If the remainders of the polynomial `f(x)` when divided by `x-1, x-2` are 2, 5 then the remainder of `f(x)` when divided by `(x-1)(x-2)` is

Options

A. 0
B. `1-x`
C. `2x-1`
D. `3x-1`

Answer

To find the remainder of the polynomial \( f(x) \) when divided by \( (x-1)(x-2) \), we can use the Remainder Theorem. According to the theorem, the remainder when a polynomial \( f(x) \) is divided by a polynomial of degree \( n \) will be a polynomial of degree less than \( n \). Since \( (x-1)(x-2) \) is a quadratic polynomial (degree 2), the remainder will be a linear polynomial of the form:

\[ R(x) = ax + b \]

### Step 1: Set up the equations using the given remainders
We know from the problem that:
- The remainder when \( f(x) \) is divided by \( x-1 \) is 2:
\[ f(1) = 2 \]
- The remainder when \( f(x) \) is divided by \( x-2 \) is 5:
\[ f(2) = 5 \]

### Step 2: Substitute into the remainder equation
Substituting \( x = 1 \) into the remainder equation \( R(x) = ax + b \):
\[ R(1) = a(1) + b = a + b = 2 \quad \text{(Equation 1)} \]

Substituting \( x = 2 \) into the remainder equation:
\[ R(2) = a(2) + b = 2a + b = 5 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of two equations:
1. \( a + b = 2 \) (Equation 1)
2. \( 2a + b = 5 \) (Equation 2)

We can solve these equations by elimination. Subtract Equation 1 from Equation 2:
\[ (2a + b) - (a + b) = 5 - 2 \]
This simplifies to:
\[ 2a + b - a - b = 3 \]
\[ a = 3 \]

### Step 4: Substitute back to find \( b \)
Now that we have \( a \), we can substitute it back into Equation 1 to find \( b \):
\[ 3 + b = 2 \]
\[ b = 2 - 3 = -1 \]

### Step 5: Write the final remainder
Now we have both \( a \) and \( b \):
\[ R(x) = ax + b = 3x - 1 \]

Thus, the remainder of \( f(x) \) when divided by \( (x-1)(x-2) \) is:
\[ \boxed{3x - 1} \]