Finding roots of $P(z)$ through $\text{gcd}(P(z),P'(z))$ 
This is an excerpt from Zorich's book. I have some issues with understanding the last paragraph of it. This is what I understood so far:
Suppose we have polynomial with complex coefficients $P(z)$. We would like to find roots and their multiplicities.
Suppose $q(z):=\text{gcd}(P(z),P'(z)).$ Suppose we know factorization of $q(z)$, i.e. $q(z)=(z-z_1)^{\alpha_1}\dots (z-z_p)^{\alpha_p}$ then somehow he deduces that roots of $P(z)$ are also $z_1,\dots,z_p$ with multiplicities $\alpha_1+1,\dots, \alpha_p+1$.
Am I right that h is doing exactly this?
If yes how to prove it? I am bit confused.
 A: Let $q$ be the monic $\gcd(P, P^\prime)$. Since $q$ is a monic divisor of $P$, then it must be of the form $$q(z) = (z - z_1)^{\alpha_1} \cdots (z - z_p)^{\alpha_p}$$ where each $\alpha_i$ is nonnegative and at most the multiplicity of the root $z_i$ in $P$.
There is an inverse statement to the corollary which is not stated that if $z_i$ is a root of $P$ of multiplicity $1$, then it is not a root of $P^\prime$.
Since $q$ is also a divisor of $P’$, there are two cases:

*

*The number $\alpha_i = 0$ and thus $z_i$ is not a root of $P^\prime$.

*The number $\alpha_i$ is positive, and thus $z_i$ is a common root of $P$ and $P^\prime$.

In the first case, the contrapositive to the corollary tells us that the multiplicity of $z_i$ in $P$ is $1$. In the second case, the inverse statement to the corollary shows that the multiplicity of $z_i$ in $P$ is greater than $1$.
In both cases, the maximal power of $(z - z_i)$ that divides $P$ and $P^\prime$ is $(z - z_i)^{k_i - 1}$, and $\alpha_i = k_i - 1$ where $k_i$ is the multiplicity of the root $z_i$ in $P$.
