# How do you solve x^3-13x-12<=0?

Nov 24, 2016

$x \le - 3 \mathmr{and} x \in \left[- 1 , 4\right]$

#### Explanation:

Let $f \left(x\right) = {x}^{3} - 13 x - 12$. The sum of the coefficients is 0. Xo, x+1 is a

factor.

Let the other factor be ${x}^{2} + a x + b$.

By comparing

the product with f(x), $a = - 1 \mathmr{and} b = - 12$.

The factors of ${x}^{2} - x - 12 a r e x + 3 \mathmr{and} x - 4$. So,

$f \left(x\right) = \left(x + 1\right) \left(x + 3\right) \left(x - 4\right) \le 0$.

One factor only or all the factors should be $\le 0$.

For $x \le - 3 \mathmr{and} x \in \left[- 1 , 4\right]$, this happens.