NCERT Textbook- Factorisation Class 8 Notes | EduRev

Mathematics (Maths) Class 8

Class 8 : NCERT Textbook- Factorisation Class 8 Notes | EduRev

 Page 1


FACTORISATION  217
14.1  Introduction
14.1.1  Factors of natural numbers
Y ou will remember what you learnt about factors in Class VI. Let us take a natural number ,
say 30, and write it as a product of other natural numbers, say
 30 = 2 × 15
= 3 × 10 = 5 × 6
Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30.
Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
A number written as a product of prime factors is said to
be in the prime factor form; for example, 30 written as
2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors.  This is
what we shall learn to do in this chapter.
14.1.2  Factors of algebraic expressions
W e have seen in Class VII that in algebraic expressions, terms are formed as products of
factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed
by the factors 5, x and y, i.e.,
5xy = y x× × 5
Observe that the factors 5, x and y of 5xy cannot further
be expressed as a product of factors. We may say that 5,
x and y are ‘prime’ factors of 5xy.  In algebraic expressions,
we use the word ‘irreducible’ in place of ‘prime’.  W e say that
5 × x × y is the irreducible form of 5xy.  Note 5 × (xy) is not
an irreducible form of  5xy, since the factor xy can be further
expressed as a product of x and y, i.e., xy = x × y.
Factorisation
CHAPTER
14
Note 1 is a factor of 5xy, since
5xy = y x× × × 5 1
In fact, 1 is a factor of every term. As
in the case of natural numbers, unless
it is specially required, we do not show
1 as a separate factor of any term.
We know that 30 can also be written as
30 = 1 × 30
Thus, 1 and 30 are also factors of 30.
You will notice that 1 is a factor of any
number.  For example, 101 = 1 × 101.
However, when we write a number as a
product of factors, we shall not write 1 as
a factor, unless it is specially required.
2019-20
Page 2


FACTORISATION  217
14.1  Introduction
14.1.1  Factors of natural numbers
Y ou will remember what you learnt about factors in Class VI. Let us take a natural number ,
say 30, and write it as a product of other natural numbers, say
 30 = 2 × 15
= 3 × 10 = 5 × 6
Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30.
Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
A number written as a product of prime factors is said to
be in the prime factor form; for example, 30 written as
2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors.  This is
what we shall learn to do in this chapter.
14.1.2  Factors of algebraic expressions
W e have seen in Class VII that in algebraic expressions, terms are formed as products of
factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed
by the factors 5, x and y, i.e.,
5xy = y x× × 5
Observe that the factors 5, x and y of 5xy cannot further
be expressed as a product of factors. We may say that 5,
x and y are ‘prime’ factors of 5xy.  In algebraic expressions,
we use the word ‘irreducible’ in place of ‘prime’.  W e say that
5 × x × y is the irreducible form of 5xy.  Note 5 × (xy) is not
an irreducible form of  5xy, since the factor xy can be further
expressed as a product of x and y, i.e., xy = x × y.
Factorisation
CHAPTER
14
Note 1 is a factor of 5xy, since
5xy = y x× × × 5 1
In fact, 1 is a factor of every term. As
in the case of natural numbers, unless
it is specially required, we do not show
1 as a separate factor of any term.
We know that 30 can also be written as
30 = 1 × 30
Thus, 1 and 30 are also factors of 30.
You will notice that 1 is a factor of any
number.  For example, 101 = 1 × 101.
However, when we write a number as a
product of factors, we shall not write 1 as
a factor, unless it is specially required.
2019-20
218  MATHEMATICS
Next consider the expression 3x (x + 2).  It can be written as a product of factors.
3, x and (x + 2)
3x(x + 2) = ( ) 2 3 + × × x x
The factors 3, x and (x +2) are irreducible factors of 3x (x + 2).
Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form
as  10x (x + 2) (y + 3) = 
( ) ( ) 2 5 2 3 x x y × × × + × + .
14.2  What is Factorisation?
When we factorise an algebraic expression, we write it as a product of factors.  These
factors may be numbers, algebraic variables or algebraic expressions.
Expressions like 3xy, y x
2
5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form.
Their factors can be just read off from them, as we already know.
On the other hand consider expressions like 2x + 4, 3x + 3y, x
2
 + 5x, x
2
 + 5x + 6.
It is not obvious what their factors are. W e need to develop systematic methods to factorise
these expressions, i.e., to find their factors. This is what we shall do now.
14.2.1  Method of common factors
• We begin with a simple example: Factorise 2x + 4.
W e shall write each term as a product of irreducible factors;
2x = 2 × x
4 = 2 × 2
Hence 2x + 4 = (2 × x) + (2 × 2)
Notice that factor 2 is common to both the terms.
Observe, by distributive law
2 × (x + 2) = (2 × x) + (2 × 2)
Therefore, we can write
2x + 4 = 2 × (x + 2) = 2 (x + 2)
Thus, the expression 2x + 4 is the same as 2 (x + 2).  Now we can read off its factors:
they are 2 and (x + 2). These factors are irreducible.
Next, factorise 5xy + 10x.
The irreducible factor forms of 5xy and 10x are respectively,
5xy = 5 × x × y
10x = 2 × 5 × x
Observe that the two terms have 5 and x as common factors.  Now,
5xy + 10x = (5 × x × y)  + (5 × x × 2)
= (5x × y)  + (5x × 2)
W e combine the two terms using the distributive law,
(5x× y) + (5x× 2) = 5x × ( y + 2)
Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.)
2019-20
Page 3


