Is $K[x]/\langle f(x) \rangle $ always the splitting field of an irreducible polynomial $f(x)$ over a field $K$ In general for $f(x)$ irreducible in $K[x]$ (for $K$ a field) we look for a field extension of $K$ having a root of $f(x)$ as $K[x]/\langle f(x) \rangle $. 
In the case of  $x^3-x+1$ over $\mathbb{F}_3$, I have found $\mathbb{F}_3[x]/(x^3-x+1)$ to be the splitting field of $x^3-x+1$.
I do not think this would be the case for all such polynomials and all such fields.
Could someone help me to know what is special in this case??
 A: Your observation is true, if $K$ is a finite field.. It is also true, if $K=\mathbb{R}$, because the only algebraic extension of that field is algebraically closed. If $\deg f(x)=2$, then it also holds because the sum of the two roots is in $K$, so if you join one, you automatically also join the other.
Over other fields it may or may not hold. For example over $\mathbb{Q}$ the claim does not hold, when $f(x)=x^3-2$, because exactly one of the roots of that polynomial is real, so joining that real root won't give you the rest. On the other hand, if $f(x)=x^3+x^2-2x-1$, then $\mathbb{Q}[x]/(f(x))$ is the splitting field of $f(x)$, because the zeros of $f(x)$ are $2\cos 2\pi/7$, $2\cos4\pi/7$ and $2\cos8\pi/7$. If you join one of those roots, you will also join the others because 
$$
\cos2\alpha=2\cos^2\alpha-1.
$$
I think that, in a sense that I cannot make precise, it is rare for a field $K$ to have this property for all irreducible polynomials $f(x)$.
A: you can show this fact for finite fields because all extensions are Galois everywhere! Namely, the splitting field of an irreducible $p$ of degree $n$ will be contained in $F_{p^n}$ since $x^{p^n}-x$ contains all irreds of degree $n$. So the splitting field of $p$ will be a subfield of $F_{p^n}$ which corresponds to a subgroup of $Z/nZ$ which is normal, hence the extension is Galois. Since in a Galois extension having one root means you have all roots, your field is splitting.
