Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible?

Question

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.

Answer 1

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.

Answer 2

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}