If $f(x)$ is a polynomial such that $f(x) = [f'(x)]^2$ and $f(2) = 0$, then $f(-2) = \ldots$

Let the degree of the polynomial $f(x)$ be $n$.

The degree of $f'(x)$ is $n-1$, so the degree of $[f'(x)]^2$ is $2(n-1)$.

Equating the degrees, $n = 2n - 2 \Rightarrow n = 2$.

Let $f(x) = ax^2 + bx + c$, where $a \neq 0$.

Then $f'(x) = 2ax + b$.

Substituting into the given equation $f(x) = [f'(x)]^2$:

$ax^2 + bx + c = (2ax + b)^2$

$ax^2 + bx + c = 4a^2x^2 + 4abx + b^2$

Comparing the coefficients of like powers of $x$:

$a = 4a^2 \Rightarrow a = \dfrac{1}{4}$ (since $a \neq 0$)

$b = 4ab \Rightarrow b = b$

$c = b^2$

Thus, the polynomial is $f(x) = \dfrac{1}{4}x^2 + bx + b^2$.

Given $f(2) = 0$:

$\dfrac{1}{4}(2)^2 + 2b + b^2 = 0$

$b^2 + 2b + 1 = 0$

$(b+1)^2 = 0 \Rightarrow b = -1$

Substituting $b = -1$ into $f(x)$:

$f(x) = \dfrac{1}{4}x^2 - x + 1$

Now, evaluating $f(-2)$:

$f(-2) = \dfrac{1}{4}(-2)^2 - (-2) + 1$

$f(-2) = 1 + 2 + 1 = 4$

Answer: $4$