How many strict local maxima the function $|p(x)|$ might have for any cubic polynomial $p:\Bbb R^n\to \Bbb R$?

Question

Instead of looking for roots of a polynomial $p:\Bbb R^n$, we might be interested about the points at which $p$ is "furthest" from having a root:

Question. What is the maximum number $N$ of strict local maxima of the function $|p(x)|$, where $p:\Bbb R^n \to \Bbb R$ is a cubic polynomial?

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The answer in case of a quadratic polynomial $p$ was trivial: At how many isolated points in $\Bbb R^n$ might the distance of a quadratic polynomial $p:\Bbb R^n\to \Bbb R$ from 0 be locally maximized? However, it seems to me that in case of a cubic polynomial we run into the same problems as those that I describe in the question: How many strict local minima a quartic polynomial in two variables might have?

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As @GerryMyerson commented, $N\leq 2^n$ because by Bézout's theorem the partial derivatives of $p$ can be zero simultaneously at no more than $2^n$ points. Can this upper bound be attained?

Answer 1

Proposition. $N = 2$.

Proof. As argued in A cubic polynomial in $n$ variables has at most one strict local minimum., a cubic polynomial has at most one strict local minimum. Applying this to $p$ and $-p$, we conclude that $|p|$ has at most two strict local maxima (one corresponding to a local minimum of $p$ and the other to local maximum of $p$). Clearly this limit is attained, for example by the polynomial $$ p(x,y) = x(x^2-3) + y (y^2-3), $$ which has a strict local maximum at the point $(-1,-1)$, a strict local minimum at the point $(1,1)$, and the two remaining critical points $(1,-1)$ and $(-1,1)$ are saddle points. $\tag*{$\Box$}$