How to prove that $X^4+8X^3+X^2+2X+5$ is irreducible? How to prove that $X^4+8X^3+X^2+2X+5$ is irreducible on $\mathbb Q[X]$?
I can try to find $a,b,c,d$ such that
$$X^4+8X^3+X^2+2X+5=(X^2+aX+b)(X^2+cX+d)$$ 
and prove that $a,b,c,d\notin\mathbb Q$ and find $a,b,c,d$ such that
$$X^4+8X^3+X^2+2X+5=(X-a)(X^3+bX^2+cX+d)$$
and prove that $a,b,c,d\notin \mathbb Q$, but it's really long and I would like to know if there is an other method. 

First, by the Rational Root Theorem, no rational number other than $\pm 1, \pm 5$ can be a root of the given polynomial $f$, and evaluating quickly shows that none of these are roots, so $f$ has no linear factors.
There are a few strategies to deal with the possibilities of quadratic factors. A monic polynomial with coefficients in $\mathbb{Z}$ is irreducible over $\mathbb{Z}$ iff it is irreducible over $\mathbb{Q}$, so the only possibilities for the constant terms $(b, d)$ (up to relabeling) in
$$X^4 + 8 X^3 + X^2 + 2 X + 5 = (X^2 + aX + b) (X^2 + cX + d)$$
are $(1, 5)$ and $(-1, -5)$. Substituting the first pair and expanding gives the system
$$\left\{\begin{array}{rcl} a + c &=& 8 \\ ac + 6 &=& 1 \\ 5 a + c &=& 2 \end{array}\right.;$$ solving the first and third equation shows that there is no solution $(a, c)$ over $\mathbb{Z}$, and we can treat the second pair similarly. So, there is no factorization over $\mathbb{Z}$, and hence over $\mathbb{Q}$.
Remark Sometimes a good technique for showing that a polynomial $f$ is irreducible is to find a prime $p$ such that $f$ is irreducible modulo $p$. In our case, this is less efficient than the above method, because the smallest such prime for this $f$ is $p = 23$.
On the other hand, in our case this method does immediately give that $f$ has no linear factors more quickly than the Rational Root Theorem, as substitution gives that $$f(0) \equiv f(1) \equiv 1 \bmod 2.$$