FACTORISATION  217
14.1  Introduction
14.1.1  Factors of natural numbers
Y ou will remember what you learnt about factors in Class VI. Let us take a natural number ,
say 30, and write it as a product of other natural numbers, say
 30 = 2 × 15
= 3 × 10 = 5 × 6
Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30.
Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
A number written as a product of prime factors is said to
be in the prime factor form; for example, 30 written as
2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors.  This is
what we shall learn to do in this chapter.
14.1.2  Factors of algebraic expressions
W e have seen in Class VII that in algebraic expressions, terms are formed as products of
factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed
by the factors 5, x and y, i.e.,
5xy = y x× × 5
Observe that the factors 5, x and y of 5xy cannot further
be expressed as a product of factors. We may say that 5,
x and y are ‘prime’ factors of 5xy.  In algebraic expressions,
we use the word ‘irreducible’ in place of ‘prime’.  W e say that
5 × x × y is the irreducible form of 5xy.  Note 5 × (xy) is not
an irreducible form of  5xy, since the factor xy can be further
expressed as a product of x and y, i.e., xy = x × y.
Factorisation
CHAPTER
14
Note 1 is a factor of 5xy, since
5xy = y x× × × 5 1
In fact, 1 is a factor of every term. As
in the case of natural numbers, unless
it is specially required, we do not show
1 as a separate factor of any term.
We know that 30 can also be written as
30 = 1 × 30
Thus, 1 and 30 are also factors of 30.
You will notice that 1 is a factor of any
number.  For example, 101 = 1 × 101.
However, when we write a number as a
product of factors, we shall not write 1 as
a factor, unless it is specially required.
2019-20
218  MATHEMATICS
Next consider the expression 3x (x + 2).  It can be written as a product of factors.
3, x and (x + 2)
3x(x + 2) = ( ) 2 3 + × × x x
The factors 3, x and (x +2) are irreducible factors of 3x (x + 2).
Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form
as  10x (x + 2) (y + 3) = 
( ) ( ) 2 5 2 3 x x y × × × + × + .
14.2  What is Factorisation?
When we factorise an algebraic expression, we write it as a product of factors.  These
factors may be numbers, algebraic variables or algebraic expressions.
Expressions like 3xy, y x
2
5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form.
Their factors can be just read off from them, as we already know.
On the other hand consider expressions like 2x + 4, 3x + 3y, x
2
 + 5x, x
2
 + 5x + 6.
It is not obvious what their factors are. W e need to develop systematic methods to factorise
these expressions, i.e., to find their factors. This is what we shall do now.
14.2.1  Method of common factors
• We begin with a simple example: Factorise 2x + 4.
W e shall write each term as a product of irreducible factors;
2x = 2 × x
4 = 2 × 2
Hence 2x + 4 = (2 × x) + (2 × 2)
Notice that factor 2 is common to both the terms.
Observe, by distributive law
2 × (x + 2) = (2 × x) + (2 × 2)
Therefore, we can write
2x + 4 = 2 × (x + 2) = 2 (x + 2)
Thus, the expression 2x + 4 is the same as 2 (x + 2).  Now we can read off its factors:
they are 2 and (x + 2). These factors are irreducible.
Next, factorise 5xy + 10x.
The irreducible factor forms of 5xy and 10x are respectively,
5xy = 5 × x × y
10x = 2 × 5 × x
Observe that the two terms have 5 and x as common factors.  Now,
5xy + 10x = (5 × x × y)  + (5 × x × 2)
= (5x × y)  + (5x × 2)
W e combine the two terms using the distributive law,
(5x× y) + (5x× 2) = 5x × ( y + 2)
Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.)
2019-20
FACTORISATION  219
TRY THESE
Example 1: Factorise  12a
2
b + 15ab
2
Solution: We have 12a
2
b = 2 × 2 × 3 × a × a × b
15ab
2
 = 3 × 5 × a × b × b
The two terms have 3, a and b as common factors.
Therefore, 12a
2
b + 15ab
2
 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
= 3 × a × b × [(2 × 2 × a) + (5 × b)]
 = 3ab × (4a + 5b)
= 3ab (4a + 5b) (required factor form)
Example 2: Factorise 10x
2
 – 18x
3
 + 14x
4
Solution: 10x
2 
= 2 × 5 × x × x
        18x
3
  = 2 × 3 × 3 × x × x × x
       14x
4
 = 2 × 7 × x × x × x × x
The common factors of the three terms are 2, x and x.
Therefore, 10x
2
 – 18x
3
 + 14x
4
 = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x)
+ (2 × x × x × 7 × x × x)
= 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)]
= 2x
2 
 × 
 
