Answer
The inequality is valid for values less than -2 and values between 1 and 3 (including them) i.e. $(-\infty,-2]\cap [1,3]$
Work Step by Step
First, we are going to find the x-intercepts by equating to zero:
$\dfrac{(x-3)(x+2)}{x-1}=0$
$\dfrac{(x-3)(x+2)}{x-1}\cdot (x-1)=0\cdot (x-1)$
$(x-3)(x+2)=0$
Note that even though we have eliminated $x-1$, the equation has the restriction $x\ne1$, therefore we are going to count it as a critical point.
$x_1=-2$
$x_2=1$
$x_3=3$
These are the critical points. We are going to take four values: one less than -2; one between -2 and 1; one between 1 and 3; and one more than 3 to test in the original equation and check if the inequality is true or not:
First test with a value less than -2:
$\dfrac{(-3-3)(-3+2)}{-3-1}\leq0$
$\dfrac{(-6)(-1)}{-4}\leq0$
$-1.5\leq0 \rightarrow \text{ TRUE}$
Second test with a value between -2 and 1:
$\dfrac{(0-3)(0+2)}{0-1}\leq0$
$\dfrac{(-3)(2)}{-1}\leq0$
$6\leq0 \rightarrow \text{ FALSE}$
Third test with a value between 1 and 3:
$\dfrac{(2-3)(2+2)}{2-1}\leq0$
$\dfrac{(-1)(4)}{1}\leq0$
$-4\leq0 \rightarrow \text{ TRUE}$
Fourth test with a value more than 3:
$\dfrac{(4-3)(4+2)}{4-1}\leq0$
$\dfrac{(1)(6)}{3}\leq0$
$2\leq0 \rightarrow \text{ FALSE}$
These tests show that the inequality $\dfrac{(x-3)(x+2)}{x-1}\leq0$ is valid for values less than -2 and values between 1 and 3 (including them) i.e. $(-\infty,-2]\cap [1,3]$
