If a polynomial is divided by x-2 the remainder is 1 and when it is divided by x-3 the remainder is 2. What will be remainder when the polynomial is divided by `x^(2) -5x +6`.

Answer

To solve the problem step by step, we will use the information given about the polynomial and the properties of polynomial division.

### Step 1: Understand the problem
We are given that when a polynomial \( f(x) \) is divided by \( x-2 \), the remainder is 1. This means:
\[ f(2) = 1 \]

When the polynomial is divided by \( x-3 \), the remainder is 2. This means:
\[ f(3) = 2 \]

We need to find the remainder when \( f(x) \) is divided by \( x^2 - 5x + 6 \).

### Step 2: Factor the divisor
The expression \( x^2 - 5x + 6 \) can be factored as:
\[ x^2 - 5x + 6 = (x-2)(x-3) \]

### Step 3: Assume the form of the remainder
When dividing by a quadratic polynomial, the remainder will be a linear polynomial. Therefore, we can assume the remainder \( R(x) \) has the form:
\[ R(x) = ax + b \]
where \( a \) and \( b \) are constants we need to determine.

### Step 4: Set up equations using known values
Since \( f(x) = (x-2)(x-3)Q(x) + R(x) \) for some polynomial \( Q(x) \), we can substitute \( x = 2 \) and \( x = 3 \) to find \( a \) and \( b \).

1. For \( x = 2 \):
\[ f(2) = 2a + b = 1 \quad \text{(Equation 1)} \]

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

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

We can subtract Equation 1 from Equation 2 to eliminate \( b \):
\[ (3a + b) - (2a + b) = 2 - 1 \]
This simplifies to:
\[ a = 1 \]

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

### Step 6: Write the final remainder
Thus, the remainder when \( f(x) \) is divided by \( x^2 - 5x + 6 \) is:
\[ R(x) = ax + b = 1x - 1 = x - 1 \]

### Conclusion
The remainder when the polynomial is divided by \( x^2 - 5x + 6 \) is:
\[ \boxed{x - 1} \]