After learning multiplicity in polynomials we were given the task of proving that:
if $ f(\alpha) = f'(\alpha) = f''(\alpha) = f'''(\alpha) =$ .... $f^{k-1}(\alpha) = 0$ and $f^{k}(\alpha) \not= 0$
then $(x-\alpha)$ is a root of $f(x)$ of multiplicity of $k$.
By logic we know this to be true however when trying to prove it I have experienced great difficulty.
I have so far tried using mathematical induction with little success and by algebraically differentiating it to ultimately show that $ (x-\alpha)^{m-k} = 0 $ where $m$ is the multiplicity of $ (x-\alpha) $ and therefore must be $k$ but do not feel like this proof is substantial enough as it relies on an undefined value.
The starting point for my proof's so far have been with $ f(x) = (x-\alpha)^m * Q(x)$.
I was hoping if someone here could show me a proof for this using one of the above methods or any others (such as contradiction).
Thanks.
It is not (necessarily) about induction...
$f(x)=(x-\alpha)g(x)$ because $f(\alpha)=0$
$f'(x)=(x-\alpha)g'(x)+g(x)$..
As $f'(\alpha)=0$ we have $g(\alpha)=0$ so, $g(x)=(x-\alpha)g_1(x)$ and so $f(x)=(x-\alpha)^2g_1(x)$...
Repeat this... Can you continue from here?