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Exam code:C300

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) + 2+ 6 

Factorise the second two terms, using 2 as a common factor 

b(a + 3) + 2(a + 3) 

(+ 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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