Which of the following statements includes a function divisible by 2x+1?
I. `f(x)=8x^(2)-2`
II. `g(x)=2x^(2)-9x+4`
III. `h(x)=4x^(3)+2x^(2)-6x-3`

Options

A. I only
B. I and II only
C. I and III only
D. I, II, and III

Answer

To determine which of the given functions is divisible by \(2x + 1\), we need to check if substituting \(x = -\frac{1}{2}\) into each function results in zero. This is based on the Factor Theorem, which states that if \(f(c) = 0\) for some polynomial \(f(x)\), then \(x - c\) is a factor of \(f(x)\).

### Step-by-Step Solution:

1. **Identify the functions:**
- \(f(x) = 8x^2 - 2\)
- \(g(x) = 2x^2 - 9x + 4\)
- \(h(x) = 4x^3 + 2x^2 - 6x - 3\)

2. **Set \(2x + 1 = 0\) to find the value of \(x\):**
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]

3. **Evaluate \(f(-\frac{1}{2})\):**
\[ f(-\frac{1}{2}) = 8\left(-\frac{1}{2}\right)^2 - 2 \]
\[ = 8 \cdot \frac{1}{4} - 2 = 2 - 2 = 0 \]
Thus, \(f(x)\) is divisible by \(2x + 1\).

4. **Evaluate \(g(-\frac{1}{2})\):**
\[ g(-\frac{1}{2}) = 2\left(-\frac{1}{2}\right)^2 - 9\left(-\frac{1}{2}\right) + 4 \]
\[ = 2 \cdot \frac{1}{4} + \frac{9}{2} + 4 = \frac{1}{2} + \frac{9}{2} + 4 = \frac{1}{2} + \frac{9}{2} + \frac{8}{2} = \frac{18}{2} = 9 \]
Thus, \(g(x)\) is not divisible by \(2x + 1\).

5. **Evaluate \(h(-\frac{1}{2})\):**
\[ h(-\frac{1}{2}) = 4\left(-\frac{1}{2}\right)^3 + 2\left(-\frac{1}{2}\right)^2 - 6\left(-\frac{1}{2}\right) - 3 \]
\[ = 4 \cdot \left(-\frac{1}{8}\right) + 2 \cdot \frac{1}{4} + 3 - 3 \]
\[ = -\frac{1}{2} + \frac{1}{2} + 3 - 3 = 0 \]
Thus, \(h(x)\) is also divisible by \(2x + 1\).

### Conclusion:
The functions that are divisible by \(2x + 1\) are \(f(x)\) and \(h(x)\). Therefore, the correct statements are:
- I. \(f(x) = 8x^2 - 2\) (Divisible)
- III. \(h(x) = 4x^3 + 2x^2 - 6x - 3\) (Divisible)

### Final Answer:
The functions divisible by \(2x + 1\) are I and III.

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