(5 – 9x + 7x
2
) = 
2 2
2 (7 9 5) x x x - +
  
Factorise: (i)  12x + 36 (ii)  22y – 33z (iii)  14pq + 35pqr
14.2.2  Factorisation by regrouping terms
Look at the expression 2xy + 2y + 3x + 3.  Y ou will notice that the first two terms have
common factors 2 and y and the last two terms have a common factor 3.  But there is no
single factor common to all the terms.  How shall we proceed?
Let us write (2xy + 2y) in the factor form:
2xy + 2y = (2 × x × y) + (2 × y)
= (2 × y × x) + (2 × y × 1)
= (2y × x) + (2y × 1) = 2y (x + 1)
Similarly ,     3x + 3 = (3 × x) + (3 × 1)
= 3 × (x + 1) = 3 ( x + 1)
Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)
Observe, now we have a common factor (x + 1) in both the terms on the right hand
side. Combining the two terms,
2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3)
The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its
factors are (x + 1) and (2y + 3). Note, these factors are irreducible.
Note, we need to
show1 as a factor
here. Why?
Do you notice that the factor
form of an expression has only
one term?
(combining the three terms)
(combining the terms)
2019-20
Page 4


FACTORISATION  217
14.1  Introduction
14.1.1  Factors of natural numbers
Y ou will remember what you learnt about factors in Class VI. Let us take a natural number ,
say 30, and write it as a product of other natural numbers, say
 30 = 2 × 15
= 3 × 10 = 5 × 6
Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30.
Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
A number written as a product of prime factors is said to
be in the prime factor form; for example, 30 written as
2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors.  This is
what we shall learn to do in this chapter.
14.1.2  Factors of algebraic expressions
W e have seen in Class VII that in algebraic expressions, terms are formed as products of
factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed
by the factors 5, x and y, i.e.,
5xy = y x× × 5
Observe that the factors 5, x and y of 5xy cannot further
be expressed as a product of factors. We may say that 5,
x and y are ‘prime’ factors of 5xy.  In algebraic expressions,
we use the word ‘irreducible’ in place of ‘prime’.  W e say that
5 × x × y is the irreducible form of 5xy.  Note 5 × (xy) is not
an irreducible form of  5xy, since the factor xy can be further
expressed as a product of x and y, i.e., xy = x × y.
Factorisation
CHAPTER
14
Note 1 is a factor of 5xy, since
5xy = y x× × × 5 1
In fact, 1 is a factor of every term. As
in the case of natural numbers, unless
it is specially required, we do not show
1 as a separate factor of any term.
We know that 30 can also be written as
30 = 1 × 30
Thus, 1 and 30 are also factors of 30.
You will notice that 1 is a factor of any
number.  For example, 101 = 1 × 101.
However, when we write a number as a
product of factors, we shall not write 1 as
a factor, unless it is specially required.
2019-20
218  MATHEMATICS
Next consider the expression 3x (x + 2).  It can be written as a product of factors.
3, x and (x + 2)
3x(x + 2) = ( ) 2 3 + × × x x
The factors 3, x and (x +2) are irreducible factors of 3x (x + 2).
Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form
as  10x (x + 2) (y + 3) = 
( ) ( ) 2 5 2 3 x x y × × × + × + .
14.2  What is Factorisation?
When we factorise an algebraic expression, we write it as a product of factors.  These
factors may be numbers, algebraic variables or algebraic expressions.
Expressions like 3xy, y x
2
5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form.
Their factors can be just read off from them, as we already know.
On the other hand consider expressions like 2x + 4, 3x + 3y, x
2
 + 5x, x
2
 + 5x + 6.
It is not obvious what their factors are. W e need to develop systematic methods to factorise
these expressions, i.e., to find their factors. This is what we shall do now.
14.2.1  Method of common factors
• We begin with a simple example: Factorise 2x + 4.
W e shall write each term as a product of irreducible factors;
2x = 2 × x
4 = 2 × 2
Hence 2x + 4 = (2 × x) + (2 × 2)
Notice that factor 2 is common to both the terms.
Observe, by distributive law
2 × (x + 2) = (2 × x) + (2 × 2)
Therefore, we can write
2x + 4 = 2 × (x + 2) = 2 (x + 2)
Thus, the expression 2x + 4 is the same as 2 (x + 2).  Now we can read off its factors:
they are 2 and (x + 2). These factors are irreducible.
Next, factorise 5xy + 10x.
The irreducible factor forms of 5xy and 10x are respectively,
5xy = 5 × x × y
10x = 2 × 5 × x
Observe that the two terms have 5 and x as common factors.  Now,
5xy + 10x = (5 × x × y)  + (5 × x × 2)
= (5x × y)  + (5x × 2)
W e combine the two terms using the distributive law,
(5x× y) + (5x× 2) = 5x × ( y + 2)
Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.)
2019-20
FACTORISATION  219
TRY THESE
Example 1: Factorise  12a
2
b + 15ab
2
Solution: We have 12a
2
b = 2 × 2 × 3 × a × a × b
15ab
2
 = 3 × 5 × a × b × b
The two terms have 3, a and b as common factors.
Therefore, 12a
2
b + 15ab
2
 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
= 3 × a × b × [(2 × 2 × a) + (5 × b)]
 = 3ab × (4a + 5b)
= 3ab (4a + 5b) (required factor form)
Example 2: Factorise 10x
2
 – 18x
3
 + 14x
4
Solution: 10x
2 
= 2 × 5 × x × x
        18x
3
  = 2 × 3 × 3 × x × x × x
       14x
4
 = 2 × 7 × x × x × x × x
The common factors of the three terms are 2, x and x.
Therefore, 10x
2
 – 18x
3
 + 14x
4
 = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x)
+ (2 × x × x × 7 × x × x)
= 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)]
= 2x
2 
 × 
 
