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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:
$$P(x) = x^d + A_d-1x^d-1 + A_d-2x^d-2 + cdots + A_0;. $$

Q. What does the set of
polynomials, with real coefficients $a_i$,
$$p(x) = x^d + a_d-1x^d-1 + a_d-2x^d-2 + cdots + a_0 ;,$$
look like (geometrically) in $mathbbR^d$,
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:

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 $D : mathrmdisc(p)=0$ 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 $varepsilon$.

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: