`f(x)=3x^(3)+kx^(2)-32x+28`
The function f is defined by the equation above where k is a nonzero constant. In the xy-plane the graph of f intersects the x-axis at three points: `(-2, 0), ((3)/(2), 0), and (c, 0)`. What is the value of k?

Options

A. `-25`
B. `-17`
C. `7`
D. `14`

Answer

To find the value of \( k \) in the function \( f(x) = 3x^3 + kx^2 - 32x + 28 \), we know that the function intersects the x-axis at the point \((-2, 0)\). This means that when we substitute \( x = -2 \) into the function, \( f(-2) \) should equal 0.

### Step-by-Step Solution:

1. **Substitute \( x = -2 \) into the function:**
\[ f(-2) = 3(-2)^3 + k(-2)^2 - 32(-2) + 28 \]

2. **Calculate \( (-2)^3 \) and \( (-2)^2 \):**
\[ (-2)^3 = -8 \quad \text{and} \quad (-2)^2 = 4 \]

3. **Substitute these values into the equation:**
\[ f(-2) = 3(-8) + k(4) - 32(-2) + 28 \]
\[ = -24 + 4k + 64 + 28 \]

4. **Combine the constant terms:**
\[ f(-2) = 4k + (-24 + 64 + 28) \]
\[ = 4k + 68 \]

5. **Set \( f(-2) = 0 \) since it intersects the x-axis:**
\[ 4k + 68 = 0 \]

6. **Solve for \( k \):**
\[ 4k = -68 \]
\[ k = -\frac{68}{4} = -17 \]

Thus, the value of \( k \) is \(-17\).