Is $X^3+2X^2-3X-2$ irreducible in $\mathbb F_7[X]$? Is $X^3+2X^2-3X-2$ irreducible in $\mathbb F_7[X]$?
I have checked for every $x \in [0,6]$, but the polynom never turns $0$.
Is there a possibility of factorization so that the polynom will be reducible?
I refer to that post - I have not seen the reducibility there, either.  
 A: When you are checking whether polynomial is irreducible in general, you want to find out whether factorization $f = gh$ is possible or not (with $g,h$ non-invertible). In the case of polynomial of degree $3$ over a field, factorization $f = gh$ implies $\deg g = 1$ or $\deg h = 1$, since $\deg g + \deg h = 3$ (make sure that you understand why $3 = 0 + 3$ doesn't count). But, that means that $f(X) = (X-\alpha)g(X)$ for some $\alpha$ from the field, which is equivalent to $f(\alpha) = 0$ (equivalence is due to Euclidean division). Thus, polynomial of degree $3$ over a field is reducible if and only if it has a root in the field.
So, in your case, check all elements of $\Bbb F_7$, as you did. If there are no roots, the polynomial is irreducible.
Unrelated note. Segment notation $[0,6]$ is very poor choice in this case. First of all, it assumes order on $\Bbb F_7$, but it won't work, since you can't make it respect multiplication. For example, $2<3$, yet $4 = 2\cdot 2 > 3\cdot 3 = 2$. On the other hand, when I see segment, I'm expecting uncountable set due to being used to working with it on $\Bbb R$.

Edit: Where does $f(X) = (X-\alpha)g(X)$ come from?
It is well known that any polynomial over $\Bbb C$ can be written as $$p(X) = a(X-\alpha_1)(X-\alpha_2)\cdots(X-\alpha_n)$$ where $a$ is $p$'s leading coefficient and $\alpha_i$ are its roots. This is motivation for doing the same in general case as well. If you have field $K$, then $K[X]$ is Euclidean domain, i.e. for any $f,g\in K[X]$ ($g\neq 0$), there exist $q, r\in K[X]$ such that $f = qg + r$ and $\deg r<\deg g$ or $r = 0$. Letting $g = X-\alpha$, we get that for any polynomial $f\in K[X]$, $f(X) = (X-\alpha)q(X) + f(\alpha)$. As you can see, $X-\alpha$ divides $f$ if and only if $\alpha$ is root of $f$. This is why we prefer $X-\alpha$ instead of $X+\alpha$ (although, there is no essential difference).
In our case, if $f$ were reducible, then there would exist $g$ and $h$ non-invertible such that $f = gh$. As we mention above, without loss of generality, $\deg h = 1$. That means that $h(X) = aX + b$ and $f(X) = (aX+b)g(X) = (X+\frac ba)(ag(X))$. Now, let $\alpha = -\frac ba$ and $\bar g(X) = ag(X)$. We get that $f(X) = (X-\alpha)\bar g(X)$. This means that it was not loss of generality to assume that $h(X) = X-\alpha$ for some $\alpha\in\Bbb F_7$ in the first place.
A: A polynomial of degree $3$ which is reducible over a field $\Bbb F$ must have a root on that field, because at least one of its factors must have degree $1$. So $f$ is irreducible.
This is not true for greater degrees. For example, a product of irreducible polynomials of degree $2$ is reducible but it has no roots.
