Question

Solve each inequality algebraically.

$(x+1)(x-3)(x-5)>0$

Short answer

Solution set and interval is:

$$\begin{gathered} \left\{x\right|-1<x<3orx>5\left\} \\ \right(-1,3)\cup (5,\infty ) \end{gathered}$$

Step-by-step solution

Step 1. Change the inequality to equal to zero.

Change the inequality $(x+1)(x-3)(x-5)>0$to equal to zero to make it easier and then get the value of x.

$$\begin{gathered} (x+1)(x-3)(x-5)=0 \\ x=-1;x=3;x=5 \end{gathered}$$

Step 2. Form the intervals

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

So the interval we have is:

$(-\infty ,-1)\cup (-1,3)\cup (3,5)\cup (5,\infty )$

Step 3. Form the table

Since, $f\left(x\right)>0$, so $f$is positive.

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

Interval	Number chosen	Resultant
localid="1646893443729" $(-\infty ,-1)$	-3	-96>0
localid="1646893451333" $(-1,3)$	2	9>0
localid="1646893456613" $(3,5)$	4	-5>0
localid="1646893495815" $(5,\infty )$	6	21>0

Step 4. Solution set and interval

Since,

$9>0$and $21>0$satisfy $f$, Thus

Solution set and interval is:

localid="1646157446539" $$\begin{gathered} \left\{x\right|-1<x<3orx>5\left\} \\ \right(-1,3)\cup (5,\infty ) \end{gathered}$$