The following exercise is from Golan's book on linear algebra.
Problem: Is the polynomial $6x^4+3x^3+6x^2+2x+5$ irreducible over $GF(7)$, the field with 7 elements.
Work so far: If the polynomial is reducible, it facts into the product of a linear term and something else, or it factors as two quadratics. The first case is easy to exclude; simply plug all seven elements of $\mathbb{Z}_7$ into the polynomial and confirm none of them are a root. The second is harder. Of course, one could just set up the systems of equations resulting from
$$(ax^2+bx+c)(dx^2+ex+f)$$
and go through all the possible values of $a,c,d,f$ and see if the resulting values of $b$ and $e$ are permissible, and while I know that would eventually give me the answer, I have no desire to do all of those computations. Is there a slicker way?
Hint $ $ As suggested you could use trial-and-error, and program a computer to test the $7^4 = 2401$ cases that arise from $4$ undetermined coefficients in a factorization into two quadratics. But, as is often true, a little insight trumps brute force. By exploiting innate symmetry, we can reduce the $2401$ cases to $2$ cases. First, shifting $\rm\:x\to x\!-\!1\:$ to kill the $\rm\:x^3\:$ term yields
$$\rm\begin{eqnarray}\rm -f(x\!-\!1) &\equiv&\rm\ \ \ x^4\ +\ 2\ x^2\ -\ 3\ x\ +\ 2\pmod 7 \\[.2em]&\equiv&\rm\ (x^2\!- a\, x + b)\ (x^2\! + a\, x + c)\\[.2em] &\equiv&\rm\ \ x^4\! + (b\!+\!c\!-\!a^2)\!\: x^2\! + a(b\!-\!c)\:\!x + bc\end{eqnarray}\qquad $$
Up to $\rm\, b,c\, $ swaps, $\rm\: bc\equiv 2\!\iff\! (b,c)\, \equiv\, \pm(2,1),\, \pm\:\!(3,3).\:$ $\rm\:b\not\equiv c\:$ else coef of $\rm\,x\,$ is $\,0\not\equiv -3$.
If $\rm\ (b,c) \equiv\ \ \: (\ 2,\ 1\ )\ $ then $\rm\:-3 \equiv a(b\!-\!c)\equiv\ \: a\:\ $ so $\rm\:b\!+\!c\!-\!a^2\equiv\ \ \ \, 2\!+\!1\!-\!(-3)^2\equiv\ \ 1\:\not\equiv 2$
If $\rm\ (b,c) \equiv (-2,\!-1)\:$ then $\rm\:-3 \equiv a(b\!-\!c)\equiv -a\:$ so $\rm\:b\!+\!c\!-\!a^2\equiv\, -2\!-\!1\!-\!(+3)^2\equiv -5 \equiv 2$
So $\rm\:a,b,c \equiv 3,-2,-1,\:$ is a solution, which yields the factorization $$\begin{align}\rm -f(x\!-\!1)\, &\equiv\, x^4 + 2\,x^2 - 3\,x+2\\[.2em] &\equiv\,\rm (x^2-3\,x-2)(x^2+3\,x-1)\pmod 7\\[.2em] \underset{x\ \to\ x+1}\Longrightarrow \rm-f(x) \,&\equiv\,\rm (x^2+6\,x+3)(x^2+5\,x+3)\pmod 7\end{align}\qquad$$
Remark $ $ Alternatively, we can apply the Euclidean algorithm to compute $\rm\,gcd(f(x\!+\!c),x^{24}\!-\!1)\,$ for random $\rm\,c,\,$ which, $\,$ for $\rm\,c=1\,$ quickly yields $\rm\,x^2\!+\!2\,|\,f(x\!+\!1),\,$ hence $\rm\,f(x)\,$ has the factor $\rm\,(x\!-\!1)^2\!+\!2\, =\, x^2-2\,x+3.\,$ This is how some factoring algorithms work.