Question

Consider the inequality x3 + 4x2 - 5x < 0.

Select all intervals for which the statement is true.
There may be more than one correct answer. Select all correct answers.

Answer 1

Interval notation is

`\left(-\infty, -5\right)\cup \left(0,1)`

Solutions:

`\left(-\infty, -5\right)`

`\left(0,1)`

Explanation:

`x^3 + 4x^2 - 5x < 0`

In this inequality, luckly we can easily factor it.

`x^3 + 4x^2 - 5x`

`x(x^2+4x-5)`

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

So we have

`x(x-1)(x+5)<0`

In exercises of this kind I usually do in my mind, but just to make it clear, let's do a table to organize. This table represents the x-intercepts in order to evaluate the inequality.

Consider
`x(x-1)(x+5)=0`. Here, those are the possible values for
`x` for each factor to be 0:

The first step to complete the table is the x value where the factor will be equal to zero.

`x<-5`
`x=5`
`-5<x<0`
`x=0`
`0<x<1`
`x=1`
`x>1`

`x` 0

`x-1` 0

`x+5` 0

Then, just consider the signal:

`x<-5`
`x=5`
`-5<x<0`
`x=0`
`0<x<1`
`x=1`
`x>1`

`x` - - - 0 + + +

`x-1` - - - - - 0 +

`x+5` - 0 + + + + +

`x(x-1)(x+5)` - 0 + 0 - 0 +

When
`x(x-1)(x+5)<0` ?

It happens when
`x<-5` and when
`0<x<1`

The solution is

`\{x \in \mathbb{R} | x<-5 \text{ or } 0<x<1 \}`

`\left(-\infty, -5\right)\cup \left(0,1)`