Question

How do you solve `p^5-p>0` using a sign chart?

Answer 1

The answer is `=p in ] -1,0 [uu ] 1,+oo [`

Explanation:

Let's factorise the equation

`p^5-p=p(p^4-1)=p(p^2+1)(p^2-1)`

`=p(p^2+1)(p+1)(p-1)`

The term `(p^2+1)>0`

Let `f(p)=p^5-p`

Let's do the sign chart

`color(white)(aaaa)` `p` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `0` `color(white)(aaaa)` `1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `p+1` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `p` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `p-1` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(p)` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

So `f(p)>0` when `p in ] -1,0 [uu ] 1,+oo [`

graph{x^5-x [-8.89, 8.89, -4.444, 4.445]}