How to prove that r is a double root if and only if it is a root of a polynomial and of its derivative. I don't know how to start the question. The title is self explanatory. How to approach and make a formal proof?
 A: Hint $\ $ Wlog, by a shift, assume the root is $\rm\: r = 0.$
Notice $\rm\ \ x^2\: |\ f(x)$
$\rm\ \iff\ \color{#0c0}{x\ |\ f(x)}\ $ and $\rm\ x\ \bigg|\ \dfrac{f(x)}{x}$
$\rm\ \iff\ \color{#0a0}{f(0) = 0}\ $ and $\rm\ x\ \bigg|\ \dfrac{f(x)-\color{#0a0}{f(0)}}x\iff \dfrac{f(x)-f(0)}x\bigg|_{\:x\:=\:0} =\: 0$
$\rm\ \iff\ f(0) = 0\ $ and $\rm\ f'(0) = 0$
Remark $\ $ It's often overlooked that many divisibility properties of numbers are specializations of this double-root criterion for (polynomial) functions, e.g. see my answer here., or below from here.
Hint: a conceptual way: we seek  $\,\color{#c00}{47^2\mid f(47)}\,$ for $\,f(x) = (x\!+\!5)^{n+1} + ((n\!-\!1)x\!-\!5)(2x\!+\!5)^n$
But $\,f(0)=0=f'(0)\,$ so by the double root test (or first two terms of the Binomial Theorem), it follows that $\,x=0\,$ is a double root so $\,f(x) = x^2 g(x)\,$ for $g$ a polynomial with integer coef's. Evaluating at $\,x=47\,$ yields $\,\color{#c00}{f(47)}= 47^2 g(47).\ $ QED
A: Hint: (a slightly different approach).
Suppose that $f(x)$ has $r$ as a root $m$ times ($m\geq 1$).  Then $f(x)=(x-r)^m g(x)$ for some polynomial $g(x)$ with $g(r)\not=0$.  What is $f'(x)$?  What has to be true about $m$ if $f'(r)=0$?
A: I was trying to solve this and came across this question. This is my solution. I hope it helps.
Suppose $P(x)$ is a polynomial of degree n and r is a root of $P(x)$ and its derivative.
$$P(x)=(x-r)Q(x)$$
$$P ^\prime(x)=(x-r)S(x)$$
where $Q(x)$ and $S(x)$ are also polynomials.
Based on the first equation, we can also write $$P ^\prime(x)=Q(x)+(x-r)Q^\prime(x)$$
For $x \neq r$
$$S(x)=\frac{P ^\prime(x)}{(x-r)}=\frac{Q(x)+(x-r)Q^\prime(x)}{x-r}$$
$$=\frac{Q(x)}{x-r}+Q^\prime(x)$$
$$=\frac{P(x)}{(x-r)^2}+Q^\prime(x)$$
$$\Rightarrow P(x)=(x-r)^2(S(x)-Q^\prime(x))$$
So, r is a double root of $P(x)$.
If r is a double root of $P(x)$, we have
$$P(x)=(x-r)^2T(x)$$
where $T(x)$ is a polynomial. We have
$$P^\prime(x)=2(x-r)T(x)+(x-r)^2T^\prime(x)$$
So, r is also a root of  the derivative of $P(x)$.
