 # Polynomial | Remainder Theorem | Factor Theorem 2/3

## B. FACTOR THEOREM

If f(x) is divided by (x – a) and the remainder is 0 ⇔ f(a)=0
⇒ (x – a) is a factor of f(x)
⇒ f(x) is exactly divisible by (x – a)

From your Secondary 2 Expansion & Factorisation chapter:

1. Expansion → remove brackets –> multiply in
2. Factorisation → add brackets –> divide by factor

Question: Factorise x2 – 5x + 6

4 Methods to find the factors of x2 – 5x + 6 –> To solve for x where f(x) or y = 0

1. Trial and Error

(I will use this method if the power of x is 3 and above, usual for polynomial questions)

2. Frame Method

(my preferred method if the power of x is 2 : quadratic equation)

3. Box Method

(I don’t like this method… just a personal opinion :p)

4. General Equation x=roots= [-b ± (b2-4ac)0.5]/(2a)

(works all the time for quadratic equations even if the roots are not integer or nice numbers)

Method 1 via Trial & Error

Choose 2 factors of the constant 6
try: 1 x 6 → 1x + 6x ≠ -5x (reject)
try: 2 x 3 → 2x + 3x ≠ -5x (reject)
try: (-2) x (-3) → -2x + (-3)x = -5x (YAY!)
⇒ cross-check: f(3) = f(2) = 0 (YAY!)
⇒ x2 – 5x + 6 = (x – 3)(x – 2)try: 1 x 6 → 1x + 6x ≠ -5x (reject)
try: 2 x 3 → 2x + 3x ≠ -5x (reject)
try: (-2) x (-3) → -2x + (-3)x = -5x (YAY!)
⇒ cross-check: f(3) = f(2) = 0 (YAY!)

factors / roots / x =2 or 3
⇒ x2 – 5x + 6 = (x – 3)(x – 2)

Method 2 via frame method

⇒ x2 – 5x + 6 = (x – 3)(x – 2)

Method 3 via box method

⇒ x2 – 5x + 6 = (x – 3)(x – 2)

Method 4 via general equation method

x = [-b ± (b2-4ac)0.5]/(2a) = 3 or 2

⇒ x2 – 5x + 6 = (x – 3)(x – 2)﻿

When you see the keyword Factorise, final answer must be in (brackets) i.e. DON’T write x = 3, 2 as the answer (unless question is asking you to solve for x / solve for the roots / solve for the factors of x2 – 5x + 6 )