Since the polynomial has integer coefficients, the rational root theorem applies. Thus any rational root must be of the form $x=\pm p/q$, where $p$ divides the constant term 100 and $q$ divides the leading coefficient 1. In this case, the only possibility for $q$ is 1. This tells you that any rational root must be a divisor of $100=2^2*5^2$. It turns out that this polynomial does have rational roots, after which you find one you can perform polynomial division to get a complete factorization.
For instance, we have the potential rational roots $x=\pm2,\pm5,\pm10,\pm20\pm25,\pm50,\pm100$. We could plug in $x=5$ and verify that this is a root. Then, $ \frac{x^4 - 10x^3 + 21x^2 + 40x - 100}{x-5} = x^3-5x^2-4x+20. $ Since all the roots are rational, repeating this process will generate all of them. Not every polynomial with integer coefficients has rational roots (for instance $x^2-2=0$), so this won't always be the case.