Solve the inequality algebraically

$\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{\le }\mathbf{0}$

we have a given inequality

$(x-1)(x-2)(x-3)\le 0$

Zeroes of inequality

$f\left(x\right)=(x-1)(x-2)(x-3)\le 0$

are,

$$\begin{gathered} x=1 \\ x=2 \\ x=3 \end{gathered}$$

Now we use the zeros to separate the real number line into intervals.

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

Now we select a test number in each interval found in Step 3 and evaluate at each number to determine if $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{=}\mathbf{0}$

is positive or negative.

In the interval $(-\infty ,1)$we chose 0, where f isnegative

In the interval$(1,2)$

we chose 1.5 , where f is positive.

In the interval (2,3) we chose 2.5 , where f is negative.

In the interval $(3,\infty )$

we chose 4 , where f ispositive.

We know that our required inequality is$f\left(x\right)\le 0$

Here the inequality is strict$(\ge or\le )$so we have to include the solutions of $f\left(x\right)=0$

in the solution set.

So we want the interval where f is negative or equals to 0.

So required solution set is$(-\infty ,1]\cup [2,3]$