What are roots of $x^{3}+3x^{2}+4x+1$? There are no divisor of 1 in this polynomial for which would be satisfied $x^{3}+3x^{2}+4x+1=0$. 
How to find roots here?

This can be done by hand with a bunch of substitutions. 
Let $y=x+1$, then
$$1+4(y-1)+3(y-1)^2+(y-1)^3=0$$
If we expand all the terms we get
$$1+4y-4+3y^2-6y+3+y^3-3y^2+3y-1=0$$
which boils down to 
$$y^3+y-1=0$$
Now we can do a change in coordinates with another substitution. 
Let $y=u+\frac{\lambda}{u}$. We will find $\lambda$ later. This yields. 
$$-1+u+\frac{\lambda}{u}+\left ( u + \frac{\lambda}{u} \right )^3=0$$
We can expand this by multiplying by $u^3$ which gives
$$u^4(3\lambda + 1) + u^2\lambda(3\lambda+1)-u^3+u^6+\lambda^3=0$$
We want the coefficients on the $u^2$ and $u^4$ terms to be $0$. So $\lambda=-\frac{1}{3}$. 
Substituting we now have
$$u^6-u^3-\frac{1}{27}=0$$
If we perform yet another substitution, we can finally get to a quadratic.
Let $z=u^3$. 
$$z^2-z+\frac{1}{27}=0$$ 
Using the quadratic formula we get
$$z=\frac{1}{18}(9+\sqrt{93})$$.
Now we climb back up our ladder of subsitutions
$$u^3=\frac{1}{18}(9+\sqrt{93})$$
$$u=\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}$$
We substitute for $u$ and get
$$y=\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}-\frac{1}{3\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}}$$
Now to get $x$ we just subtract 1. 
$$x=\left( \frac{1}{18}(9+\sqrt{93})\right )^{1/3} - \frac{1}{3}\left( \frac{1}{18}(9+\sqrt{93})\right )^{-1/3}-1$$
