How to Solve: Remainder Theorem

How to Solve: Remainder Theorem

(For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script)

Introduction

"Ever wondered how to find the remainder of a polynomial division in seconds—without doing long division? The Remainder Theorem is your shortcut, and it’s a guaranteed question on your exam. Master this, and you’ll save time, avoid mistakes, and pick up easy marks."

What You Need To Know First

Before diving into the Remainder Theorem, you must understand: 1. Polynomials – Expressions like 3x² + 2x – 5 (terms with variables raised to whole-number powers). 2. Evaluating a polynomial – Substituting a value for x (e.g., f(2) means replace x with 2). 3. Division of polynomials – The idea that dividend = divisor × quotient + remainder (like 17 ÷ 5 = 3 with remainder 2).

If any of these are unclear, review them first—this guide assumes you’re solid on them.

Key Vocabulary

Formulas To Know

1. The Remainder Theorem

Formula: If a polynomial f(x) is divided by (x – a), the remainder is f(a).

What it means: - f(x) = your polynomial (e.g., 2x³ – 5x + 1). - (x – a) = the linear divisor (e.g., (x – 4) means a = 4). - The remainder = f(a) (plug a into f(x)).

Memorise This. It’s not usually given on exam sheets.

Example: If f(x) = x² – 3x + 2 is divided by (x – 1), the remainder is f(1) = (1)² – 3(1) + 2 = 0.

2. Extended Remainder Theorem (for (x + a) divisors)

Formula: If f(x) is divided by (x + a), the remainder is f(–a).

What it means: - (x + a) is the same as (x – (–a)), so a in the theorem becomes –a. - Just plug –a into f(x).

Memorise This. It’s a direct extension of the first formula.

Example: If f(x) = x³ + 2x² – x is divided by (x + 2), the remainder is f(–2) = (–2)³ + 2(–2)² – (–2) = –8 + 8 + 2 = 2.

Step-by-Step Method

Follow these steps exactly for every problem.

1. Identify the polynomial f(x) – Write down the dividend (the big polynomial).
2. Identify the divisor (x – a) – Check if it’s (x – a) or (x + a).
3. If it’s (x + a), rewrite it as (x – (–a)) to find a.
4. Find the value of a – This is the number you’ll plug into f(x).
5. Evaluate f(a) – Substitute a (or –a) into f(x) and simplify.
6. State the remainder – The simplified result is the remainder.

Worked Example Using the Steps

Problem: Find the remainder when f(x) = 2x³ – 5x² + 3x – 1 is divided by (x – 2).

Step 1: Identify f(x). - f(x) = 2x³ – 5x² + 3x – 1.

Step 2: Identify the divisor. - Divisor is (x – 2), so a = 2.

Step 3: Find a. - a = 2.

Step 4: Evaluate f(a). - f(2) = 2(2)³ – 5(2)² + 3(2) – 1 - f(2) = 2(8) – 5(4) + 6 – 1 - f(2) = 16 – 20 + 6 – 1 - f(2) = 1.

Step 5: State the remainder. - The remainder is 1.

Worked Examples

Example 1 – Basic

Problem: Find the remainder when f(x) = x² – 4x + 5 is divided by (x – 3).

Solution: 1. f(x) = x² – 4x + 5. 2. Divisor is (x – 3), so a = 3. 3. Evaluate f(3):
- f(3) = (3)² – 4(3) + 5 = 9 – 12 + 5 = 2. 4. Remainder = 2.

What we did and why: We used the Remainder Theorem to avoid long division. Since the divisor is (x – 3), we plugged 3 into f(x) to find the remainder.

Example 2 – Medium (Divisor is (x + a))

Problem: Find the remainder when f(x) = x³ + 2x² – x + 4 is divided by (x + 1).

Solution: 1. f(x) = x³ + 2x² – x + 4. 2. Divisor is (x + 1), which is (x – (–1)), so a = –1. 3. Evaluate f(–1):
- f(–1) = (–1)³ + 2(–1)² – (–1) + 4 = –1 + 2(1) + 1 + 4 = –1 + 2 + 1 + 4 = 6. 4. Remainder = 6.

What we did and why: The divisor was (x + 1), so we rewrote it as (x – (–1)) and plugged –1 into f(x). This is the same as the Remainder Theorem but with a negative a.

Example 3 – Exam Style (Disguised Problem)

Problem: The polynomial p(x) = 3x³ – kx² + 5x – 2 leaves a remainder of 10 when divided by (x – 2). Find the value of k.

Solution: 1. p(x) = 3x³ – kx² + 5x – 2. 2. Divisor is (x – 2), so a = 2. 3. Remainder is p(2) = 10. 4. Evaluate p(2):
- p(2) = 3(2)³ – k(2)² + 5(2) – 2 = 24 – 4k + 10 – 2 = 32 – 4k. 5. Set equal to the given remainder:
- 32 – 4k = 10. 6. Solve for k:
- –4k = 10 – 32
- –4k = –22
- k = 5.5.

What we did and why: The problem gave the remainder and asked for a coefficient. We used the Remainder Theorem to set up an equation (p(2) = 10) and solved for k. This is a common exam trick—don’t panic if the question looks different!

Common Mistakes

Exam Traps

1-Minute Recap

"Okay, let’s lock this in. The Remainder Theorem is your secret weapon for finding remainders fast. Here’s the deal:

1. If you’re dividing a polynomial f(x) by (x – a), the remainder is just f(a). Plug a into the polynomial and simplify.
2. If the divisor is (x + a), rewrite it as (x – (–a)) and plug in –a.
3. Watch out for tricks: negative signs, unknown coefficients, or disguised divisors. Always rewrite the divisor in the form (x – a) first.
4. Double-check your arithmetic—exam questions love to test this!

That’s it. No long division, no stress. Just plug and play. Now go ace that exam!