Factoring polynomials of degree $a p^b$ over extension fields. Let $f(x)$ be an irreducible polynomial with integer coefficients, which is irreducible over $\mathbb{Z}$ and has degree $a p^b > p$ with $p,a,b>0$ and $p$ a prime.
It appears that $f(x)$ factors in at least $b$ ways over extensions of degree $p$ that do not belong to the zeros of $f(x)$ or their extensions.
For clarity we count 1 way per extension , even if there are more for that extension.
How to prove or disprove this ?
 A: I think that the question is asking about the following. I'm prepared for this to be wrong, but I want to open up the discussion with this "guess".
Let $L$ be the splitting field of $F$. Then we can ask, whether there exist intermediate fields $K$, $\mathbb{Q}\subseteq K\subseteq L$, such that A) $f(x)$ factors in a non-trivial way  in $K[X]$, B) $K$ does not contain any of the zeros of $f(x)$, and C) the degree condition $[K:\mathbb{Q}]=p$ is satisfied.
The OP then asks, whether there are always at least the prescribed number of such intermediate fields $K$.
I lead off by proffering a counterexample. Let 
$$
f(x)=x^4+x^3+x^2+x+1=(x-\zeta)(x-\zeta^2)(x-\overline{\zeta})(x-\overline{\zeta}^2),
$$
where $\zeta=e^{2\pi i/5}$.Then the splitting field of $f(x)$ is the fifth cyclotomic field
$L=\mathbb{Q}(\zeta)$. We have $\deg f(x)=4=2^2$ as well as $[L:\mathbb{Q}]=4$, so $p=2$, $b=2$, and $a=1$.
But in this case there exists only one intermediate field $K=\mathbb{Q}(\sqrt5).$ This follows from Galois theory. The Galois group $Gal(L/\mathbb{Q})$ is cyclic of order four, so it has a unique subgroup of index two, and the Galois correspondence associates the subfield $K$ with that subgroup.
The corresponding factorization is
$$
f(x)=f_1(x)f_2(x),
$$
where
$$
f_1(x)=(x-\zeta)(x-\overline{\zeta})=x^2-2(\cos\frac{2\pi}5)\,x+1
$$
and
$$
f_2(x)=(x-\zeta^2)(x-\overline{\zeta^2})=x^2-2(\cos\frac{4\pi}5)\,x+1.
$$
Here the coefficients of the linear term are in the subfield $K$ as
$$
\cos\frac{2\pi}5=\frac{\sqrt5 -1}4\qquad\text{and}\qquad\cos\frac{4\pi}5=-\frac{\sqrt5 +1}4
$$
are clearly elements of $K$. Galois theory tells us that the coefficients of other potential factors such as $(x-\zeta)(x-\zeta^2)$ necessarily generate all of $L$.
Cyclotomic polynomials of order $17$, $257$ or $65537$ yield similar polynomials of respective degrees $16$, $256$ and $65536$ that only have a single such field $K$.

Galois theory will say a lot about the problem. An intermediate field $K$ will satisfy condition C) iff the subgroup $H=Gal(L/K)$ is of index $p$ in $G=Gal(L/\mathbb{Q})$. It will satisfy condition B) iff the subgroup $H$ has no fixed points among the roots of $f(x)$. It will satisfy condition A) iff the action of $H$ on the roots of $f(x)$ is not transitive. I dare not suggest the most general case, when all these would hold for a large number of subgroups $H$. When $G$ is elementary $p$-abelian, it seems to be easy to produce examples of several such subgroups $H$. In a comment the OP himself exhibited the
example $f(x)=x^4+1$. In this case $G\simeq C_2\times C_2$ is elementary $2$-abelian, and the splitting field $L=\mathbb{Q}(i,\sqrt2)$ has three subfields like $K$ corresponding to the cyclic subgroups generated by the permutations of the roots $\sigma_1=(12)(34)$,
$\sigma_2=(13)(24)$ and $\sigma_3=(14)(23)$ respectively. The corresponding subfields are (in some order) $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt{-2})$.
