Factorising (1 Bracket)

Understanding

It is best to understand how to expand and simplify brackets first, before understanding how to factorise expressions. Check the Expanding Brackets Part 1 page and understand that, because the opposite of expanding is factorising.
Lets say we have the we to expand the expression 2 (a + b), we know how to do this
2 x a = 2a2 x b = 2b
So 2(a + b) = 2a + 2b (So we have expanded 2(a + b) to get 2a + 2b)
What if the question was factorise 2a + 2b? In other words how to get from 2a + 2b to 2(a + b). THIS IS CALLED FACTORISATION.
So the way we factorise is by looking at the terms of the expression. Do the terms have a common factor which can be factored out to be placed outside the bracket.
The number 2 is a common factor in 2a and 2b.
So we have we have to take out the the highest common factor of all the terms, and place them outside the bracket. The highest common factor of 2a and 2b would be 2.
So we can write 2( ).
Now what do we put inside the bracket to get 2a + 2b, when 2 is multiplied by all the terms inside the bracket.
Lets look at the first term which is 2a.
2a = 2 x a, we have 2 already outside the bracket, so to get 2a, the term inside the bracket should be a.
So we have currently 2(a )
Lets look at the second term now which is 2b.
2b = 2 x b, we have 2 already outside the bracket, so to get 2b, the term inside the bracket should be b. Remember the term inside the bracket is what is multiplied by the outside the term.
So we have 2(a + b) = 2a + 2b
Lets check,
does 2 x a = 2a YESdoes 2 x b = 2b YES
So lets add the products 2a + 2b = 2a + 2b
ONE POINT THAT IS CRUCIAL TO UNDERSTAND IS THAT WE MUST TAKE OUT THE HIGHEST COMMON FACTOR OF ALL TERMS. THIS HIGHEST COMMON FACTOR IS WHAT GOES OUTSIDE THE BRACKET.
Question 1: Factorise the expression 2a² + a
By looking at both terms, we see that both terms have the unknown number a, in the first term a is squared and the second term itself is a.
2a² + a = 2a² + 1a¹
We need to find the highest common factor between the two terms we have. The first term is 2a² and a¹
Let's factorise by taking a step by step approach.
2a² + 1a¹
In purple are the coefficient of both terms (which include the unknown number a).
What is the highest common factor of 2 and 1. It is 1, so this number would be use to go outside the bracket.
What is the highest common factor of a² and a¹?
When we answer this question, we always choose the a term with the lowest power, in this case it is a¹. So a¹ would go outside the bracket
So we can currently write
1a¹( ) = a( ), because 1 x a¹ = a
What do we do next?
So a multiplied with the terms inside the bracket need to equal a² + a.
Lets look at the first term which is a².
a² = a x a, we already have an a outside the bracket, so to get a², the term inside the bracket is a.
So we currently have a(a ).
Lets now look at the second term which is a. a = a, we already have an a outside the bracket, so to get a, the term inside the bracket needs to be 1.
Remember any number multiplied by 1 is just the number itself.
So
a(a + 1) = a² + a
Lets check
a x a = a²a x 1 = a
Add the products
a² + a
So the expression a² + a factorised is equal to a(a + 1).
Let's try another question so can understand better on how we separate the coefficients from their terms to factorise.
Question: Factorise the expression 3b² - 6b.
3b² - 6b = 3b² - 12b¹
What is the highest common factor of 3 and 12. It is 3( this number will go outside the bracket.
What is the highest common factor of b² and b. Remember we always choose the one with smallest power. In this case it is b. (this will go outside the bracket).
so we can write
3b( ) = 3b² - 6b
So 3b multiplied by something is equal to 3b², that something is b.
3b multiplied by something is equal to -6b, that something is -2.
Now we can write
3b(b - 2) = 3b² - 6b
Therefore, 3b(b - 2) is the answer.
Question 2: Factorise the expression 2ab - 3ab + b² .
We can see that the expression can be simplified, once we simplify this makes the factorisation a simpler process.
So 2ab - 3ab + b² = -ab + b²
So what we need to factorise is -ab + b²
-ab + b² = -1a¹b¹ + 1a⁰b²
Why did we write a⁰? Anything to the power of 0 is equal to 1, so a⁰ = 1, no matter what number a is. So that is why -ab + b² equals -1a¹b¹ + 1a⁰b² . Why we wrote a⁰ this will become clear now.
The highest common factor between 1 and 1 is just 1 (this number will go outside the bracket).
The highest common factor between a⁰ and a¹ is a⁰ (remember we always choose the one with the smallest power), so a⁰ will go outside the bracket.
The highest common factor between b¹ and b² is b¹ (remember we always choose the one with the smallest power), so b¹ will go outside the bracket.

So we can write 1a⁰b¹( ) = b( ), because 1a⁰b¹ = 1 x 1 x b = b
so
b( ) = -ab + b²
Now what do we put in the bracket to get -ab + b²?
Lets look at the first term which is -ab.
-ab = -a x b = b x -a, we already have b outside the bracket, so the term inside the bracket needs to be -a.
So we currently have b(-a )
Lets look at the second term which is b².
b² = b x b, we already have a b outside the bracket, so the term inside the bracket needs to be b.
This gives us b(-a + b) = -a + b²
Lets check
b x -a = -ab
b x b = b²
Add the products to get
-ab + b²
So the expression 2ab - 3ab + b² factorised is b(-a + b).

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