Answer
The inequality is valid for x-values less than -5 and x-values between -2 and 0.5 (not including them) i.e. $(-\infty,-5)\cap (-2,0.5)$.
Work Step by Step
First, we are going to find the x-intercepts by equating them to zero:
$(2x-1)(x+2)(x+5)=0$
$x_1=-5$
$x_2=-2$
$x_3=0.5$
These are the critical points. We are going to take four values: one less than -5; one between -5 and -2; one between -2 and 0.5; and one more than 0.5 to test in the original equation and check if the inequality is true or not:
First test with a value less than -5:
$(2(-6)-1)(-6+2)(-6+5)<0$
$(-13)(-4)(-1)<0$
$-52<0 \rightarrow \text{ TRUE}$
Second test with a value between -5 and -2:
$(2(-3)-1)(-3+2)(-3+5)<0$
$(-7)(-1)(2)<0$
$14<0 \rightarrow \text{ FALSE}$
Third test with a value between -2 and 0.5:
$(2(0)-1)(0+2)(0+5)<0$
$(-1)(2)(5)<0$
$-10<0 \rightarrow \text{ TRUE}$
Fourth test with a value more than 0.5:
$(2(1)-1)(1+2)(1+5)<0$
$(1)(3)(6)<0$
$18<0 \rightarrow \text{ FALSE}$
These tests show that the inequality $(2x-1)(x+2)(x+5)<0$ is valid for values less than -5 and values between -2 and 0.5 (not including them) i.e. $(-\infty,-5)\cap (-2,0.5)$
