If `S` is the set of all real `x` such that `(2x-1)/(2x^(3)+3x^(2)+x)` is positive then S cantains

Options

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

Explanation

To solve the problem, we need to determine the set \( S \) of all real numbers \( x \) such that the expression

\[ \frac{2x - 1}{2x^3 + 3x^2 + x} \]

is positive.

### Step-by-step Solution:

1. **Identify the Numerator and Denominator**:
- The numerator is \( 2x - 1 \).
- The denominator is \( 2x^3 + 3x^2 + x \).

2. **Set the Numerator Greater than Zero**:
- For the fraction to be positive, both the numerator and denominator must be either both positive or both negative.
- First, solve \( 2x - 1 > 0 \):
\[ 2x > 1 \implies x > \frac{1}{2} \]

3. **Set the Denominator Greater than Zero**:
- Now, we need to find when \( 2x^3 + 3x^2 + x > 0 \).
- Factor the denominator:
\[ x(2x^2 + 3x + 1) > 0 \]
- Find the roots of \( 2x^2 + 3x + 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{-3 \pm \sqrt{1}}{4} = \frac{-3 \pm 1}{4} \]
- This gives us the roots:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -1 \]

4. **Determine Intervals**:
- The critical points from the numerator and denominator are \( -1, -\frac{1}{2}, \frac{1}{2}, 0 \).
- We will test the sign of the expression in the intervals:
- \( (-\infty, -1) \)
- \( (-1, -\frac{1}{2}) \)
- \( (-\frac{1}{2}, 0) \)
- \( (0, \frac{1}{2}) \)
- \( (\frac{1}{2}, \infty) \)

5. **Test Each Interval**:
- **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[ \frac{2(-2) - 1}{2(-2)^3 + 3(-2)^2 + (-2)} = \frac{-4 - 1}{-16 + 12 - 2} = \frac{-5}{-6} > 0 \]
- **Interval \( (-1, -\frac{1}{2}) \)**: Choose \( x = -0.75 \):
\[ \frac{2(-0.75) - 1}{2(-0.75)^3 + 3(-0.75)^2 + (-0.75)} = \frac{-1.5 - 1}{-0.84375 + 1.6875 - 0.75} < 0 \]
- **Interval \( (-\frac{1}{2}, 0) \)**: Choose \( x = -0.25 \):
\[ \frac{2(-0.25) - 1}{2(-0.25)^3 + 3(-0.25)^2 + (-0.25)} = \frac{-0.5 - 1}{-0.03125 + 0.1875 - 0.25} < 0 \]
- **Interval \( (0, \frac{1}{2}) \)**: Choose \( x = 0.25 \):
\[ \frac{2(0.25) - 1}{2(0.25)^3 + 3(0.25)^2 + (0.25)} = \frac{0.5 - 1}{0.03125 + 0.1875 + 0.25} < 0 \]
- **Interval \( (\frac{1}{2}, \infty) \)**: Choose \( x = 1 \):
\[ \frac{2(1) - 1}{2(1)^3 + 3(1)^2 + (1)} = \frac{2 - 1}{2 + 3 + 1} = \frac{1}{6} > 0 \]

6. **Combine Results**:
- The expression is positive in the intervals \( (-\infty, -1) \) and \( (\frac{1}{2}, \infty) \).
- Exclude points where the denominator is zero: \( x = -1, -\frac{1}{2}, 0 \).

### Final Set \( S \):
\[ S = (-\infty, -1) \cup \left(-\frac{1}{2}, 0\right) \cup \left(\frac{1}{2}, \infty\right) \]