Question

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

Answer 1

The answer is `x in [-8,-2] uu [3, +oo[`

Explanation:

Let `f(x)=(x+8)(x+2)(x-3)`

Let's do the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-8` `color(white)(aaaa)` `-2` `color(white)(aaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+8` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+2` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)>=0`, when `x in [-8,-2] uu [3, +oo[`

graph{(x+8)(x+2)(x-3) [-154.2, 146, -60.4, 89.8]}