I was thinking this has a chance in $\mathbb R$ but once you step in $\mathbb R^2$ there should be counterexamples. Let us have a look at the polynomial function: $f\colon (x,y)\in \mathbb R^2 \longmapsto x^4 + 12 x^2 y^2 + y^4 + (x+2y)^2$. You can factor it as $f(x,y) = (x^2+y^2)^2 + 10x^2y^2 + (x+2y)^2$, this establishes that $f(0,0) = 0$ is the unique minimum.
We should note that $f$ is not locally convex around $(0,0)$. For example you can check that,
\begin{equation}
\begin{bmatrix} 2 \\ -1 \end{bmatrix}^\top \nabla^2 f(t,t) \begin{bmatrix} 2 \\ -1 \end{bmatrix} = -12t^2 < 0,
\end{equation}
for all $t \neq 0$, no matter how small.
It seemed that there existed a unique minimum for a small enough linear perturbation, however upon closer inspection the gradient map is not injective. I have plotted (rotated and resized) $\nabla f(r\cos \theta, r\sin\theta)$ with $r$ fixed smaller and smaller, and with $\theta$ varying from $0$ to $2 \pi$. It seems to describe a deformed circle around $(0,0)$ (it actually becomes close to a segment) but those circles actually overlap (see the attached plot). In any case the equation is,
\begin{equation}
\begin{bmatrix}
4 x^3 + 24 x y^2 + 2 (x + 2 y) \\ 24 x^2 y + 4 y^3 + 4 (x + 2 y)
\end{bmatrix}
=
\begin{bmatrix}
-u \\ -v
\end{bmatrix}.
\end{equation}
Arcs of the gradient are drawn, $\nabla f(r \cos t, r \sin t)$, for $r=10^{-7}(1+n/4)$ with $n=1,\dots,10$ and $t = 0 \dots 2\pi$. The picture has been rotated and stretched to show clearly the overlap.
Indeed the solution in the video, which is quite similar to your work, is somewhat hand wavy. The key expression used later on comes out of nowhere and some of the things are not explained right.
Given data is: $f'(\alpha) = f'(\beta) = 0$ and $f(\alpha) = f(\beta) = M$
Now since, $x> a$ we can write $f(x) = \frac{g(x)}{(x-a)}$.
Using quotient rule, we get $$f'(x) = \frac{g'(x)(x-a) - g(x)}{(x-a)^2}$$
Now using the given data, we get \begin{align}
g'(\alpha) = \frac{g(\alpha)}{(\alpha-a)} = f(\alpha) \qquad
g'(\beta) = \frac{g(\beta)}{(\beta-a)} = f(\beta) \tag{1}
\end{align}
which means both $g'(\alpha)$ and $g'(\beta)$ are the local maximums $``M"$.
Now¹ if you consider $f(\beta) - f(\alpha)$ and use $(1)$ with the given data and that $\beta - \alpha = 6 \sqrt{3} \implies \beta = 6 \sqrt{3} + \alpha$ we get $$\frac{g(\alpha) - g(\beta)}{\alpha - \beta} = \frac{g(\alpha)}{\alpha - a}$$
Which you should recognize the R.H.S as $g'(\alpha)$ or $f(\alpha)$ or $M$.
Some words —
Note that at this point we only have to worry about $g(x)$ and its relation with $M$. For the motivation for the rest of the solution, also note that the L.H.S is a difference quotient which is sufficient evidence to look into the graph of $g(x)$ and the tangents at $x = \alpha$ & $x = \beta$. This is also sufficient to say that the difference quotient ($``M"$) is the tangent
for two points $(\alpha , g(\alpha))$ and $(\beta, g(\beta))$ framing a triangle on a graph $x \mapsto x$ and $y \mapsto g(x)$.
It should be noted that $f(x)$ is not a cubic as suggested by (a) alone. From (b) it says that $f(x)$ has two maximums (implying a flex between the two maximums) which would be strange if $f(x)$ was a cubic since it can only have $3-1 = 2$ waves at most, contradicting (b).
So far we have only used conditions (a) and (b) thus we now also turn our attention to (c). It simply means that the number of local extremas are $\leq 2$. Also given is $g(x)$ is a quartic which means there is symmetry and it has a coefficient -1 implying both "hands" go to $-\infty$. The curve will have $4-1 = 3$ waves hence one of the points is a point of inflection.
We have $g(\alpha) = f(\alpha)(\alpha - a)$ and $g(\beta) = f(\beta)(\beta - a)$ from $(x-a)f(x) = g(x)$ $\qquad \ldots(\lambda)$
Hence, using two-point form for tangent equation gives $$g(x) - g(\beta) = M \cdot (x - \beta).$$ Again using $g(\beta) = f(\beta)(\beta - a) = M(\beta - a)$ gives $$g(x) = M(x - \beta) + M(\beta - a) = M(x - a)$$ Note that although this form is reminiscent of the slope intercept form, x-intercept variant, directly trying to use that initially fails.
We have $g(x) = M(x-a) \implies g(x) - M(x-a) = 0$.
