Use the given zero to find the remaining zeros of the function. \( f(x)=x^{3}-7 x^{2}+9 x-63 ; \) zero: \( -3 i \) The remaining zero(s) of \( f \) is(are) (Use a comma to separate answers as needed.)

To find the remaining zeros of the function \( f(x) = x^3 - 7x^2 + 9x - 63 \) given that one zero is \( -3i \), we can use the fact that the coefficients of the polynomial are real numbers. This means that the complex zeros must occur in conjugate pairs. Therefore, if \( -3i \) is a zero, then its conjugate \( 3i \) is also a zero. Next, we can find the remaining zero by performing polynomial long division or synthetic division to divide \( f(x) \) by the quadratic polynomial formed by the two complex zeros, which is \( (x + 3i)(x - 3i) = x^2 + 9 \). Let's perform the division of \( f(x) \) by \( x^2 + 9 \) to find the remaining zero. 1. **Divide \( f(x) \) by \( x^2 + 9 \)**: - The first term of \( f(x) \) is \( x^3 \), and the first term of \( x^2 + 9 \) is \( x^2 \). We divide \( x^3 \) by \( x^2 \) to get \( x \). - Multiply \( x \) by \( x^2 + 9 \) to get \( x^3 + 9x \). - Subtract this from \( f(x) \): \[ (x^3 - 7x^2 + 9x - 63) - (x^3 + 9x) = -7x^2 - 63 \] - Now, divide \( -7x^2 \) by \( x^2 \) to get \( -7 \). - Multiply \( -7 \) by \( x^2 + 9 \) to get \( -7x^2 - 63 \). - Subtract this from the previous result: \[ (-7x^2 - 63) - (-7x^2 - 63) = 0 \] The result of the division is \( x - 7 \). 2. **Set the quotient equal to zero to find the remaining zero**: \[ x - 7 = 0 \implies x = 7 \] Thus, the remaining zeros of the function \( f(x) \) are \( 3i, -3i, 7 \). The remaining zero(s) of \( f \) is(are) \( 7 \).