Question

Solve each inequality algebraically.

$\frac{(x-3)(x+2)}{(x-1)}\le 0$

Short answer

Solution to the inequality $\frac{(x-3)(x+2)}{(x-1)}\le 0$is $(-\infty ,-2]\cup (1,3]$.

Step-by-step solution

Step 1. Given information 

We have been given an inequality $\frac{(x-3)(x+2)}{(x-1)}\le 0$.

We have to solve this inequality algebraically.

Step 2. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined.  

Assume $f\left(x\right)=\frac{(x-3)(x+2)}{x-1}$

So, $$\begin{gathered} (x-3)(x+2)=0 \\ x=-2,3 \end{gathered}$$

Also,$$\begin{gathered} x-1=0 \\ x=1 \end{gathered}$$

Step 3. Form the intervals  

Using the values of x found in previous step, we can divide the real numbers in the intervals:

$(-\infty ,-2)\cup (-2,1)\cup (1,3)\cup (3,\infty )$

Step 4. Select a number in each interval and evaluate f at the number  

Create the following table:

Interval	$(-\infty ,-2)$	$(-2,1)$	$(1,3)$	$(3,\infty )$
Number chosen	$-4$	$-1$	2	4
Value of f	$f(-4)=-\frac{14}{5}$	$f(-1)=2$	$f\left(2\right)=-4$	$f\left(4\right)=2$
Conclusion	negative	positive	negative	positive

Step 5. Identify the interval 

Since we want to know where f is negative or zero, we conclude that $f\left(x\right)\le 0$in the interval$(-\infty ,-2]\cup (1,3]$.