Remainder Theorem - Algebra I | Luna Learn 24

Remainder Theorem - Algebra I

Grade: 9 | Subject: Algebra I | Topic: Remainder Theorem

1. Introduction

The Remainder Theorem is a fundamental concept in algebra that allows you to find the remainder when a polynomial is divided by a linear divisor. This theorem simplifies polynomial division and is a key tool in factorization and solving polynomial equations.

2. Lesson Outline

Definition of the Remainder Theorem
Understanding Polynomial Division
How to Apply the Remainder Theorem
Examples and Applications
Exercises with Solutions
Homework and Revision

3. What is the Remainder Theorem?

The Remainder Theorem states that if a polynomial \( f(x) \) is divided by a linear divisor of the form \( x - c \), the remainder of this division is equal to \( f(c) \). In simpler terms, the remainder when dividing a polynomial by \( x - c \) is the value of the polynomial evaluated at \( c \).

4. Examples

Example 1: Find the remainder when \( f(x) = x^3 - 6x^2 + 11x - 6 \) is divided by \( x - 2 \).

Solution: Using the Remainder Theorem, we evaluate the polynomial at \( x = 2 \).
\( f(2) = 2^3 - 6(2)^2 + 11(2) - 6 = 8 - 24 + 22 - 6 = 0 \).
Therefore, the remainder is 0.

Example 2: Find the remainder when \( f(x) = x^2 - 4x + 3 \) is divided by \( x - 1 \).

Solution: Evaluate the polynomial at \( x = 1 \).
\( f(1) = 1^2 - 4(1) + 3 = 1 - 4 + 3 = 0 \).
Therefore, the remainder is 0.

Example 3: Find the remainder when \( f(x) = 2x^3 + 3x^2 - 5x + 7 \) is divided by \( x + 1 \).

Solution: We need to evaluate \( f(-1) \).
\( f(-1) = 2(-1)^3 + 3(-1)^2 - 5(-1) + 7 = -2 + 3 + 5 + 7 = 13 \).
Therefore, the remainder is 13.

Example 4: Find the remainder when \( f(x) = x^4 - 2x^3 + x^2 - x + 5 \) is divided by \( x - 2 \).

Solution: Evaluate the polynomial at \( x = 2 \).
\( f(2) = 2^4 - 2(2)^3 + 2^2 - 2 + 5 = 16 - 16 + 4 - 2 + 5 = 7 \).
Therefore, the remainder is 7.

Example 5: Find the remainder when \( f(x) = 3x^2 - 5x + 2 \) is divided by \( x - 3 \).

Solution: Evaluate the polynomial at \( x = 3 \).
\( f(3) = 3(3)^2 - 5(3) + 2 = 27 - 15 + 2 = 14 \).
Therefore, the remainder is 14.

Example 6: Find the remainder when \( f(x) = 4x^3 - x^2 + 2x - 5 \) is divided by \( x + 2 \).

Solution: We need to evaluate \( f(-2) \).
\( f(-2) = 4(-2)^3 - (-2)^2 + 2(-2) - 5 = -32 - 4 - 4 - 5 = -45 \).
Therefore, the remainder is -45.

5. Exercises

Try solving the following problems. Solutions are provided below.

Find the remainder when \( f(x) = x^2 + 2x + 1 \) is divided by \( x - 1 \).
Find the remainder when \( f(x) = 2x^3 + 3x^2 - x + 4 \) is divided by \( x + 1 \).
Find the remainder when \( f(x) = x^4 - 3x^3 + 2x^2 - x + 6 \) is divided by \( x - 2 \).