Why can't polynomials have negative exponents or division by a variable Why can't:
$$2x^{-3} - 3x$$
or
$$\frac{1}{2x}$$
be polynomials too?
Why have a definition that excludes these algebraic forms?
 A: Here’s my understanding of the reason for the restrictiveness of the definition of “polynomial”, in response to @QuoraFea’s suggestion.
I’ll try to make my explanation as elementary as I can, with parenthetical expansions for the more technically inclined.
The set (ring, actually) of polynomials with real coefficients (more generally with coefficients in any commutative ring) has a “universal” property that the larger sets do not. Just to speak only of polynomials in one variable, the set of all such, $\mathbb R[x]$ ($R[x]$ for a general ring), has the property that the variable $x$ may be evaluated to any real number $a$ (to any element $\alpha$ of an algebra $A$ over the base ring $R$) so that this “evaluation mapping” can be applied to any polynomial at all (to any element of $R[x]$ at all). That is, once you’ve chosen $a$, then $f(a)$ makes sense as a real number (as an element of $A$) no matter what polynomial $\,f$ you look at.
Furthermore, the choice of $a$, once it’s done, gives you a function from polynomials to constants, I’ll call it $e_a$, namely $e_a(f)=f(a)$. And this function is both additive and multiplicative (it’s a ring morphism from $R[x]$ to $A$). That is, it satisfies $e_a(f+g)=e_a(f)+e_a(g)$ and $e_a(fg)=e_a(f)e_a(g)$, as well as $e_a(\mathbf1)=1$, where the bold-face $\mathbf1$ is the constant polynomial $1$, and the other $1$ is the ordinary unit element of $\mathbb R$. The function $e_a$ may be called the evaluation morphism. I stress that the domain is the set of polynomials, and the target space, also known as codomain, is $\mathbb R$ (the algebra $A$ in the general case).
There are, indeed, other algebraic structures that allow the evaluation functions to constants [for instance the set of all entire analytic functions on the line], but the ring of real polynomials $R[x]$ is the only one that allows the variable $x$ to be evaluated to an element of any larger structure containing $\mathbb R$ (to any $R$-algebra).
A: Polynomials are defined as they are for a few distinct reasons: (1) because polynomials as functions have certain properties that your 'polynomials with division' don't have, and (2) because there are other terms for more generalized algebraic forms.
First, the properties of polynomials: unlike e.g., $2x^{-3}+3x$, polynomials have no poles; there's no place where a polynomial 'blows up', where it goes to infinity.  In particular, this means that the domain of a polynomial as a function is all of $\mathbb{R}$ — or, alternately, all of $\mathbb{C}$).  In fact, polynomials in the complex plane have even nicer properties - they're analytic functions, which means that not only are they well-defined everywhere, but all of their derivatives are well-defined everywhere.  This means that polynomials have a lot of structural properties that make them 'nice' objects of study in ways that your expressions aren't.
Secondly, there's also a notion that (roughly) corresponds to your 'extended' version of polynomials, with its own set of nice properties: the ring of rational functions, which includes not just items like $x^2+\frac1x+5$ but also terms like $\dfrac{x^3+2}{x+5}$ that your formulation doesn't at least at first glance suggest.  These represent a useful extension of polynomials for study because they're closed under the operation of division; for any two rational functions $f(x)$ and $g(x)$, the function $h(x) = \frac{f(x)}{g(x)}$ is also a rational function.  This property doesn't hold for your 'Lambert polynomials', because there's no finite expression in positive and/or negative powers of $x$ that corresponds to the function $\frac1{x+1}$.
What's more, there's also an object occasionally studied that more directly corresponds to your notion: the notion of Laurent Polynomial.  As that article suggests, they're of particular importance and interest for their connections with the field of Hopf Algebras (and by extension, quantum groups).  I would opine that the reason that they're not the primary object of study is because they're not the 'simplest' structure of interest among any of their peers, and fundamentally the most important structures in mathematics tend to be the simplest structures exhibiting some given property.
A: I don't know if there is a "good" reason. Polynomial is just a name for a certain kind of structure. Practically speaking you frequently want to work with negative powers of $x$ as well as positive powers (e.g. Laurent Series), but then you just wouldn't call it a polynomial.
There are good pedagogical reasons to teach polynomials with just positive powers. There are a lot of rules that apply to the polynomial family as a whole which you would lose if you included negative powers into the definition. Polynomials are probably the most well behaved functions there are (no singularities, continuous and differentiable everywhere, etc.) it is helpful to study them separately first. Then when the student is comfortable with polynomials adding the more complicated negative powers is just an easy modification of what has already been learned.
There is a brief investigation of the origin of the word "Polynomial" here : Link
There is one quotation from there that suggests to me the original meaning of polynomial  may have been: Any string of mathematical expressions connected by addition and subtraction. I have no hard evidence of this so take it worth a grain of salt, but it would be more consistent with your suggested definition.
A: You'll be glad to know that as Steven remarked, "Laurent polynomials" include both positive and negative exponents. If $R$ is a ring, like the complex numbers, you can consider such expressions with the coefficients in $R$ and they are used frequently in higher algebra. The notation for them is $R[x,x^{-1}]$.
Here is something to think about if you have seen quotient rings before (or when you do see them): the Laurent polynomial ring is just the quotient of another polynomial ring: in fact, $R[x,x^{-1}]$ is isomorphic to $R[x,y]/(xy - 1)$. Even if you haven't seen quotients before, this latter ring is just the set of all polynomials in two variables with the relation that $xy = 1$. So Laurent polynomials are just a natural extension of the idea of polynomials.
A: Some reasons: they have domain $\mathbb{R}$, the family of polynomials is  closed under addition, substraction, multiplication, differentiation, composition, and integration, and any continuous function on a closed and bounded interval can be approximated as closely as you want by polynomials.  That's quite a lot of nice properties.  There are good reasons beyond pedagogical reasons (especially the fact that the domain is $\mathbb{R}$).  Note that antiderivatives (integrals) of rational functions are not necessarily rational functions.
A: I would have added this as a comment to Newb's answer but for some reason I can't add a comment today. But in answer to Quora Fea's question in the comments of that answer - is it possibly because a 'polynomial' with negative exponents could be converted into a valid one with positive exponents by multiplying by some x^n. The roots would be the same (with the exception of an extra root at x=0 in the converted polynomial). 
