If the remainder of the polynomial `f(x)` when divided by `x+1` and `x-1` are 7, 3 then the remainder of `f(x)` when devided by `x^(2)-1` is

Options

A. `3x+5`
B. `2x+7`
C. `2x+5`
D. `3x+7`

Answer

To find the remainder of the polynomial \( f(x) \) when divided by \( x^2 - 1 \), we can use the Remainder Theorem and the information given about the remainders when \( f(x) \) is divided by \( x + 1 \) and \( x - 1 \).

### Step-by-Step Solution

1. **Understanding the Problem**:
We know that the remainder of \( f(x) \) when divided by \( x + 1 \) is 7, and when divided by \( x - 1 \) is 3. This means:
\[ f(-1) = 7 \quad \text{and} \quad f(1) = 3 \]

2. **Form of the Remainder**:
Since we are dividing by \( x^2 - 1 \), which can be factored as \( (x + 1)(x - 1) \), the remainder \( R(x) \) when dividing \( f(x) \) by \( x^2 - 1 \) will be a linear polynomial:
\[ R(x) = ax + b \]

3. **Setting Up Equations**:
We will use the values of \( f(-1) \) and \( f(1) \) to find \( a \) and \( b \):
- For \( x = -1 \):
\[ R(-1) = -a + b = 7 \quad \text{(1)} \]
- For \( x = 1 \):
\[ R(1) = a + b = 3 \quad \text{(2)} \]

4. **Solving the Equations**:
We have the system of equations:
\[ -a + b = 7 \quad \text{(1)} \]
\[ a + b = 3 \quad \text{(2)} \]
We can solve these equations simultaneously.

From equation (2), we can express \( b \) in terms of \( a \):
\[ b = 3 - a \quad \text{(3)} \]

Now substitute equation (3) into equation (1):
\[ -a + (3 - a) = 7 \]
Simplifying this gives:
\[ -2a + 3 = 7 \]
\[ -2a = 4 \quad \Rightarrow \quad a = -2 \]

Now substitute \( a = -2 \) back into equation (3) to find \( b \):
\[ b = 3 - (-2) = 3 + 2 = 5 \]

5. **Final Remainder**:
Thus, the remainder \( R(x) \) is:
\[ R(x) = -2x + 5 \]

### Conclusion:
The remainder of \( f(x) \) when divided by \( x^2 - 1 \) is:
\[ R(x) = 2x + 5 \]