How do I find out if a polynomial is irreducible?

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.

There are algorithms, e.g. Kronecker's, that will find factors over $\mathbb Q$ if they exist. These are implemented in the various Computer Algebra systems, so in practice you would just ask Maple or Wolfram Alpha.