Polynomials that share at least one root This is a generalization of an MSE question,
Polynomials that share at least one root.
Let $P(x)$ be a specific polynomial of degree $d$, with given
real coefficients $A_i$ ($A_d=1$), and real roots:
$$P(x) = x^d + A_{d-1}x^{d-1} + A_{d-2}x^{d-2} + \cdots + A_{0}\;. $$

Q. What does the set of
  polynomials, with real coefficients $a_i$,
  $$p(x) = x^d + a_{d-1}x^{d-1} + a_{d-2}x^{d-2} + \cdots + a_{0} \;,$$
  look like (geometrically) in $\mathbb{R}^d$,
  when each $p(x)$ shares at least one root with 
  $P(x)$?

I am seeking a description of this set in the space of the $d$ coefficients, $(a_0,\ldots,a_{d-1})$.
The reason there is hope for a nice description, is that it makes
a pretty picture for $d=2$.
Let $P(x) = x^2 + A_1x + A_0$, with $A_1=3$ and $A_0=-1$.
The plot in the $a_0 a_1$-plane 
of all other $p(x)= x^2 + a_1x + a_0$ that share a root with $P(x)$,
looks like this:

          


          

Lines intersect at $(a_0,a_1)=(-1,3)$. The discriminant is $a_1^2 = 4 a_0$.


All those $(a_0,a_1)$ on the two lines share a root with 
$x^2 + 3x -1$. The lines are tangent to the discriminant parabola.
 A: The locus of real polynomials $p(x)$ sharing a root with $P(x)$ is the union of the hyperplanes $H_\alpha : p(\alpha)=0$, where $\alpha$ runs over the roots of $P(x)$. This is an arrangement of hyperplanes.
The intersection of $H_\alpha$ with the discriminant variety $D : \mathrm{disc}(p)=0$ contains the subspace $H^{(1)}_{\alpha} : p(\alpha)=p'(\alpha)=0$. In fact $H_\alpha$ and $D$ are tangent along $H^{(1)}_\alpha$. To see this, take a generic polynomial $p_0(x)=(x-\alpha)^2 q_0(x)$ in $H^{(1)}_\alpha$. Near $p_0$, polynomials in $D$ will be of the form $p(x)=(x-\alpha+\varepsilon)^2 q(x)$ where $q$ is near $q_0$. Then $p(\alpha)=\varepsilon^2 q(\alpha)$, while the distance from $p_0$ to $p$ is of first order with respect to $\varepsilon$.
A: I tried to make a 3D image for 
$P(x)=x^3+3 x^2-2 x-1$. The set consists of three planes,
each tangent to the discriminant surface.
But it became too visually complex, partly because 
the discriminant
is complicated.
For what it's worth:

          


          

Discriminant surface: blue.


Perhaps someone can do better...
