How to Solve: Factor Theorem

How to Solve: Factor Theorem

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

Introduction

"Ever spent 10 minutes trying to factor a cubic polynomial, only to hit a dead end? The Factor Theorem is your shortcut—it tells you instantly whether (x – a) is a factor, saving you time and stress on exam day."

What You Need To Know First

Before diving into the Factor Theorem, ensure you understand: 1. Polynomials: Expressions like 3x³ – 2x² + 5x – 7 (terms with variables raised to whole-number powers). 2. Evaluating Polynomials: Substituting a value (e.g., x = 2) into a polynomial to find its output. 3. Factors of a Polynomial: If (x – a) is a factor, the polynomial equals zero when x = a.

Key Vocabulary

Formulas To Know

1. Factor Theorem

Formula: If f(a) = 0, then (x – a) is a factor of f(x). Variables: - f(x) = any polynomial (e.g., 2x³ – x + 1). - a = a number (e.g., a = 2). MEMORISE THIS – It’s the core of the topic.

2. Remainder Theorem (Bonus)

Formula: If f(x) is divided by (x – a), the remainder is f(a). Variables: - Same as above. Given on exam sheet (but understanding it helps with the Factor Theorem).

Step-by-Step Method

Goal: Determine if (x – a) is a factor of f(x) and use it to factor the polynomial.

Steps:

1. Identify a:
2. Look at the potential factor (x – a). The number a is the value to test.

3. Example: For (x – 3), a = 3.

4. Evaluate f(a):

5. Substitute x = a into f(x).

6. Calculate the result. If f(a) = 0, (x – a) is a factor.

7. Check for Zero:

8. If f(a) = 0 → (x – a) is a factor.

9. If f(a) ≠ 0 → (x – a) is not a factor.

10. Factor the Polynomial (if applicable):

11. If (x – a) is a factor, use polynomial division (long or synthetic) to divide f(x) by (x – a).

12. The quotient is the other factor(s).

13. Repeat (if needed):

14. For higher-degree polynomials, test other possible roots (e.g., from the Rational Root Theorem).

Worked Example Using the Steps

Problem: Is (x – 2) a factor of f(x) = x³ – 4x² + 5x – 2? If yes, factor f(x) completely.

Step 1: Identify a. - Potential factor: (x – 2) → a = 2.

Step 2: Evaluate f(2). - f(2) = (2)³ – 4(2)² + 5(2) – 2 - = 8 – 16 + 10 – 2 - = 0

Step 3: Check for zero. - f(2) = 0 → (x – 2) is a factor.

Step 4: Factor the polynomial. - Divide f(x) by (x – 2) using synthetic division: 2 | 1 -4 5 -2
| 2 -4 2
----------------
1 -2 1 0 - Quotient: x² – 2x + 1. - So, f(x) = (x – 2)(x² – 2x + 1).

Step 5: Factor further (if possible). - x² – 2x + 1 = (x – 1)². - Final factorisation: f(x) = (x – 2)(x – 1)².

What we did and why: - We tested x = 2 because (x – 2) was the suspected factor. - Since f(2) = 0, we confirmed it was a factor and used division to break down the polynomial.

Worked Examples

Example 1 – Basic

Problem: Show that (x + 1) is a factor of f(x) = x³ + 2x² – 5x – 6.

Step 1: Identify a. - (x + 1) = (x – (–1)) → a = –1.

Step 2: Evaluate f(–1). - f(–1) = (–1)³ + 2(–1)² – 5(–1) – 6 - = –1 + 2 + 5 – 6 - = 0

Step 3: Check for zero. - f(–1) = 0 → (x + 1) is a factor.

What we did and why: - We rewrote (x + 1) as (x – (–1)) to match the Factor Theorem form. - Substituting x = –1 gave zero, confirming the factor.

Example 2 – Medium

Problem: Factor f(x) = 2x³ – 3x² – 8x + 12 completely, given that (x – 2) is a factor.

Step 1: Verify (x – 2) is a factor. - f(2) = 2(8) – 3(4) – 8(2) + 12 = 16 – 12 – 16 + 12 = 0 → Confirmed.

Step 2: Divide f(x) by (x – 2) using synthetic division. 2 | 2 -3 -8 12
| 4 2 -12
----------------
2 1 -6 0 - Quotient: 2x² + x – 6.

Step 3: Factor the quotient. - 2x² + x – 6 = (2x – 3)(x + 2).

Step 4: Write the complete factorisation. - f(x) = (x – 2)(2x – 3)(x + 2).

What we did and why: - We used the given factor (x – 2) to break down the cubic polynomial. - Synthetic division gave a quadratic, which we factored further.

Example 3 – Exam Style

Problem: The polynomial f(x) = x³ + kx² – 4x – 12 has a factor of (x + 3). Find the value of k and factor f(x) completely.

Step 1: Use the Factor Theorem. - (x + 3) is a factor → f(–3) = 0.

Step 2: Substitute x = –3 and solve for k. - f(–3) = (–3)³ + k(–3)² – 4(–3) – 12 = 0 - –27 + 9k + 12 – 12 = 0 - 9k – 27 = 0 - k = 3.

Step 3: Rewrite f(x) with k = 3. - f(x) = x³ + 3x² – 4x – 12.

Step 4: Factor using (x + 3). - Synthetic division: -3 | 1 3 -4 -12
| -3 0 12
----------------
1 0 -4 0 - Quotient: x² – 4.

Step 5: Factor the quadratic. - x² – 4 = (x – 2)(x + 2).

Step 6: Write the complete factorisation. - f(x) = (x + 3)(x – 2)(x + 2).

What we did and why: - We used the Factor Theorem to find k by setting f(–3) = 0. - Then, we factored the polynomial step by step.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in for exam day. The Factor Theorem is your secret weapon: if plugging x = a into f(x) gives zero, then (x – a) is a factor. Here’s the drill: 1. Spot the a: For (x – 5), a = 5. For (x + 2), a = –2. 2. Plug and pray: Substitute x = a into f(x). If it’s zero, you’ve got a factor. 3. Divide and conquer: Use synthetic division to break down the polynomial. Repeat until you’ve factored it completely. 4. Watch the traps: Check for disguised factors like (2x – 1), and don’t stop until the polynomial is fully factored. Practice with past papers—this is a 3-mark question you cannot afford to lose. You’ve got this!