# How do you solve (x+8)(x+2)(x-3)>=0?

Jun 28, 2016

$- 8 \le x \le - 2$ and $3 \le x$

#### Explanation:

Zeros of the function $\left(x + 8\right) \left(x + 2\right) \left(x - 3\right)$ are $\left\{- 8 , - 2 , 3\right\}$ and they divide the real number line in four parts.

$x < - 8$ - In this region all binomials $\left(x + 8\right)$, $\left(x + 2\right)$ and $\left(x - 3\right)$ are negative and hence their product is negative. So this region is not a solution.

$- 8 \le x \le - 2$ - In this region while $\left(x + 8\right)$ is positive, $\left(x + 2\right)$ and $\left(x - 3\right)$ are negative and hence their product is positive. So this region forms a solution.

$- 2 < x < 3$ - In this region while $\left(x + 8\right)$ and $\left(x + 2\right)$ is positive, $\left(x - 3\right)$ is negative and hence their product is negative. So this region does not form a solution.

$3 \le x$ - In this region all binomials $\left(x + 8\right)$, $\left(x + 2\right)$ and $\left(x - 3\right)$ are positive and hence their product is positive. So this region forms a solution.

This can also be checked from the following graph.

graph{(x+8)(x+2)(x-3) [-12.38, 7.62, -100, 100]}