Question

How do you solve `((x+3)(x+5))/(x+2)>=0`?

Answer 1

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

Explanation:

We solve this inequality with a sign chart

Let `f(x)=((x+3)(x+5))/(x+2)`

The domain of `f(x)` is `D_f(x)=RR-{-2}`

Let build the sign chart

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

`color(white)(aaaa)` `x+5` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+3` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+`

Therefore,

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