# How do you solve x^3+3x^2+x+3<=0?

Apr 23, 2017

$x \le 3$

#### Explanation:

${x}^{3} + 3 {x}^{2} + x + 3 \le 0$

$\Leftrightarrow {x}^{2} \left(x + 3\right) + 1 \left(x + 3\right) \le 0$

or $\left({x}^{2} + 1\right) \left(x + 3\right) \le 0$

As ${x}^{2} + 1$ is always positive,

dividing by $\left({x}^{2} + 1\right)$, we get $x + 3 \le 0$ or $x \le - 3$