A polynomial in x of degree greater than three, leaves remainders 1, -2 and-1 when divided, respectively, by (x-l ), (x + 2) and (x + 1 ). What will be the remainder when is divided by (x - 1) (x +2) (x + 1).

Options

A. (7x)^2 + (3/2)x - (2/3)
B. `-2x`
C. x
D. `-x`

Explanation

To solve the problem, we need to find the remainder of a polynomial \( F(x) \) when it is divided by \( (x - 1)(x + 2)(x + 1) \). We know that the polynomial leaves specific remainders when divided by \( (x - 1) \), \( (x + 2) \), and \( (x + 1) \).

### Step-by-Step Solution:

1. **Understanding the Polynomial Form**:
Since \( F(x) \) is a polynomial of degree greater than 3, we can express it as:
\[ F(x) = (x - 1)(x + 2)(x + 1)Q(x) + Ax^2 + Bx + C \]
where \( Q(x) \) is the quotient and \( Ax^2 + Bx + C \) is the remainder we need to find.

2. **Using Remainders**:
We know:
- \( F(1) = 2 \)
- \( F(-2) = 1 \)
- \( F(-1) = -1 \)

We can substitute these values into the polynomial form to create equations.

3. **Setting Up the Equations**:
- For \( F(1) = 2 \):
\[ A(1)^2 + B(1) + C = 2 \implies A + B + C = 2 \quad \text{(Equation 1)} \]
- For \( F(-2) = 1 \):
\[ A(-2)^2 + B(-2) + C = 1 \implies 4A - 2B + C = 1 \quad \text{(Equation 2)} \]
- For \( F(-1) = -1 \):
\[ A(-1)^2 + B(-1) + C = -1 \implies A - B + C = -1 \quad \text{(Equation 3)} \]

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

From Equation (1), we can express \( C \):
\[ C = 2 - A - B \]

Substituting \( C \) into Equations (2) and (3):
- Substituting into Equation (2):
\[ 4A - 2B + (2 - A - B) = 1 \implies 3A - 3B + 2 = 1 \implies 3A - 3B = -1 \implies A - B = -\frac{1}{3} \quad \text{(Equation 4)} \]
- Substituting into Equation (3):
\[ A - B + (2 - A - B) = -1 \implies -2B + 2 = -1 \implies -2B = -3 \implies B = \frac{3}{2} \]

5. **Finding A and C**:
Substitute \( B = \frac{3}{2} \) back into Equation (4):
\[ A - \frac{3}{2} = -\frac{1}{3} \implies A = -\frac{1}{3} + \frac{3}{2} = -\frac{1}{3} + \frac{9}{6} = \frac{7}{6} \]

Now substitute \( A \) and \( B \) back into Equation (1) to find \( C \):
\[ \frac{7}{6} + \frac{3}{2} + C = 2 \implies C = 2 - \frac{7}{6} - \frac{9}{6} = 2 - \frac{16}{6} = 2 - \frac{8}{3} = \frac{6}{3} - \frac{8}{3} = -\frac{2}{3} \]

6. **Final Remainder**:
Now we have:
\[ A = \frac{7}{6}, \quad B = \frac{3}{2}, \quad C = -\frac{2}{3} \]
Therefore, the remainder when \( F(x) \) is divided by \( (x - 1)(x + 2)(x + 1) \) is:
\[ R(x) = \frac{7}{6}x^2 + \frac{3}{2}x - \frac{2}{3} \]