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July 30, 2014

July 30, 2014

Posted by **Jill** on Tuesday, April 9, 2013 at 3:21pm.

f(x)=x^3-2x^2+9x-18;zero;3i

Enter the remaining zeros of f

=

- use the given zero -
**Reiny**, Tuesday, April 9, 2013 at 6:40pmThe 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

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