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

Apr 9, 2016

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

#### Explanation:

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

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

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

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

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