The remainder of the polynomial `f(x)` when divided by `x+1, x+2, x-2` are 6, 15, 3 the remainder of `f(x)` when divided by `(x+1)(x+2)(x-2)` is

Options

A. `2x^(2)-3x+1`
B. `3x^(2)-2x+1`
C. `2x^(2)-x-3`
D. `3x^(2)-2x+1`

Answer

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

### Step 1: Set Up the Remainder
Since the degree of the polynomial \( f(x) \) is greater than or equal to the degree of the divisor \( (x+1)(x+2)(x-2) \), the remainder \( R(x) \) must be a polynomial of degree less than 3. Thus, we can express the remainder as:
\[ R(x) = ax^2 + bx + c \]
where \( a, b, c \) are constants we need to determine.

### Step 2: Use the Remainder Theorem
According to the problem, we have the following conditions based on the remainders:
1. \( f(-1) = 6 \)
2. \( f(-2) = 15 \)
3. \( f(2) = 3 \)

### Step 3: Set Up the Equations
We can substitute the values of \( x \) into \( R(x) \) to create a system of equations.

1. For \( x = -1 \):
\[ R(-1) = a(-1)^2 + b(-1) + c = a - b + c = 6 \quad \text{(Equation 1)} \]

2. For \( x = -2 \):
\[ R(-2) = a(-2)^2 + b(-2) + c = 4a - 2b + c = 15 \quad \text{(Equation 2)} \]

3. For \( x = 2 \):
\[ R(2) = a(2)^2 + b(2) + c = 4a + 2b + c = 3 \quad \text{(Equation 3)} \]

### Step 4: Solve the System of Equations
We now have the following system of equations:
1. \( a - b + c = 6 \) (Equation 1)
2. \( 4a - 2b + c = 15 \) (Equation 2)
3. \( 4a + 2b + c = 3 \) (Equation 3)

#### Step 4.1: Eliminate \( c \)
Subtract Equation 1 from Equation 2:
\[ (4a - 2b + c) - (a - b + c) = 15 - 6 \]
This simplifies to:
\[ 3a - b = 9 \quad \text{(Equation 4)} \]

Subtract Equation 1 from Equation 3:
\[ (4a + 2b + c) - (a - b + c) = 3 - 6 \]
This simplifies to:
\[ 3a + 3b = -3 \quad \Rightarrow \quad a + b = -1 \quad \text{(Equation 5)} \]

#### Step 4.2: Solve for \( a \) and \( b \)
Now we have two equations:
1. \( 3a - b = 9 \) (Equation 4)
2. \( a + b = -1 \) (Equation 5)

From Equation 5, we can express \( b \) in terms of \( a \):
\[ b = -1 - a \]
Substituting this into Equation 4:
\[ 3a - (-1 - a) = 9 \]
This simplifies to:
\[ 3a + 1 + a = 9 \quad \Rightarrow \quad 4a = 8 \quad \Rightarrow \quad a = 2 \]

Now substituting \( a = 2 \) back into Equation 5:
\[ 2 + b = -1 \quad \Rightarrow \quad b = -3 \]

#### Step 4.3: Find \( c \)
Substituting \( a \) and \( b \) back into Equation 1:
\[ 2 - (-3) + c = 6 \quad \Rightarrow \quad 2 + 3 + c = 6 \quad \Rightarrow \quad c = 1 \]

### Step 5: Write the Remainder
Now we have:
\[ a = 2, \quad b = -3, \quad c = 1 \]
Thus, the remainder is:
\[ R(x) = 2x^2 - 3x + 1 \]

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