Question

How do you solve `x^3-5x^2<x-5` using a sign chart?

Answer 1

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

Explanation:

Let's rewrite the inequality

`x^3-5x^2 < x-5`

`x^3-5x^2-x+5<0`

Let `f(x)=x^3-5x^2-x+5`

`f(1)=1-5-1+5 =0`

so,

`(x-1)` is a factor of `f(x)`

`f(-1)=-1-5+1+5=0`

so,

`(x+1)` is a factor of `f(x)`

So,

`(x+1)(x-1)=x^2-1` is a factor of `f(x)`

We perform a long division to find the 3rd factor

`color(white)(aaaa)` `x^3-5x^2-x+5` `color(white)(aaaa)` `|` `x^2-1`

`color(white)(aaaa)` `x^3` `color(white)(aaaaaa)` `-x` `color(white)(aaaaaaaa)` `|` `x-5`

`color(white)(aaaa)` `0-5x^2` `color(white)(aaa)` `0+5`

`color(white)(aaaaaa)` `-5x^2` `color(white)(aaaaa)` `+5`

`color(white)(aaaaaaaaa)` `0` `color(white)(aaaaa)` `+0`

Therefore,

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

Let's build the sign chart

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

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

Therefore,

`f(x)<0` when `x in ]-oo, -1 [uu]1,5[`