How to factorized this 4th degree polynomial?

I am not sure what you mean about being irreducible over $\mathbb Q$ - it may have no roots over $\mathbb Q$, which is a different thing.

The general process for finding the roots of a quartic starts by trying to express it as a difference of two squares, and solving a cubic for the parameter involved (you can pair the four roots of the quartic in three different ways, which represent three different factorisations into two quadratics).

Often trying to spot a difference of two squares is a good method with quartics anyway, if you are stuck.

Here I spotted $x^4+2x^3+x^2-9x^2-6x-1$ as a possible way forward - and this leads to a factorisation into quadratics with integer coefficients.

You can always find the roots of a fourth degree polynomial. The formulas are not pretty, but they do exist.

Using the formulas, or, like me, using Wolfram Alpha, can show you that the roots of the polynomial are $1\pm\sqrt{2}$ and $-2\pm \sqrt{3}$ so the polynomial is

$$\begin{align}f(x)&=(x-1-\sqrt{2})(x-1+\sqrt{2})(x+2-\sqrt{3})(x+2+\sqrt{3})\\&=(x^2-2x-1)(x^2+4x+1)\end{align}$$