Let $f = 15x^4+22x^3-x=0$ a polynomial in $\mathbb Z_p[x]$, find the first prime $p$ value that will make $f$ result in being grade 3 and monic. Then factorize $f$ in $\mathbb Z_3[x]$ as product of irreducible factors.
Find the $p$ value
In order for $f$ to be monic and grade 3, I need to find a $p$ value so that:
$\begin{aligned} 15 \equiv_p 0 \end{aligned}$ $\begin{aligned} 22 \equiv_p 1 \end{aligned}$
it's easy to see that $p = 3$, so I will be working in $\mathbb Z_3[x]$ and $f = x^3+2x$.
Factorizing $f$
I need to find all the solutions to $x^3+2x=0$ in order to get it factorized as product of irreducible factors.
$\begin{aligned} x^3+2x = x(x^2+2) = 0 \Leftrightarrow x = 0 \vee (x^2+2) = 0 \end{aligned}$
Let's examine $\Delta = b^2-4ac$ of $(x^2+2)$
$\begin{aligned} \Delta = 0 - 8 = -8 \equiv_3 = 1 \Rightarrow \end{aligned}$ $\begin{aligned} \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{0 \pm \sqrt{1}}{2} \Rightarrow x_1 = +\frac{1}{2} \equiv_3 2 \text{, } x_2=-\frac{1}{2} \equiv_3 1 \end{aligned}$
So $f = x^3+2x= x(x+2)(x+1)$.
As my algebra exam day is fast approaching I really am in need of knowing if I've got this right or not. Sorry if I incuded trivial calculations in it, but I have to be extra sure I have understood everything. I have a question, though. Is it wrong to assume $x^3 = 1$ as we are dealing with $\mathbb Z_3$ elements? Doesn't the modulo apply to the exponents as well?