What is a polynomial and how is it different from a function? I have a problem that asks me to find a polynomial $P(x)$ so that $P(3)$ is 9.
Now I can say with certainty that $P(x)$ can be $x^2$. This is a second degree polynomial.
But what about functions such as $\frac{1}{x^2+3}$, are these not polynomials? If not, why?
P.S. I seem to be having troubles with the math tags. Some help on that would be appreciated as well.
 A: A polynomial (in one variable, over the real numbers) is by definition an expression of the form
$f(x) = a_0 + a_1x+a_2x^2 + \cdots + a_nx^n$
for some non-negative integer $n$ and real numbers $a_i,\ i=0,1,\dotsc, n$.
Note that the variable $x$ must have non-negative powers; things like $r(x) = \frac{1}{x^2+3}$ are, by definition, not polynomials in $x$ since they include negative powers of $x$. (Properly speaking, it contains a negative power of the polynomial $x^2-3$, and hence cannot be written in the same form as $f(x)$ above.)
Of course, the numerator and denominator of $r$ are both polynomials; expressions of this form $\frac{p}{q}$ where $p,q$ are polynomials are called rational functions.
A: Polynomials are finite formal sums $\sum_{i = 0}^n a_i x^i$, where $a_i$ belongs to some ring (if you need some idea of what a ring can be, the real numbers are a ring, the integers are a ring, and the rationals are a ring). By definition, there are no negative powers of the indeterminate $x$ involved (likewise, there are no reciprocals of expressions involving $x$).
Technically, a polynomial is not a function, although you can consider it to be one if you take an element $r$ in the original ring and "plug it in" (let $x = r$ and compute). However, if you interpret a polynomial as a function in that manner, two different polynomials can give the same function. For example, consider the field $\Bbb Z/p\Bbb Z$ (i.e. integers modulo $p$). The polynomials $0$ and $x^p - x$ define the same function by Fermat's little theorem, but they are not the same polynomial.
