Degree 3 polynomial with integer coefficients with zeros -3i and (9/7).

For a degree 3 polynomial with integer coefficients, the given zeros are \( -3i \) and \( \frac{9}{7} \). Here's how we construct the polynomial step by step:

1. **Use the fact that the coefficients are integers**: Since the polynomial has real coefficients, the complex root \( -3i \) must have its conjugate \( 3i \) as a root as well. Therefore, the roots are:
   \[
   -3i, \ 3i, \ \frac{9}{7}
   \]

2. **Form factors from the roots**:
   - From the roots \( -3i \) and \( 3i \), the corresponding factor is:
     \[
     (x + 3i)(x - 3i) = x^2 + 9
     \]
   - From the root \( \frac{9}{7} \), the corresponding factor is:
     \[
     \left( x - \frac{9}{7} \right)
     \]
     To get integer coefficients, multiply by 7 to eliminate the fraction:
     \[
     7 \left( x - \frac{9}{7} \right) = 7x - 9
     \]

3. **Form the polynomial**:
   Multiply the factors \( (x^2 + 9) \) and \( (7x - 9) \) to obtain the polynomial:
   \[
   P(x) = (x^2 + 9)(7x - 9)
   \]
   Expand this expression:
   \[
   P(x) = (x^2 + 9)(7x - 9) = x^2(7x - 9) + 9(7x - 9)
   \]
   Simplify:
   \[
   P(x) = 7x^3 - 9x^2 + 63x - 81
   \]

Thus, the degree 3 polynomial with integer coefficients is:
\[
P(x) = 7x^3 - 9x^2 + 63x - 81
\]

Would you like further explanation, or do you have any questions about the process?

Here are 5 related questions to expand on this concept:
1. How do complex conjugates affect the formation of polynomials with real coefficients?
2. Can you derive a polynomial if only two roots (one real, one complex) are given?
3. What is the relationship between roots and factors of a polynomial?
4. How would the polynomial change if all the roots were rational numbers?
5. What if you were asked for a polynomial of degree 4 with one real and one complex root?

**Tip**: When dealing with polynomials with complex roots, always include the conjugate of the complex root to ensure the coefficients remain real.

Learn how to construct a degree 3 polynomial with integer coefficients given the zeros -3i and 9/7. This problem uses complex conjugates and rational roots to form the polynomial P(x) = 7x^3 - 9x^2 + 63x - 81. Suitable for advanced algebra students in Grades 11-12.

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