If `S` is the set of all real `x` such that `(2x-1)(x^(3)+2x^(2)+x) gt 0`, then `S` contains which of the following intervals :

Options

A. `((-3)/(2),(1)/(2))`
B. `((-3)/(2),(-1)/(4))`
C. `((-1)/(4),(1)/(2))`
D. `((1)/(2),3)`

Answer

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:
\[ (2x-1)(x^3 + 2x^2 + x) \]
We can factor \(x^3 + 2x^2 + x\) by taking \(x\) common:
\[ x(x^2 + 2x + 1) = x(x + 1)^2 \]
Thus, we can rewrite the original expression as:
\[ (2x - 1)(x(x + 1)^2) > 0 \]

### Step 2: Identify the critical points
Next, we find the critical points by setting each factor to zero:
1. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
2. \(x = 0\)
3. \(x + 1 = 0 \Rightarrow x = -1\)

The critical points are \(x = -1\), \(x = 0\), and \(x = \frac{1}{2}\).

### Step 3: Create a number line
We will place the critical points on a number line:
\[ -\infty \quad -1 \quad 0 \quad \frac{1}{2} \quad +\infty \]

### Step 4: Test intervals
We will test the sign of the expression in each of the intervals defined by the critical points:
1. **Interval \((- \infty, -1)\)**: Choose \(x = -2\)
\[ (2(-2) - 1)(-2(-2 + 1)^2) = (-4 - 1)(-2(1)) = (-5)(-2) = 10 > 0 \]
So, this interval is valid.

2. **Interval \((-1, 0)\)**: Choose \(x = -0.5\)
\[ (2(-0.5) - 1)(-0.5(-0.5 + 1)^2) = (-1 - 1)(-0.5(0.25)) = (-2)(-0.125) = 0.25 > 0 \]
So, this interval is valid.

3. **Interval \((0, \frac{1}{2})\)**: Choose \(x = 0.25\)
\[ (2(0.25) - 1)(0.25(0.25 + 1)^2) = (0.5 - 1)(0.25(1.5625)) = (-0.5)(0.390625) < 0 \]
So, this interval is not valid.

4. **Interval \((\frac{1}{2}, +\infty)\)**: Choose \(x = 1\)
\[ (2(1) - 1)(1(1 + 1)^2) = (2 - 1)(1(4)) = (1)(4) = 4 > 0 \]
So, this interval is valid.

### Step 5: Combine valid intervals
The valid intervals for which the expression is greater than zero are:
\[ (-\infty, -1) \cup (-1, 0) \cup \left(\frac{1}{2}, +\infty\right) \]

### Conclusion
Thus, the set \(S\) contains the intervals:
- \((- \infty, -1)\)
- \((-1, 0)\)
- \((\frac{1}{2}, +\infty)\)