(5 – 9x + 7x
2
) = 
2 2
2 (7 9 5) x x x - +
  
Factorise: (i)  12x + 36 (ii)  22y – 33z (iii)  14pq + 35pqr
14.2.2  Factorisation by regrouping terms
Look at the expression 2xy + 2y + 3x + 3.  Y ou will notice that the first two terms have
common factors 2 and y and the last two terms have a common factor 3.  But there is no
single factor common to all the terms.  How shall we proceed?
Let us write (2xy + 2y) in the factor form:
2xy + 2y = (2 × x × y) + (2 × y)
= (2 × y × x) + (2 × y × 1)
= (2y × x) + (2y × 1) = 2y (x + 1)
Similarly ,     3x + 3 = (3 × x) + (3 × 1)
= 3 × (x + 1) = 3 ( x + 1)
Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)
Observe, now we have a common factor (x + 1) in both the terms on the right hand
side. Combining the two terms,
2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3)
The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its
factors are (x + 1) and (2y + 3). Note, these factors are irreducible.
Note, we need to
show1 as a factor
here. Why?
Do you notice that the factor
form of an expression has only
one term?
(combining the three terms)
(combining the terms)
2019-20
220  MATHEMATICS
What is regrouping?
Suppose, the above expression was given as 2xy + 3 + 2y + 3x; then it will not be easy to
see the factorisation. Rearranging the expression, as 2xy + 2y + 3x + 3, allows us to form
groups (2xy + 2y) and (3x + 3) leading to factorisation. This is regrouping.
Regrouping may be possible in more than one ways. Suppose, we regroup the
expression as: 2xy + 3x + 2y + 3. This will also lead to factors. Let us try:
2xy + 3x + 2y + 3 = 2 × x × y + 3 × x + 2 × y + 3
= x × (2y + 3) + 1 × (2y + 3)
           = (2y + 3) (x + 1)
The factors are the same (as they have to be), although they appear in different order.
Example 3: Factorise  6xy – 4y + 6 – 9x.
Solution:
Step 1 Check if there is a common factor among all terms.  There is none.
Step 2 Think of grouping.  Notice that first two terms have a common factor 2y;
6xy – 4y = 2y (3x – 2) (a)
What about the last two terms? Observe them.  If you change their order to
– 9x + 6, the factor ( 3x – 2) will come out;
–9x + 6 = –3 (3x) + 3 (2)
= – 3 (3x – 2) (b)
Step 3 Putting (a) and (b) together,
6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)
The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3).
EXERCISE 14.1
1. Find the common factors of the given terms.
(i) 12x, 36 (ii) 2y, 22xy (iii) 14 pq, 28p
2
q
2
(iv) 2x, 3x
2
, 4 (v) 6 abc, 24ab
2
, 12 a
2
b
(vi) 16 x
3
, – 4x
2
, 32x (vii) 10 pq, 20qr, 30rp
(viii) 3x
2
 y
3
, 10x
3
 y
2
,6 x
2
 y
2
z
2. Factorise the  following expressions.
(i) 7x – 42 (ii) 6p – 12q (iii) 7a
2
 + 14a
(iv) – 16 z + 20 z
3
(v) 20 l
2
 m + 30 a l m
(vi) 5 x
2
 y – 15 xy
2
(vii) 10 a
2
 – 15 b
2
 + 20 c
2
(viii) – 4 a
2
 + 4 ab – 4 ca (ix) x
2
 y z +  x y
2
z + x y z
2
(x) a x
2
 y + b x y
2
 + c x y z
3. Factorise.
(i) x
2
 +  x y + 8x + 8y (ii) 15 xy – 6x + 5y – 2
2019-20
Page 5


