To form a polynomial $f(x)$ with real coefficients of degree 5, given the zeros and the requirement for real coefficients, let's analyze the details:

Given Information:

• Degree: 5
• Zeros: 7, $-i$, and $9 + i$

Since the polynomial must have real coefficients, the complex zeros must come in conjugate pairs. Therefore, the conjugates $i$ (the conjugate of $-i$) and $9 - i$ (the conjugate of $9 + i$) are also zeros of the polynomial.

Steps to Form the Polynomial:

1. List All Zeros: The zeros of $f(x)$ are $7$, $-i$, $i$, $9 + i$, and $9 - i$.

2. Form Factors Corresponding to Zeros: Each zero $r$ gives a factor of $(x - r)$ in the polynomial.

   • For zero $7$: the factor is $(x - 7)$.
   • For zeros $-i$ and $i$: the factors are $(x + i)(x - i) = x^2 + 1$.
   • For zeros $9 + i$ and $9 - i$: the factors are $(x - (9 + i))(x - (9 - i)) = (x - 9)^2 + 1 = x^2 - 18x + 82$.

3. Combine All Factors: The polynomial $f(x)$ will be: $f(x) = a(x - 7)(x^2 + 1)(x^2 - 18x + 82)$ where $a$ represents the leading coefficient.

Final Expression:

$f(x) = a(x - 7)(x^2 + 1)(x^2 - 18x + 82)$

This is the polynomial with real coefficients, of degree 5, and with the specified zeros.

Would you like more details on any part of this solution?

Follow-up Questions

1. What happens if the leading coefficient $a$ is specified?
2. How do complex conjugate pairs ensure real coefficients in polynomials?
3. How can we expand $f(x)$ to get a simplified form of the polynomial?
4. What is the role of the degree in determining the form of the polynomial?
5. How can we verify that the given zeros satisfy the polynomial equation?

Tip

Remember that for polynomials with real coefficients, complex roots always appear in conjugate pairs.