Find the value of k, if (x+2) exactly divides `x^(3)+6x^(2)+4x+k`.

Options

A. 4
B. 6
C. `-8`
D. `-10`

Answer

To find the value of \( k \) such that \( (x + 2) \) exactly divides \( x^3 + 6x^2 + 4x + k \), we can use the Remainder Theorem. According to this theorem, if a polynomial \( f(x) \) is divided by \( (x - a) \), the remainder is \( f(a) \). If \( (x + 2) \) divides the polynomial exactly, then substituting \( x = -2 \) into the polynomial should yield a result of zero.

### Step-by-Step Solution:

1. **Identify the Polynomial**:
We have the polynomial \( f(x) = x^3 + 6x^2 + 4x + k \).

2. **Substitute \( x = -2 \)**:
Since \( (x + 2) \) divides the polynomial, we substitute \( x = -2 \):
\[ f(-2) = (-2)^3 + 6(-2)^2 + 4(-2) + k \]

3. **Calculate Each Term**:
- \( (-2)^3 = -8 \)
- \( 6(-2)^2 = 6 \times 4 = 24 \)
- \( 4(-2) = -8 \)

Now substituting these values into the equation:
\[ f(-2) = -8 + 24 - 8 + k \]

4. **Combine Like Terms**:
Combine the constants:
\[ f(-2) = (-8 - 8 + 24) + k = 8 + k \]

5. **Set the Equation to Zero**:
Since \( (x + 2) \) divides \( f(x) \) exactly, we set \( f(-2) = 0 \):
\[ 8 + k = 0 \]

6. **Solve for \( k \)**:
Rearranging gives:
\[ k = -8 \]

### Final Answer:
The value of \( k \) is \( -8 \).

---