Factorising by Grouping (Cambridge (CIE) IGCSE Maths)
Revision Note
Written by: Mark Curtis
Reviewed by: Dan Finlay
Factorising by Grouping
How do I factorise expressions with a common bracket?
Look at the expression 3x(t + 4) + 2(t + 4)
Both terms have a common bracket, (t + 4)
The whole bracket, (t + 4), can be "taken out" like a common factor:
(t + 4)(3x + 2)
This is like factorising 3xy + 2y to get y(3x + 2)
y represents (t + 4) above
How do I factorise by grouping?
Some questions may require you to form a common bracket yourself
For example xy + 3x + 5y + 15
The first two terms have a common factor of x
The second two terms have a common factor of 5
Factorising fully the first pair of terms, and the last pair of terms:
x(y + 3) + 5(y + 3)
You can now spot a common bracket of (y + 3)
(y + 3)(x + 5)
This is called factorising by grouping
Does it matter what order I group in?
You can often rearrange terms to factorise in a different order
Rewriting the same example, xy + 3x + 5y + 15, but in a different order:
xy + 5y + 3x + 15
The first pair of terms have a common factor of y
The second pair of terms have a common factor of 3
Factorising gives y(x + 5) + 3(x + 5)
You can now spot a common bracket, this time of (x + 5)
(x+5)(y+3)
This gives the same result as found previously
Some rearrangements cannot be factorised as "first pair" then "second pair"
For example, rewriting the above example as xy + 15 + 3x + 5y
Examiner Tips and Tricks
Once you have factorised something, expand it by hand to check your answer is correct.
Worked Example
Factorise ab + 3b + 2a + 6.
Method 1:
Notice that ab and 3b have a common factor of b
Notice that 2a and 6 have a common factor of 2
Factorise the first two terms, using b as a common factor
b(a + 3) + 2a + 6
Factorise the second two terms, using 2 as a common factor
b(a + 3) + 2(a + 3)
(a + 3) is a common bracket
We can now factorise out the bracket (a + 3)
(a + 3)(b + 2)
Method 2:
Notice that ab and 2a have a common factor of a
Notice that 3b and 6 have a common factor of 3
Rewrite the expression, grouping these terms together
ab + 2a + 3b + 6
Factorise the first two terms, using a as a common factor
a(b + 2) + 3b + 6
Factorise the second two terms, using 3 as a common factor
a(b + 2) + 3(b + 2)
(b + 2) is a common bracket
We can now factorise out the bracket (b + 2)
(b + 2)(a + 3)
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