We are given the function:

$h(x) = 3x^4 + 7x^3 + 6x^2 + 28x - 24$

with one known zero $-2i$.

Step 1: Use Complex Conjugate Root Theorem

The coefficients of the polynomial are real, meaning that if $-2i$ is a zero, its conjugate $2i$ must also be a zero.

So far, we have two zeros: $-2i \quad \text{and} \quad 2i.$

Step 2: Factor out the quadratic corresponding to the complex roots

The factor that corresponds to the complex roots $-2i$ and $2i$ is: $(x - (-2i))(x - 2i) = (x + 2i)(x - 2i) = x^2 + 4.$

Thus, $h(x)$ can be factored as: $h(x) = (x^2 + 4)q(x),$ where $q(x)$ is the remaining quadratic factor that we need to determine.

Step 3: Polynomial Division

We will now divide $h(x)$ by $x^2 + 4$ to find $q(x)$.

Using synthetic or long division of the polynomial $3x^4 + 7x^3 + 6x^2 + 28x - 24$ by $x^2 + 4$, we get:

$q(x) = 3x^2 + 7x - 6.$

Step 4: Solve for the remaining zeros

Now, we solve the quadratic $3x^2 + 7x - 6 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 3$, $b = 7$, and $c = -6$.

$x = \frac{-7 \pm \sqrt{7^2 - 4(3)(-6)}}{2(3)} = \frac{-7 \pm \sqrt{49 + 72}}{6} = \frac{-7 \pm \sqrt{121}}{6} = \frac{-7 \pm 11}{6}.$

Thus, the two solutions are: $x = \frac{-7 + 11}{6} = \frac{4}{6} = \frac{2}{3}$ and $x = \frac{-7 - 11}{6} = \frac{-18}{6} = -3.$

Step 5: Final zeros

The zeros of the polynomial $h(x)$ are: $-2i, \, 2i, \, \frac{2}{3}, \, -3.$

Would you like more details or have any questions?

Here are five related questions for further practice:

1. What are the roots of $4x^4 - 64x^2 + 256 = 0$?
2. How does the complex conjugate root theorem apply to real-coefficient polynomials?
3. How do you factor a polynomial given complex roots?
4. What is the general approach to solving higher-degree polynomials?
5. How can synthetic division be applied to find polynomial quotients efficiently?

Tip: When solving polynomial equations, always check for complex conjugate pairs if the coefficients are real, as this can simplify the process.