Chapter 4.6, Problem 26AYU

In Problems 19-48, solve each inequality algebraically.

${\left(x\text{ - 5}\right)}^2\left(x\text{ + 2}\right)\text{ < 0}$

To determine

To find: The inequality algebraically.

Answer to Problem 26AYU

$\left\{x|-\infty <x\le -3,-2\le x\le -1\right\}$

Explanation of Solution

Given:

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

The zero of $f\left(x\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)$ is $x=-1,-2,-3$.

Use the zeros to separate the real number line into intervals.

$\left(-\infty ,-3\right],\left[-2,-1\right],\left(-1,\infty \right)$

Select a test number in each interval found in above and evaluate $f\left(x\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)$ each number to determine if $f\left(x\right)$ is positive or negative.

Interval	$\left(-\infty ,-3\right]$	$\left[-2,-1\right]$	$\left(-1,\infty \right)$
Number chosen	$-5$	$-2$	0
Value of $f$	$-24$	0	6
Conclusion	Negative	Zero	Positive

Since we want to know where $f\left(x\right)$ is negative and zero, conclude that $f\left(x\right)\le 0$. The solution set is $\left\{x|-\infty <x\le -3,-2\le x\le -1\right\}$ or, using interval notation $\left(-\infty ,-3\right]\cup \left[-3,-2\right]$.