Why $P(x) = 0$ is equivalent to $\dfrac{P(x)}{P'(x)} = 0$ for every polynomial $P$?
On can see the following:
Claim 1: Iff a polynomial $P(x)$ has a root $r$ with multiplicity $k>0$ i.e., $(x-r)^k$ divides $P$ but $(x-r)^{k+1}$ does not, then $r$ is a root of $P'(x)$ with multiplicity exactly $k-1$.
[Indeed, let us write $P(x) = (x-r)^kQ(x)$, where $Q(x)$ is a polynomial that is nonzero at $x=r$. Then note that $$P'(x) = k(x-r)^{k-1}Q(x) + (x-r)^kQ'(x),$$ where $Q'(x)$ is the derivative of $Q(x)$ with respect to $x$. As $(x-r)^k$ divides $(x-r)^kQ'(x)$ but not $k(x-r)^{k-1}Q(x)$, it follows that $(x-r)^k$ does not divide $P'(x)$. Can you see that on the other hand, $(x-r)^{k-1}$ does divide $P'(x)$?]
So if $P(x) = \prod_{i=1}^m (x-r_i)^{k_i}$, where $r_1,\ldots, r_m$ are the distinct roots of $P(x)$ and each $k_i$ is a positive integer, then for some polynomial $p(x)$ that is nonzero at each of $r_1,\ldots, r_m$, the derivative $P'(x)$ of $P(x)$ with respect to $x$ can be written $$P'(x) = p(x)\prod_{i=1}^m (x-r_i)^{k_i-1}.$$
Can you take it from here. In particular, let $R(x)$ be the rational function $$R(x) \doteq \frac{1}{p(x)} × \prod_{i=1}^m (x-r_i)^{k_i-k'_i},$$ where for each $i=1,\ldots, m$, the integer $k'_i$ is defined to be the largest integer such that $(x-r_i)^{k'_i}$ divides $P'(x)$. Then your question asks to prove the following: No matter the choice of non-constant polynomial $P(x)$, this rational function $R(x)$ is indeed defined and takes the value $0$ for $x=r_i$ for each $i=1,\ldots, m$. However, to this end, first use the fact established above that $k_i-k'_i$ is precisely $1$ for each such $i$. Next, note that $p(x)$ is nonzero at each of $r_1,\ldots, r_m$....
You can take it from there.
The division $$ P(x)=P'(x)Q(x)+R(x) $$ may yield a nonzero remainder, but it's not a real problem. As an easy example, the derivative of $x^2+1$ is $2x$ and we cannot do exact division.
The correct statement needs to use the greatest common divisor (which over polynomials is only defined up to a nonzero multiplicative constant).
Theorem. If $D(x)$ is the (monic) greatest common divisor of $P(x)$ and $P'(x)$ and $P(x)=D(x)Q(x)$, then $Q(x)$ has the same roots as $P(x)$ and all roots of $Q(x)$ are simple.
I assume that the base field is $\mathbb{C}$, but the argument works with any base field of characteristic $0$. With $x$ I denote an indeterminate; other lowercase letters denote elements of the base field; uppercase letters denote polynomials.
First a definition: $r$ is said to be a root of multiplicity $k$ of the polynomial $P(x), where $k$ is a nonnegative integer, if
- $(x-r)^k$ divides $P(x)$, and
- $(x-r)^{k+1}$ does not divide $P(x)$.
Of course, saying that $r$ is a root of multiplicity $0$ means it's not a root at all, but it's convenient for the sequel. A simple root is a root of multiplicity $1$.
The derivative satisfies the standard properties, in particular the product rule, and we have, if $P(x)=(x-r)^kQ(x)$, $$ P'(x)=k(x-r)^{k-1}Q(x)+(x-r)^kQ(x) $$ It follows that, if $r$ is a root of multiplicity $k$ of $P(x)$, then $r$ is a root of multiplicity $k-1$ of $P'(x)$ (exercise).
If $r_1,r_2,\dots,r_m$ is the complete set of distinct roots of $P(x)$, with respective multiplicities $k_1,k_2,\dots,k_m$, then $$ P(x)=(x-r_1)^{k_1}(x-r_2)^{k_2}\dotsm(x-r_m)^{k_m}A(x) $$ where $A(x)$ has no roots. Applying the argument above, we see that $$ P'(x)=(x-r_1)^{k_1-1}(x-r_2)^{k_2-1}\dotsm(x-r_m-1)^{k_m-1}B(x) $$ where none of $r_1,r_2,\dots,r_m$ is a root of $B(x)$. Now it's clear that we have, for the greatest common divisor, $$ D(x)=(x-r_1)^{k_1-1}(x-r_2)^{k_2-1}\dotsm(x-r_m)^{k_m-1}C(x) $$ and $C(x)$ is the greatest common divisor of $A(x)$ and $B(x)$, so it cannot have roots. Therefore we have proved that $$ Q(x)=(x-r_1)(x-r_2)\dotsm(x-r_m)Q_0(x) $$ where $Q_0(x)$ has no roots.
Now if we consider polynomials over $\mathbb{C}$ as functions, we have $$ f(x)=\frac{P(x)}{P'(x)}=(x-r_1)(x-r_2)\dotsm(x-r_m)\frac{A(x)}{B(x)} $$ which is defined where $B(x)\ne0$ and $A(x)$ has no roots, so the zeros of the function $f$ are the same as the zeros of $P(x)$.