Answer
The inequality is valid for x-values between -1 and 3 and x-values more than 5 (not including them) i.e. $(-1,3)\cap (5,\infty)$
Work Step by Step
First, we are going to find the x-intercepts by equating to zero:
$(x+1)(x-3)(x-5)=0$
$x_1=-1$
$x_2=3$
$x_3=5$
These are the critical points. We are going to take four values: one less than -1; one between -1 and 3; one between 3 and 5; and one more than 5 to test in the original equation and check if the inequality is true or not:
First test with a value less than -1:
$(-2+1)(-2-3)(-2-5)>0$
$(-1)(-5)(-7)>0$
$-35>0 \rightarrow \text{ FALSE}$
Second test with a value between -1 and 3:
$(0+1)(0-3)(0-5)>0$
$(1)(-3)(-5)>0$
$15>0 \rightarrow \text{ TRUE}$
Third test with a value between 3 and 5:
$(4+1)(4-3)(4-5)>0$
$(5)(1)(-1)>0$
$-5>0 \rightarrow \text{ FALSE}$
Fourth test with a value more than 5:
$(6+1)(6-3)(6-5)>0$
$(7)(3)(1)>0$
$21>0 \rightarrow \text{ TRUE}$
These tests show that the inequality $(x+1)(x-3)(x-5)>0$ is valid for values between -1 and 3 and values more than 5 (not including them) i.e. $(-1,3)\cap (5,\infty)$
