Show that $2x^5-4x^2-5$ is irreducible over $\mathbb{Q}$ The excercise is to show that $2x^5-4x^2-5$ is irreducible over $\mathbb{Q}$.
I actually managed to solve this, but using some luck and previous excercises, so I wonder if you can think of a better way.
My way is as follows:


*

*The given polynomial is irreducible iff the polynomial obtained by the substitution of $x$ by $x+1$ is irrdeucible (why $x+1$? a lucky guess). 
This is a previous excercise.
The result is $$ 2x^5+10x^4+20x^3+16x^2+2x-5 $$

*Another excercise says that a polynomial is irreducible iff the polynomial obtained by "flipping" the coefficients is irreducible. That is:
$$ -5x^5+2x^4+16x^3+20x^2+10x+2 $$

*Apply eisenstein's criterion for $p=2$ to deduce irreducibillity.


Do you have any other, maybe more natural solutions?
 A: The polynomial is irreducible modulo $13$, and hence irreducible over $\mathbb{Z}$, and hence over $\mathbb{Q}$. If the polynomial had a non-trivial factorization over $\mathbb{Z}$, then it had also one modulo $p$.
Irreducibility over a finite field $\mathbb{F}_p$ is much easier than over $\mathbb{Z}$, either by using the Berlekamp algorithm, or by assuming
$$
f(x)=(2x^3+ax^2+bx+c)(x^2+dx+e)
$$
and solving the equations over the finite field (which is easy).
A: One way to do it would be using Cohn's criterion, which states that a polynomial $p$ is irreducible over the integers if its coefficients are non-negative, and if $p(b)$ is prime for some integer $b > \max\{a_n,\cdots,a_0,2\}$.
Note that $p(x)$ is irreducible if and only if $p(-x)$ is. So let us look at the irreducibility of 
$$p(x) = 2x^5 + 4x^2 + 5.$$
This is a form to which we can (try to) apply Cohn' criterion. 
To that end, we can just quickly try some integers $b>5$ to see if $p(b)$ is prime. After a few trials, we find that $p(11)=322591$, which is indeed a prime number, so the polynomial is irreducible over $\mathbb{Z}$.
