What is meant by "the coefficients of a polynomial function $f(x)$ are symmetric functions of its roots"? I am reading Rotman's Introduction to Group Theory. 
One of his first remarks is that:

By the middle of the eighteenth
  century, it was realized that permutations of the roots of a polynomial $f(x)$
  were important; for example, it was known that the coefficients of $f(x)$ are
  "symmetric functions" of its roots.

Aren't the coefficients of a polynomial function constants? How can they then be functions, let alone "symmetric functions"?
 A: Consider the cubic of the form $x^3+Px^2+Qx+R$, and suppose it has roots $a$, $b$, and $c$. Then:
\begin{align}
P&=-a-b-c\\
Q&=ab+ac+bc\\
R&=-abc
\end{align}
These are called "symmetric" functions of $a$, $b$, and $c$, because switching them around doesn't change the value of the function.
Proof: This follows from noting that the cubic is $(x-a)(x-b)(x-c)$. Just multiply it out and equate coefficients.
A: If you know what the roots of a polynomial are, you can use those roots to compute the coefficients, up to a constant factor. This means that, for example for monic fifth-degree polynomials, there is a function that takes the five roots as inputs and produces, say, the coefficient of $x^2$ in the polynomial. (To emphasize, this is one function that works for all degree-5 polynomials).
It is this function that is symmetric, meaning that the coefficient it produces does not depend on which order you feed the various roots to it. (Moreover the coefficient-generating functions are themselves polynomials in several variables).
A: But that's the ingenuity that revolutionized the theory of solving polynomial equations - it's sort of the reverse of the usual situation! Normally we're given an actual polynomial and want to find its roots. 
However, that doesn't stop us from deciding that we want certain roots and asking, "What kind of polynomial can I write down with these roots?" (By the way, questions like this arise quite naturally when you're writing quizzes and tests for students of the kind of algebra you learned years ago!)
Let's say we want a quadratic with roots $x = r_1$ and $x = r_2$. In this case, any quadratic of the form
$$p(x) = a(x - r_1)(x - r_2) = ax^2 -a(r_1 + r_2)x + ar_1r_2$$
has the roots we want, $r_1$ and $r_2$. So our coefficients for $x^2,\, x,$ and $1$ are respectively 


*

*$a$ (which is completely independent of both $r_1$ and $r_2$; we could swap them, and it wouldn't change $a$)

*$-a(r_1 + r_2)$ which again remains invariant under permuting the two roots, and

*$ar_1r_2$, another function of $r_1$ and $r_2$ that doesn't depend on which is which.
This is an example of the sort of thing they mean.
A: Let's introduce a bit of notation first: let $f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_1 x + a_0$ where each $a_i \in \mathbb{C}$ (if the coefficient of $x^n$ is not $1$, factor it out of the polynomial).  Then the fundamental theorem of algebra tells us that $f$ has $n$ (not necessarily distinct roots) $\omega_1,\ldots,\omega_n$.  Then we can also write $f(x) = (x - \omega_1)(x - \omega_2)\cdots (x - \omega_n)$.  If we multiply this all out and equate coefficients, we can express each $a_i$ in terms of the roots; in this sense, the $a_i$'s are functions of the roots, and we can write $a_i(\omega_1,\ldots,\omega_n)$.  If you were to multiply it out, this $a_i$ is going to be a polynomial in $n$ variables
Now, given a function of $n$ variables, $g(x_1,x_2,\ldots,x_n)$, we say that $g$ is a symmetric function iff $g(x_1,x_2,\ldots,x_n) = g(x_{\sigma(1)},x_{\sigma(2)},\ldots,x_{\sigma(n)})$ for every $\sigma \in S_n$.  That is, $g$ is a symmetric function if we can permute the variables without changing the function.  We then see that the $a_i$'s can be viewed as symmetric functions of the roots in that we can permute the order of the roots without changing the coefficients. That is, the function $a_i(\omega_1,\ldots,\omega_n)$ is a symmetric polynomial in $n$  variables. 
A: Consider a quadratic polynomial with roots $r_1$ and $r_2$. If you expand $(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$ you can see that the coefficients of this quadratic are related to its roots in a way that’s independent of the order in which you list the roots. Similarly, for a cubic, its coefficients are $-(r_1+r_2+r_3)$, $(r_1r_2+r_1r_3+r_2r_3)$ and $-(r_1r_2r_3)$. These values, too, are clearly independent of the order in which you list the three roots. This sort of symmetry holds for higher-degree polynomials as well.
