Answer
The inequality is valid for values less than -3 and values between -2 and -1 (including them) i.e. $(-\infty,-3]\cap [-2,-1]$
Work Step by Step
First, we are going to find the x-intercepts by equating them to zero:
$(x+1)(x+2)(x+3)=0$
$x_1=-3$
$x_2=-2$
$x_3=-1$
These are the critical points. We are going to take four values: one less than -3; one between -3 and -2; one between -2 and -1; and one more than -1 to test in the original equation and check if the inequality is true or not:
First test with a value less than -3:
$(-4+1)(-4+2)(-4+3)\leq0$
$(-3)(-2)(-1)\leq0$
$-6\leq0 \rightarrow \text{ TRUE}$
Second test with a value between -3 and -2:
$(-2.5+1)(-2.5+2)(-2.5+3)\leq0$
$(-1.5)(-0.5)(0.5)\leq0$
$0.375\leq0 \rightarrow \text{ FALSE}$
Third test with a value between -2 and -1:
$(-1.5+1)(-1.5+2)(-1.5+3)\leq0$
$(-0.5)(0.5)(1.5)\leq0$
$-0.375\leq0 \rightarrow \text{ TRUE}$
Fourth test with a value greater than -1:
$(0+1)(0+2)(0+3)\leq0$
$(1)(2)(3)\leq0$
$6\leq0 \rightarrow \text{ FALSE}$
These tests show that the inequality $(x+1)(x+2)(x+3)\leq0$ is valid for values less than -3 and values between -2 and -1 (including them) i.e. $(-\infty,-3]\cap [-2,-1]$
