TITLE: Octic family with Galois group of order 1344?
QUESTION [5 upvotes]: Does the octic,
$\tag{1} x^8+3x^7-15x^6-29x^5+79x^4+61x^3+29x+16 = nx^2$
for any constant n have Galois group of order 1344? Its discriminant D is a perfect square,
$D = (1728n^4-341901n^3-11560361n^2+3383044089n+28121497213)^2$
Surely (1) is not an isolated result. How easy is it to find another family with the same Galois group and same form,
$\tag{2} \text{octic poly in}\ x = nx^2$ 
Perusing Kluener's "A Database For Number Fields" this seems to be the only one. (Though I was able to find a 2-parameter family of a different form.)

REPLY [3 votes]: In arxiv.org/abs/1209.5300 I give the following polynomial with the same Galois group over Q(t): $(y+1)(y^7-y^6-11y^5+y^4+41y^3+25y^2-34y-29)-t(2y+3)^2$. This has $(6912t^4-3456t^3-95472t^2+23976t-1417)^2$ for a discriminant. This doesn't quite answer the second question, but substituting $y \mapsto (x-3)/2$, $t \mapsto s/256$ gives $(x-1) (x^7-23 x^6+181 x^5-547 x^4+515 x^3-325 x^2+71 x-1)-s x^2$, with discriminant $256 (27 s^4-3456 s^3-24440832 s^2+1571291136 s-23773315072)^2$, which does.