<s>[INST] Factorise $f(x)=x^5-x$ into a product of irreducibles in $\mathbb Z_{5}[x]$ So plugging in $1$ gives $f(1) = 0$ which means $1$ is a root and $f$ has a factor $(x-1)$ which is $\equiv (x+4)$ in $\mathbb Z_{5}[x]$ ?
I then divide $f(x)$ by $(x+4)$ using polynomial long division and I get 
$x^4-4x^3+16x^2-64x+255$ with remainder $-1020$ 
And that's looking kinda strange.. not sure what I'm doing at this point
I notice $f(0)$ also $= 0$ so maybe if I divide $f(x)$ by $x$ instead I would get $(x^4-1)$ but then I'm still stuck and don't know what else to do 
 [/INST] $f(x) = x (x-1)(x^3+x^2 + x+1)=x(x-1)(x+1)(x^2+1)$
$(x+a)(x+b)=x^2 + (a+b)x + ab = x^2 +1$ in ${\bf Z}_5[x]$
So $a+b =0(5)$ and $ab=1(5)$ Hence $a=2$ and $b=3$
$ f(x) =x(x-1)(x+1)(x+2)(x+3)$
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