However, $g(x) - M(x-a)$ itself is interesting, since from the data we have $\alpha, \beta$ to be roots of $g(x) - M(x-a)$. $\underline{\text{And this is very surprising !}}$ To see this let $g(x) - M(x-a) = t(x)$ then we have $$t(\alpha) = g(\alpha) - M(\alpha - a) = f(\alpha)(\alpha - a) - f(\alpha)(\alpha - a) = 0$$ where we use $\ldots (\lambda)$ and the given data. Similarly $\beta$ is also a root. But also we know that $M(x-a)$ is the tangent hence it must be so that $g(\alpha)$ and $g(\beta)$ merely touch the tangent at a point and are double roots.
Therefore, using canonical form of a polynomial gives $$t(x) = k(x-\alpha)^2(x-\beta)^2.$$ To figure out $k$ we again turn to the graph of $g$ and the tangent, and because both hands go to $-\infty$ we have that $k=-1.$ Now we re-orient the graph to treat the tangent as the axis to get $$g(x) = M(x-a) - (x-\alpha)^2(x-\beta)^2$$
Here we are interested in finding the least position in the vertical direction so that when we translate back to the initial graph of $g(x)$ we have $M$ to be minimum which is what we want.
Thus taking the derivative one time gives $g'(x) = -2(x-\alpha)(x - \beta) (2x - (\alpha + \beta)) +M$. Now for the graph of the derivative, it is easy to see that $\alpha, \beta$ are roots (indeed this is also true by Double Root Theorem) of $g'(x) - M$.
Hence the mid point is $\frac{\alpha + \beta}{2}$ by Intermediate Value Theorem gives the value of $g'(x) - M = 0$. Note that the $- M$ only dictates the vertical shift. Now this also means the graph of $g(x) - M$ is symmetric and thus we can divide the $\beta - \alpha = 6 \sqrt{3}$ into two parts with $\beta = 3\sqrt{3}$ and $\alpha = -3\sqrt{3}$ and plugging this into the expression for $g'(x)$ gives $g'(x) = -4x(x^2 - 27) + M$. From the graph of $g'(x) - M$ we can get a minimum (viz. $-3$). To see this you can also employ second derivative and equate to $0$ and get two $x \in \{-3, 3\}$ (But in this way you have to reason why $x = 3$ wouldn't work). Plugging it to find $g'(x)$ gives the it to be $-216 + M.$ To make $g'(x)$ a minimum, we equate $-216 + M = 0$ which means $M = \boxed{216}$ $\blacksquare$
¹ .- There is ugly algebra involved which I didn't write. If required I will add it. Added.
As requested, the "ugly" algebra.
\begin{align}
f(\beta) - f(\alpha) = 0 \\
&\implies f(\beta) - f(\alpha) = \frac{g(\alpha)}{\alpha - a} - \frac{g(\beta)}{\beta - \alpha} \\
&\implies \frac{g(\beta)}{\beta - \alpha} = \frac{g(\alpha)}{\alpha - a}\\
&\implies \frac{g(6\sqrt{3} + \alpha)}{6\sqrt{3} + \alpha - a} = \frac{g(\alpha)}{\alpha - a}\\
&\implies g(\beta)\alpha - g(\beta)a = g(\alpha) 6 \sqrt{3} + g(\alpha)\alpha - g(\alpha)a \\
&\implies g(\alpha)a - g(\beta)a = g(\alpha) 6 \sqrt{3} + g(\alpha) \alpha - g(\beta) \alpha \\
&\implies a(g(\alpha) - g(\beta)) = g(\alpha)6\sqrt{3} + \alpha(g(\alpha) - g(\beta)) \\
&\implies a(g(\alpha) - g(\beta)) = g(\alpha)(\beta - \alpha) + \alpha(g(\alpha) - g(\beta)) \\
&\implies a(g(\alpha) - g(\beta)) - \alpha(g(\alpha) - g(\beta)) = g(\alpha)(\beta - \alpha) \\
&\implies (g(\alpha) - g(\beta))(a - \alpha) = g(\alpha)(\beta - \alpha) \\
&\implies (g(\alpha) - g(\beta))(\alpha - a) = g(\alpha)(\alpha - \beta) \\
&\implies \frac{g(\alpha) - g(\beta)}{\alpha - \beta} = \frac{g(\alpha)}{\alpha - a}
\end{align}
This is the what I did originally purely using $(1)$ and the given information that $\beta - \alpha = 6 \sqrt{3}$. Much cleaner way is to use $\ldots(\lambda)$ as shown here, however we only arrive at $\ldots(\lambda)$ after the consideration of the desired equation hence the above way is much preferred (by me).
\begin{align}
&g(\alpha) - g(\beta) = f(\alpha)(\alpha - a) - f(\beta)(\beta - a) \\
&\implies g(\alpha) - g(\beta) = M(\alpha - a - \beta + a) \\
&\implies \frac{g(\alpha) - g(\beta)}{\alpha - \beta} = M
\end{align}
Best Answer
If $x_0, v_0 \in \mathbb{R}^n$, then $t \mapsto p(x_0+v_0t) : \mathbb{R} \to \mathbb{R}$ is a polynomial of degree at most $3$, so it has at most one isolated minimum.
Now assume $p$ has more than one isolated minimum and restrict to the line through any two of them.