# How do you find all the complex roots of x^3+6x^2+11x+6?

Jan 11, 2016

There are no complex roots.

#### Explanation:

Try to divide the polynomial, either through polynomial long division or synthetic division.

The factors you should try are $\pm 1 , \pm 2 , \pm 3 , \pm 6$.

The first root I encountered was $- 1$, or $\left(x + 1\right)$.

$\frac{{x}^{3} + 6 {x}^{2} + 11 x + 6}{x + 1} = {x}^{2} + 5 x + 6$

This is easily factorable into $\left(x + 2\right) \left(x + 3\right)$, revealing the final two roots of $- 2$ and $- 3$.

Thus, this cubic has three real roots, $- 1 , - 2$, and $- 3$ and no complex roots.

graph{x^3+6x^2+11x+6 [-11.29, 8.71, -4, 6]}