How do you find all the zeros of F(x)=x3−3x2+9x+13?

1 Answer
George C.
Mar 5, 2016

Identify x=−1 as a zero by examining the coefficients, separate out the corresponding factor (x+1), then solve the remaining quadratic factor by completing the square.

Explanation:

First notice that reversing the signs of the coefficients of the terms with odd degree results in coefficients that sum to zero.

That is −1−3−9+13=0.

So F(−1)=0 and (x+1) is a factor:

x3−3x2+9x+13

=(x+1)(x2−4x+13)

=(x+1)(x2−4x+4+9)

=(x+1)((x−2)2−(3i)2)

=(x+1)((x−2)−3i)((x−2)+3i)

=(x+1)(x−2−3i)(x−2+3i)

So the zeros of F(x) are x=−1 and x=2±3i

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