Construct a field of 25 elements. We need to construct a field of 25 elements.

By using a result : For a prime p and a monic irreducible polynomial $p(x)$ in $\Bbb F_p [x]$ of degree $n$ , then the ring $(\Bbb F_p [x] / &lt;p(x)&gt;)$ is a field of order $p^{n}$.

So , for the question problem above , we need a irreducible polynomial of degree $2$ in $\Bbb Z_5 [x]$.

Solution says , $x^{2} + x +1$ does our job , since it's irreducible in $\Bbb Z_5 [x]$.

My question : Can't we take $ x^{2} + 2$ , it has no zeroes in $\Bbb Z_5$ , and is irreducible. In every solution I've gone through , the polynomial $x^{2} + x+1$ is taken. Can't we consider $ x^{2} + 2$ ?
Yes, you can take $x^2+2$: any degree two irreducible polynomial would do.

If you take $K=\mathbb{F}_p[x]/\langle x^2+2\rangle$ and denote by $u$ an element such that $u^2=3$, you can find in it a root of $x^2+x+1$:
$$
(a+bu)^2+(a+bu)+1=0
$$
is the same as
$$
a^2+3b^2+a+1=0,\qquad 2ab+b=0
$$
so we must take $a=-1/2=2$ and so $3b^2=3$; thus $2+u$ and $2-u$ are the solutions.
there is a theorem that says that if you have a field $F$ and a polynomial $f(x)$ of degree 2 or 3 in $F[x]$ that does not have zeros in $F$, then $f(x)$ is irreducible in $F$, therefore $x^2+2$ is irreducible in $\mathbb{Z}_5$.