A question (2)

We have f(x)=(x-a)^k P(x) ; f'(x)=(x-a)^k Q(x)

On the other hand f'(x)=k(x-a)^(k-1) P(x) + (x-a)^k P'(x)
=(x-a)^(k-1) [kP(x)+(x-a)P'(x)]

Compare with x-a)^k Q(x) => P(x) has a factor (x-a)
So f(x)=(x-a)^(k+1) R(x) where P(x)=(x-a)R(x) and a is a (k + 1) - multiple root of equation f(x) = 0

2009-12-26 13:24:07 補充：
(x-a)^k Q(x) = (x-a)^(k-1) [kP(x)+(x-a)P'(x)]