Complex Polynomial Roots Suppose $p$ is a polynomial with coefficients in $\mathbb{C}$ and it has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p′$ have no roots in common.
 A: Let $r$ be the highest power of $x-a$ that divides $f(x),$
we can write $f(x)=(x-a)^rg(x)$ where $1\le r\le m$ so that $g(x)$ is a polynomial of  degree $m-r$ with $(x-a)\not\mid g(x)$
$f'(x)=(x-a)^rg'(x)+r(x-a)^{r-1}g(x)$
If $a$ is a root of $f'(x)=0,f'(a)=0\iff r-1\ge1$ as  $(x-a)\not\mid g(x)$
$\implies r\ge 2$ using Polynomial remainder theorem.
Alternatively,
if $f(x)=(x-a)g(x)\implies f(a)=0\iff (x-a)\mid f(x)$
Now, $f'(x)=(x-a)g'(x)+g(x)$
If $a$ is a root of $f'(x)=0\iff f'(a)=0$
$\implies  g(a)=0\implies (x-a)\mid g(x)\implies (x-a)^2\mid f(x)$
A: Let $p(x)$ be a polynomial with complex coefficients, of degree $m\ge 1$. The hard part, which is essentially the Fundamental Theorem of Algebra, is that we can express $p(x)$ in the form
$$p(x)=a(x-r_1)(x-r_2)\cdots (x-r_m),$$
where the $r_i$ are complex numbers, and $a$ is a non-zero complex number. The solutions of the equation $p(x)=0$ are then $r_1,\dots,r_m$. 
We show (A) If the $r_i$ are not all distinct, then $p(x)$ and $p'(x)$ have a common root and (B) If $f(x)$ and $f'(x)$ have a common root, then the $r_i$ are not all distinct.
Proof of A: Suppose the roots are not all distinct. So suppose for example that $r_1=r_2=r$. Then $p(x)=(x-r)^2q(x)$ for some $q(x)$. Differentiate, We get 
$p'(x)=(x-r)^2q'(x)+2(x-r)q(x)$. It follows that $r$ is a root $p'(x)$, and hence is a common root of $p(x)$ and $p'(x)$.
Proof of B: Let $r$ be a common root of $p(x)$ and $p'(x)$. We show that $(x-r)^2$ divides $p(x)$. This will show that $r$ is a root of $p(x)$ of multiplicity $\gt 1$, meaning that the roots of $p(x)$ are not all distinct.
Because $r$ is a root of $p(x)$, by the Factor Theorem we have that $x-r$ divides $p(x)$, and therefore $p(x)=(x-r)q(x)$ for some polynomial $q(x)$. 
Differentiate. We get $p'(x)=(x-r)q'(x)+q(x)$. But $r$ is a root of $p'(x)$. So $p;(r)=(r-r)q'(r)+q(r)$. It follows that $q(r)=0$. So again by the Factor Theorem, $x-r$ divides $q(x)$. 
Suppose that $q(x)=(x-r)s(x)$. Then $p(x)=(x-r)q(x)=(x-r)^2s(x)$, and therefore $r$ is a root of $p(x)$ of multiplitity $\gt 1$.
Remark: The Fundamental Theorem is needed for the proof of the result in the OP.  We could prove a weaker result without using the Fundamental Theorem. 
Without the Fundamental Theorem, we can show, as in the argument above,  that if $p(x)$ has a root of multiplicity $\gt 1$, then it is a common root of $p(x0$ and $p'(x)$. One can also show that if $p(x)$ and $p'(x)$ have a common root, then $p(x)$ has a root of multiplicity $\gt 1$.
But one cannot show that there are (counting multiplicity) $m$ roots. I suspect the wording of the problem is not quite what was intended.  
A: Hint $\  $ It follows immediately by the double root test, one proof of which is sketched below.
$$\rm\begin{eqnarray} &&\rm\!\! (x\!-\!a)^2 |\ p(x)\!\!\!\!\!\!\!\\
\iff\ &&\rm x\!-\!a\ |\ p(x)\  &\rm and\ \ &\rm  x\!-\!a\ \bigg|\ \dfrac{p(x)}{x\!-\!a}\\
\\
\iff\ &&\rm p(a) = 0 &\rm and&\rm x\!-\!a\ \bigg|\ \dfrac{p(x)-p(a)}{x\!-\!a}\ \ \left[\!\iff \color{#C00}{\dfrac{p(x)-p(a)}{x\!-\!a}\Bigg|_{\large\:x\:=\:a}} \!=\: 0\ \right] \\
\\
\iff\ &&\rm p(a) = 0 &\rm and&\rm \color{#C00}{p'(a)} = 0\end{eqnarray}$$
