Polynomial `P (x)=x ^(2)-ax +5 and Q(x)=2x ^(3)+5x-(a-3)` when divided by `x-2` have same remainders, then 'a' is equal to

Options

A. 10
B. `-10`
C. 20
D. `-20`

Answer

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

### Step-by-Step Solution:

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

2. **Evaluate \( P(2) \)**:
\[ P(2) = 2^2 - a(2) + 5 \]
\[ P(2) = 4 - 2a + 5 = 9 - 2a \]

3. **Evaluate \( Q(2) \)**:
\[ Q(2) = 2(2^3) + 5(2) - (a - 3) \]
\[ Q(2) = 2(8) + 10 - (a - 3) \]
\[ Q(2) = 16 + 10 - a + 3 = 29 - a \]

4. **Set the Remainders Equal**: Since the remainders are the same, we set \( P(2) \) equal to \( Q(2) \):
\[ 9 - 2a = 29 - a \]

5. **Solve for \( a \)**:
- Rearranging gives:
\[ 9 - 29 = -a + 2a \]
\[ -20 = a \]

6. **Conclusion**: Thus, the value of \( a \) is:
\[ a = -20 \]