We are given the function:

$h(x) = 3x^4 + 10x^3 + 19x^2 + 90x - 72$

and the zero $-3i$.

Step 1: Complex Conjugate Root Theorem

Since the coefficients of the function are real, the complex zeros must come in conjugate pairs. This means that if $-3i$ is a zero, then its conjugate $3i$ must also be a zero of the function.

Thus, we know two zeros of $h(x)$:

• $-3i$
• $3i$

Step 2: Form the quadratic factor

The two complex zeros $-3i$ and $3i$ can be used to form a quadratic factor. Using the fact that the roots $-3i$ and $3i$ correspond to the factor:

$(x - (-3i))(x - 3i) = (x + 3i)(x - 3i)$

This simplifies to:

$(x + 3i)(x - 3i) = x^2 + 9$

Step 3: Divide the original polynomial by the quadratic factor

Now that we have the quadratic factor $x^2 + 9$, we need to divide $h(x)$ by $x^2 + 9$ to find the remaining quadratic factor.

Perform polynomial division:

$h(x) \div (x^2 + 9)$

Let's do this step-by-step:

1. Divide the leading term of $h(x)$ (which is $3x^4$) by the leading term of $x^2 + 9$ (which is $x^2$) to get $3x^2$.
2. Multiply $3x^2$ by $x^2 + 9$, which gives $3x^4 + 27x^2$.
3. Subtract $3x^4 + 27x^2$ from $h(x)$, leaving $10x^3 + (19x^2 - 27x^2) + 90x - 72$, which simplifies to $10x^3 - 8x^2 + 90x - 72$.
4. Now, divide the leading term $10x^3$ by $x^2$, which gives $10x$.
5. Multiply $10x$ by $x^2 + 9$, which gives $10x^3 + 90x$.
6. Subtract $10x^3 + 90x$, leaving $-8x^2 - 72$.
7. Divide the leading term $-8x^2$ by $x^2$, which gives $-8$.
8. Multiply $-8$ by $x^2 + 9$, which gives $-8x^2 - 72$.
9. Subtract $-8x^2 - 72$, leaving a remainder of 0.

Thus, after dividing, we have:

$h(x) = (x^2 + 9)(3x^2 + 10x - 8)$

Step 4: Find the remaining zeros

Now, to find the remaining zeros, solve the quadratic equation:

$3x^2 + 10x - 8 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For $a = 3$, $b = 10$, and $c = -8$:

$x = \frac{-10 \pm \sqrt{10^2 - 4(3)(-8)}}{2(3)}$ $x = \frac{-10 \pm \sqrt{100 + 96}}{6}$ $x = \frac{-10 \pm \sqrt{196}}{6}$ $x = \frac{-10 \pm 14}{6}$

So, the two solutions are:

$x = \frac{-10 + 14}{6} = \frac{4}{6} = \frac{2}{3}$ $x = \frac{-10 - 14}{6} = \frac{-24}{6} = -4$

Final Answer:

The remaining zeros of $h(x)$ are $\frac{2}{3}$ and $-4$.

Recap of the zeros:

• $-3i$
• $3i$
• $\frac{2}{3}$
• $-4$

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Would you like a further breakdown of any step, or do you have additional questions?

Here are 5 related questions you might be interested in:

1. What is the connection between the Complex Conjugate Root Theorem and real-coefficient polynomials?
2. How do you perform synthetic division instead of polynomial long division?
3. Can all polynomials with real coefficients be factored using complex numbers?
4. What would the graph of this polynomial look like given these zeros?
5. How do you apply the quadratic formula to complex numbers?

Tip: When dividing polynomials, always check for the possibility of factoring out simple common terms before attempting long division.