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Let's analyze the given problem.

We need to solve for a polynomial \( f(x) \) of degree four that has critical points at \( x = -2, 0, \) and \( 2 \). Since these are critical points, the derivative \( f'(x) \) will have these as its roots. Moreover, for a polynomial of degree four, its derivative \( f'(x) \) will be of degree three. Hence, we can express \( f'(x) \) as:

\[ f'(x) = k(x + 2)x(x - 2) \]

Where \( k \) is a constant. Integrating \( f'(x) \), we get:

\[ f(x) = k\left(\frac{x^3}{3} + x^2 - \frac{4x}{3}\right) + C \]

The condition \( f(x) = f(2\sqrt{2}) \) defines the set \( P \). Since \( f'(x) = 0 \) at \( x = -2, 0, 2 \), these values are candidates for \( P \) if \( f(x) = f(2\sqrt{2}) \) for these \( x \). We need to verify these values to see if they produce the same \( f(x) \) value as \( f(2\sqrt{2}) \). 

First, let's calculate \( f(2\sqrt{2}) \):

\[ f(2\sqrt{2}) = k\left(\frac{(2\sqrt{2})^3}{3} + (2\sqrt{2})^2 - \frac{4(2\sqrt{2})}{3}\right) + C \]

\[ f(2\sqrt{2}) = k\left(\frac{16\sqrt{2}}{3} + 8 - \frac{8\sqrt{2}}{3}\right) + C \]

\[ f(2\sqrt{2}) = k\left(\frac{8\sqrt{2}}{3} + 8\right) + C \]

Now, calculate \( f(-2) \), \( f(0) \), and \( f(2) \) to verify if any of these are equal to \( f(2\sqrt{2}) \):

\[ f(-2) = k\left(\frac{(-2)^3}{3} + (-2)^2 - \frac{4(-2)}{3}\right) + C = k\left(-\frac{8}{3} + 4 + \frac{8}{3}\right) + C = 4k + C \]
\[ f(0) = k\left(0 + 0 - 0\right) + C = C \]
\[ f(2) = k\left(\frac{2^3}{3} + 2^2 - \frac{4 \times 2}{3}\right) + C = k\left(\frac{8}{3} + 4 - \frac{8}{3}\right) + C = 4k + C \]

Notice \( f(-2) \) and \( f(2) \) are the same and could be equal to \( f(2\sqrt{2}) \) if constants align. Since we only know that critical points produce the same value under \( f(x) \), assume \( x = -2, 0, 2, 2\sqrt{2} \) are potential solutions. We now find \( x \) such that \( f(x) = f(2\sqrt{2}) \) for other values:

Given that the problem likely seeks whether \( f(2\sqrt{2}) \) repeats among \( -2, 0, 2 \), and the sum of squares of all elements in \( P \) is likely considering all possible \( x \), if \( f(x) \) value is unique at \( x = 2\sqrt{2} \) then \( P = \{-2, 0, 2, 2\sqrt{2}\} \).

Sum of squares:
\[ (-2)^2 + 0^2 + 2^2 + (2\sqrt{2})^2 = 4 + 0 + 4 + 8 = 16 \]

The sum of squares of all elements of \( P \) is \( 16 \).

Do you need any more details on this problem? Here are some additional queries you might consider:
1. How do we find the constant \( k \) and \( C \) in the polynomial?
2. What are other properties of polynomials with specific critical points?
3. Can we verify if other values of \( x \) belong to \( P \) using a different approach?
4. What happens if the polynomial degree changes?
5. How does the choice of \( k \) affect the graph of \( f(x) \)?

Tip: When dealing with polynomials, always consider their derivatives to find critical points and possible inflection points.

If f(x) is a polynomial of degree four having critical points at -2, 0, 2. If P = {x ∈ R | f(x) = f(2√2)}, then the sum of squares of all the elements of P is:

Solve the problem of a degree four polynomial f(x) with critical points at -2, 0, and 2, and find the sum of squares of the elements of P where f(x) = f(2√2). Suitable for students in Grades 11-12, covering key concepts in calculus and polynomials.

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