(1)的右方: 16f(x) + g(x) = (x - α)Q(x)，Q(x)為一多項式

由於α (不等於-a / 4)是f(x) = 0的根，所以式(1)可表達成：

f'(x) = (x - α)Q(x) / (4x + a)，x =/= -a/4

= (x - α)P(x)，P(x)為另外的多項式

所以證明了α是f(x) = 0的根，亦是f'(x) = 0的根

因此，α是f(x) = 0的二重根。



2009-03-28 17:28:39 補充：
我相信你有能力把我的中文解答轉成英文吧...

" only if " part :

let α be the repeated root of f(x)

s.t. f(α)= 0 and f'(α)=

16f(x)=(4x+a)f'(x)-g(x)

=>16f(α)=(4α+a)f'(α)-g(α)

=>g(α)=o

so α is the root of g(x)

" if " part

Let α be the root of g(x) and f(x)

s.t. g(α)=f(α)=0

16 f(x)=(4x+a)f'(x)-g(x)

=> 16 f(α) = (4α+a) f'(α) -g(α)

=> f'(α)=0

so α is a repeated root of f(x)=0

α is a repeated root of f(x) = 0 <=> α is root of g(x)=0

Use @ to represent alpha.

Suppose @ is a root of g(x) = 0,

By (b),
16f(@) = (4@ + a)f'(@) - g(@)
0 = (4@ + a)f'(@) [ since @ is a root of f(x) = 0 and g(x) = 0 ]
f'(@) = 0 [ since @ <> -a/4 ]

Therefore @ is a repeated root of f(x)