Question

How do you solve `(x + 5) ( 3x + 9) ( 2x - 4) \geq 0`?

Answer 1

The solution is `x in [-5,-3]uu[2,+oo)`

Explanation:

Let `f(x)=6(x+5)(x+3)(x-2)`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaa)` `-5` `color(white)(aaaaaa)` `-3` `color(white)(aaaaaaaaa)` `2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+5` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `0` `color(white)(aaa)` `+` `color(white)(aa)` `0` `color(white)(aaaa)` `+` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `x+3` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `0` `color(white)(aaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `+` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `x-2` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `0` `color(white)(aaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaaa)` `0` `color(white)(aaa)` `+` `color(white)(aa)` `0` `color(white)(aaaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

Therefore,

`f(x)>=0` when `x in [-5,-3]uu[2,+oo)`

graph{(x+5)(3x+9)(2x-4) [-41.1, 41.08, -20.54, 20.57]}