If the polynomial `x^(4)-3x^(3)-6x^(2)+kx-16` is exactly divisible by `x^(2)-3x+2`, then find the value of `k`.

Answer

To solve the problem, we need to find the value of \( k \) such that the polynomial \( P(x) = x^4 - 3x^3 - 6x^2 + kx - 16 \) is exactly divisible by \( g(x) = x^2 - 3x + 2 \).

### Step-by-Step Solution:

1. **Identify the Roots of \( g(x) \)**:
The polynomial \( g(x) = x^2 - 3x + 2 \) can be factored as:
\[ g(x) = (x - 1)(x - 2) \]
This means the roots are \( x = 1 \) and \( x = 2 \).

2. **Use the Remainder Theorem**:
According to the Remainder Theorem, if \( P(x) \) is divisible by \( g(x) \), then \( P(1) = 0 \) and \( P(2) = 0 \).

3. **Calculate \( P(1) \)**:
Substitute \( x = 1 \) into \( P(x) \):
\[ P(1) = 1^4 - 3(1^3) - 6(1^2) + k(1) - 16 \]
Simplifying this gives:
\[ P(1) = 1 - 3 - 6 + k - 16 = k - 24 \]
Setting this equal to zero for divisibility:
\[ k - 24 = 0 \implies k = 24 \]

4. **Calculate \( P(2) \)** (to verify):
Substitute \( x = 2 \) into \( P(x) \):
\[ P(2) = 2^4 - 3(2^3) - 6(2^2) + k(2) - 16 \]
Simplifying this gives:
\[ P(2) = 16 - 24 - 24 + 2k - 16 = 2k - 48 \]
Setting this equal to zero for divisibility:
\[ 2k - 48 = 0 \implies 2k = 48 \implies k = 24 \]

5. **Conclusion**:
Both calculations confirm that the value of \( k \) is:
\[ \boxed{24} \]