Here is Example 4.33 (pages 55-56) from http://jmilne.org/math/CourseNotes/FT.pdf:
Select separable monic polynomials of degree $n$, $f_1,f_2,f_3$ with coefficients in $\mathbb{Z}$ such that:
(1) $f_1$ is irreducible modulo 2;
(2) $f_2=(\textrm{degree 1})(\textrm{irreducible of degree }n-1)$ mod 3;
(3) $f_3=(\textrm{irreducible of degree 2})(\textrm{product of 1 or 2 irreducible polynomials of odd degree}$) mod 5.
Take $$f=-15f_1+10f_2+6f_3.$$ Then
(i) $G_f$ is transitive (it contains an $n$-cycle because $f\equiv f_1$ mod 2);
(ii) $G_f$ contains a cycle of length $n-1$ (because $f\equiv f_2$ mod 3);
(iii) $G_f$ contains a transposition (because $f\equiv f_3$ mod 5, and so it contains the product of a transposition with a commuting element of odd order; on raising this to an appropriate odd power, we are left with the transposition). Hence $G_f$ is $S_n$.
My question is: Why must $f_3$ have only 1 or 2 factors of odd degree? In particular, I don't see why there can't be more than 2; it seems to me as if the same argument would hold if (for example) $f_3$ had 3 (irreducible) factors of odd degree.