Question

How do you solve `(x+8)^2(x+5)(x+7)^2>=0` using a sign chart?

Answer 1

The solution is `x in {-8} uu {-7}uu[-5,+oo)`

Explanation:

The inequality is

`(x+8)^2(x+5)(x+7)^2>=0`

Let `f(x)=(x+8)^2(x+5)(x+7)^2`

Let 's build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `-8` `color(white)(aaaaaa)` `-7` `color(white)(aaaaa)` `-5` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `(x+8)^2` `color(white)(aaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `(x+7)^2` `color(white)(aaaa)` `+` `color(white)(aaaa)` ###color(white)(aaaa)#`+` `color(white)(aa)` `0` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+5` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` ###color(white)(aaaa)#`-` `color(white)(aa)` ###color(white)(aaaa)#`-` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `-` `color(white)(a)` `0` `color(white)(aa)` `+`

Therefore,

`f(x)>=0` when `x in {-8} uu {-7}uu[-5,+oo)`

graph{(x+8)^2(x+5)(x+7)^2 [-9.692, -3.533, -1.228, 1.85]}