Perhaps by looking at an example you can understand what we mean by the splitting field. Suppose you look at $x^3 - 2$ over $\Bbb{Q}$. Then this polynomial is irreducible by Eisenstein's Criterion with $p = 2$. Now I claim that in the field $\Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2})$ where $\omega = e^{2\pi i/3}$ our polynomial $x^3 - 2$ factors into linears. But this is clear from the fact that $x^3 - 2 = (x - \sqrt[3]{2})(x - \omega\sqrt[3]{2})(x - \omega^2 \sqrt[3]{2}).$
Hence we can make an ansatz that in general if a field extension $E/F$ contains all the roots of some polynomial $f(x) \in F[x]$, then we should be able to factorise $f(x)$ over $E$ into linears. Now several people have given you the definition of a splitting field of a polynomial. Let us see why $\Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2})$ is a splitting field for $x^3 - 2$. On one hand by minimality of the splitting field $K$, we must have that $K \subset \Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2})$.
On the other,because each term that we adjoined to $\Bbb{Q}$ to get our extension is a root of $x^3 - 2$, they are all in the splitting field and hence $\Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}) \subset K$. It follows that
$\Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}) = K.$
Now if we view everything as sitting inside the big algebraically closed field $\Bbb{C}$, one can speak of the splitting field. Otherwise, if you are over a finite field, it is not so easy to see what big algebraically closed field all the finite fields sit inside. In fact if we wanted to even speak of the algebraic closure of a finite field, we need to set up some directed system and take a direct limit. So in this case, by the splitting field we have to mean the one upto isomorphism.
Let's turn our attention to what happens over finite fields. By Lagrange's theorem, the polynomial $x^{p^n}- x$ splits completely over the finite field $\Bbb{F}_{p^n}$. Since any two splitting fields are isomorphic, we conclude that any two finite fields with the same number of elements are isomorphic. For example, weird as it may seem consider the polynomials
$x^4 + x^3 + 1, \hspace{2mm} x^4 + x + 1$
over $\Bbb{F}_2$ that are irreducible. Then we have a very interesting phenomena: In $\Bbb{F}_2[x]/(x^4 + x^3 + 1)$, the polynomial $x^4 + x^3 +1$ splits completely! In fact this always happens: If you have an extension $\Bbb{F}_{p^n}(a)$ obtained by adjoining some root $a$ of $f(x) \in \Bbb{F}_{p^n}[x]$, then suppose that $\deg f = n$. Then
$ \Bbb{F}_{p^n}(a) \cong \Bbb{F}_{p^{n^2}}.$
But the guy on the right is a separable extension and by what I said above is also a normal extension so is Galois. Hence $\Bbb{F}_{p^n}(a)$ is a Galois extension of $\Bbb{F}_{p^n}$, so in particular is normal so that $f(x)$ splits completely in here! Similary we see that $x^4 + x + 1$ splits completely in $\Bbb{F}_2[x]/(x^4 + x + 1)$ so that since
$\Bbb{F}_2[x]/(x^4 + x + 1) \cong \Bbb{F}_2[x]/(x^4 + x^3 + 1),$
both of these are splitting fields for $x^4 + x + 1$ and $x^4 + x^3 + 1$!