Generator of the multiplicative groups of units in $\mathbb{Z_5[x]}/(x^{2}+3x+3)$ I would like to find a generator of the multiplicative groups of units in $\mathbb{Z_5[x]}/(x^{2}+3x+3)$. This is a field since $x^{2}+3x+3$ is irreducible, so every coset with $bx+a\not=0$ as a representative should be a unit... 
I do not understand how to go from here though... Since $b\not=0$ we have $4$ diffrent choices for $b$ and $5$ different options for coefficient $a$ hence $24$ elements in the multiplicative group of units. 
Should any $bx+a$ with order $24$ be a generator? How do I go from here? 

It suffices to rule out the maximal factors of $24$, i.e. $8=24/3$ and $12=24/2$ as possible orders. The (coset of) $x$ fits the bill, because 
$$
x^4=(-x^2)^2\equiv (3x+3)^2=4(x^2+2x+1)=-x^2+3x+4\equiv6x+7=x+2,
$$
and therefore neither
$$
x^8=(x^4)^2\equiv(x+2)^2=x^2+4x+4\equiv x+1
$$
not
$$
x^{12}=x^8\cdot x^4\equiv(x+2)(x+1)=x^2+3x+2\equiv-1
$$
are equal to $1$.
The point is that other possible orders ($2,3,4,6$) that are less than $24$ are factors of either $8$or $12$.
Now that we settled that $x$ is a generator, the other generators can be found by reducing the powers $x^j$, $0<j<24, \gcd(j,24)=1$ modulo the generator $x^2+3x+3$.
