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The problem from the image asks about a polynomial function \( f(x) \) with real coefficients. It provides the degree, zeros, and a solution point:

- **Degree**: 3
- **Zeros**: \( -2 \), \( 2 \), and \( 2 + 2\sqrt{2}i \)
- **Solution Point**: \( f(-1) = -68 \)

### Analysis:

1. Since the polynomial has real coefficients, the complex zero \( 2 + 2\sqrt{2}i \) must have a conjugate pair: \( 2 - 2\sqrt{2}i \). So, the polynomial has zeros at \( -2 \), \( 2 \), \( 2 + 2\sqrt{2}i \), and \( 2 - 2\sqrt{2}i \).

2. The general form of the polynomial is:
   \[
   f(x) = a(x + 2)(x - 2)\left((x - (2 + 2\sqrt{2}i))(x - (2 - 2\sqrt{2}i))\right)
   \]

3. Simplify the complex pair:
   \[
   (x - (2 + 2\sqrt{2}i))(x - (2 - 2\sqrt{2}i)) = (x - 2)^2 + (2\sqrt{2})^2 = (x - 2)^2 + 8
   \]

4. Now, the polynomial becomes:
   \[
   f(x) = a(x + 2)(x - 2)((x - 2)^2 + 8)
   \]

5. To find the constant \( a \), use the given point \( f(-1) = -68 \):
   \[
   f(-1) = a(-1 + 2)(-1 - 2)((-1 - 2)^2 + 8)
   \]
   \[
   f(-1) = a(1)(-3)(9 + 8) = a(1)(-3)(17) = -51a
   \]
   Set \( f(-1) = -68 \):
   \[
   -51a = -68 \implies a = \frac{-68}{-51} = \frac{68}{51} = \frac{4}{3}
   \]

6. Therefore, the polynomial is:
   \[
   f(x) = \frac{4}{3}(x + 2)(x - 2)((x - 2)^2 + 8)
   \]

Would you like more details on any part of this solution? 

Here are 5 related questions to expand on this:

1. How would you derive the polynomial if only the degree and real zeros were given?
2. Why does the complex conjugate rule apply when a polynomial has real coefficients?
3. How do we handle imaginary roots in polynomial equations?
4. What is the significance of the solution point in finding the constant \( a \)?
5. How can this approach be used for higher-degree polynomials?

**Tip:** Always look for symmetry when working with polynomials, especially when complex numbers are involved.

A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point:
Degree: 3
Zeros: -2, 2, 2 + 2√2i
Solution Point: f(-1) = -68

Explore a degree 3 polynomial function with real coefficients, zeros at -2, 2, and 2 + 2√2i, and a solution point f(-1) = -68. Learn how to apply the conjugate root theorem and solve for the constant a in the polynomial.

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