The polynomials `ax^(3)+3x^(2)-3 and 2x^3- 5x +a,` when divided by x - 4, leave the same remainder in each case. Find the value of a.

Answer

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

### Step-by-Step Solution:

1. **Identify the Polynomials:**
\[ f(x) = ax^3 + 3x^2 - 3 \]
\[ q(x) = 2x^3 - 5x + a \]

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

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

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

5. **Set the Remainders Equal:**
Since the problem states that both polynomials leave the same remainder when divided by \( x - 4 \), we can set the two expressions equal to each other:
\[ 64a + 45 = 108 + a \]

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

### Final Answer:
The value of \( a \) is \( 1 \).

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