Use the given zero to find the remaining zeros of the function.

f(x)=x^3-2x^2+9x-18;zero;3i
Enter the remaining zeros of f
=

The function is actually quite easy to factor using grouping

f(x_ = x^3 - 2x^2 + 9x - 18
= x^2(x-2) + 9(x-2)
= (x-2)(x^2+9)

so x^2 + 9 = 0 ----> x = ± 3i
and x-2 = 0 ---> x = 2

the other way:
since complex roots always come in conjugate pairs
the other one had to be -3i
and its corresponding factor would have been x^2 = 9
dividing this into the f(x) function would have given us the other remaining factor of (x-2) for the third root of 2

To find the remaining zeros of the function f(x) = x^3 - 2x^2 + 9x - 18, where one zero is given as 3i, we can make use of the complex conjugate theorem. This theorem states that if a polynomial has coefficients that are real numbers, and it has a complex zero a + bi, then its conjugate a - bi is also a zero.

Given that one zero is 3i, the complex conjugate would be -3i, which means that -3i is also a zero of the function.

Therefore, the remaining zeros of f(x) are 3i and -3i.