How can the Poles of the Root Locus be negative?
If the denominator polynomial is given by

$$Q(s)=(s-s_0)(s-s_1)\ldots (s-s_{n-1})\tag{1}$$

then the corresponding time domain function is made up of a sum of scaled exponentials of the form $e^{s_it}u(t)$, where $u(t)$ is the unit step function. These terms decay only if $\textrm{Re}\{{s_i}\}&lt;0$, so for stability the zeros of the denominator polynomial (i.e., the poles of the system function) must lie in the left half-plane.

Let $h(t)$ be the impulse response of a causal LTI system. The transfer function is the Laplace transform of $h(t)$:

$$H(s)=\int_{0}^{\infty}h(t)e^{-st}dt\tag{2}$$

This integral only converges in a region $\textrm{Re}\{s\}&gt;c$ ("region of convergence"), with some real-valued constant $c$. That constant depends on the properties of $h(t)$. If $h(t)$ is a weighted sum of exponentials $e^{s_it}u(t)$ then $c$ equals the most positive real part of the complex numbers $s_i$:

$$c=\max_i\big\{\textrm{Re}\{s_i\}\big\}\tag{3}$$

This makes sure that the damping by the term $e^{-st}$ in $(2)$ is enough to make $h(t)e^{-st}$ decay sufficiently such that the integral converges.

For stability we require $c&lt;0$, i.e., all poles lie in the left half-plane, and the imaginary axis of the $s$-plane $s=j\omega$ lies inside the region of convergence.