Show that $x^3 + 3x+2$ is irreducible in $\mathbb{Z}[5]$ I am reading Fraleigh in which Fraleigh proves that $f(x) = x^3 + 3x+ 2$ is irreducible over $\mathbb{Z}[5]$. He explains that the if $f(x) = x^3 + 3x+ 2$ were reducible over $\mathbb{Z}[5]$, then there would exist ar least one linear factor of $f(x)$ of the form $x-a$ for some $a \in \mathbb{Z}[5]$. Then, $f(a)$ would be $0$. However, $f(0) = 2, f(1) = 1, f(2) = 1, f(-1) = -2,f(-2) = -2$. 
I don't understand why Fraleigh didn't analyze $f(4)$ and $f(3)$ instead of $f(-1)$ and $f(-2)$ respectively. Can someone please explain? Thanks!
 A: The set $\{0,1,2,-1,-2\}$ forms a complete residue system modulo $5$, just like $\{0,1,2,3,4\}$. This means all possibilities for the linear factor of $f$ are still covered with the former set as they are with the latter set.
It is just a matter of convention to adopt one system or the other. The more appropriate choice may also depend on context – calculations related to quadratic reciprocity, for example, heavily rely on manipulating negative numbers (e.g. is $-1$ a square modulo $n$ or not).
A: Just check the polynomial's value $\pmod 5$ at each of $\{0,1,2,3,4\}=\Bbb Z_5$.
Oh, and in answer to your question, we have $-1\equiv 4\pmod5$ and $ -2\equiv 3\pmod5$.
A: When multiplying modulo $n$ for relatively small $n$ if you choose to use negative numbers instead of higher positive numbers you keep the numbers small at the expense of having to keep track of the sign. This is often easier when using brute force to check for irreducibly of a polynomial and other such computations. For example it can make proofs of quadratic reciprocity easier to follow. 
