To solve this, we start by analyzing the given information:

Function:

$h(x) = 4x^4 + 5x^3 + 94x^2 + 125x - 150$

Known zero:

One of the zeros is $-5i$.

Since the coefficients of the polynomial are real, the complex zeros must occur in conjugate pairs. Thus, if $-5i$ is a zero, its conjugate $5i$ is also a zero.

This gives us two zeros: $x_1 = -5i, \quad x_2 = 5i$

Step 1: Form a quadratic factor

The product of $(x - (-5i))(x - 5i)$ gives: $(x + 5i)(x - 5i) = x^2 - (5i)^2 = x^2 + 25$

Thus, $x^2 + 25$ is a factor of the polynomial.

Step 2: Polynomial division

We now need to divide $h(x)$ by the quadratic factor $x^2 + 25$ to find the other factor. This will help identify the remaining zeros.

Performing synthetic or long division:

$h(x) \div (x^2 + 25)$

The result of the division is: $4x^2 + 5x - 6$

Step 3: Solve for remaining zeros

We now solve the quadratic $4x^2 + 5x - 6 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$, $b = 5$, and $c = -6$.

$x = \frac{-5 \pm \sqrt{5^2 - 4(4)(-6)}}{2(4)}$ $x = \frac{-5 \pm \sqrt{25 + 96}}{8}$ $x = \frac{-5 \pm \sqrt{121}}{8}$ $x = \frac{-5 \pm 11}{8}$

Thus, the two real roots are: $x_1 = \frac{-5 + 11}{8} = \frac{6}{8} = \frac{3}{4}, \quad x_2 = \frac{-5 - 11}{8} = \frac{-16}{8} = -2$

Final answer:

The remaining zeros of $h(x)$ are: $\frac{3}{4}, \, -2, \, 5i, \, -5i$

Would you like further clarification on any part of this solution?

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Here are five related questions:

1. What is the general form of the quadratic formula?
2. How do you find the conjugate of a complex number?
3. What is synthetic division and when is it used?
4. How can you factor a cubic polynomial?
5. What are the conditions for a polynomial to have complex roots?

Tip: When a polynomial has real coefficients, remember that complex roots always come in conjugate pairs!