If the polynomials `2x^(3)+ax^(2)+3x-5` and `x^(3)+x^(2)-4x+a` leave the same remainder when divided by `x-2` , then the value of a is

Options

A. `13//3`
B. `-13//3`
C. `26//3`
D. `-26//3`

Answer

To solve the problem, we need to find the value of \( a \) such that the two polynomials \( P(x) = 2x^3 + ax^2 + 3x - 5 \) and \( Q(x) = x^3 + x^2 - 4x + a \) leave the same remainder when divided by \( x - 2 \).

### Step-by-Step Solution:

1. **Evaluate the Remainder for \( P(x) \)**:
To find the remainder when \( P(x) \) is divided by \( x - 2 \), we substitute \( x = 2 \) into \( P(x) \):
\[ P(2) = 2(2^3) + a(2^2) + 3(2) - 5 \]
\[ = 2(8) + a(4) + 6 - 5 \]
\[ = 16 + 4a + 6 - 5 \]
\[ = 17 + 4a \]

2. **Evaluate the Remainder for \( Q(x) \)**:
Similarly, we find the remainder when \( Q(x) \) is divided by \( x - 2 \) by substituting \( x = 2 \) into \( Q(x) \):
\[ Q(2) = (2^3) + (2^2) - 4(2) + a \]
\[ = 8 + 4 - 8 + a \]
\[ = 4 + a \]

3. **Set the Remainders Equal**:
Since the remainders are equal, we set \( P(2) \) equal to \( Q(2) \):
\[ 17 + 4a = 4 + a \]

4. **Solve for \( a \)**:
Rearranging the equation gives:
\[ 17 + 4a - a = 4 \]
\[ 17 + 3a = 4 \]
\[ 3a = 4 - 17 \]
\[ 3a = -13 \]
\[ a = -\frac{13}{3} \]

Thus, the value of \( a \) is \( -\frac{13}{3} \).