How/Why does factoring the linear/differential operator suggest a specific change of variables? 
Find a solution of the PDE $u_{tt} - c^2 u_{xx} = 0$, (where $c$ is a constant) in the half plane $t > 0$ with initial conditions $u(x, 0) = g_0(x)$ and $u_t(x, 0) = g_1(x)$. 

$$\therefore \mathcal{L} = \partial^2_t - c^2 \partial^2_x = (\partial_t + c \partial_x)(\partial_t - c \partial_x)$$ (Linear/Differential Operator.)
Apparently, this factorisation suggests the change of variables 
$$y = x - ct, s = x + ct$$
How/Why does factoring the linear/differential operator suggest a specific change of variables? I see the obvious similarity between the change of variables above and the linear operator of the wave equation, but, speaking in terms of the general concept/theory, this was not explained before this example.
Thank you.
Postscript: Please note that this is within the context of a section on characteristics of second-order PDEs (method of characteristics for Second-Order PDEs).
 A: Just give it a try:
$$
(\partial_t+c \partial_x)y = 0 = (\partial_t -c \partial_x)s 
$$
$$
(\partial_t - c \partial_x)y = -2c = -(\partial_t + c \partial_x)s 
$$
which means
$$
(\partial_t - c \partial_x) = -2c \partial_y \text{ and } (\partial_t + c \partial_x) = 2c \partial_s.
$$
This will strongly simplify your differential equation:
$$
\mathcal{L} = -4 c^2 \partial_y\partial_s
$$
which means g only depends on s or y, if $\mathcal{L}g = 0$.
In general, having
$$
\mathcal{L} = (a_{11}\partial_x + a_{12}\partial_t)(a_{21}\partial_x+a_{22}\partial_t)
$$
by the ansatz
$$
y = b_{11}x + b_{21}t \text{ and } s = b_{12}x + b_{22}t
$$
you set
$$
(a_{11}\partial_x + a_{12}\partial_t)y = 1
$$
$$
(a_{21}\partial_x + a_{22}\partial_t)y = 0
$$
$$
(a_{11}\partial_x + a_{12}\partial_t)s = 0
$$
$$
(a_{21}\partial_x + a_{22}\partial_t)s = 1
$$
This is equivalent to
$$
\left(\matrix{a_{11} && a_{12}\\a_{21}&&a_{22}}\right) \left(\matrix{b_{11}\\b_{21}} \right)= \left(\matrix{1\\0} \right)
$$
and
$$
\left(\matrix{a_{11} && a_{12}\\a_{21}&&a_{22}}\right) \left(\matrix{b_{12}\\b_{22}} \right)= \left(\matrix{0\\1} \right)
$$
In other words $(b_{ij})_{ij} = (a_{ij})^{-1}_{ij}$.
Then the transform $y = b_{11}x + b_{21}t$ and $s = b_{12}x + b_{22}t$ will simplify your eqation as above.
A: From your comments it seems like you're more interested in some intuition about why to make that change of variables aside from the similarity of the two? Forgive me if I have misinterpreted. Also the answer of denklo is the more complete one, as I'm really just saying in words what that last eigensystem is saying.
The most intuitive way I grasp it is from a physics perspective: that factorization suggests a type of symmetry, that what happens going leftwards is the same as that going rightwards.
Changing to $x \pm c t$ is to follow the information packets traveling left and rightwards with speed $c$. In physical terms, it's a change of frame of reference to that of the waveform, which because the system has no diffusion or dispersion, must retain its structure.
