Polynomial Remainder Theorem Explained

This problem involves finding the remainder of a polynomial division using the Remainder Theorem. We are given information about the remainders when the polynomial $p(x)$ is divided by simpler linear factors, $(x-1)$ and $(x-2)$. We need to determine the remainder when $p(x)$ is divided by the product of these factors, $(x-1)(x-2)$.

Remainder Theorem Application

The Remainder Theorem states that if a polynomial $p(x)$ is divided by a linear factor $(x-a)$, the remainder is $p(a)$.

• We are given that when $p(x)$ is divided by $(x-1)$, the remainder is 2. According to the Remainder Theorem, this means:

  $ p(1) = 2 $

• We are also given that when $p(x)$ is divided by $(x-2)$, the remainder is 1. This implies:

  $ p(2) = 1 $

Remainder Form for Quadratic Divisor

When a polynomial $p(x)$ is divided by another polynomial, the degree of the remainder must be less than the degree of the divisor.

• In this case, the divisor is $(x-1)(x-2)$, which is a quadratic polynomial (degree 2).
• Therefore, the remainder, let's call it $R(x)$, must be a polynomial of degree less than 2. This means the remainder can be represented in the form $ax + b$, where $a$ and $b$ are constants.

Polynomial Division Equation Setup

We can express the polynomial $p(x)$ using the division algorithm:

$ p(x) = Q(x) \cdot (x-1)(x-2) + R(x) $

Where $Q(x)$ is the quotient and $R(x)$ is the remainder. Substituting $R(x) = ax + b$, we get:

$ p(x) = Q(x) \cdot (x-1)(x-2) + (ax + b) $

Solving Remainder Coefficients

Now, we use the information we have about $p(1)$ and $p(2)$:

• Substitute $x=1$ into the equation:

  $ p(1) = Q(1) \cdot (1-1)(1-2) + (a(1) + b) $

  $ p(1) = Q(1) \cdot (0)(-1) + a + b $

  $ p(1) = 0 + a + b $

  Since we know $p(1) = 2$, we have:

  $ a + b = 2 \quad \quad (1) $

• Substitute $x=2$ into the equation:

  $ p(2) = Q(2) \cdot (2-1)(2-2) + (a(2) + b) $

  $ p(2) = Q(2) \cdot (1)(0) + 2a + b $

  $ p(2) = 0 + 2a + b $

  Since we know $p(2) = 1$, we have:

  $ 2a + b = 1 \quad \quad (2) $

System of Linear Equations Solution

We now have a system of two linear equations with two variables, $a$ and $b$:

To solve this system, we can subtract Equation (1) from Equation (2):

$ (2a + b) - (a + b) = 1 - 2 $

$ 2a + b - a - b = -1 $

$ a = -1 $

Now, substitute the value of $a$ back into Equation (1):

$ (-1) + b = 2 $

$ b = 2 + 1 $

$ b = 3 $

Final Remainder Calculation

We found the coefficients $a = -1$ and $b = 3$. The remainder $R(x)$ was defined as $ax + b$. Therefore, the remainder is:

$ R(x) = (-1)x + 3 $

$ R(x) = 3 - x $

So, when the polynomial $p(x)$ is divided by $(x-1)(x-2)$, the remainder is $3-x$.