It looks like there's some formatting confusion in the polynomial expression provided. Based on the elements present, I assume you meant to write something like:

$f(x) = x^3 + 4x^2 + 2x + 8$

with $x = 2i$ being a zero of the polynomial. If this is correct, I'll proceed to find all zeros of this polynomial under the assumption that $2i$ is indeed a root.

Step-by-Step Solution

1. Identify Complex Conjugate Pair:

   • Since the coefficients of the polynomial are real, the complex roots must come in conjugate pairs. Therefore, if $x = 2i$ is a root, then $x = -2i$ must also be a root.

2. Form the Quadratic Factor:

   • Given $x = 2i$ and $x = -2i$, we can find a factor of $f(x)$ by multiplying $(x - 2i)(x + 2i)$: $(x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 + 4$
   • Thus, $x^2 + 4$ is a factor of $f(x)$.

3. Divide $f(x)$ by $x^2 + 4$:

   • To find the remaining factor, we divide $f(x) = x^3 + 4x^2 + 2x + 8$ by $x^2 + 4$.
   • Performing the division: $\frac{x^3 + 4x^2 + 2x + 8}{x^2 + 4} = x + 2$
   • So, we have: $f(x) = (x^2 + 4)(x + 2)$

4. Find All Zeros:

   • Setting each factor equal to zero:
     • For $x^2 + 4 = 0$: $x = \pm 2i$
     • For $x + 2 = 0$: $x = -2$

5. Conclusion:

   • The zeros of $f(x)$ are $x = 2i$, $x = -2i$, and $x = -2$.

Would you like a breakdown of the division step or have any other questions?

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Here are some related questions that expand on this problem:

1. How do we verify if a complex root is part of a complex conjugate pair?
2. What if the polynomial had only real coefficients but no imaginary roots?
3. How does synthetic division work for polynomials with complex roots?
4. How do higher-order polynomials factor when given a complex zero?
5. How can we determine the multiplicity of each root?

Tip: Always remember, real polynomials with complex roots will have those roots in conjugate pairs!