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I've been struggling to prove the general formula for how the polynomial in the title factors mod $p$, for some arbitrary prime $p$. This is how it must factors, although I should mention this "guess" wasn't formulated by factoring for some primes and then assuming any pattern notice held;

$x^4-2x^3-2x+1=Q_{1}\cdot L_{1} \cdot L_{2}$ if $p \equiv 11 \pmod{12}$

$x^4-2x^3-2x+1=Q_{1}\cdot Q_{2}$ if $p \equiv 7 \pmod{12}$

$x^4-2x^3-2x+1=Qu_{1}$ if $p \equiv 5 \pmod{12}$

$x^4-2x^3-2x+1=L_{1} \cdot L_{2} \cdot L_{3} \cdot L_{4}$ if $p = a^2+36 b^2$

$x^4-2x^3-2x+1=Q_{1}\cdot Q_{2}$ if $p=4a^2+9b^2$

Where $Q, L, Qu$ are irreducible in $\mathbb{F}_{p}$, and $Q_{i}, L_{j}, Qu_{k}$ are quadratic, linear, and quartic polynomials respectively. I'm working on various other polynomials too of varying degree as well, all with solvable Galois Groups. Any help as to how to factor them like this formula would be much appreciated.

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    Have you computed the complex roots of such a polynomial? If you perform that, it is clear that such a polynomial splits in $\mathbb{F}_p$ according to $3$ being (or not) a quartic residue and $2$ being (or not) a quadratic residue. – 2017-01-27
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    Im relatively aware and comfortable with solving for the roots. As a matter of fact, it is much easier to solve for the quartic polynomial whose roots are the square of the roots of $x^4-2x^3-2x+1$. I have learned quadratic residues but nothing about quartic. – 2017-01-27

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The polynomial is reciprocal, so there is an easy way of solving the corresponding equation: dividing through by $x^2$ gives $$ x^2 - 2x - \frac{2}{x} + \frac1{x^2} = \Big(x + \frac1x\Big)^2 - 2 \Big(x + \frac1x\Big) - 2 = 0,$$ hence $$ x + \frac1x = 1 \pm \sqrt{3}. $$ Multiplying through by $x$ gives the quadratic $$ x^2 - \omega x + 1 = 0, $$ where $\omega = 1 \pm \sqrt{3}$. This shows that the field generated by the roots of the polynomial is ${\mathbb Q}(\sqrt{2\sqrt{3}}) = {\mathbb Q}(\sqrt[4]{12})$.

The remaining calculations involve quadratic and quartic reciprocity for describing the splitting of primes in pure quartic extensions.

As an example, look at the normal closure of ${\mathbb Q}(\sqrt[4]{12})$, which is $L = {\mathbb Q}(i, \sqrt[4]{-3})$. Its maximal abelian subfield $F$ is the field of $12$-th roots of unity, hence exactly the primes $p \equiv 1 \bmod 12$ split in $F$. Among these primes, the ones splitting completely in $L$ are exactly those for which $-3$ is a quartic residue modulo $p$. By classical criteria due to Gauss this happens if and only if $p = a^2 + 4b^2$ for $b$ divisible by $3$.

A direct approach for describing the splitting of primes in dihedral extensions using binary quadratic forms requires the concept of ring class fields. You might want to look into Cox's book on primes of the form $x^2 + ny^2$, where this is explained in detail.

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    Great work on calculating the roots, much appreciated. I don't understand reciprocity and am only familiar with quadratic. I am still quite confused as to how to compute the the factors in $\mathbb{F}_{p}$. – 2017-01-28
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    The splitting field is not $\mathbb{Q}(\sqrt[4]{12})$; it is $\mathbb{Q}(\sqrt[4]{12}, i)$, since the splitting field has all four roots $\pm\sqrt{\pm 2 \sqrt{3}}$. $\mathbb{Q}(\sqrt[4]{12})$ is merely (isomorphic to) the field you get by adjoining a single root. – 2017-01-28