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.

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.
