The polynomials `2x^3- 7x^2 + ax -6` and `x^3-8x^2 +(2a + 1)x- 16` leave the same remainder when divided by x- 2. Find the value of a

Answer

To solve the problem, we need to find the value of \( a \) such that the two given polynomials leave the same remainder when divided by \( x - 2 \).

### Step-by-step Solution:

1. **Identify the Polynomials**:
Let \( f(x) = 2x^3 - 7x^2 + ax - 6 \) and \( g(x) = x^3 - 8x^2 + (2a + 1)x - 16 \).

2. **Find the Remainder When Divided by \( x - 2 \)**:
According to the Remainder Theorem, the remainder of a polynomial \( f(x) \) when divided by \( x - c \) is \( f(c) \). Here, we need to evaluate both polynomials at \( x = 2 \).

3. **Calculate \( f(2) \)**:
\[ f(2) = 2(2^3) - 7(2^2) + a(2) - 6 \]
\[ = 2(8) - 7(4) + 2a - 6 \]
\[ = 16 - 28 + 2a - 6 \]
\[ = 2a - 18 \]

4. **Calculate \( g(2) \)**:
\[ g(2) = (2^3) - 8(2^2) + (2a + 1)(2) - 16 \]
\[ = 8 - 8(4) + (4a + 2) - 16 \]
\[ = 8 - 32 + 4a + 2 - 16 \]
\[ = 4a - 38 \]

5. **Set the Remainders Equal**:
Since both polynomials leave the same remainder when divided by \( x - 2 \), we set \( f(2) \) equal to \( g(2) \):
\[ 2a - 18 = 4a - 38 \]

6. **Solve for \( a \)**:
Rearranging the equation:
\[ 38 - 18 = 4a - 2a \]
\[ 20 = 2a \]
\[ a = \frac{20}{2} = 10 \]

### Final Answer:
The value of \( a \) is \( 10 \).