Showing that $f(a)=f'(a) = 0$ if and only if $f = (x-a)^2q$. This was problem that I just cannot figure out:  
Let $F$ be a field, $a\in F$ and $f\in F[x]$.  Show that $f(a)=f'(a)=0$ if and only if $f=(x-a)^2q$ for some $q\in F[x]$.  And $f'$ refers to the derivative of $f$.

The if part is obvious.
For the only if part, since $f(a)=0$, there exist $g\in F[x]$ with $f=(x-a)g$. Now, $f'(x)=g(x)+(x-a)g'(x)$. So, $f'(a)=0$ implies $g(a)=0$. Thus again there exist $q \in F[x]$ with $g(x)=(x-a)q(x)$. Therefore, $f=(x-a)^2q$.
