If the remainders obtained when a polynomial `f(x)` is divided with `x, x-1, x+1` respectively are 1, -2, 3 then find the remainder when `f(x)` is divided with `x^(3)-x` ?

Answer

To solve the problem, we need to find the remainder when the polynomial \( f(x) \) is divided by \( x^3 - x \). We know that the remainders when \( f(x) \) is divided by \( x \), \( x-1 \), and \( x+1 \) are 1, -2, and 3, respectively.

### Step-by-Step Solution:

1. **Understanding the Division**:
The polynomial \( f(x) \) can be expressed in terms of its remainders when divided by its factors. Since \( x^3 - x = x(x-1)(x+1) \), the remainder \( R(x) \) when \( f(x) \) is divided by \( x^3 - x \) can be expressed as a polynomial of degree less than 3:
\[ R(x) = Ax^2 + Bx + C \]

2. **Using the Remainders**:
We know:
- \( f(0) = 1 \) (remainder when divided by \( x \))
- \( f(1) = -2 \) (remainder when divided by \( x-1 \))
- \( f(-1) = 3 \) (remainder when divided by \( x+1 \))

3. **Setting Up the Equations**:
- From \( f(0) = 1 \):
\[ R(0) = A(0)^2 + B(0) + C = C = 1 \]
- From \( f(1) = -2 \):
\[ R(1) = A(1)^2 + B(1) + C = A + B + 1 = -2 \implies A + B = -3 \quad \text{(Equation 1)} \]
- From \( f(-1) = 3 \):
\[ R(-1) = A(-1)^2 + B(-1) + C = A - B + 1 = 3 \implies A - B = 2 \quad \text{(Equation 2)} \]

4. **Solving the System of Equations**:
We have two equations:
- \( A + B = -3 \) (Equation 1)
- \( A - B = 2 \) (Equation 2)

Adding these two equations:
\[ (A + B) + (A - B) = -3 + 2 \implies 2A = -1 \implies A = -\frac{1}{2} \]

Substituting \( A \) back into Equation 1:
\[ -\frac{1}{2} + B = -3 \implies B = -3 + \frac{1}{2} = -\frac{6}{2} + \frac{1}{2} = -\frac{5}{2} \]

5. **Finding the Values**:
We already found \( C = 1 \). Thus, we have:
\[ A = -\frac{1}{2}, \quad B = -\frac{5}{2}, \quad C = 1 \]

6. **Writing the Remainder**:
The remainder \( R(x) \) is:
\[ R(x) = -\frac{1}{2}x^2 - \frac{5}{2}x + 1 \]

### Final Answer:
The remainder when \( f(x) \) is divided by \( x^3 - x \) is:
\[ R(x) = -\frac{1}{2}x^2 - \frac{5}{2}x + 1 \]