Question

Solve each inequality algebraically.

$(2x-1)(x+2)(x+5)<0$

Short answer

Solution set and the interval is:

$$\begin{gathered} \left\{x\right|x<-5or-2<x<\frac{1}{2}\left\} \\ \right(-\infty ,-5)\cup (-2,\frac{1}{2}) \end{gathered}$$

Step-by-step solution

Step 1. Change the inequality to equal to zero.

Change the inequality equal to zero to make it easier and then get the value of x.

$$\begin{gathered} (2x-1)(x+2)(x+5)=0 \\ 2x-1=0;x+2=0x+5=0 \\ x=\frac{1}{2};x=-2;x=-5 \end{gathered}$$

Step 2. Form the intervals

From the obtained values of x, we can form the interval,

So the interval we have is:

$(-\infty ,-5)\cup (-5,-2)\cup (-2,\frac{1}{2})\cup (\frac{1}{2},\infty )$

Step 3. Form the table

Since, $f\left(x\right)<0$, so is negative.

check a number in each interval and evaluate the function to see whether it is satisfying the function.

Interval	Number chosen	Resultant
$(-\infty ,-5)$	-6	-52<0
$(-5,-2)$	-3	14<0
$(-2,\frac{1}{2})$	0	-10<0
$(\frac{1}{2},\infty )$	1	18<0

Step 4. Solution set and interval

Since,

$-52<0$and $-10<0$satisfy $f$Thus

Solution set and the interval is:

localid="1646158528553" $$\begin{gathered} \left\{x\right|x<-5or-2<x<\frac{1}{2}\left\} \\ \right(-\infty ,-5)\cup (-2,\frac{1}{2}) \end{gathered}$$