Prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$ 
How to prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$?

I tried Eisenstein criteria on $f(x+n)$ with $n$ ranging from $-10$ to $10$. None can be applied. I tried factoring over mod $p$ for primes up to $1223$. $f(x)$ is always reducible over these. 
$f(x)$ has roots $\pm\sqrt{\left(2\pm\sqrt{2}\right) \left(3\pm\sqrt{3}\right)}$, and according to computation by PARI, should have Galois group isomorphic to the quaternion group. The splitting field of $f(x)$ is $\mathbb{Q}(\sqrt{\left(2+\sqrt{2}\right) \left(3+\sqrt{3}\right)})$, and it contains $\sqrt{2}, \sqrt{3}, \sqrt{6}$, so we know $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is in the splitting field, so $\mathbb{Q}(\sqrt{\left(2+\sqrt{2}\right) \left(3+\sqrt{3}\right)})$ has degree $4$ or $8$.
I tried showing the degree is $8$ by showing that $(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})^2=(2+\sqrt{2})(3+\sqrt{3})$ cannot have a solution with $a, b, c, d \in \mathbb{Q}$, and got these equations:
$$a^2+2b^2+3c^2+6d^2=6$$
$$2ab + 6cd = 3$$
$$ac+2bd = 1$$
$$2ad+2bc = 1$$
which I'm unable to handle.
Addendum: now that I've solved this problem thanks to the answers, I found some additional related information: 
In A Rational Polynomial whose Group is the Quaternions, a very similar polynomial, $$f(x)=x^8 - 72 x^6 + 180 x^4 - 144 x^2 + 36$$ is studied and its Galois group is proven to be the quaternion group. I subjected this polynomial, as well as two related ones: $f(\sqrt{x})$, $f(6\sqrt{x})/36$, to the same battery of tests (Eisenstein; mod p) and these tests also failed to show them to be irreducible. Maybe there's something common about these polynomials.
So I subjected $f(x)$ to the prime numbers test demonstrated by Robert Israel, and found that it is prime at $\pm\{7, 13, 23, 25, 49, 53, 55, 79, 91, 127, 139, 145, 151, 181, 239, 251, 277, 283, 319, 355, 379, 403, 413, 425, 473, 485, 595, 607, 623, 679, 733, 743, 779, 827, 851, 923, 965, ...\}$ and thus $f(x)$ is irreducible.

Assume that $f(x) = f_1(x) f_2(x)$.
Note that $f(1) = - 23$, so one of $f_1(1)$, $f_2(1) = \pm 1$, and the other $\mp 23$.
The polynomial factors $\mod 5$ as $(x^4+2)(x^4 + x^2 + 2)$ ( see Jyrki's answer). Say we have $f_1(x)\equiv x^4 + 2$ and $f_2(x) \equiv x^4 + x^2 + 2 \mod 5$.  Now, $f_1(1) \equiv 1^4 + 2 = 3 \mod 5$, and $f_2(1) \equiv 1^4 + 1^2 + 2 \equiv -1 \mod 5$. We conclude that $f_1(1) = 23$, and $f_2(1) = -1$.
Now, $f(x)$ factors $\mod 3$ as $x^8$. We  have $f_1(x) \equiv x^4$, and $f_2(x) \equiv x^4 \mod 3$. This implies $f_1(1), f_2(1) \equiv 1 \mod 3$.  We got a contradiction
$\bf{Added:}$ Another proof:  Assume that $(2+\sqrt{2})(3+ \sqrt{3})$ is  a square in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. Taking the norm to $\mathbb{Q}(\sqrt{2})$ we get
$(2+\sqrt{2})^2 (3+ \sqrt{3})(3-\sqrt{3})$ is a square in $\mathbb{Q}(\sqrt{2})$, so $6$ is, contradiction.
