Question

How do you find all the zeros of `h(x) = x³ - 3x² + 4x - 2`?

Answer 1

`x=1` or `x = 1+-i`

Explanation:

`h(x) = x^3-3x^2+4x-2`

Note that the sum of the coefficients is `0`. That is:

`1-3+4-2 = 0`

So `h(1) = 0`, `x=1` is a zero and `(x-1)` is a factor:

`x^3-3x^2+4x-2 = (x-1)(x^2-2x+2)`

The remaining quadratic has negative discriminant, so only Complex zeros, but we can complete the square and use the difference of squares identity to find them:

`x^2-2x+2`

`= x^2-2x+1+1`

`= (x-1)^2-i^2`

`= ((x-1)-i)((x-1)+i)`

`= (x-1-i)(x-1+i)`

So the other two zeros are `x = 1+-i`