SOLUTION: What are the real and imaginary zeros in this equation: f(x)=x^3-6x^2+12x-9

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Question 445393: What are the real and imaginary zeros in this equation: f(x)=x^3-6x^2+12x-9
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
f(x)=x^3-6x^2+12x-9
The function can be factored as [confirm this yourself]
(x-3)(x^2-3x+3). To find the zeros, set f(x) = 0:
(x-3)(x^2-3x+3) = 0
The first factor gives us x = 3
Using the quadratic formula on the 2nd factor gives:
x = (3 +- sqrt(9 - 12))/2
x = 3/2 +- sqrt(-3)/2 -> x+=+3%2F2+%2B-+sqrt%283%29%2F2i
So the zeros are: 3,3%2F2+%2B-+sqrt%283%29%2F2i