Degree 3 polynomial with integer coefficients with zero 4i and (8/7)

A degree 3 polynomial with integer coefficients must have conjugate pairs of complex roots if it includes any complex roots. Given that \(4i\) is a zero, its conjugate \(-4i\) must also be a zero.

Thus, the polynomial will have three roots:
1. \( 4i \)
2. \( -4i \)
3. \( \frac{8}{7} \)

Now, let's construct the polynomial.

1. The factor corresponding to the root \( 4i \) is \( (x - 4i) \).
2. The factor corresponding to the root \( -4i \) is \( (x + 4i) \).
3. The factor corresponding to the root \( \frac{8}{7} \) is \( \left(7x - 8\right) \) because we need integer coefficients.

First, combine the factors corresponding to \( 4i \) and \( -4i \):

\[
(x - 4i)(x + 4i) = x^2 - (4i)^2 = x^2 - (-16) = x^2 + 16
\]

Now, the polynomial is:

\[
(x^2 + 16)(7x - 8)
\]

Let's expand this:

\[
(x^2 + 16)(7x - 8) = 7x^3 - 8x^2 + 112x - 128
\]

Thus, the degree 3 polynomial is:

\[
7x^3 - 8x^2 + 112x - 128
\]

### Would you like more details or have any questions?

Here are some related questions you might find useful:

1. How do you find the conjugate of a complex root?
2. What is the reasoning behind multiplying polynomials with rational roots?
3. How can you verify if a polynomial has integer coefficients?
4. What are the conditions for a polynomial to have complex roots?
5. Can a polynomial of degree 3 have only real roots?

**Tip:** Whenever a polynomial has a complex root, its conjugate must also be a root if the polynomial has real (or integer) coefficients.

Explore how to construct a degree 3 polynomial with integer coefficients given the roots 4i, -4i, and 8/7. This solution covers important concepts such as complex numbers, conjugate roots, and polynomial expansion. Suitable for advanced algebra students in Grades 11-12.

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