Irreducible polynomial in $\mathbb Q[x]$ of degree $n$ having exactly $n-2$ real roots.

I need to show that for any $n$ there is an irreducible polynomial in $\mathbb Q[x]$ of degree $n$ having exactly $n-2$ real roots.

I know that from a previous exercise that if $f(x) \in \mathbb{R}[x]$ is any polynomial having exactly $k$ distinct real roots, there exists $\epsilon > 0$ for which $f(x) +a$ has exactly $k$ real roots, for all $a\in \mathbb{R}$ with $|a|<\epsilon$.

Then by starting with any polynomial $f(x) \in \mathbb Q[x]$ with exactly $n-2$ distinct real roots, and using the paragraph above $f(x)+a$ has the same property for infinitely many $a\in \mathbb Q$. Now, how can use the Eisenstein irreducibility criterion for $f(x) \in \mathbb Z[x]$ and $a \in \mathbb Q$ to prove my initial statement?

Thanks

• See this related answer for a more general result. Listing this, because the method closely follows what you seem to be trying. The twist in getting the resulting polynomial to be irreducible is kinda neat IMHO. Commented Mar 18, 2015 at 13:56

I'd start with some polynomial $f\in\mathbb Z[X]$, say $$f(X)=(X^2+m)(X-k_1)\cdots(X-k_{n-2})$$ with $m,k_1,\dots,k_{n-2}$ positive even integers, and $k_1<\cdots<k_{n-2}$. Now use your previous post which says that $f(X)+\dfrac a b$ has also exactly $n-2$ real roots for all $a,b\in\mathbb Z$, $b\ne0$ with $\bigg\vert\dfrac ab\bigg\vert<\epsilon$ for some $\epsilon>0$. Then $bf(X)+a$ has the same property. If $$f(X)=X^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$$ then $$bf(X)+a=bX^n+(ba_{n-1})X^{n-1}+\cdots+(ba_1)X+(ba_0+a).$$ Now chose $b$ odd, $a=2$, and use Eisenstein's Criterion for $p=2$.