Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible? Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible?
What is the best/simplest/elementary way to approach this? Any solutions or hints are greatly appreciated.
 A: No, it is not. If we set $p(x)=6x^4+3x^3+6x^2+2x+5$, over $\mathbb{F}_7$ we have:
$$ p(x+2) = -\left(x^4-2x^3-2x+1\right) $$
that is a palyndromic polynomial, from which:
$$\frac{p(x+2)}{x^2} = -\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right) = -\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}\right)+2$$
and:
$$ p(x+2) = -(x^2+2x+3)(x^2+3x-2),$$
since the previous line gives:
$$\begin{eqnarray*} p(x+2)=-(x^2+1)^2+2x(x^2+1)+2x^2&=&-(x^2+1-x)^2+3x^2\\&=&-(x^2+1-4x)^2+2(x+2)^2\end{eqnarray*} $$
and the RHS is now the difference of two squares.
A viable alternative is to notice, through Stickelberger criterion, that since the discriminant of $p$, $\Delta=-1728=-12^3$, is a quadratic residue $\pmod{7}$, $p$ splits as the product of an even number of irreducible polynomials, hence $p$ cannot be irreducible.
A: Because the modulus is so small, we can do this the hard way.
All arithmetic will be done in $\mathbb Z_7.$
$p(x)=6x^4+3x^3+6x^2+2x+5$
$-p(x) = x^4 + 4x^3 + x^2 + 5x + 2$
$-p(x-1) = x^4 + 2x^2 + 4x + 2$
Now we suppose that
\begin{align}
   -p(x-1)
      &= (x^2 + ax + b)(x^2 + cx + d)\\
      &= x^4 + (a+c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd
\end{align}
We end up with the equations
\begin{align}
   a+c &= 0\\
   ac + b + d &= 2\\
   ad+bc &= 4\\
   bd &= 2
\end{align}
Iterating through $b$, we finally get $(a,b,c,d) = (4,1,3,2)$.
So 
\begin{align}
   -p(x-1) &= (x^2 + 4x + 1)(x^2 + 3x + 2)\\
    p(x-1) &= -(x^2 + 4x + 1)(x^2 + 3x + 2)\\
      p(x) &= -(x^2 + 5x + 3)(x^2 + 6x + 3)\\
\end{align}
