A prime polynomial $f$ in $K[X]$ (where $K$ is a field, e.g. $k=\Bbb R$) is the same as an irreducible polynomial (because $K[X]$ is a UFD [= unique factorization domain]).
A polynomial $f$ is irreducible in $K[X]$ if the following holds : whenever you can write $f$ as a product $g \cdot h$, one of the two factors $g$ or $h$ is a non-zero constant.
For instance, $X^2+1$ is irreducible in $\Bbb R[X]$, but $x^4+4=(x^2-2 x+2) (x^2+2 x+2)$ is not.
Factorising polynomials can be difficult in general. Over finite fields, there are Berkelamp's algorithm and Cantor-Zassenhaus algorithm. A polynomial of degree $≤3$ is irreducible if and only if it has no roots.
Some examples: $x^2+x+1 \in \Bbb F_2[X]$ is irreducible, because you can't write it as $(x-a)(x-b)$ for some $a,b \in \Bbb F_2[X]$. On the other hand, $x^4+x^2+1 \in \Bbb F_2[X]$ is not irreducible, since you can factor it as $(x^2+x+1)^2 =x^4+2 x^3+3 x^2+2 x+1$ (recall that $2=0$ in $\Bbb F_2$).
Let $\Bbb F_{p^n} = \text{GF}(p^n)$ be the field with $p^n$ elements, which is an extension of the field $\Bbb F_p$.
You may know that its multiplicative group is a cyclic group, say generated by some element $a$ : we have $(\Bbb F_{p^n})^* = \langle a \rangle$, so that $\Bbb F_{p^n} = \Bbb F_p(a)$.
Then a primitive polynomial $f \in \Bbb F_p[X]$, in that context, is the minimal polynomial of $a$ over $\Bbb F_p$.
In particular, a primitive polynomial is irreducible. However, the converse doesn't hold: for instance, $x^4+x^3+x^2+x+1$ is irreducible in $\Bbb F_2$ but is not a primitive polynomial.
For examples about primitive polynomials, you can have a look at this question or this one.
