Answer:
Interval notation is
[tex]\left(-\infty, -5\right)\cup \left(0,1)[/tex]
Solutions:
[tex]\left(-\infty, -5\right)[/tex]
[tex]\left(0,1)[/tex]
Step-by-step explanation:
[tex]x^3 + 4x^2 - 5x < 0[/tex]
In this inequality, luckly we can easily factor it.
[tex]x^3 + 4x^2 - 5x[/tex]
[tex]x(x^2+4x-5)[/tex]
[tex]x(x-1)(x+5)[/tex]
So we have
[tex]x(x-1)(x+5)<0[/tex]
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 [tex]x(x-1)(x+5)=0[/tex]. Here, those are the possible values for [tex]x[/tex] 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.
[tex]x<-5[/tex] [tex]x=5[/tex] [tex]-5<x<0[/tex] [tex]x=0[/tex] [tex]0<x<1[/tex] [tex]x=1[/tex] [tex]x>1[/tex]
[tex]x[/tex] 0
[tex]x-1[/tex] 0
[tex]x+5[/tex] 0
Then, just consider the signal:
[tex]x<-5[/tex] [tex]x=5[/tex] [tex]-5<x<0[/tex] [tex]x=0[/tex] [tex]0<x<1[/tex] [tex]x=1[/tex] [tex]x>1[/tex]
[tex]x[/tex] - - - 0 + + +
[tex]x-1[/tex] - - - - - 0 +
[tex]x+5[/tex] - 0 + + + + +
[tex]x(x-1)(x+5)[/tex] - 0 + 0 - 0 +
When [tex]x(x-1)(x+5)<0[/tex] ?
It happens when [tex]x<-5[/tex] and when [tex]0<x<1[/tex]
The solution is
[tex]\{x \in \mathbb{R} | x<-5 \text{ or } 0<x<1 \}[/tex]
[tex]\left(-\infty, -5\right)\cup \left(0,1)[/tex]
