While reading a chapter entitled "Functional equations for polynomials" in the book "Polynomials" by Victor Prasolov, he states that
Every polynomial $f$ of degree $n+1$ satisfies the identity $$ f(x) = f(y) +(x-y)f'(y)+\cdots+(x-y)^{n+1}\frac{f^{(n+1)}(y)}{(n+1)!}$$
He essentially transcribes a given polynomial into a functional form. I'm having a little difficulty understanding what's going on here. Clearly, the functional form resembles a Taylor series, but what does $y$ represent here? Could someone give me an example of how a polynomial would look under this transformation?
This literally is just the Taylor series. Intuitively, you think about it as taking a Taylor expansion around $y$ to find the value at $x$. Normally, you expect Taylor series to give you an approximation close to $y$, but since all of the higher order terms disappear and polynomials are analytic function, you get exact equality.
The beauty of this is that you can define a formal derivative for polynomials over any ring, rather that just for polynomials with coefficients in $\mathbb{R}$ or $\mathbb{C}$. You can then check that the usual identities still hold (although, in the present case, to be able to write a Taylor series as usual for any polynomial, you should request that the characteristic be 0).
If you are wondering about the definition of formal derivative, it is exactly what you would expect: if you have a polynomial $p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$, then $p'(x) = a_1 + 2 a_2 x + \ldots + n a_n x^{n - 1}$.