Question

How do you find all the zeros of `F(x)=x^3-3x^2+9x+13`?

Answer 1

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:

`x^3-3x^2+9x+13`

`= (x+1)(x^2-4x+13)`

`=(x+1)(x^2-4x+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`