Question

How do you solve `m^3-2m^2-15m>0` using a sign chart?

Answer 1

The solution is `m in ]-3,0 [uu]5, +oo[`

Explanation:

Let's factorise the inequality

`m^3-2m^2-15m>0`

`m(m^2-2m-15)>0`

`m(m+3)(m-5)>0`

Let `f(m)=m(m+3)(m-5)`

Now, we can build the sign chart

`color(white)(aaaa)` `m` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `0` `color(white)(aaaaa)` `5` `color(white)(aaaaa)` `+oo`

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

`color(white)(aaaa)` `m` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `m-5` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

Therefore,

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