In Jacobson's Basic Algebra Vol. 1, there is a construction of a rational irreducible polynomial with exactly two complex roots. In brief, its as follows. Let $k$ be odd integer with $k\ge 5$. $$f(x)=(x^2+m)(x-n_1)(x-n_2)\cdots (x-n_{k-2})$$ where $n_1<n_2<\cdots < n_{k-2}$ are even integers and $m$ is positive even integer. Then, for suitable choice of $m$, the polynomial $g(x)=f(x)-2$ has exactly $k-2$ real roots (and is irreducible). I proceeded in following way (little different from Jacobsons's; I want to check if its correct) On can skip proofs of Claims and see statements of claims only.
Claim 1. $n_1,n_2,\cdots, n_{k-2}$ are simple roots of $f(x)$. [Clear: since they are distinct]
Claim 2. In intervals $(n_i,n_{i+1})$, $f(x)$ is either positive or negative.
If $f$ takes positive as well as negative value in $(n_i,n_{i+1})$, it should take (intermediate) zero value, i.e. it will have root in $(n_i,n_{i+1})$, not possible (since only roots of $f(x)$ are $n_1<n_2<\cdots < n_{k-2}$).
Claim 3. On $[n_i,n_{i+1}]$ if $f\ge 0$, then $f$ attains maximum value which is $>2$.
For, if $a$ is odd integer (between $n_i$ and $n_{i+1}$), then $|a-n_j|\ge 1$ for all $j$, and $|a^2+m|>m\ge 2$.
Claim 4. If $f\ge 0$ on $[n_i,n_{i+1}]$, then $g(x)=f(x)-2$ has at least $2$ distinct roots in $[n_i,n_{i+1}]$.
For, let $\beta$ be a point in $(n_i,n_{i+1})$ at which $f$ attains maximum; we saw $f(\beta)>2$, so $g(\beta)>0$ and $g(n_i)=g(n_{i+1})=-2<0$. Apply intermediate value theorem to $g$ on $[n_i,\beta]$ and $[\beta,n_{i+1}]$.
Claim 5. If $f\le 0$ on $[n_i,n_{i+1}]$, then $g(x)=f(x)-2$ has no roots in $[n_i,n_{i+1}]$.
Claim 6. Consider $k-3$ (which is even) successive intervals $[n_1,n_2]\cup [n_2,n_3]\cup \cdots \cup [n_{k-3},n_{k-2}]$; on half of them, $g(x)$ has at least two roots, and on remaining half, $g(x)$ has no root. (Thus, $g(x)$ has at least $k-3$ real roots.)
Since $n_i$ is simple root of $f$, so in small neighbourhood of $n_i$, $f$ is monotonic. Thus, if $f\ge 0$ on $[n_i,n_{i+1}]$ then $f\le 0$ on next (and previous) interval. Now apply previous two claims.
Claim 7. $g(x)$ has at least $k-2$ real roots (for every choice of positive even integer $m$).
We found at least $k-3$ real roots for $g(x)$ in previous claim. One more root can be found in $(n_{k-3},\infty)$. For this, first, if $a\in (n_{k-3},n_{k-2})$ then all the factors in expression of $f$ take positive value at $a$ except the last one, hence $f(a)<0$. On the other hand, $f(x)\rightarrow +\infty$ as $x\rightarrow +\infty$. Hence $f(x)$ (and hence $g(x)=f(x)-2$) has a root in $(n_{k-2},\infty)$.
After this, I think, we should choose $m$ cleverly so that $g(x)$ will have exactly $k-2$ real roots. Is this justification correct? The arguments in Jacobson I couldn't understand and I thought there was some flaw in justification (in the sense, the statements of individual claims there were shuffled). I thought it in above way, and I want to know if Claims (and proofs are correct). One may suggest shortcuts in above arguments if any.