transforming $\,x^3 + x^2 + 1 \implies x\left(x^2 + x + \dfrac{1}{x}\right) \implies\,$ div by $\,0\,$?! EDIT: Sorry. I basically was confused by that just valid mathematical transforming could lead into a undefined behavior. 
I have to admit, my question has not much to do with the zero of a function, i've just used one method of the "zero calculating tool" set to reshape my function. (i explain that a bit more below)
I have a function $\,f(x)=x^3 + x^2 + 1\,$. I can now reshape $\,1\,$ to $\,x^0\,$:
\begin{align}
    f(x)&= x^3 + x^2 + x^0
\end{align}
and factor one x out to 
\begin{align}
    x\left(x^2 + x + x^{-1}\right)
\end{align}
A product of zero remains zero. Therefore, either
\begin{align}
    x=0
\end{align}
or 
\begin{align}
     x^2 + x + \dfrac{1}{x} = 0
\end{align}
But $\,\dfrac{1}{x}\,$ is not defined, as it could be $0$ and divide through $0$ is not valid.  My confusion arises primarily because reformatting/reshaping a function normally doesn't change its value. It's simply another way to write it, hence it should be the exact same. But as i showed above, it is not. The beginning function is a normal grade 3 function. But simply transforming can obviously change the whole validity of a function. That's what i don't understand.
My Question: 


*

*Where is my mistake?

*Is it expected that applying valid mathematical methods to a function correctly can result in not defined things?


The function
$$f(x)= x^3 + x^2 + 1$$ 
has domain $x \in \mathbb{R}$ when otherwise unspecified. It is not the same function as
$$g(x)= x \left(x^2 + x + \frac 1x\right) $$
whose domain is considered to be $x \in \mathbb{R}\setminus\{0\}$.
So $f(x) = g(x)$ if $x\ne0$.
If you wish to utilize the form of $f(x)$ where you factor out an $x$ but retain the original mapping, you could employ some extra notation... Perhaps...
$$f(x) = \begin{cases}
x \left(x^2 + x + \frac 1x\right) & x \ne 0 \\
 1 & x = 0 \\
 \end{cases}$$
It appears that in your post, you've manipulated the function to a case that is valid on only part of the original domain of definition. That is, when you created the $1/x$ term, you made the assumption that $x\ne 0$.
To avoid future confusion, be 1) aware of the domain and 2) aware of any assumptions you make during algebraic manipulations. 
