Factor $x^8 + x^2 + 13$ into irreducible polynomials in $\Bbb F_{23} [x]$. I want to factor  $x^8+ x^2 + 13$ into irreducible polynomials in $\Bbb F_{23} [x]$. I am trying using the method given in this link but not able to find its factors. Any help would be appreciated.
Thanks.
Factor $X^4 + 3$ into irreducible factors in $F_7[X]$
 A: I'd start by viewing this as $y^4+y+13$, where $y=x^2$. Does that factor? Search for linear factors by testing numbers as roots. It takes a little patience (or a calculator) but $y=8$ is a root. So you have $$(y-8)\left(y^3+8y^2-5y+7\right)$$
Does this factor further? Only if the cubic has a root. Checking for roots, it does not. So you have:
$$(x^2-8)\left(x^6+8x^4-5x^2+7\right)$$
Does $x^2-8$ factor? Again checking for linear factors, we find a pair.
$$(x-10)(x-13)\left(x^6+8x^4-5x^2+7\right)$$
Could that sixth degree polynomial factor further?

*

*There are no linear factors. (If it had a root, then the cubic from before would have had a root.)

*It could factor into a quadratic and an irreducible quartic. But for the original polynomial $f$, $f(x)=f(-x)$. So the map $x\to -x$ must permute the factors (while negating any odd-degree factor). In this case, the trivial permutation. So that quadratic would have to have no linear term, and the quartic would have no linear term nor a cubic term. This would make our cubic in $y$ from earlier factorable, which we know it is not.

*It could factor into three quadratics. Again, the map $x\to-x$ might preserve all three quadratics, but that means none of them have linear terms and our cubic in $y$ would have been factorable. The other possibility is that it only preserves one quadratic, and we have $$\begin{align}x^6+8x^4-5x^2+7&=(x^2+a)(x^2+bx+c)(x^2-bx+c)\\&=x^6+(2c+a)x^4+(c^2+2ac)x^2+ac^2\end{align}$$ from which we deduce $a=8-2c$, so $$\begin{align}x^6+8x^4-5x^2+7&=x^6+8x^4+(c^2+2(8-2c)c)x^2+(8-2c)c^2\\&=x^6+8x^4+(16-3c^2)x^2+(8-2c)c^2\end{align}$$ Can $16-3c^2$ equal $-5$? Then $3c^2=21$, and $c^2=7$. But $7$ is not a square mod $23$, which can be checked by inspection.

*It could factor into two cubics. Again , $x\to -x$ permutes the factors (and since both are odd-degree, negates them at the same time).  It is impossible for the map $x\to-x$ to preserve the cubics while negating them, or each factor is of the form $x^3+ax$, and $0$ would be a root. It follows that they are negated and nontrivially permuted. So $$\begin{align}x^6+8x^4-5x^2+7&=(x^3+ax^2+bx+c)(x^3-ax^2+bx-c)\\&=x^6+(2b-a^2)x^4+(b^2-2ac)x^2-c^2\end{align}$$ Since $-7$ is a square, we can deduce wlog that $c=4$. So $$\begin{align}x^6+8x^4-5x^2+7&=x^6+(2b-a^2)x^4+(b^2-8a)x^2+7\end{align}$$ Now can we solve $2b-a^2\equiv8$ with $b^2-8a\equiv-5$? You can try each value of $a$ in the first relation to get a value of $b$, and try that in the second. You find $a\equiv1$, $b\equiv16$ works. So we have a factorization: $$\begin{align}x^6+8x^4-5x^2+7&=(x^3+x^2+16x+4)(x^3-x^2+16x-4)\end{align}$$
And then we conclude with: $$(x-10)(x-13)(x^3+x^2+16x+4)(x^3-x^2+16x-4)$$
A: Using a CAS, we get $f(x) = (x - 10)(x + 10)(x^3 - x^2 - 7x - 4)(x^3 + x^2 - 7x + 4).$
Looks pretty painful to figure this out by hand, in particular the cubics.
