Polynomial is irreducible over $\mathbb{Q}$ I've been scratching my head lately over this problem in my textbook:

Prove that the polynomial $x^4 + x^3 +x^2 +x +1$ is irreducible over $\mathbb{Q}$.

I've done some research and found this link, but they talk about Eisenstein's criterion, which we haven't covered in our math class just yet. Is there a general strategy where we can show whether a polynomial is irreducible over a field? 
My textbook really didn't go into depth on the topic of irreducibility of polynomials, but this Wiki link somewhat helps. Perhaps the rational root theorem may be helpful here, but how would I go about starting this proof?
EDIT
Please see the first comment for my initial strategy at showing the required result.

There exists a simple general method for monic quartic polynomials in $\mathbb{Z}[x]$. Suppose you have verified that the polynomial
$$p(x) = x^4 + a x^3 + b x^2 + c x + d$$
has no rational roots. Then we want to see if it has quadratic factors, and if so we want to factor it. Also we don't want to use very elaborate trial and error methods. The first test that needs to be performed is to see whether p(x) is the square of a quadratic polynomial, that's easy to see by computing the GCD of $p(x)$ with its derivative. If a nontrivial GCD, then you're done. If not, then we proceed as follows. We reduce $p(x)$ modulo $x^2 - p x - q$, this yields:
$$\left(p^3 + a p^2+ 2 p q + a q+ b p+c\right) x + p^2 q + q^2 +a p q+ b q+d$$
Then if $p(x)$ has a quadratic factor, this has to be identically zero. We thus need to equate the coefficient of $x$ and the constant term to zero and solve the two equations for $p$ and $q$. It's then convenient to start with eliminating the highest powers of $p$ in favor of lower powers until $p$ has been completely eliminated in favor of $q$. We then end up with an equation for $q$, and an expression for $p$ in terms of $q$. Then since $q$ had to divide $d$, you only have a few cases to check. If none of them work then $p(x)$ is irreducible. If $p(x)$ is reducible you'll find the factorization, unless both factor have the same value for $q$. In the latter case the single solution for $q$ obviously cannot tell you what both values for $p$ are (and they are different because $p(x)$ was verified to be square-free). If this exceptional case does not occur, then we have:
$$p = \frac{a q^2+c q}{d-q^2}\tag{1}$$
and
$$q^6 +b q^5 +(a c-d)q^4 + \left(a^2 d-2 b d+c^2\right)q^3 +\left(a c d-d^2\right)q^2 +b d^2 q+d^3 = 0$$
So, $q$ must be an integer that divides $d$ that satisfies this equation. If two such values for $q$ are found then you have found the factorization, the corresponding values for $p$ follow from Eq. (1). If only one solution for $q$ is found, then Eq. (1) will be singular, the two values for $p$ are then solutions of the equation:
$$a p^2 + a^2 p+a b-2 c = 0$$
