$f(x)$ is irreducible over finite field $F_p$ , does $f(x)$ split over $F_p / \langle f(x) \rangle$? I want to know whether it is possible to generalize the next thing exercise.
$x^4 + x + 1$ is irreducible over $F_2[x]$ , so $F_2[x]/\langle x^4 +x+1\rangle = span_{F_2} \{1,\bar{x} , \bar{x}^2,\bar{x}^3\} \cong F_{2^4}$. So $\bar{x}^4 + \bar{x} +1 = 0$ in $F_2[x]/\langle x^4+x+1\rangle$ and due to the fact that in $F_2$ $(a+b)^2 = a^2 + b^2$ ,we get that $\bar{x}^8 + \bar{x}^2 +1 = 0=(\bar{x}^4 + \bar{x} +1)^2 = 0$ so $\bar{x}^2$ is also a root. And similarly $\bar{x}^4 , \bar{x}^8$ are roots. A simple check gives that those are 4 different roots, thus indeed the polynomial splits. 
So my question,
is there a way to use the fact that $F_{p^n}$ is the splliting field of $x^{p^n}-x$ over $F_p$ and the fact that for every $f(x)$ irreducible polynomial over $F_p[x]$ from degree $n$ , $F_p[x]/ \langle p(x) \rangle \cong F_{p^n}$ to show that $f(x)$  splits over $F_p[x]/ \langle f(x) \rangle$ ?
 A: Yes, your conjecture is true and can be proven without Galois theory (@commenters):

Theorem 1. Let $K$ be a finite field. Let $f \in K\left[x\right]$ be an irreducible polynomial. Let $L$ be the field $K\left[t\right] / \left(f\left(t\right)\right)$. Then, $f$ factors into linear polynomials over the field $L$.

Note that I have used two different indeterminates $x$ and $t$ here in order to avoid the awkwardness of having a polynomial in $x$ over a ring that already sort-of contains an $x$ in it. This is either necessary or redundant depending on the pedantic minutiae of your definition of polynomial rings; in either case, it does not hurt.
Proof of Theorem 1. Let $q = \left|K\right|$. It is well-known that for any commutative $K$-algebra $A$, the map $A \to A,\ a \mapsto a^q$ is a $K$-algebra endomorphism of $A$ (since any $a, b \in A$ satisfy $\left(a+b\right)^q = a^q + b^q$ and $\left(ab\right)^q = a^q b^q$, and since each $\lambda \in K$ satisfies $\lambda^q = \lambda$). This endomorphism will be denoted by $F_A$, and is called the Frobenius endomorphism of $A$.
Let $n$ be the degree of $f$. For each polynomial $g \in K\left[t\right]$, let $\overline{g}$ be the projection of $g \in K\left[t\right]$ onto the quotient ring $K\left[t\right] / \left(f\left(t\right)\right) = L$. Let $u = \overline{t} \in L$. Let $F$ be the map $F_L$. Thus, $F$ sends each $a \in L$ to $a^q$ (by the definition of $F_L$). Hence, for each nonnegative integer $i$ and each $a \in L$, we have
\begin{align}
F^i\left(a\right) = a^{q^i}
\label{darij1.pf.t1.Fi}
\tag{1}
\end{align}
(indeed, this can be proven by induction on $i$, using the previous sentence in the induction step).
We shall show that $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ are $n$ distinct roots of $f$ in $L$. This will quickly yield that $f$ factors into linear polynomials over $L$. Let us proceed in three steps:

Claim 1: The $n$ elements $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ of $L$ are distinct.

