Being a quartic, this polynomial is reducible if and only if it has a linear or quadratic factor with integer coefficients.
A linear factor implies an integer root. The only possible roots are $1,-1,5,-5$. Checking, none of them works. Note that for $x=\pm5$ you don't need to do all the calculations since it is easy to see that then
$$x^4+x^3-4x^2-5x-5\equiv-5\pmod{25}$$
and so LHS${}\ne0$.
So try
$$x^4+x^3-4x^2-5x-5=(x^2+ax+b)(x^2+cx+d)\ .$$
Expanding and equating coefficients,
$$a+c=1\ ,\quad ac+b+d=-4\ ,\quad ad+bc=-5\ ,\quad bd=-5\ .$$
The last equation gives four possibilities for $b$ and $d$, (in fact only two, as we may assume by symmetry that $b=\pm1$ and $d=\pm5$) and then it's easy to find the other coefficients:
- $b=-1$, $d=5$, $a=(b+5)/(b-d)=-\frac23$, didn't work;
- $b=1$, $d=-5$, $a=(b+5)/(b-d)=1$, $c=1-a=0$,
and we then check that
$$x^4+x^3-4x^2-5x-5=(x^2+x+1)(x^2-5)\ .$$
Note that if our second attempt above had failed, this would be enough to conclude that the polynomial is irreducible.
It's not the roots, it's the "$2$"!
A polynomial is irreducible over a ring if it cannot be written as a product of two non-invertible polynomials. In $\mathbb{Z}$, "$2$" is noninvertible, so $(x^2+2)2$ is an appropriately "nontrivial" factorization.
Meanwhile, over in $\mathbb{Q}$, the polynomial "$2$" is invertible, since ${1\over 2}$ is rational (proof: exercise :P). So the factoriztion $(x^2+2)2$ is "trivial" in the context of $\mathbb{Q}$, since we can always extract a factor of $2$ from any polynomial.
EDIT: Think of it this way: saying that a polynomial is irreducible over a ring means it has no "nontrivial" factorizations. Now, when we make the ring bigger (e.g. pass from $\mathbb{Z}$ to $\mathbb{Q}$) two things happen:
So even though your first instinct might be "polynomials will only go from "irreducible" to "reducible" as the ring gets bigger," actually the opposite can happen!
In fact, here's a good exercise:
Can you find a polynomial $p\in\mathbb{Z}[x]$ which is irreducible over $\mathbb{Z}$ but reducible over $\mathbb{Q}$?
Note that the definition of reducibility over a field may sound different:
For $F$ a field, a polynomial $p\in F[x]$ is irreducible if $p$ cannot be written as the product of two nonconstant polynomials.
But this is actually equivalent to the definition I gave above, in case we're over a field: the noninvertible elements of $F[x]$ are precisely the nonconstant polynomials!
Best Answer
In this case you can just look at $f$ in $\mathbb F_2[x]$. The only irreducible quadratic polynomial is $x^2+x+1$ and it doesn't divide $x^4+x+1=x^2(x^2+1)+(x^2+x+1)$.