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
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
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