Polynomials `ax^(3)+3x^(2)-3 " and" 2x^(3)-5x+a` when divided by (x-4) leave the same remainder. Find the value of a.

Answer

To solve the problem, we need to find the value of \( a \) such that the polynomials \( p_1(x) = ax^3 + 3x^2 - 3 \) and \( p_2(x) = 2x^3 - 5x + a \) leave the same remainder when divided by \( x - 4 \).

### Step-by-Step Solution:

1. **Identify the Polynomials and the Divisor**:
- The first polynomial is \( p_1(x) = ax^3 + 3x^2 - 3 \).
- The second polynomial is \( p_2(x) = 2x^3 - 5x + a \).
- The divisor is \( x - 4 \).

2. **Use the Remainder Theorem**:
- According to the Remainder Theorem, the remainder of a polynomial \( f(x) \) when divided by \( x - c \) is \( f(c) \).
- Here, we will evaluate both polynomials at \( x = 4 \).

3. **Calculate the Remainder for \( p_1(4) \)**:
\[ p_1(4) = a(4^3) + 3(4^2) - 3 \]
\[ = a(64) + 3(16) - 3 \]
\[ = 64a + 48 - 3 \]
\[ = 64a + 45 \]

4. **Calculate the Remainder for \( p_2(4) \)**:
\[ p_2(4) = 2(4^3) - 5(4) + a \]
\[ = 2(64) - 20 + a \]
\[ = 128 - 20 + a \]
\[ = 108 + a \]

5. **Set the Remainders Equal**:
Since both remainders are equal, we set them equal to each other:
\[ 64a + 45 = 108 + a \]

6. **Solve for \( a \)**:
- Rearranging the equation:
\[ 64a - a = 108 - 45 \]
\[ 63a = 63 \]
\[ a = \frac{63}{63} = 1 \]

7. **Conclusion**:
The value of \( a \) is \( 1 \).