We are given the polynomial function:

$P(x) = x^3 + 9x^2 + x + 9$

and one of its zeros, $x = -i$.

Step 1: Complex Conjugate

Since the coefficients of the polynomial are real, if $-i$ is a zero, its complex conjugate $i$ is also a zero. Therefore, $x = i$ is another zero.

Step 2: Factorization Using Known Zeros

We know that $x = -i$ and $x = i$ are zeros, so the polynomial can be divided by $(x + i)(x - i)$, which simplifies to $x^2 + 1$.

Now, we can factor the polynomial by dividing $P(x)$ by $x^2 + 1$.

Step 3: Polynomial Division

We divide:

$\frac{x^3 + 9x^2 + x + 9}{x^2 + 1}$

Performing the division step-by-step:

1. $x^3 \div x^2 = x$
2. Multiply $x \times (x^2 + 1) = x^3 + x$
3. Subtract: $(x^3 + 9x^2 + x + 9) - (x^3 + x) = 9x^2 + 9$
4. $9x^2 \div x^2 = 9$
5. Multiply $9 \times (x^2 + 1) = 9x^2 + 9$
6. Subtract: $(9x^2 + 9) - (9x^2 + 9) = 0$

So, the quotient is $x + 9$, meaning:

$P(x) = (x^2 + 1)(x + 9)$

Step 4: Find the Remaining Zero

The remaining factor is $x + 9$, so the remaining zero is $x = -9$.

Final Answer:

The zeros of the polynomial are: $x = -i, i, -9$

Would you like more details or explanations about any step? Here are five related questions:

1. How do you perform polynomial division in more complex cases?
2. Why must the complex conjugate $i$ also be a zero if $-i$ is?
3. Can you explain the significance of real and complex zeros in polynomials?
4. What happens if a polynomial has only real coefficients but a zero is complex?
5. How does the degree of a polynomial relate to the number of its zeros?

Tip: When working with polynomials with real coefficients, always check if complex zeros occur in conjugate pairs.