I'd like to show that the following two propositions are equivalent:
(1) $f \in \mathbb{R}[x]$ has a multiple Null, so it's $\ge 2$.
(2) $f$ and $f'$ have a common Null, whereas $f'$ describes the derivative of $f$.
$(1) \to (2)$
If $f$ has a multiple Null/Zero, then $\exists r \in \mathbb{R}$ and $s \in \mathbb{R}[x] \backslash 0$ so that
(1.1) $f = (x - r)^2 \cdot s$ [I'll proof this later]
Then it is $f' = 2(x-r) \cdot s + (x-r)^2 \cdot s'$
Thus $f'(r) = 0$.
$(2) \to (1)$
Let $r \in \mathbb{R}$ be the common Zero/Null.
We then have a definite $s \in \mathbb{R}[x] \backslash 0$ with
$f = (x-r)\cdot s$ and follows $f' = s + (x-r)\cdot s'$.
I know want to show: we have a specific form of $s$ so that $s = 0$ because of $r$ being the zero. Something like $ s = (x-r)\cdot s_2$ ?!
I hope that both steps are okay so far. If there are any problems, please tell me.
Now (1.1) and its proof:
Let $k \in K$ be a Zero of $f \in K[t]$, then there exists a specific $g \in K[t]$ so that $f = (t- k)\cdot g$.
Proof: Do a division with remainder for $f$ divided by $(t-k)$. There are now specific $g,r \in K[t]$ with $f = (t-k)\cdot g + r$ and $deg r < deg (t-k) = 1$.
Therefore $r = a_0, a_0 \in K$.
Since $f(k)=0$ we put in $k$ and get $0 = (k-k)\cdot g(k) + r = 0 + a_0$,
so $a_0 = r = 0$.
This is what I wanted to show.