# Existence of irreducible polynomial over $\mathbb Q$, which has exactly two non-real roots

I know the following statement:

If $f$ is an irreducible polynomial over $\mathbb Q$ of prime degree (let $\deg f=p$) and if $f$ has exactly two non-real roots in $\mathbb C$, then the Galois group of $f$ is the symmetric group $S_p$.

But, I can't show the existence of such polynomial $f$ for each prime $p$. How can I find it?

Start with $Q(X) = (X^2+1) \times \prod_{i=1}^{p-2} (X-i)$. Then $Q(X)$ has exactly two non-real roots.
For $k \in \mathbb{N}$, let $P_k(X) := Q(X) + \frac{X^p+p}{kp^2} \in \mathbb{Q}[X]$. Then $P_k(X)$ is irreducible (apply Eisenstein criterion to $kp^2 P$). By Rouché's theorem for some $k>>1$, $P_k$ has a root in each disk $D(z,1/2)$ where $z \in {\rm Root}(Q(X))=\{1, 2, \dots, p-2 , i ,-i\}$. Since $P_k$ has real coefficients, each root in $D(y,1/2)$ ($y \in \{1, 2, \dots, p-2 \}$) are real.
Hence $P_k$ is irreducible and has only two non-real roots (namely the ones in $D(i,1/2)$ and $D(-i,1/2)$).