Polynomial or not? My question is whether $x(x+{1\over x}+2)$ the same as $x^2+2x+1$ ?
And if so then: 
-Why doesn't 0 belong to the domain of the first but it belongs to the second ?
-Is the first function a polynomial ?
 A: The question "is this function a polynomial" can be answered at various different levels of sophistication and depends on what we mean by a function as well as what we mean by a polynomial.
For example, we can think of a function as a collection of ordered pairs $(x, f(x))$ independent of how the function is expressed.
Here there is no indication what kind of mathematical object $x$ is - for this purpose I will assume $x$ is a real number.
Then the function defined by the first expression also has an expression as a polynomial on any subset of its domain of definition (which excludes $x=0$). Also as a function $\mathbb R\setminus \{0\} \to \mathbb R$ it can be extended to a function represented by a polynomial at $x=0$. But the expression you are given is not a polynomial.
Probably these technicalities are not meant in your context and may seem over pedantic - but it is sometimes important to distinguish between a function and how it is expressed. Sometimes a function cannot be expressed nicely in its whole domain, but can be patched together from nice pieces (here it can be easily repaired at $0$). Sometimes this can be done in such a way that the niceness of the pieces can be exploited. And other times the niceness of the pieces suggests a different point of view which is mathematically simpler (the Riemann Sphere - which represents the complex plane with the addition of a point at infinity is an example).
A: formally they are not so because every function comes with a domain and a range and those functions do not have the same domain, if you consider them in the same domain (i.e. $\Bbb R\setminus\{0\}$) if they would be the same.
A: No, first function is not a polynomial since $0$ is not in it domain. 
Why? Because we can not divide by $0$. 
So the functions are not the same since the other one is a polynomial of a degree 2 and the $0$ is in it domain.