FACTORISATION  217
14.1  Introduction
14.1.1  Factors of natural numbers
Y ou will remember what you learnt about factors in Class VI. Let us take a natural number ,
say 30, and write it as a product of other natural numbers, say
 30 = 2 × 15
= 3 × 10 = 5 × 6
Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30.
Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
A number written as a product of prime factors is said to
be in the prime factor form; for example, 30 written as
2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors.  This is
what we shall learn to do in this chapter.
14.1.2  Factors of algebraic expressions
W e have seen in Class VII that in algebraic expressions, terms are formed as products of
factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed
by the factors 5, x and y, i.e.,
5xy = y x× × 5
Observe that the factors 5, x and y of 5xy cannot further
be expressed as a product of factors. We may say that 5,
x and y are ‘prime’ factors of 5xy.  In algebraic expressions,
we use the word ‘irreducible’ in place of ‘prime’.  W e say that
5 × x × y is the irreducible form of 5xy.  Note 5 × (xy) is not
an irreducible form of  5xy, since the factor xy can be further
expressed as a product of x and y, i.e., xy = x × y.
Factorisation
CHAPTER
14
Note 1 is a factor of 5xy, since
5xy = y x× × × 5 1
In fact, 1 is a factor of every term. As
in the case of natural numbers, unless
it is specially required, we do not show
1 as a separate factor of any term.
We know that 30 can also be written as
30 = 1 × 30
Thus, 1 and 30 are also factors of 30.
You will notice that 1 is a factor of any
number.  For example, 101 = 1 × 101.
However, when we write a number as a
product of factors, we shall not write 1 as
a factor, unless it is specially required.
2019-20
218  MATHEMATICS
Next consider the expression 3x (x + 2).  It can be written as a product of factors.
3, x and (x + 2)
3x(x + 2) = ( ) 2 3 + × × x x
The factors 3, x and (x +2) are irreducible factors of 3x (x + 2).
Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form
as  10x (x + 2) (y + 3) = 
( ) ( ) 2 5 2 3 x x y × × × + × + .
14.2  What is Factorisation?
When we factorise an algebraic expression, we write it as a product of factors.  These
factors may be numbers, algebraic variables or algebraic expressions.
Expressions like 3xy, y x
2
5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form.
Their factors can be just read off from them, as we already know.
On the other hand consider expressions like 2x + 4, 3x + 3y, x
2
 + 5x, x
2
 + 5x + 6.
It is not obvious what their factors are. W e need to develop systematic methods to factorise
these expressions, i.e., to find their factors. This is what we shall do now.
14.2.1  Method of common factors
• We begin with a simple example: Factorise 2x + 4.
W e shall write each term as a product of irreducible factors;
2x = 2 × x
4 = 2 × 2
Hence 2x + 4 = (2 × x) + (2 × 2)
Notice that factor 2 is common to both the terms.
Observe, by distributive law
2 × (x + 2) = (2 × x) + (2 × 2)
Therefore, we can write
2x + 4 = 2 × (x + 2) = 2 (x + 2)
Thus, the expression 2x + 4 is the same as 2 (x + 2).  Now we can read off its factors:
they are 2 and (x + 2). These factors are irreducible.
Next, factorise 5xy + 10x.
The irreducible factor forms of 5xy and 10x are respectively,
5xy = 5 × x × y
10x = 2 × 5 × x
Observe that the two terms have 5 and x as common factors.  Now,
5xy + 10x = (5 × x × y)  + (5 × x × 2)
= (5x × y)  + (5x × 2)
W e combine the two terms using the distributive law,
(5x× y) + (5x× 2) = 5x × ( y + 2)
Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.)
2019-20
FACTORISATION  219
TRY THESE
Example 1: Factorise  12a
2
b + 15ab
2
Solution: We have 12a
2
b = 2 × 2 × 3 × a × a × b
15ab
2
 = 3 × 5 × a × b × b
The two terms have 3, a and b as common factors.
Therefore, 12a
2
b + 15ab
2
 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
= 3 × a × b × [(2 × 2 × a) + (5 × b)]
 = 3ab × (4a + 5b)
= 3ab (4a + 5b) (required factor form)
Example 2: Factorise 10x
2
 – 18x
3
 + 14x
4
Solution: 10x
2 
= 2 × 5 × x × x
        18x
3
  = 2 × 3 × 3 × x × x × x
       14x
4
 = 2 × 7 × x × x × x × x
The common factors of the three terms are 2, x and x.
Therefore, 10x
2
 – 18x
3
 + 14x
4
 = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x)
+ (2 × x × x × 7 × x × x)
= 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)]
= 2x
2 
 × 
 
