$f(x)=3+2x+\alpha^2x^3+\beta x^4\in \mathbb{Z}_5[x]$ Find $\alpha,\beta,\lambda^{625}$ I have the following question :
$K=\frac{\mathbb{Z}_5[x]}{(f)}$ is a field with 125 elements.
$\lambda=x+(f)\in K$
$f(x)=3+2x+\alpha^2x^3+\beta x^4\in \mathbb{Z}_5[x]$ Find $\alpha,\beta,$ and $\lambda^{625}$ express $\lambda^{625}\in K$ in the following form $a_0+a_1 \lambda+a_2\lambda ^2$ while $a_0,a_1,a_2\in \mathbb{Z}_5$
I did manage to find $\beta$ which is $0$ since there's a theorem that say $q^n=125$ (125-number of elements) so the max power should be $3$, Yet I don't know how to find $\alpha$ and $\lambda^{625}$ 
Any help will be appreciated.

We have $\frac{\mathbb Z_5[x]}{(f)}\cong \mathbb Z_5(a)$ then number of the element in $\mathbb Z_5(a)=125=5^3$ then $[\mathbb Z_5(a):\mathbb Z_5]=3$ then $f(x)$ is a polynomial of degree $3$ and donot have  roots in $\mathbb Z_5$, then $\beta=0$.
we have $\alpha\in \mathbb Z_5 $,then $\alpha^2=0,1,4 $
if $\alpha^2=0$ then $f(x)=3+2x$,then $deg(f)<3$ contradiction
if $\alpha^2=1$ then $f(x)=3+2x+x^3 $, then $2$ root of $f(x)$ contradiction
if $\alpha^2=4$ then $f(x)=3+2x+4x^3$ and donnot have a root in $\mathbb Z_5$
Let $a$ root of $f(x)$ in $ \mathbb Z_5(a)$ then $\lambda=x+(f)$ root in $\mathbb Z_5[x]/(f)$, for calculate $\lambda^{625}$ we must to calculate $a^{625}$
we have $4a^3+2a+3=0\Rightarrow a^3=2a+3\Rightarrow a^4=2a^2+3a \Rightarrow a^5=3a^2+4a+1\Rightarrow a^6=4a^2+2a+4 $, so $a^{10}=3a+3\Rightarrow a^{20}=4a^2+3a+4\Rightarrow a^{25}=2a^2+4\Rightarrow a^{50}=4a^2+2a+1 $, so $  a^{50}=a^6+2 \Rightarrow a^{100}=(a^6+2)^2=a^{12}+4a^6+4=4a^2+4a+4\Rightarrow a^{100}=a^6+2a $,so $a^{200}=(a^6+a)^2=4a+2 \Rightarrow a^{400}=a^2+a+4\Rightarrow a^{600}=a^2+a$.
Therefor $a^{625}=a^{600}a^{25}=3a^2+4a+1=a^5$
