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 remainders of the polynomials \( P(x) = x^2 - ax + 5 \) and \( Q(x) = 2x^3 + 5x - (a - 3) \) when divided by \( x - 2 \) are equal.

### Step-by-Step Solution:

1. **Find the remainder of \( P(x) \) when divided by \( x - 2 \)**:
- According to the Remainder Theorem, the remainder of a polynomial \( P(x) \) when divided by \( x - c \) is \( P(c) \).
- Here, we need to calculate \( P(2) \):
\[ P(2) = 2^2 - a(2) + 5 = 4 - 2a + 5 = 9 - 2a \]
- So, the remainder \( R_1 \) of \( P(x) \) when divided by \( x - 2 \) is:
\[ R_1 = 9 - 2a \]

2. **Find the remainder of \( Q(x) \) when divided by \( x - 2 \)**:
- We will calculate \( Q(2) \):
\[ Q(2) = 2(2^3) + 5(2) - (a - 3) = 2(8) + 10 - (a - 3) = 16 + 10 - a + 3 = 29 - a \]
- So, the remainder \( R_2 \) of \( Q(x) \) when divided by \( x - 2 \) is:
\[ R_2 = 29 - a \]

3. **Set the remainders equal to each other**:
- According to the problem, the remainders are equal:
\[ R_1 = R_2 \]
- Therefore, we have:
\[ 9 - 2a = 29 - a \]

4. **Solve for 'a'**:
- Rearranging the equation, we get:
\[ 9 - 29 = -a + 2a \]
\[ -20 = a \]
- Thus, the value of \( a \) is:
\[ a = -20 \]

### Final Answer:
The value of \( a \) is \( -20 \).