To solve the inequality [tex]\((x+1)(x-3)^2 > 0\)[/tex], let's follow a detailed, step-by-step approach.
### Step 1: Determine the critical points
First, find the values of [tex]\(x\)[/tex] that make either [tex]\(x+1 = 0\)[/tex] or [tex]\((x-3)^2 = 0\)[/tex]:
- [tex]\(x+1 = 0 \implies x = -1\)[/tex]
- [tex]\((x-3)^2 = 0 \implies x = 3\)[/tex]
These are the critical points: [tex]\(x = -1\)[/tex] and [tex]\(x = 3\)[/tex].
### Step 2: Analyze the intervals around the critical points
The critical points divide the number line into three intervals. We need to test the sign of the expression [tex]\((x+1)(x-3)^2\)[/tex] in each of these intervals:
1. [tex]\(x < -1\)[/tex]
2. [tex]\(-1 < x < 3\)[/tex]
3. [tex]\(x > 3\)[/tex]
### Step 3: Test each interval
Choose a test point in each interval and substitute it into the inequality [tex]\((x+1)(x-3)^2 > 0\)[/tex]:
#### Interval 1: [tex]\(x < -1\)[/tex]
Let's choose [tex]\(x = -2\)[/tex]:
[tex]\[
(x+1)(x-3)^2 = (-2+1)(-2-3)^2 = (-1)(-5)^2 = (-1)(25) = -25
\][/tex]
Since [tex]\(-25 < 0\)[/tex], [tex]\((x+1)(x-3)^2 < 0\)[/tex] in this interval.
#### Interval 2: [tex]\(-1 < x < 3\)[/tex]
Let's choose [tex]\(x = 0\)[/tex]:
[tex]\[
(x+1)(x-3)^2 = (0+1)(0-3)^2 = (1)(-3)^2 = (1)(9) = 9
\][/tex]
Since [tex]\(9 > 0\)[/tex], [tex]\((x+1)(x-3)^2 > 0\)[/tex] in this interval.
#### Interval 3: [tex]\(x > 3\)[/tex]
Let's choose [tex]\(x = 4\)[/tex]:
[tex]\[
(x+1)(x-3)^2 = (4+1)(4-3)^2 = (5)(1)^2 = (5)(1) = 5
\][/tex]
Since [tex]\(5 > 0\)[/tex], [tex]\((x+1)(x-3)^2 > 0\)[/tex] in this interval.
### Step 4: Checking the critical points
- At [tex]\(x = -1\)[/tex]:
[tex]\[(x+1)(x-3)^2 = (0)(4) = 0\][/tex]
- At [tex]\(x = 3\)[/tex]:
[tex]\[(x+1)(x-3)^2 = (4)(0) = 0\][/tex]
So, in these intervals, the equality holds, not the inequality.
### Step 5: Combine the results
[tex]\((x+1)(x-3)^2 > 0\)[/tex] holds for:
1. [tex]\( -1 < x < 3 \)[/tex]
2. [tex]\( x > 3 \)[/tex]
Therefore, the solution to the inequality [tex]\((x+1)(x-3)^2 > 0\)[/tex] is:
[tex]\[
x > -1 \text{ and } x \neq 3
\][/tex]
In interval notation, this is:
[tex]\[
(-1, 3) \cup (3, \infty)
\][/tex]
