The expression $ \frac{x^2 -4}{x-2}$ doesn’t looks like a polynomial because of $ x $ in the denominator. But it can be factorized. After factorizing and canceling the common factor $$\frac{x^2 -4}{x-2} = \frac{(x+2)(x-2)}{x-2} = x+2, $$ and $ x+2 $ is a polynomial which satisfies the definition of the polynomial.
Is the original expression $ \frac{x^2 -4}{x-2} $ is a polynomial or not?
The answer to your question depends a great deal upon context. Other answers have discussed contexts which depend on particular branches of mathematics—I want to focus on contexts related to the educational environment. This answer is likely somewhat more localized than the others, as it is based on the American curriculum, but there is likely a somewhat more universal flavor to it (as the curricula in the rest of the world probably don't vary too much).
"Yes" and/or "No".
In the American system, students are usually guided along a path of courses with titles such as "Algebra", "Precalculus", and "Calculus" (in that sequence). The usual goal of these courses is to introduce students to some basic analysis (estimation, differentiation, integration, etc). In the modern setting, analysis is built on the study of functions, hence the fundamental objects of study are functions. In this setting, a common definition is a follows:
Definition: A polynomial is a function $p : \mathbb{R} \to \mathbb{R}$ which can be written in the form $$ p(x) = a_n x^n + a_{n-1}x^{n-1} + \dotsb + a_2 x^2 + a_1 x + a_0, $$ where $n$ is a nonnegative integer and $a_1, a_2, \dotsc, a_n \in \mathbb{R}$.
In this setting, the function defined by the formula $$ r(x) = \frac{x^2-4}{x-2} $$ is not a polynomial. This issue here is a little subtle—a function is defined by three data: the domain of the function, the codomain of the function, and the rule or law which pairs an element of the domain with an element of the codomain. Two functions with different domains are not equal. While no domain for $r$ has been given, the formula defining $r$ is not defined when $x=2$, which implies that the domain of $r$ is $\mathbb{R}\setminus\{2\}$. The domain of a polynomial is all real numbers, while the domain of $r$ is something less than this. Therefore $r$ is not a polynomial.
That being said, the "hole" in $r$ can be filled by defining a new function $\overline{r} : \mathbb{R} \to \mathbb{R}$ by the formula $$ \overline{r}(x) = \begin{cases} r(x) & \text{if $x \ne 2$, and} \\ 4 & \text{if $x = 2$.} \end{cases} $$ This function $\overline{r}$ is an extension of $r$ to a larger domain which has some nice properties. In particular, $\overline{r}$ is continuous (analytic, even), and for any $x \in \mathbb{R}$ $$ \overline{r}(x) = x+2. $$ Hence $\overline{r}$ is a polynomial. In this setting, we often say that $r$ has a removable singularity, and can be extended to a polynomial on $\mathbb{R}$.
Students who are intending to complete an undergraduate degree in mathematics are often required to take a course with a title like "Modern Algebra" or "Abstract Algebra". The focus of such a course is on the structures created when a set is endowed with certain operations (such as addition or multiplication). There are a lot of technicalities here (and, as I am analyst, I am likely to gloss over a lot of this), but the basic idea is something like the following:
Definition: Let $R$ be a field[1] (a set endowed with two operations which "play nice"—just take $R$ to be the set of real numbers with the usual addition and multiplication). The ring of polynomials with coefficients in $R$, denoted by $R[x]$, is the ring with elements of the form $$ a_n x^n + a_{n+1} x^{n-1} + \dotsb + a_2 x^2 + a_1 x + a_0, $$ where $n$ is a nonnegative integer, $a_1, a_2, \dotsc, a_n \in R$, and $x, x^2, \dotsc$ are abstract symbols which represent "powers" of $x$. The ring of polynomials are endowed with an addition and multiplication operation. Addition is defined so that (for example) $$ ( a_n x^n + \dotsb a_1 x + a_0) + (b_n x^n + \dotsb + b_1 x + b_0) = (a_n + b_n) x^n + \dotsb (a_1 + b_1) x + (a_0 + b_0), $$ and multiplication is defined so that (for example) $$ (a_1 x + a_0) \cdot (b_1 x + b_0) = (a_1 b_1) x^2 + (a_1 b_0 + a_0 b_1) x + (a_0 b_0). $$ (That is, addition and multiplication of polynomials is defined so that they behave in exactly the way we would expect them to behave based on our previous experience with polynomials.)
A ring of polynomials is closed with respect to addition (and subtraction), and with respect to multiplication. This means that if we add, subtract, or multiply two polynomials, the result is a polynomial. However, most polynomials do not have a "multiplicative inverse"—given an arbitrary polynomial $p$, there almost certainly isn't another polynomial $p^{-1}$ such that $p\cdot p^{-1} = 1$. Because of this, it doesn't make sense to define division hence an expression such as $p/q$ (where $p$ and $q$ are polynomials) is simply nonsense.
By way of analogy, if $a$ and $b$ are both integers, then (for most values of $a$ and $b$), the expression $a/b$ is not an integer. If you are working with integers (and can't work with rational numbers), then the expression $a/b$ is simply undefined—it does not exist in the universe being considered.
In this setting, both $x^2 - 4$ and $x - 2$ are polynomials in $\mathbb{R}[x]$, but the expression $$ \frac{x^2-4}{x-2} $$ is not a polynomial. It does not exist in this universe.
