# How do you find all the real and complex roots of x^3 + 4x^2 + x + 4=0?

Feb 3, 2016

Factor by grouping to find roots:

$x = - 4$, $x = i$, $x = - i$

#### Explanation:

Factor by grouping:

${x}^{3} + 4 {x}^{2} + x + 4$

$= \left({x}^{3} + 4 {x}^{2}\right) + \left(x + 4\right)$

$= {x}^{2} \left(x + 4\right) + 1 \left(x + 4\right)$

$= \left({x}^{2} + 1\right) \left(x + 4\right)$

$\left(x + 4\right)$ is zero when $x = - 4$

$\left({x}^{2} + 1\right)$ is zero when $x = \pm i$

So the roots of ${x}^{3} + 4 {x}^{2} + x + 4 = 0$ are $x = - 4$ and $x = \pm i$