To solve the inequality \( (2x - 1)(x^3 + 2x^2 + x) > 0 \), we will follow these steps:
### Step 1: Factor the expression
We start with the expression:
\[
(2x - 1)(x^3 + 2x^2 + x)
\]
First, we can factor out \( x \) from the second term:
\[
x^3 + 2x^2 + x = x(x^2 + 2x + 1) = x(x + 1)^2
\]
Thus, the inequality becomes:
\[
(2x - 1)(x(x + 1)^2) > 0
\]
### Step 2: Identify the critical points
Next, we need to find the critical points where the expression equals zero:
1. \( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \)
2. \( x = 0 \)
3. \( (x + 1)^2 = 0 \) gives \( x = -1 \)
The critical points are \( x = -1, 0, \frac{1}{2} \).
### Step 3: Test intervals
We will test the sign of the expression in the intervals defined by the critical points:
1. \( (-\infty, -1) \)
2. \( (-1, 0) \)
3. \( (0, \frac{1}{2}) \)
4. \( (\frac{1}{2}, \infty) \)
**Interval 1: \( (-\infty, -1) \)**
Choose \( x = -2 \):
\[
(2(-2) - 1)(-2(-2 + 1)^2) = (-4 - 1)(-2(1)) = (-5)(-2) = 10 > 0
\]
**Interval 2: \( (-1, 0) \)**
Choose \( x = -0.5 \):
\[
(2(-0.5) - 1)(-0.5(-0.5 + 1)^2) = (-1 - 1)(-0.5(0.5)^2) = (-2)(-0.5 \cdot 0.25) = 0.25 > 0
\]
**Interval 3: \( (0, \frac{1}{2}) \)**
Choose \( x = 0.25 \):
\[
(2(0.25) - 1)(0.25(0.25 + 1)^2) = (0.5 - 1)(0.25(1.25)^2) = (-0.5)(0.25 \cdot 1.5625) < 0
\]
**Interval 4: \( (\frac{1}{2}, \infty) \)**
Choose \( x = 1 \):
\[
(2(1) - 1)(1(1 + 1)^2) = (2 - 1)(1(2)^2) = (1)(4) = 4 > 0
\]
### Step 4: Combine results
From our tests, we find:
- The expression is positive in the intervals \( (-\infty, -1) \) and \( (-1, 0) \).
- The expression is negative in the interval \( (0, \frac{1}{2}) \).
- The expression is positive in the interval \( (\frac{1}{2}, \infty) \).
### Step 5: Write the solution set
Thus, the solution set \( S \) is:
\[
S = (-\infty, -1) \cup (0, \frac{1}{2}) \cup (\frac{1}{2}, \infty)
\]
### Conclusion
The intervals that \( S \) contains are \( (-\infty, -1) \) and \( (\frac{1}{2}, \infty) \).
