First, pull out a factor of \(x\).
\(x^6+3x^5+x^4-5x^3-6x^2-2x=x(x^5+3x^4+x^3-5x^2-6x-2)\)
Notice that when \(x=-1\) (which you can arrive at via the rational root theorem), you have
\((-1)^5+3(-1)^4+(-1)^3-5(-1)^2-6(-1)-2=-1+3-1-5+6-2=0\)
which means you can pull out a factor of \(x+1\). Upon dividing you get
\(\dfrac{x^5+3x^4+x^3-5x^2-6x-2}{x+1}=x^4+2x^3-x^2-4x-2\)
The rational root theorem will come in handy again, suggesting that \(x=-1\) appears a second time as a root, which means
\(\dfrac{x^4+2x^3-x^2-4x-2}{x+1}=x^3+x^2-2x-2\)
Now this is more readily factored without having to resort to the rational root theorem. You have
\(x^3+x^2-2x-2=x^2(x+1)-2(x+1)=(x^2-2)(x+1)\)
so in fact, \(x=-1\) shows up as a root for a third time.
So, you have
\(x^6+3x^5+x^4-5x^3-6x^2-2x=x(x+1)^3(x^2-2)=0\)
Two roots are obvious, \(x=0\) and \(x=-1\) (with multiplicity 3). The remaining two are given by
\(x^2-2=0\implies x^2=2\implies x=\pm\sqrt2\)
