Question

How do you find all the zeros of `x^3+x^2-4x+6`?

Answer 1

The only real zero of `x^3+x^2-4x+6` is `-3`. However, if we extend the domain to Complex numbers `1+2i` and `1-2i` are other two zeros.

Explanation:

In `x^3+x^2-4x+6`, zeros of the function would be factors of `6` i.e. among `{1,-1,2,-2,3,-3,6,-6}`.

It is apparent that one such zero is `-3` as putting `x=-3` in `x^3+x^2-4x+6`, we get `(-3)^3+(-3)^2-4(-3)+6=-27+9+12+6=0`.

Hence `(x+3)` is a factor of `x^3+x^2-4x+6` and dividing latter by former we get `x^2-2x+2`, for which discriminant is `(-2)^2-4*1*2=-4` and hence cannot be factorized with real coefficients.

Hence, the only zero of `x^3+x^2-4x+6` is `-3`.

However, if domain of `x` is extended to complex numbers, we will also have `(2+-sqrt(-4))/2` (as they are zeros of `x^2-2x+2` using quadratic formula) or `1+2i` and `1-2i`.