Over $\mathbb{Z}[x]$, both $2$ and $x^3-5$ are irreducible, so you can't factor further than $$
2x^3 - 10 = 2(x^3 - 5) \text{.}
$$
The reason you can't factor $x^3 - 5$ further is that if it had any more factors, one of them would need to have degree $1$, meaning $x^3 - 5$ would have a zero in $\mathbb{Z}$, which it doesn't.
The same argument works over $\mathbb{Q}[x]$, but you now need to use that $\sqrt[3]{5}$ isn't rational.
You are mixing up lots of matters in your question.
The polynomials that you are considering have their coefficients in
the finite field $\mathbb F_5$ or GF$(5)$ while their roots lie in
a larger field $\mathbb F_{5^n}$ where $n$ is the degree of the polynomials ($2$ in this instance).
A polynomial with coefficients in a field $\mathbb F$ is said to
irreducible over that field $\mathbb F$ if the polynomial cannot
be factored into two (or more) polynomials of smaller degrees with
coefficients in that field $\mathbb F$. The polynomial might well
factor into polynomials of smaller degree over a larger field, but
that is not relevant to irreducubility over the given field. For
example, $x^2+1$ is irreducible over the real number field $\mathbb R$
but factors into $(x+i)(x-i)$ over the larger field $\mathbb C$.
Your lecture notes may have a typo, or you may have transcribed them
incorrectly into your question above, but the definition of primitive
polynomial should mention the field to which the coefficients belong.
$f(x)$, an irreducible polynomial of degree $n$ with coefficients
in $\mathbb F_p$, (e.g. $\mathbb F_5$) is called a
primitive polynomial over $\mathbb F_p$ (by coding theorists) if
the roots of $f(x)$ are
primitive elements of the field $\mathbb F_{p^n}$. More
generally, $g(x)$, an irreducible polynomial of degree $n$ with
coefficients in $\mathbb F_q$, is called a
primitive polynomial over $\mathbb F_q$ if its roots are
primitive elements of the field $\mathbb F_{q^n}$. Here,
of course, $q$ must be a prime $p$ or a prime power $p^m$.
So, let us take up $f_1(x) = x^2+2$ with coefficients in $\mathbb F_5$.
Its roots lie in the field $\mathbb F_{5^2}$ which has 25 elements
in it, and whose primitive elements have order $24$.
If $\alpha$ is a root of $x^2+2$, then
$$\alpha^2+2 = 0 \implies \alpha^2=-2 \implies
\alpha^8 = (-2)^4 = 1.$$ Thus, $\alpha \in \mathbb F_{5^2}$
is of order $8$ and so $\alpha$ is not a primitive
element of $\mathbb F_{5^2}$, and $x^2+2$ is not a primitive polynomial over
$\mathbb F_{5}$.
On the other hand, if $\beta$ is a root of $f_2(x) = x^2+x+2$,
then your calculations show that none of $\beta, \beta^2, \beta^3,
\beta^4, \beta^6, \beta^8$ equal $1$. But, the multiplicative
group of $\mathbb F_{5^2}$ is a cyclic group of order $24$ and
so the elements of the group can have orders $24$ or a divisor
thereof. You have shown that $\beta$ is not an element of order $1,2,3,4,6,8$. The only remaining possibilities are $\beta^{12}=1$
or $\beta^{24} = 1$, and we can eliminate $\beta^{12} = (\beta^6)^2
= 2^2 = 4 \neq 1$ from consideration as well. Hence, $\beta$
is a primitive element of $\mathbb F_{5^2}$ and $x^2+x+2$ (of
which $\beta$ is a root), is a primitive polynomial of degree $2$
over $\mathbb F_5$.
Bear in mind that $x^2+x+2$ is neither primitive
nor irreducible when viewed as a polynomial with coefficients
in $\mathbb F_{5^2}$: over $\mathbb F_{5^2}$, $x^2+x+2$ factors
into two linear terms $(x-\beta)(x-\beta^5)$.
Best Answer
Theorem: If $p(x)$ is irreducible polynomial in $F[x]$ and $K/F$ is some extension s.t. $\alpha \in K$ is a root of $p$, i.e. $p(\alpha) = 0$, then $F[x]/(p(x)) \cong F(\alpha)$.
For a proof see e.g. see Dummit and Foote 13.1.
This should give you a way to find the right polynomials.