Proof of Claim 1. Assume the contrary. Thus, there exist two elements $i$ and $j$ of $\left\{0,1,\ldots,n-1\right\}$ such that $i<j$ and $F^i\left(u\right) = F^j\left(u\right)$. Consider these $i$ and $j$. We have $i < j$, thus $q^i < q^j$ (since $q = \left|K\right| > 1$).
Let $a \in L$ be arbitrary. We are going to prove that $a^{q^j} - a^{q^i} = 0$.
Indeed, we have $a \in L = K\left[t\right] / \left(f\left(t\right)\right)$; in other words, $a = \overline{s}$ for some polynomial $s \in K\left[t\right]$. Consider this $s$. But $s = s\left(t\right)$, so that $\overline{s} = \overline{s\left(t\right)} = s\left(\overline{t}\right)$ (since $s$ is a polynomial with coefficients in $K$). In view of $u = \overline{t}$, this rewrites as $\overline{s} = s\left(u\right)$. Hence, $a = \overline{s} = s\left(u\right)$. But $F^i$ is a $K$-algebra homomorphism (since $F$ is a $K$-algebra homomorphism), and thus commutes with the application of any polynomial over $K$. Hence, in particular, $F^i$ commutes with $s$. Thus, $F^i\left(s\left(u\right)\right) = s\left(F^i\left(u\right)\right)$. Now, applying $F^i$ to both sides of the equality $a = s\left(u\right)$, we find
\begin{align}
F^i\left(a\right) = F^i\left(s\left(u\right)\right) = s\left(F^i\left(u\right)\right) .
\end{align}
Comparing this with \eqref{darij1.pf.t1.Fi}, we obtain $a^{q^i} = s\left(F^i\left(u\right)\right)$.
The same argument (applied to $j$ instead of $i$) yields $a^{q^j} = s\left(F^j\left(u\right)\right)$. Hence,
\begin{align}
a^{q^i} = s\left(\underbrace{F^i\left(u\right)}_{= F^j\left(u\right)}\right)
= s\left(F^j\left(u\right)\right) = a^{q^j} .
\end{align}
In other words, $a^{q^j} - a^{q^i} = 0$.
Now, forget that we fixed $a$. We thus have shown that $a^{q^j} - a^{q^i} = 0$ for each $a \in L$. In other words, each $a \in L$ is a root of the polynomial $x^{q^j} - x^{q^i} \in K\left[x\right]$. This polynomial is nonzero (since $q^i < q^j$, so that the terms $x^{q^j}$ and $x^{q^i}$ do not cancel) and has degree $q^j$ (since $q^i < q^j$), and thus has at most $q^j$ many roots in $L$ (since any nonzero polynomial over $L$ of degree $d$ has at most $d$ many roots in $L$). Since each $a \in L$ is a root of this polynomial, we thus conclude that there are at most $q^j$ many $a \in L$. In other words, $\left|L\right| \leq q^j$.
But $L = K\left[t\right] / \left(f\left(t\right)\right)$ is a $K$-vector space of dimension $n$ (since the polynomial $f$ has degree $n$), and thus has size $\left|K\right|^n = q^n$ (since $\left|K\right| = q$). Hence, $\left|L\right| = q^n$, so that $q^n = \left|L\right| \leq q^j$ and therefore $n \leq j$. This contradicts $j \in \left\{0,1,\ldots,n-1\right\}$. This contradiction shows that our assumption was wrong. Hence, Claim 1 is proven.

Claim 2: The $n$ elements $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ of $L$ are roots of $f$.

Proof of Claim 2. From $u = \overline{t}$, we obtain $f\left(u\right) = f\left(\overline{t}\right) = \overline{f\left(t\right)}$ (since $f$ is a polynomial with coefficients in $K$). Thus, $f\left(u\right) = \overline{f\left(t\right)} = 0$ (since $f\left(t\right) \equiv 0 \mod f\left(t\right)$).
Now, let $i \in \left\{0,1,\ldots,n-1\right\}$ be arbitrary. Then, $F^i$ is a $K$-algebra homomorphism (since $F$ is a $K$-algebra homomorphism), and thus commutes with the application of any polynomial over $K$. Hence, in particular, $F^i$ commutes with $f$. Thus, $F^i\left(f\left(u\right)\right) = f\left(F^i\left(u\right)\right)$. Hence, $f\left(F^i\left(u\right)\right) = F^i\left(\underbrace{f\left(u\right)}_{=0}\right) = F^i\left(0\right) = 0$ (again since $F^i$ is a $K$-algebra homomorphism). In other words, $F^i\left(u\right)$ is a root of $f$.
Now, forget that we fixed $i$. We thus have shown that $F^i\left(u\right)$ is a root of $f$ for each $i \in \left\{0,1,\ldots,n-1\right\}$. In other words, the $n$ elements $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ of $L$ are roots of $f$. This proves Claim 2.
We are now almost done proving Theorem 1. Claim 2 shows that the $n$ elements $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ of $L$ are roots of $f$. Since these $n$ elements are distinct (by Claim 1), we thus conclude that the polynomial $f$ has (at least) $n$ distinct roots over $L$. But this polynomial $f$ has degree $n$; hence, if it has $n$ distinct roots over $L$, then it factors into linear polynomials over $L$. So we have proven that $f$ factors into linear polynomials over $L$. This proves Theorem 1. $\blacksquare$
