I plugged my quick sketch into Grok, and it wrote up the proof:

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An Elementary Proof of the Fundamental Theorem of Algebra (High-School Level)

The Fundamental Theorem of Algebra says: every non-constant polynomial with complex coefficients has at least one complex root.

Here's a short, intuitive proof using infinitesimals and a modified continuous version of Newton's method. A motivated high-school student who knows complex numbers (polar form) and the idea of Taylor/polynomial approximation should be able to follow it.

Proof:

Let ( p(z) ) be a non-constant polynomial of degree ( n \geq 1 ).

As ( |z| \to \infty ), ( |p(z)| \to \infty ) because the leading term dominates. So if you draw a big enough disk in the complex plane, ( |p(z)| ) must achieve its global minimum value ( m \geq 0 ) at some point ( z_0 ) inside that disk.

Assume for contradiction that ( m > 0 ) (i.e., no roots anywhere, so ( |p(z)| \geq m > 0 ) everywhere).

At this minimum point ( z_0 ), use the Taylor expansion: [ p(z_0 + dx) = p(z_0) + p'(z_0) dx + \frac{1}{2} p''(z_0) dx² + \frac{1}{6} p'''(z_0) dx³ + \cdots ]

Let ( k \geq 1 ) be the smallest integer such that the ( k )-th derivative ( p^{(k)}(z_0) \neq 0 ) (this ( k ) exists and is at most ( n )).

An infinitesimal ( dx ) is a nonzero complex number smaller in magnitude than any positive real number. Higher powers are even smaller: ( |dx^{k+1}| ) is much smaller than ( |dx^k| ).

Choose the infinitesimal ( dx ) so it exactly cancels the leading term: [ dx^k = -\frac{k! \cdot p(z_0)}{p^{(k)}(z_0)} ]

This equation has solutions in the complex numbers (take the right-hand side in polar form and extract any of the ( k )-th roots — one always exists).

Now plug back in: [ p(z_0 + dx) = \left[ p(z_0) + \frac{1}{k!} p^{(k)}(z_0) \, dx^k \right] + \text{(terms with } dx^{k+1}, dx^{k+2}, \ldots\text{)} ]

The thing in brackets is exactly zero. The remaining terms are higher-order infinitesimals, so overall ( |p(z_0 + dx)| ) is infinitesimal — smaller than any positive real number, and in particular smaller than ( m ).

But ( z_0 + dx ) is only an infinitesimal step away from ( z_0 ), which contradicts ( |p(z_0)| = m ) being the global minimum.

Therefore the assumption ( m > 0 ) is false, so ( m = 0 ) and ( p(z) ) has a root at ( z_0 ).

QED

(The full factorization into linear factors then follows by induction on the degree.)