mathematics

1. Factorisation of quadratic
expressions

2. A. Factorisation by extracting common factors
𝒂𝒃 + 𝒂𝒄 = 𝒂 𝒃 + 𝒄
Factorisation is the reverse of expansion:
𝑎 𝑏 + 𝑐 𝑎𝑏 + 𝑎𝑐
Expand
Factorise

3. • When extracting negative common factors such as −1, we have to
change the sign inside the brackets as shown below:
−𝑥 − 𝑦 = −(𝑥 + 𝑦)
Factorisation is the reverse of Expansion.
Consider the reverse procedure of the distributive law:
change sign
Factor out -1
( )
ab ac a b c
  

4. Factorising linear expressions
Example
Factorise each of the following expressions completely
a. 12𝑥 + 18
b. −15𝑥 − 6
c. −10𝑎𝑥 + 25𝑎𝑦
Solution
a. 𝟏𝟐𝒙 + 𝟏𝟖 = 𝟔(𝟐𝒙 + 𝟑)
The factors of 12 are: 𝟏, 𝟐, 𝟑, 𝟒, 𝟔, 𝟏𝟐 HCF of 12 𝑎𝑛𝑑 18 = 6
The factors of 18 are: 𝟏, 𝟐, 𝟑, 𝟔, 𝟗, 𝟏𝟖

5. The factors of 15 are: 𝟏, 𝟑, 𝟓, 𝟏𝟓
The factors of 3 are: 𝟏, 𝟑
HCF of 15 𝑎𝑛𝑑 3 = 3, so -3 is a common factor
b. −15𝑥 − 6 = −3(5𝑥 + 2)
c. −10𝑎𝑥 + 25𝑎𝑦 = 25𝑎𝑦 − 10𝑎𝑦 rearrange order of terms
= 5𝑎(5𝑦 − 2𝑥) common factors are 5 & a
Factor out -3
change sign

6. Factorising expression involving squares and cubes
Factorise each of the following expressions completely
• 10𝑥2 + 8𝑥 = 2𝑥(5𝑥 + 4) HCF of 10 and 8 = 2
HCF of 𝑥2and 𝑥 = 𝑥
• −49𝑏 − 28𝑏2 = −7𝑏(7 + 4𝑏) HCF of 49 and 28 = 7
HCF of 𝑏2and b = b
• 𝑐2𝑑3 + 𝑐3𝑑2 − 𝑐2𝑑2 = 𝑐2𝑑2 𝑑 + 𝑐 − 1 HCF of 𝑐2 and 𝑐3 = 𝑐2
HCF of 𝑑2 and 𝑑3 = 𝑑2

7. Factorisation of quadratic expressions of the form 𝒙𝟐 + 𝒃𝒙 + 𝒄, 𝒘𝒉𝒆𝒓𝒆 𝒄 >
𝟎 (𝒂𝒏𝒅 𝒃 > 𝟎)
• Factorise 𝑥2
+ 5𝑥 + 6 using multiplication frame.
1. write the term in 𝑥2
and the constant term in the multiplication frame
2. Consider all the possible pairs of factors of the constant term 6 whose product is 6, i.e.
6 = 1𝑥6, 2𝑥3
Choose the pair of factors of 6 with a sum of 5 ( which is the coefficient of 5𝑥). Since 3𝑥 + 2𝑥 =
5𝑥 𝑎𝑛𝑑 𝑥 + 6𝑥 ≠ 5𝑥, we write 2𝑥 𝑎𝑛𝑑 3𝑥 in the remaining two regions.
x
𝒙𝟐
+𝟔
x
𝒙𝟐 3𝑥
2𝑥 +𝟔

8. x 𝑥 +3
𝑥 𝒙𝟐 3𝑥
+2 2𝑥 +𝟔
HCF𝒙𝟐
and 𝟐𝒙 = 𝒙
HCF 𝟑𝒙 and 𝟔 = 𝟑
HCF𝟐𝒙and 6= 𝒙
HCF𝒙𝟐
and 3𝒙 = 𝒙
∴ 𝑥2
+ 5𝑥 + 6 = (𝑥 + 2)(𝑥 + 3)
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 𝑥 + 𝑚 𝑥 + 𝑛
Where: 𝑚𝑥𝑛 = 𝑐,
𝑛 + 𝑚 = 𝑏

9. Factorisation of quadratic expressions of the form 𝒙𝟐 + 𝒃𝒙 + 𝒄, 𝒘𝒉𝒆𝒓𝒆 𝒄 >
𝟎 (𝒂𝒏𝒅 𝒃 > 𝟎) or 𝒙𝟐 − 𝒃𝒙 + 𝒄
• Factorise 𝑥2
− 5𝑥 + 6 using multiplication frame.
Solution
From 𝑥2
− 5𝑥 + 6 , 𝑎 = 1, 𝑏 = −5, 𝑐 = 6
Because 𝒃 = −𝟓 is negative, we need to consider both positive and negative
factors of the constant term 6, i.e.
6 = 1𝑥6 = −1 𝑥 −6
= 2𝑥3 = −2 𝑥 −3
𝒙𝟐 − 𝟓𝒙 + 𝟔 = (𝒙 − 𝟐)(𝒙 − 𝟑)
x
𝑥2 ?
? +6
𝑥2 −3𝑥
−2𝑥 +6
𝑥 −3
𝑥 𝑥2 −3𝑥
−2 −2𝑥 +6

10. • https://www.geogebra.org/m/WgE3VWQb

11. HCF ( )
x y
 
2 ( ) 3 ( )
a x y b x y
  
(c)
2 ( ) 3 ( )
a x y b x y
   
( )(2 3 )
x y a b
  

12. The next example involves changing
the sign before a bracket before
taking out the common bracket.
Before discussing the next example,
consider the expressions:
(2 )
2
2
b a
b a
a b
 
  
  
( 2 )
2
2
b a
b a
a b
  
 
  

13. Therefore it is true that:
(2 ) ( 2 )
b a b a
     
You can write this as:
(2 ) ( 2 )
b a a b
    
For example:
(3 ) ( 3 ) ( 3 )
y x y x x y
        
(3 ) ( 3 ) ( 3 )
y x y x x y
        
( 3 ) ( 3 ) ( 3 )
y x y x x y
         

14. Therefore, if you change the sign
before the bracket, change the signs
of the terms when writing them in
the brackets.

15. 2 (3 2 ) 5 (2 3 )
a x y b y x
  
EXAMPLE
Factorise:
(a)
2 (3 2 ) 5 ( 2 3 )
a x y b y x
    
2 (3 2 ) 5 (3 2 )
a x y b x y
   
(3 2 )(2 5 )
x y a b
  

16. 3 (5 ) ( 5 )
a x y x y
   
(b)
3 (5 ) ( 5 )
a x y x y
    
3 (5 ) (5 )
a x y x y
   
(5 )(3 1)
x y a
  

17. DIFFERENCE OF TWO SQUARES
Consider the product
2 2
( )( )
a b a b a b
   
The reverse process is called the
factorisation of the difference of
two squares.
2 2
( )( )
a b a b a b
   

18. Another way of seeing this type of
factorisation is:
2 2 2 2 2 2
( )( )
a b a b a b
   
( )( )
a b a b
  
0, 0
a b
 

19. EXAMPLE
Factorise fully:
2
9
x 
(a)
2 2
( 9)( 9)
x x
  
( 3)( 3)
x x
  

20. 4 2
49 64
x y

(b)
2 2
(7 8 )(7 8 )
x y x y
  
8 8
8 8
a b

(c)
8 8
8( )
a b
 
4 4 2 2
8( )( )( )( )
a b a b a b a b
    
4 4 4 4
8( )( )
a b a b
  
4 4 2 2 2 2
8( )( )( )
a b a b a b
   

21. 2
4 ( )
x y
 
(d)
  
2 ( ) 2 ( )
x y x y
    
(2 )(2 )
x y x y
    

22. QUADRATIC TRINOMIALS
Consider the product ( )( )
x a x b
 
By multiplying out, it is clear that
this product will become:
( )( )
x a x b
 
2
x ax bx a b
    
2
( )
x a b x a b
    

23. So the expression
can be factorised as ( )( )
x a x b
 
2
( )
x a b x a b
   
For example, the trinomial
can be factorised as follows:
2
6 8
x x
 
Write the last term, 8, as the product
of two numbers ( )
a b

The options are: 1 8
 4 2


24. The middle term is now obtained
by adding the numbers of one of
the above options. The obvious
choice will be the option
because the sum of the numbers 4
and 2 gives 6. Therefore:
2
(4 2) (4 2)
x x
    
4 2

2
6 8
x x
 
( 4)( 2)
x x
  

25. So the trick to factorising trinomials is
as follows:
• Write down the last term as
the product of two numbers.
• Find the two numbers (using
the appropriate numbers from
one of the products) which
gets the middle term by adding
or subtracting.

26. • Check that when you multiply
these numbers you get the last
term.