(5 – 9x + 7x
2
) = 
2 2
2 (7 9 5) x x x - +
  
Factorise: (i)  12x + 36 (ii)  22y – 33z (iii)  14pq + 35pqr
14.2.2  Factorisation by regrouping terms
Look at the expression 2xy + 2y + 3x + 3.  Y ou will notice that the first two terms have
common factors 2 and y and the last two terms have a common factor 3.  But there is no
single factor common to all the terms.  How shall we proceed?
Let us write (2xy + 2y) in the factor form:
2xy + 2y = (2 × x × y) + (2 × y)
= (2 × y × x) + (2 × y × 1)
= (2y × x) + (2y × 1) = 2y (x + 1)
Similarly ,     3x + 3 = (3 × x) + (3 × 1)
= 3 × (x + 1) = 3 ( x + 1)
Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)
Observe, now we have a common factor (x + 1) in both the terms on the right hand
side. Combining the two terms,
2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3)
The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its
factors are (x + 1) and (2y + 3). Note, these factors are irreducible.
Note, we need to
show1 as a factor
here. Why?
Do you notice that the factor
form of an expression has only
one term?
(combining the three terms)
(combining the terms)
2019-20
220  MATHEMATICS
What is regrouping?
Suppose, the above expression was given as 2xy + 3 + 2y + 3x; then it will not be easy to
see the factorisation. Rearranging the expression, as 2xy + 2y + 3x + 3, allows us to form
groups (2xy + 2y) and (3x + 3) leading to factorisation. This is regrouping.
Regrouping may be possible in more than one ways. Suppose, we regroup the
expression as: 2xy + 3x + 2y + 3. This will also lead to factors. Let us try:
2xy + 3x + 2y + 3 = 2 × x × y + 3 × x + 2 × y + 3
= x × (2y + 3) + 1 × (2y + 3)
           = (2y + 3) (x + 1)
The factors are the same (as they have to be), although they appear in different order.
Example 3: Factorise  6xy – 4y + 6 – 9x.
Solution:
Step 1 Check if there is a common factor among all terms.  There is none.
Step 2 Think of grouping.  Notice that first two terms have a common factor 2y;
6xy – 4y = 2y (3x – 2) (a)
What about the last two terms? Observe them.  If you change their order to
– 9x + 6, the factor ( 3x – 2) will come out;
–9x + 6 = –3 (3x) + 3 (2)
= – 3 (3x – 2) (b)
Step 3 Putting (a) and (b) together,
6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)
The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3).
EXERCISE 14.1
1. Find the common factors of the given terms.
(i) 12x, 36 (ii) 2y, 22xy (iii) 14 pq, 28p
2
q
2
(iv) 2x, 3x
2
, 4 (v) 6 abc, 24ab
2
, 12 a
2
b
(vi) 16 x
3
, – 4x
2
, 32x (vii) 10 pq, 20qr, 30rp
(viii) 3x
2
 y
3
, 10x
3
 y
2
,6 x
2
 y
2
z
2. Factorise the  following expressions.
