Chapter A, Problem 22E

To determine

To solve the below inequality in terms of intervals and illustrate the solution set on the real number line -

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

Answer to Problem 22E

The solution set of the inequality is $-3\le x\le -1$ or $x\ge 2$ and the solution set on a real number line -

  

Explanation of Solution

Given: Inequality: $\left(x+1\right)\left(x-2\right)\left(x+3\right)\ge 0$

Formula Used:

An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value.

Real number line is the line whose points are the real numbers. 

Calculation:

Given: Inequality equation is $\left(x+1\right)\left(x-2\right)\left(x+3\right)\ge 0$

Simplifying the above inequality, we have:

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

To solve the above inequalities, we need to find the different intervals for which the inequality gives a value less than or equal to $0$ .

When $x<-3$ :

  $\left(x+1\right)$ is negative, $\left(x-2\right)$ is negative and $\left(x+3\right)$ is negative

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

So, $x<-3$ isnot one of the solutions.

When $x=-3$ :

  $\left(x+1\right)$ is negative, $\left(x-2\right)$ is negative and $\left(x+3\right)$ is Zero

Thus, $\left(x+1\right)\left(x-2\right)\left(x+3\right)=0$

So, $x=-3$ is one of the solutions.

When $-3<x<-1$ :

  $\left(x+1\right)$ is negative, $\left(x-2\right)$ is negative and $\left(x+3\right)$ is positive

Thus, $\left(x+1\right)\left(x-2\right)\left(x+3\right)>0$

So, $-3<x<-1$ is one of the solutions.

When $x=-1$ :

  $\left(x+1\right)$ is Zero, $\left(x-2\right)$ is negative and $\left(x+3\right)$ is positive

Thus, $\left(x+1\right)\left(x-2\right)\left(x+3\right)=0$

So, $x=-1$ is one of the solutions.

When $-1<x<2$ :

  $\left(x+1\right)$ is positive, $\left(x-2\right)$ is negative and $\left(x+3\right)$ is positive

Thus, $\left(x+1\right)\left(x-2\right)\left(x+3\right)<0$

So, $-1<x<2$ is not one of the solutions.

When $x=2$ :

  $\left(x+1\right)$ is positive, $\left(x-2\right)$ is zero and $\left(x+3\right)$ is positive

Thus, $\left(x+1\right)\left(x-2\right)\left(x+3\right)=0$

So, $x=2$ is one of the solutions.

When $x>2$ :

  $\left(x+1\right)$ is positive, $\left(x-2\right)$ is positive and $\left(x+3\right)$ is positive

Thus, $\left(x+1\right)\left(x-2\right)\left(x+3\right)>0$

So, $x>2$ is one of the solutions.

Combining all the solutions, we have:

  $x=-3$ or $-3<x<-1$ or $x=-1$ or $x=2$ or $x>2$

Thus, the solution set is $-3\le x\le -1$ or $x\ge 2$

Drawing the above inequality on a real number line, we have:

  

Conclusion:

Hence, the solution set of the inequality is $-3\le x\le -1$ or $x\ge 2$ and the solution set on a real number line -