Question 796639: Find the complex zeros of the polynomial function. Write f in a factored form.
f(x)=x^3-7x^2+25x-39
Use the complex zeros to write f in a factored form.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! I am not sure what is expected, which would be based on what has been recently studies in class.
The rational root theorem tells us that the possible rational zeros are
-13, -3, 3, and 13.
Substitution, or division tell us that x=3 is a zero, so (x-3) should be a factor.
Division shows us that
f(x) = x^3-7x^2+25x-39 = (x-3)(x^2-4x+13)
That is as far as we can factor without going into complex numbers.
x^2-4x+13 has no real zeros, but it has, as expected, two conjugate complex zeros, which we can find using the quadratic formula or completing the square.
Completing the square:
x^2-4x+13 = 0
x^2-4x+4 = -13+4
(x-2)^2 = -9
That leads to
 --->
So we could factor x^2-4x+13 as
x^2-4x+13 = [x-(2-3i)] [x-(2+3i)] = (x-2+3i)(x-2-3i)
Then we can write
f(x) = x^3-7x^2+25x-39 = (x-3)(x-2+3i)(x-2-3i)
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