If `kx^(3) + 4x^(2) + 3x - 4 ` and `x^(2) - 4x + k` leave the same remainder when divided by `(x - 3)` then the value of k is .

Options

A. 0
B. `-22/13`
C. 1
D. 2

Answer

To solve the problem, we need to find the value of \( k \) such that the polynomials \( kx^3 + 4x^2 + 3x - 4 \) and \( x^2 - 4x + k \) leave the same remainder when divided by \( x - 3 \).

### Step-by-Step Solution:

1. **Identify the Remainder Theorem**:
According to the Remainder Theorem, the remainder of a polynomial \( P(x) \) when divided by \( x - a \) is \( P(a) \). In this case, we will evaluate both polynomials at \( x = 3 \).

2. **Evaluate the first polynomial**:
\[ P_1(x) = kx^3 + 4x^2 + 3x - 4 \]
Substituting \( x = 3 \):
\[ P_1(3) = k(3^3) + 4(3^2) + 3(3) - 4 \]
\[ = k(27) + 4(9) + 9 - 4 \]
\[ = 27k + 36 + 9 - 4 \]
\[ = 27k + 41 \]

3. **Evaluate the second polynomial**:
\[ P_2(x) = x^2 - 4x + k \]
Substituting \( x = 3 \):
\[ P_2(3) = (3^2) - 4(3) + k \]
\[ = 9 - 12 + k \]
\[ = -3 + k \]

4. **Set the two remainders equal**:
Since both polynomials leave the same remainder when divided by \( x - 3 \):
\[ 27k + 41 = -3 + k \]

5. **Solve for \( k \)**:
Rearranging the equation:
\[ 27k - k = -3 - 41 \]
\[ 26k = -44 \]
\[ k = \frac{-44}{26} \]
Simplifying:
\[ k = \frac{-22}{13} \]

### Final Answer:
The value of \( k \) is \( \frac{-22}{13} \).