Find the remainder when `x^(23)` is divided by `x^(2) -3x + 2`

Answer

To find the remainder when \( x^{23} \) is divided by \( x^2 - 3x + 2 \), we can follow these steps:

### Step 1: Factor the Quadratic Polynomial
First, we need to factor the quadratic polynomial \( x^2 - 3x + 2 \).

The factors of \( x^2 - 3x + 2 \) can be found by looking for two numbers that multiply to \( 2 \) (the constant term) and add to \( -3 \) (the coefficient of \( x \)). The numbers \( -1 \) and \( -2 \) work, so we can factor it as:
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

### Step 2: Use the Remainder Theorem
According to the Remainder Theorem, the remainder of a polynomial \( f(x) \) when divided by \( (x - a) \) is \( f(a) \). In this case, we will evaluate \( x^{23} \) at the roots of the factors we found: \( x = 1 \) and \( x = 2 \).

### Step 3: Evaluate \( x^{23} \) at \( x = 1 \) and \( x = 2 \)
1. For \( x = 1 \):
\[ f(1) = 1^{23} = 1 \]

2. For \( x = 2 \):
\[ f(2) = 2^{23} \]

### Step 4: Write the Remainder as a Linear Combination
Since the divisor is a quadratic polynomial, the remainder will be of the form \( ax + b \). We can set up a system of equations using the values we found:
\[ \begin{align*} a(1) + b &= 1 \quad \text{(from } x = 1\text{)} \\ a(2) + b &= 2^{23} \quad \text{(from } x = 2\text{)} \end{align*} \]

### Step 5: Solve the System of Equations
From the first equation:
\[ a + b = 1 \quad \text{(1)} \]
From the second equation:
\[ 2a + b = 2^{23} \quad \text{(2)} \]

Now, we can subtract equation (1) from equation (2):
\[ (2a + b) - (a + b) = 2^{23} - 1 \]
This simplifies to:
\[ a = 2^{23} - 1 \]

Substituting \( a \) back into equation (1):
\[ (2^{23} - 1) + b = 1 \\ b = 1 - (2^{23} - 1) \\ b = 2 - 2^{23} \]

### Step 6: Write the Final Remainder
Thus, the remainder when \( x^{23} \) is divided by \( x^2 - 3x + 2 \) is:
\[ R(x) = (2^{23} - 1)x + (2 - 2^{23}) \]

### Summary
The remainder when \( x^{23} \) is divided by \( x^2 - 3x + 2 \) is:
\[ R(x) = (2^{23} - 1)x + (2 - 2^{23}) \]