If `f(x)=2x^(3)+3x^(2)-x+1` is divided with `x+1` and `x-1` and the respective remainders are 5 and -1 then find the remainder when `f(x)` is divided with `x^(2)-1` ?

Answer

To find the remainder when \( f(x) = 2x^3 + 3x^2 - x + 1 \) is divided by \( x^2 - 1 \), we can use the information provided about the remainders when \( f(x) \) is divided by \( x + 1 \) and \( x - 1 \).

### Step 1: Understand the Remainder Theorem
According to the Remainder Theorem, when a polynomial \( f(x) \) is divided by \( x - a \), the remainder is \( f(a) \).

### Step 2: Calculate \( f(-1) \) and \( f(1) \)
We know:
- The remainder when \( f(x) \) is divided by \( x + 1 \) (which is \( x - (-1) \)) is 5.
- The remainder when \( f(x) \) is divided by \( x - 1 \) is -1.

Thus:
- \( f(-1) = 5 \)
- \( f(1) = -1 \)

### Step 3: Set Up the Remainder for \( x^2 - 1 \)
Since \( x^2 - 1 = (x - 1)(x + 1) \), the remainder when dividing \( f(x) \) by \( x^2 - 1 \) will be a linear polynomial of the form \( R(x) = ax + b \).

### Step 4: Create a System of Equations
We can use the values of \( f(-1) \) and \( f(1) \) to set up equations:
1. \( R(-1) = a(-1) + b = -a + b = 5 \)
2. \( R(1) = a(1) + b = a + b = -1 \)

### Step 5: Solve the System of Equations
We now have the following system of equations:
1. \( -a + b = 5 \) (Equation 1)
2. \( a + b = -1 \) (Equation 2)

From Equation 2, we can express \( b \) in terms of \( a \):
\[ b = -1 - a \]

Now substitute \( b \) into Equation 1:
\[ -a + (-1 - a) = 5 \]
\[ -a - 1 - a = 5 \]
\[ -2a - 1 = 5 \]
\[ -2a = 6 \]
\[ a = -3 \]

Now substitute \( a \) back into the expression for \( b \):
\[ b = -1 - (-3) = -1 + 3 = 2 \]

### Step 6: Write the Remainder
Thus, the remainder \( R(x) \) when \( f(x) \) is divided by \( x^2 - 1 \) is:
\[ R(x) = -3x + 2 \]

### Final Answer
The remainder when \( f(x) \) is divided by \( x^2 - 1 \) is:
\[ \boxed{-3x + 2} \]

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