Question

How do you solve `(x+5)(x-2)(x-1)(x+1)<0` using a sign chart?

Answer 1

The answer is `x in ]-5,-1 [uu ]1 ,2[`

Explanation:

Let `f(x)=(x+5)(x-2)(x-1)(x+1)`

The domain of `f(x)` is `D_f(x)=RR`

Let's build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `1` `color(white)(aaaa)` `2` `color(white)(aaaaaa)` `+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+1` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-1` `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,-1 [uu ]1 ,2[`