Is polynomial $x^{2n-1}-4nx^{2n-2}-2\binom{2n}{3}x^{2n-4}-\dots-2\binom{2n}{3}x^{2}-4n$ irreducible? 
Consider following polynomials:
  \begin{align}
P_n(x)&=x^{2n-1}-\sum_{i=0}^{n-1}2\binom{2n}{2i+1}x^{2i}
\end{align}
  for $n\geq 1$. Can we prove they are irreducible over $\mathbb{Q}$?

So first couple are 
$$x-4$$
$$ x^3-8x^2-8$$
$$ x^5-12x^4-40x^2-12$$
$$ \dots$$
For small specific values of $n$ we can prove it, for example for $x^3-8x^2-8$ we only need to check for linear factor, specifically for roots $\pm 1,\pm2,\pm4\pm 8$. With Maple I've verified more than first $600$ values of $n$ so far, all appear to be irreducible (using the irreduc function). 
However for general $n$ I have no idea how to go about this, the Eisenstein criterion cannot be directly used, so I've tried substitutions like $P_n(x\pm 1)$, but it does not help. 
Edit: Just out of curiosity, I have tried to change the polynomial a bit to see when it stops to be irreducible (again for small values of $n$, let's say to $100$) and I found it interesting that you can do quite a lot changes and it still remains irreducible. For example replacing the multiple of $2$ by other integers, or changing signs of the individual terms in sum, or even replace $2i+1$ in binomial coefficients with $2i$. What is going on? 
 A: For odd $n$ the irreducibility of these polynomials follows from a generalization of Eisenstein's criterion known as Newton's polygon applied to the prime $p=2$.
When $n$ is odd the constant term $4n$ is divisible by four but not divisible by eight. Furthermore, Lucas' theorem tells us that all the binomial coefficients $\binom {2n}{2i+1}$ are even, implying that all the other coefficients, save for the leading $1$, are also all divisible by four (and possibly by a higher power of two). These imply that the Newton's polygon consists of a single line segment from $(0,2)$ to $(2n-1,0)$. The slope of this line is $-2/(2n-1)$ implying that the line segment does not pass through any integer points other than its endpoints. By $2$-adic theory this guarantees that $P_n(x)$ is irreducible over the $2$-adic field $\Bbb{Q}_2$ and hence also over the subfield $\Bbb{Q}$.
The case of an even $n$ is more complicated but it may be possible study the divisibility of the coefficients by powers of two and again settle the claim.
For example the Newton's polygon of
$$
P_4(x)=x^7-16(x^6+7x^4+7x^2+1)
$$
is the line segment from $(0,4)$ to $(7,0)$, has slope $-4/7$, and the same argument goes through.
A: Based on @JyrkiLahtonen's answer which proves the irreducibility for odd $n$ and suggests path or even $n$, I was able to prove the statement for even $n$'s for which the Newton polygon's method works, although it still leaves infinity of even $n$'s to check (hopefully someone can bring another idea to solve these as well).
The Newton polygons method seems to work for even $n$'s for which $(2n-1,v_2(4n))=1$, so $
n=4,6,10,12,16,18,20,22,\dots
$.
First, we need to show that the Newton's polygon for $P_n(x)$ is indeed a line segment between the end points. For this it is sufficient to show that $v_2(2\binom{2n}{2i+1}) \geq v_2(4n)$ for $i=0,1,\dots,n-1$ (in other words every $y$ coordinate in the Newton polygon will be at least as large as the $y$ coordinate corresponding to the constant coefficient $4n$) . This can be written as $v_2(\binom{2n}{2i+1}) \geq v_2(2n)$. But this follows almost immediately if we notice that  $(2i+1)\binom{2n}{2i+1}=2n\binom{2n-1}{2i}$, and so by applying the $v_2$ on both sides (using that $2i+1$ is odd) we get
$$
v_2\left(\binom{2n}{2i+1}\right) = v_2\left(2n\binom{2n-1}{2i}\right) \geq v_2(2n).
$$
Finally because the constant coefficient is $-4n$ and leading coefficient is $1$, the end points of the Newton polygon are $(0,v_2(4n))$ and $(2n-1,0)$, and so the method will guarantee irreducibility for $(2n-1,v_2(4n))=1$.
