Question

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

Answer 1

`-8 <= x <= -2` and `3<=x`

Explanation:

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

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

`-8 <= x <= -2` - In this region while `(x+8)` is positive, `(x+2)` and `(x-3)` are negative and hence their product is positive. So this region forms a solution.

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

`3<=x` - In this region all binomials `(x+8)`, `(x+2)` and `(x-3)` 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]}