Solve the inequality algebraically

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

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.

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

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

is positive or negative.

In the interval $(-\infty ,-3)$ we chose -4, where f isnegative

In the interval $(-3,-2)$

we chose -2.5 , where f is positive.

In the interval $(-2,-1)$ we chose 1.5 , where f is negative.

In the interval $(-1,\infty )$

we chose 4 , where f is positive

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

Here the inequality is strict localid="1646106552197" $(\ge or\le )$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$(-\infty ,-3]\cup [-2,-1]$