A general way to solve $p(x)\cdot T(x)=1$ distributional equation? 
DISCLAMER : I first should apologize, it could be that my question does not make much sense or could be imprecise. Just take this question as a naive question from a physics guy who is trying to understand what he is doing...

Is there a generic way to solve distributional equations like :
$$
p(x)\cdot T(x)=1
$$
where $p$ would be a second order polynome $p(x)=(x-z_1)(x-z_2)$ for instance ?
I do know that the general solution to the distributional equation $x\cdot T(x)=1$ reads :
$$
T(x)=\text{vp}\frac{1}{x}+\alpha\,\delta(x)\;,\quad\alpha\in\mathbb{C}
$$
For instance, is there a way to treat the equation $p\cdot T=1$ "locally around the roots" of $p$ as we would treat the equation $x\cdot T=1$ and give a solution which would look like :
$$
T(x)=\text{vp}\frac{1}{x-z_1}+\alpha_1\,\delta(x-z_1)+\text{vp}\frac{1}{x-z_2}+\alpha_2\,\delta(x-z_2)
$$
Any advice/ressources is appreciated. Thanks by advance.

If the roots of $p$ are real and distinct, then the intuition suggests that $$T" =" \frac{1}{(x-x_1)(x-x_2)}.$$ We can rewrite it as $$T(x) "=" \frac{1}{x_2-x_1}\left(\frac{1}{x-x_1} - \frac{1}{x-x_2}\right).$$ Now we apply a standard workaround to avoid non-local-integrability of terms $\frac{1}{x-x_i}$, we consider principal values (and add add delta-functions arising from the non-uniqueness of the solution):
$$T = \frac{1}{x_2-x_1}\left(PV\left(\frac{1}{x-x_1}\right) - PV\left(\frac{1}{x-x_2}\right)\right) + c_1\delta_{x_1} + c_2\delta_{x_2}.$$
If a root, say, $x_1$ has a non-zero imaginary part, then the function $x\to x-x_1$ is $C^\infty$ on $\Bbb R$ and is never zero there, so we can safely divide both parts of the equation (and therefore reduce our problem to a well-known one):
$$(x-x_2) T = \frac{1}{x-x_1}.$$
Finally, if $x_1=x_2\in\Bbb R$, then neither of the cases above apply. Let us take for simplicity $x_1=x_2=0$. We need to guess a solution of the equation $x^2T=1$. Obviously, we need to play around $PV(1/x)$. Let us take $G = -\left(PV(1/x)\right)'$, then
$$\langle x^2 G,\phi\rangle  = \langle  G,x^2\phi\rangle = \langle  PV(1/x),x^2\phi'+2x\phi\rangle = $$
$$=\lim_{\varepsilon\to 0}\left(  \int_{\varepsilon}^{+\infty}\frac{x^2\phi'(x)+2x\phi(x)}{x}dx+\int_{-\infty}^{\varepsilon}\frac{x^2\phi'(x)+2x\phi(x)}{x}dx\right)$$
$$ =  \int_{\Bbb R}(x\phi'(x)+2\phi(x))dx=\int_{\Bbb R}\left( \left( x\phi(x) \right)'+\phi(x) \right)dx=\langle 1,\varphi\rangle,$$
hence $G$ is a solution of $x^2T=1$. We can conclude 
$$T = -\left(PV(1/x)\right)' + c_0\delta_0 + c_1\delta_{0}' .$$