27. EXAMPLE
Factorise:
2
7 60
x x
 
(a)
The last term can be written as
the following products:
1 60, 30 2, 15 4,
10 6, 12 5, 20 3
  
  

28. We now need to get from one
of the options above.
7

Using will enable us to get
since
12 5

7

12 5 7
    (middle term)
and ( 12)(5) 60
   (last term)

29. Therefore:
2
7 60
x x
 
( 12)( 5)
x x
  

30. 2
5 6
x x
 
(b)
The last term can be written as
the following products:
3 2, 1 6
 
We now need to get from
one of the options above.
5

Try the option 3 2


31. Clearly which is the
middle term and
which is the last term.
Notice that the option
will not work because even
though
is the middle term,
is not the last term.
1 6

3 2 5
   
3 2 6
    
1 6 5
   
1 6 6
   

32. Therefore:
2
5 6
x x
  ( 3)( 2)
x x
  
Notice:
• If the sign of the last term of
a trinomial is negative, the
signs in the brackets are
different (see example 8a).
2
7 60 ( 12)( 5)
x x x x
    

33. • If the sign of the last term of
a trinomial is positive, the
signs in the brackets are the
same i.e. both positive or
both negative
2
5 6
x x
  ( 3)( 2)
x x
  

34. 2
3 21 24
a a
 
(c)
2
3 21 24
a a
 
Here it is necessary to first
take out the highest
common factor:
2
3( 7 8)
a a
  
3( 1)( 8)
a a
  

35. 2
12
x x
 
(d)
( 4)( 3)
x x
  
2
10 24
x x
 
(e)
( 6)( 4)
x x
  

36. In summary then, apply the following
procedure when factorising
trinomials:
• Take out the highest common
factor if necessary.
• Write down the last term as the
product of two numbers.

37. • Find the two numbers (using the
appropriate numbers from one
of the products) which gets the
middle term by adding or
subtracting.
• Check that when you multiply
these numbers you get the last
term.

38. • If the sign of the last term of a
trinomial is positive, the signs in
the brackets are the same (both
positive or both negative).
• If the sign of the last term of a
trinomial is negative, the signs in
the brackets are different.

39. MORE ADVANCED TRINOMIALS
Consider the trinomial
2
21 25 4
x x
 
The method to factorise this
trinomial is a little more involved
than with the previous
trinomials. A suggested method
is as follows.
EXAMPLE

40. Step 1
Write down the product options
of the first and last terms
2
21 : 1 21 , 7 3
4 : 1 4, 2 2
x x x x x
 
 

41. Step 2
Write the product options in a
table format as follows:
First term Last term
21x
1x
7x
3x
1
4 1
4 2
2
The product option for the
last term must also be written in
reverse order as
1 4

4 1


42. Step 3
Select any product option from the
first term and last term and write
these options using what is called
the “cross method”:
7x
3x
1
4

43. 7x
3x
1
4
Step 4
Now multiply as follows:
3x
28x

44. Step 5
The trick is to now get the middle
term, , from and
using different signs (because the
sign of the last term of the
trinomial is negative). Insert the
signs as follows:
25x
 3x 28x

45. 7x
3x
1
4
3x

28x

25x


46. 7x
3x
1
4
3x

28x

Step 6
The factors are now obtained by
reading off horizontally:
7 1
x 
3 4
x 
2
21 25 4
x x
 
(7 1)(3 4)
x x
  

47. Factorise: 2
12 11 2
x x
 
The product options of the first
and last terms:
EXAMPLE
2
12 : 1 12 , 4 3 , 6 2
2 : 1 2, 2 1
x x x x x x x
  
 

48. First term Last term
1x
12x
4x
3x
1
1
2
2
The signs in the brackets must be
the same because the sign of the
last term of the trinomial is
positive.
6x
2x

49. 4x
3x
1
2
3x

8x

11x

2
12 11 2
x x
 
(4 1)(3 2)
x x
  

50. Before discussing the next examples,
we need to deal with the concept of
“taking out a negative”. Consider the
following expressions:
( )
x y
x y
 
  
( )
x y
x y
 
  
It is clear from the above that:
( )
x y x y
    
( )
x y x y
    

51. Therefore, whenever you “take out a
negative sign” when factorising an
expression, the middle sign will
always change in the brackets.
For example:
3 ( 3)
x x
    
middle sign is positive and changes
to a negative in the brackets

52. 3 ( 3)
x x
    
middle sign is negative and changes
to a positive in the brackets
2 8 2( 4)
x x
    
middle sign is negative and changes
to a positive in the brackets

53. SIMPLIFICATION OF ALGEBRAIC
FRACTIONS
MULTIPLICATION AND DIVISION
EXAMPLE 15
Simplify the following expressions:
3 4
5
12
18
x y z
x yz
(a)

