Let $h(x)=x^4+12x^3+14x^2-12x+1$, and let $p>5$ be a prime.


I want to show $h(x)$ factors into 2 quadratics mod $p$ if $p \equiv 9,11$ mod 20, while
$h(x)$ factors mod $p$ into 4 linear factors if $p \equiv 1,19$ mod 20. 
I can show $h(x)$ is irreducible if $p \equiv 3,7$ mod 10.



Factoring a certain quartic mod primes
It's been noted already that in fact $h \bmod p$ has 
four linear factors iff $p \equiv \pm 1 \bmod 30$,
and is a product of two quadratics iff $p \equiv \pm 11 \bmod 30$.
This can be checked by identifying the splitting field of $h$
with the real subfield of the $15$-th cyclotomic field,
generated by $c := e^{2\pi i/15} + e^{-2\pi i/15} = 2 \cos (2\pi/15)$
which is a root of $c^4 - c^3 - 4c^2 + 4c + 1 = 0$; indeed
$1 + 2(c-c^2)$ is a root of $h$. The desired result soon follows from
the fact that Frobenius takes $e^{2\pi i/15}$ to $e^{2p\pi i/15}$.
This is consistent with a cyclic Galois group (**not** the Klein 4-group 
as some have claimed), since
${\rm Gal}({\bf Q}(c)/{\bf Q})$ is the cyclic group
$({\bf Z} / 15 {\bf Z})^\* / \lbrace \pm1 \rbrace$.


If for some reason you do need a quartic of this form
$x^4 + ax^3 + bx^2 - ax + 1$ (i.e. with a symmetry $x \leftrightarrow -1/x$)
that splits completely mod $p$ iff $p \equiv \pm 1 \bmod 20$,
the first few possibilities are $(a,b) = (\pm 2,-6)$, $\pm(22, -6)$,
and $(\pm 18,74)$ if I computed correctly in **gp**.