That being said, there is a way that we can construct a universe where this expression does exist:
Definition: Let $R$ be a "sufficiently nice" ring.[2] The field of fractions of $R$, denoted by $\DeclareMathOperator{Frac}{Frac}\Frac(R)$ can be constructed as follows: the underlying set consists of equivalence classes of ordered pairs of elements of $R$, where two ordered pairs $(a,b)$ and $(c,d)$ are equivalent if $$ ad - bc = 0. $$ Addition and multiplication are defined by $$ [(a,b)] + [(c,d)] = [(ad+bc, db)] \qquad\text{and}\qquad [(a,b)]\cdot[(c,d)] = [(ac, bd)], $$ where $[(a,b)]$ denotes the equivalence class of the ordered pair $(a,b)$.
This is rather complicated, so let's start with a relatively simple example: $\Frac(\mathbb{Z})$. An element of the field of fractions of $\Frac(\mathbb{Z})$ is a collection of ordered pairs of integers, where all of these ordered pairs are equivalent to each other. For example, the equivalence class of $(1,2)$ is the collection of all ordered pairs $(a,b)$ of integers such that $$1b - 2a = 0. $$ Manipulating this formally, this means that $$ 1b-2a = 0 \implies 1b = 2a \implies \frac{1}{2} = \frac{a}{b}. $$ Thus the equivalence class of $(1,2)$ is the set of all fractions which are equivalent to a half (that is, $1/2$).
In general, there is a very nice way of embedding an element of $\Frac(R)$ into $R$: the equivalence class of $(r,1)$ can be identified with the element $r \in R$. Again, looking at $\Frac(\mathbb{Z})$, think $$ [(5,1)] = \frac{5}{1} = 5. $$ That is, the equivalence class of $(5,1)$ is essentially the fraction $5/1$, which is basically just the integer $5$. From some point of view, $[(5,1)] \in \Frac(\mathbb{Z})$ is not the same object as $5 \in \mathbb{Z}$, but they are very closely related to each other, so it is not unreasonable to say that $[(5,1)]$ is an integer in $\Frac(\mathbb{Z})$. Since $\Frac(\mathbb{Z})$ is the field of rational numbers $\mathbb{Q}$, we might say that $[(5,1)]$ is the rational integer $5$.
In $\Frac(\mathbb{R}[x])$, we can play a similar game. Note that $$ 1(x^2 - 4) - (x+2)(x-2) = 0.$$ This means that $$ (x^2-4, x-2) \qquad\text{is equivalent to}\qquad (x+2, 1). $$ Using other notation, we might reasonably write $$ \frac{x^2 - 4}{x-2} = \frac{x+2}{1} = x+2, $$ which could be stated as "the rational expression $(x^2-4)/(x-2)$ is the (rational) polynomial $x+2$ in the field of fractions $\Frac(\mathbb{R}[x])$." This could very easily be shortened to "the expression $(x^2-4)/(x-2)$ is a polynomial".
In courses like those listed above (which often appear as introductory graduate level courses in the US—they are, at the very least, taught to quite advanced undergraduates), one of the major emphases is on integration. I am going to treat this section of the discourse very roughly and imprecisely—developing the theory correctly requires a lot of work, but I think that an intuition can be obtained quickly in a less rigorous manner.
While the fundamental objects of study are still functions, two functions are considered to be equivalent (essentially the same) if they always integrate to the same thing. Roughly speaking, $f$ and $g$ can be treated as the same object if $f(x) = g(x)$ for all $x$ except possibly for $x$ contained in a set of measure zero. Alternatively, $f = g$ if $$\int_{E} (f(x) - g(x)) \,\mathrm{d}x = 0 $$ for all measurable sets $E$. There are fancier (and more precise) ways of stating this—e.g. in terms of $L^p$ theory—but the essential idea is that two functions which integrate in the same way should be thought of as representing the same object.
Since $$ f(x) = \frac{x^2-4}{x-2} \qquad\text{and}\qquad g(x) = x+2 $$ only differ from each other at a single point, we will always have $$ \int_E (f(x) - g(x)) \,\mathrm{d}x = 0. $$ Hence, thought of as measurable functions, these two functions are equivalent to each other, and so it would not be unreasonable to say that $f$ is a polynomial (as it is equivalent to a polynomial in a manner meaningful to the context).
[1] For the pedants: I know that $R$ could be taken to be a commutative ring, or, perhaps, something even more general. However, for the purposes of this discussion, I don't think that it helps to discuss that level of generality.
[2] My recollection is that being an integral domain is "sufficiently nice" (zero divisors are bad), but the details here don't really matter.
It is not because it is not equivalent to $x+2$. You can evaluate $x+2$ for any real $x$ if you are working in the reals. You cannot evaluate $\frac {x^2-4}{x-2}$ at $x=2$ because of the division by $0$
Well, in terms of functions: The function $f:\mathbb{R}-\{2\} \to \mathbb{R}$ given by$f(x)=\frac{x^2-4}{x-2}$ is not equal to the function $g:\mathbb{R} \to \mathbb{R}$ given by $g(x)=x+2$ because $f$ is not even defined at $x=2;$ however, you can express $f$ at every point $x\neq 2$ as $f(x)=g(x)$. But it is important the fact that this is possible only if $x\neq 2,$ because, again, $f$ is not even defined at $x=2$.
Hence $f$ is not a polynomial function; since its domain is not all $\mathbb{R}$.
Now, in terms of abstract polynomials, say, $\mathbb{R}[x]$, the answer should be a bit different. The ring $\mathbb{R}[x]$ is not a division ring, because not all polynomials has an inverse. For instance, there's no such thing as $(x-2)^{-1}$ in $\mathbb{R}[x].$ Hence, $\frac{x^2-4}{x-2}$ has no sense on $\mathbb{R}[x]$.
