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