How to choose keywords for papers?

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.

Any with _DHE_ in them. Those are the ones that use Ephemeral Diffie-Hellman, _ECDHE_ is safe.

NOTE: I am doing $x^4 + 12 x^3 + 14 x^2 - 12 x + 1.$

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10},$ after $$(x+1)^4 \pmod 2, \; \; (x^2 + 1)^2 \pmod 3, \; \; (x+3)^4 \pmod 5. $$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$