54. 3 4
5
12
18
x y z
x yz
3 4
5 1
12
18
x y z
x y z

3
2
2
3
y
x


55. 2 7
3
12 24
5 25
a b a b
c c

(b)
2 3
7
12 25
5 24
a b c
c a b
 
2
5
5
2
c
a

2 3
7
12 25
5 24
a bc
a bc




56. 2
2
6 12
6
x x
x

(c)
2
6 ( 2)
6
x x
x


2
x
x



57. EXAMPLE
Simplify the following expressions:
2
2
12 27
4 12
x x
x x
 

(a)
  
 
9 3
4 3
x x
x x
 


9
4
x
x



58. 2
2 2
2 5
4 5 4
x x x
x x x
 

  
(b)
  
 
  
5
2
5 1 2 2
x x
x
x x x x


 
   
  
1 2
x
x x

 

59. 2 2
2 3
2 2 1
2
1 8
yx y y x
x
x x
 
 

 
(c)
2
2 2 3
2 2 2 1
1 8
yx y x x
x y x
  
  
 
 
     
2
2 2
2 1 2 1
1 1 2 2 4
y x x x
x x y x x x
  
  
    
 
  
2
2
2 1
1 2 4
x
y x x x


  

60. EXAMPLE 17
 
x y y x
   
Simplify:
3 6
2
x
x


(a)
 
3 2
2
x
x



 
 
3 2
2
x
x


 
3
 

61. EXAMPLE 18
Simplify:
2 3
1 2 1
1
2
4 3 xy
x x
  
3
LCD 12x y

2 3
1 1 2 1
1 2
4 3 xy
x x
   

62. 2 3
1 1 2 1
1 2
4 3 xy
x x
   
3 2
3
12 3 8 6
12
x y xy y x
x y
  

3
3 2 2
3
2
1 1 2 1
1 2
12 3 4 6
3 4
12 6
4 3
x y xy y x
x xy
x x
y y
x y x
 
     
3 2
3 3 3 3
12 3 8 6
12 12 12 12
x y xy y x
x y x y x y x y
   

63. 1 2 3
2 3 4
x x x
  
 
( 1) ( 2) ( 3)
2 3
6 4 3
6 4 3
4
x x x
  
     
EXAMPLE 19
Simplify:
LCD 12

6( 1) 4( 2) 3( 3)
12 12 12
x x x
  
  

64. 2 1
3 3
x x
  
 
  
  
2 1
3 3
x x
  
 
  
  
EXAMPLE 20
Simplify:
Method 1
3 2 3 1
3 3
x x
 
  
   
  

65. 3 2 3 1
3 3
x x
 
  
   
  
2
9 3 2
9
x x
 

(3 2)(3 1)
9
x x
 


66. 2 1
3 3
x x
  
 
  
  
Method 2
2 1 2 2
3 3 9
x x x
   
2
9 3 6 2
9
x x x
  

2
9 3 2
9
x x
 


67. EXAMPLE
Simplify:
1 2 1
2
x x
x x
 


(a)  
LCD 2
x x
 
 
1 (2 1)
( 2 (
)
( 2)
2)
x x x
x x
x
x
x


  
 


 
1 (2 1)( 2)
( 2) ( 2)
x x x x
x x x x
  
 
 

68.  
1 (2 1)( 2)
( 2) ( 2)
x x x x
x x x x
  
 
 
    
1 (2 1) 2
( 2)
x x x x
x x
   


2 2
(2 4 2)
( 2)
x x x x x
x x
    



69. 2 2
2 4 2
( 2)
x x x x x
x x
    


2
6 2
( 2)
x x
x x
  



70. EXAMPLE
Simplify:
2
3 2 2 5
2 2
4
x x
x x
x

 
 

(a)
  
3 2 2 5
2 2 2 2
x x
x x x x

  
   

71.       
( 2)
(3 ( 2)
( 2)
2) 2 5
2)
2 2 (
2 2
x
x x
x x x
x
x x
x
 


    
  
 
   
  
3 2 2 2 5 2
2 2
x x x x
x x
    

 
  
2
3 2 2 4 5 10
2 2
x x x x
x x
    

 
  
2
2 6 12
2 2
x x
x x
 

 

72. EXAMPLE
Simplify:
2
5 3
3
5 6
x
x
x x


 
(a)
  
5 3
2 3 3
x
x x x
 
  
    
5 3
2 3 3
x
x x x
 
   

73.     
5 3
2 3 3
x
x x x
 
  
    
5 3
2 3 )
3
( 2)
( 2
x
x x x
x
x
  
  


 
  
5 3 2
2 3
x x
x x
 

 
  
2
5 3 6
2 3
x x
x x
 

    
2
3 6 5
2 3
x x
x x
  

 

74. THE END