(i) 7x – 42 (ii) 6p – 12q (iii) 7a
2
 + 14a
(iv) – 16 z + 20 z
3
(v) 20 l
2
 m + 30 a l m
(vi) 5 x
2
 y – 15 xy
2
(vii) 10 a
2
 – 15 b
2
 + 20 c
2
(viii) – 4 a
2
 + 4 ab – 4 ca (ix) x
2
 y z +  x y
2
z + x y z
2
(x) a x
2
 y + b x y
2
 + c x y z
3. Factorise.
(i) x
2
 +  x y + 8x + 8y (ii) 15 xy – 6x + 5y – 2
2019-20
FACTORISATION  221
(iii) ax + bx – ay – by (iv) 15 pq + 15 + 9q + 25p
(v) z – 7 + 7 x y – x y z
14.2.3  Factorisation using identities
We know that (a + b)
2
  = a
2
 + 2ab + b
2
(I)
(a – b)
2
  = a
2 
– 2ab + b
2
(II)
(a + b) (a – b)  = a
2 
– b
2
(III)
The following solved examples illustrate how to use these identities for factorisation. What
we do is to observe the given expression. If it has a form that fits the right hand side of one
of the identities, then the expression corresponding to the left hand side of the identity
gives the desired factorisation.
Example 4: Factorise x
2 
+ 8x + 16
Solution: Observe the expression; it has three terms. Therefore, it does not fit
Identity III. Also, it’s first and third terms are perfect squares with a positive sign before
the middle term. So, it is of the form a
2
 + 2ab + b
2
 where a = x and b = 4
such that a
2 
+ 2ab + b
2
 = x
2
 + 2 (x) (4) + 4
2
= x
2
 + 8x + 16
Since a
2 
+ 2ab + b
2
 = (a + b)
2
,
by comparison x
2
 + 8x + 16 = ( x + 4)
2
    (the required factorisation)
Example 5: Factorise 4y
2 
– 12y + 9
Solution: Observe 4y
2
 = (2y)
2
, 9 = 3
2
 and 12y = 2 × 3 × (2y)
Therefore, 4y
2 
– 12y + 9 = (2y)
2
 – 2 × 3 × (2y) + (3)
2
= ( 2y – 3)
2
(required factorisation)
Example 6: Factorise  49p
2
 – 36
Solution: There are two terms; both are squares and the second is negative. The
expression is of the form (a
2
 – b
2
). Identity III is applicable here;
   49p
2
  – 36 = (7p)
2
 – ( 6 )
2
= (7p – 6 ) ( 7p + 6) (required factorisation)
Example 7: Factorise a
2 
– 2ab + b
2
 – c
2
Solution: The first three terms of the given expression form (a – b)
2
.  The fourth term
is a square.  So the expression can be reduced to a difference of two squares.
Thus, a
2 
– 2ab + b
2 
– c
2
 = (a – b)
2
– c
2
(Applying Identity II)
      = [(a – b) – c) ((a – b) + c)] (Applying Identity III)
= (a – b – c) (a – b + c) (required factorisation)
Notice, how we applied two identities one after the other to obtain the required factorisation.
Example 8: Factorise   m
4
 – 256
Solution: We note m
4 
= (m
2
)
2
 and 256 = (16)
 2
Observe here the given
expression is of the form
a
2
 – 2ab + b
2
.
Where a = 2y, and b = 3
with 2ab = 2 × 2y × 3 = 12y.
2019-20
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