What's the benefit to use Lyapunov stability analysis instead of checking the poles? Assume that we have a dynamic system which is described in two forms:
$$G (s) = \frac {\text{num} (s)}{\text{den} (s)} $$
And
$$\dot {x} = Ax +Bu $$ $$ y =Cx +Du$$
Then we can find if the system is stable if we find the poles from:
$$0= \text{den} (s) $$
Or 
$$\text{det} (sI-A) =0$$
If the polynomial has one or several positive roots, then it's unstable. So why should I use Lyapunov stability instead?
 A: The poles of a system are only defined for linear time-invariant (LTI) system. There are examples of linear time-variant (LTV) systems that have negative eigenvalues but still are unstable, hence the usage of eigenvalues for stability analysis is not possible for even LTV systems. 
That means if you want to study the stability of LTV or nonlinear systems in order to design nonlinear controllers or use linear controllers for nonlinear plants then a very common method is to use Lyapunov functions. 
Additionally to the advantage that Lyapunov's direct method is a general method for stability analysis it can also be used for linear control design for nonlinear plants. By using Lyapunov functions it is possible to estimate the region of attraction for an equilibrium point. Hence, it is possible to estimate the region in which the linearized model is still appropriate to describe the dynamics of the nonlinear system.
As an example look at
$$(1)\qquad \dot{x} = -x^3$$
$$(2)\qquad \dot{x} = +x^3$$
it is not possible to draw any conclusions by linearizing the systems at the equilibrium point $x=0$ as this results in the linearized system
$$\Delta \dot{x}=0$$
in both cases.
By using the Lyapunov function (which is positive definite and radially unbounded)
$$V(x)=x^2 \implies \dot{V}=2x\dot{x}$$
$$(1)\qquad \dot{V}=-2x^4<0$$
$$(2)\qquad \dot{V}=+2x^4>0.$$
By using this Lyapunov function we showed that the origin of (1) is globally asymptotically stable and the origin of (2) is unstable.
A: There are different measures of stability for systems that develop over time. A couple of common ones are:


*

*Lyapunov stability

*Exponential stability

*BIBO stability
If the system dynamics can be described as
$$
\matrix{\dot{x} = f(t, x, u) \\ y = h(t, x, u)} \tag{1}
$$
with $x$ the state vector of the system, $t$ time, $u$ the input of the system and $y$ the output of the system. Lyapunov stability essentially says that $(1)$ with $u=0$ is Lyapunov stable if $\|x\|$ remains bounded by a finite constant for all $t>t_0$. Exponential stability essentially says that $(1)$ with $u=0$ is exponential stable if $\|x\|$ remains bounded by some exponentially decaying function for all $t>t_0$. BIBO stability essentially says that $(1)$ with $u$ bounded by some constant and $x(t_0)=0$ is BIBO stable if $\|y\|$ is bounded by some other constant as well.
If a system satisfies a certain measure of stability, then that does not imply it would also satisfy the others as well. For example
$$
\matrix{
\dot{x} = u \\
y = x}
$$
is Lyapunov but not exponential or BIBO stable, while
$$
\matrix{
\dot{x} = 
\begin{bmatrix}
-1 & 0 \\ 0 & 1
\end{bmatrix} x + 
\begin{bmatrix}
1 \\ 0
\end{bmatrix} u \\
y = 
\begin{bmatrix}
1 & 0
\end{bmatrix} x
}
$$
is BIBO but not Lyapunov or exponential stable.
As MrYouMath mentioned these criteria can still be applied to for example nonlinear or time varying systems.

Also your statement: "If the polynomial has one or several positive roots, then it's unstable" is not entirely correct. For example the system with the transfer function $\frac{1}{s^2}$ is not stable according to any of the stability measures I mentioned in this answer.
