Let `a!=0a n dp(x)` be a polynomial of degree greater than 2. If `p(x)` leaves reminders `aa n d a` when divided respectively, by `+aa n dx-a ,` the remainder when `p(x)` is divided by `x^2-a^2` is `2x` b. `-2x` c. `x` d. ` x`

Options

A. `2x`
B. `-2x`
C. `x`
D. `-x`

Answer

To solve the problem, we need to find the remainder when the polynomial \( p(x) \) is divided by \( x^2 - a^2 \). We know that \( p(x) \) leaves specific remainders when divided by \( x + a \) and \( x - a \).

### Step 1: Understand the given conditions
We are given:
- \( p(x) \) leaves a remainder \( a \) when divided by \( x + a \).
- \( p(x) \) leaves a remainder \( -a \) when divided by \( x - a \).

This means:
- \( p(-a) = a \)
- \( p(a) = -a \)

### Step 2: Set up the remainder for \( x^2 - a^2 \)
Since \( x^2 - a^2 = (x - a)(x + a) \), the remainder when \( p(x) \) is divided by \( x^2 - a^2 \) can be expressed as a linear polynomial:
\[ R(x) = Ax + B \]
where \( A \) and \( B \) are constants we need to determine.

### Step 3: Use the conditions to find \( A \) and \( B \)
We can use the values of \( p(a) \) and \( p(-a) \) to set up equations:
1. From \( p(a) = -a \):
\[ R(a) = Aa + B = -a \quad \text{(1)} \]

2. From \( p(-a) = a \):
\[ R(-a) = -Aa + B = a \quad \text{(2)} \]

### Step 4: Solve the system of equations
We have the following system of equations:
1. \( Aa + B = -a \)
2. \( -Aa + B = a \)

Now, let's subtract equation (1) from equation (2):
\[ (-Aa + B) - (Aa + B) = a - (-a) \]
This simplifies to:
\[ -2Aa = 2a \]
Dividing both sides by \( -2a \) (assuming \( a \neq 0 \)):
\[ A = -1 \]

Now substitute \( A = -1 \) back into equation (1):
\[ -a + B = -a \]
This implies:
\[ B = 0 \]

### Step 5: Write the remainder
Now we have determined that:
\[ R(x) = -x + 0 = -x \]

### Conclusion
Thus, the remainder when \( p(x) \) is divided by \( x^2 - a^2 \) is:
\[ \boxed{-x} \]