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 \(3-x\).