For a polynomial `p(x)` of degree `ge1, p(a)=0` , where a is a real number, then `(x-a)` is a factor of the polynomial `p(x)`
For what value of k, the polynomial `2x^(4)+3x^(3)+2kx^(2)+3x+6` is exactly divisible by `(x+2)` ?

Options

A. `0`
B. `-1`
C. `1`
D. `2`

Answer

To find the value of \( k \) for which the polynomial \( p(x) = 2x^4 + 3x^3 + 2kx^2 + 3x + 6 \) is exactly divisible by \( (x + 2) \), we will use the fact that if \( (x + 2) \) is a factor of \( p(x) \), then \( p(-2) = 0 \).

### Step-by-Step Solution:

1. **Substitute \( x = -2 \) into the polynomial**:
\[ p(-2) = 2(-2)^4 + 3(-2)^3 + 2k(-2)^2 + 3(-2) + 6 \]

2. **Calculate each term**:
- \( 2(-2)^4 = 2 \times 16 = 32 \)
- \( 3(-2)^3 = 3 \times (-8) = -24 \)
- \( 2k(-2)^2 = 2k \times 4 = 8k \)
- \( 3(-2) = -6 \)
- The constant term is \( 6 \).

3. **Combine all the terms**:
\[ p(-2) = 32 - 24 + 8k - 6 + 6 \]

4. **Simplify the expression**:
\[ p(-2) = 32 - 24 - 6 + 6 + 8k = 32 - 24 = 8 + 8k \]
So, we have:
\[ p(-2) = 8 + 8k \]

5. **Set the polynomial equal to zero** (since \( p(-2) = 0 \)):
\[ 8 + 8k = 0 \]

6. **Solve for \( k \)**:
\[ 8k = -8 \implies k = -1 \]

### Final Answer:
The value of \( k \) for which the polynomial \( 2x^4 + 3x^3 + 2kx^2 + 3x + 6 \) is exactly divisible by \( (x + 2) \) is \( k = -1 \).