FACTORISATION
What is factorisation ?
• The process of expressing an algebraic expression as a product of its
irreducible factors is known as Factorisation .
• We have learnt in the previous classes that for a constant
12 = 2 × 6 → REDUCIBLE FACTORS
= 2 × 2 × 3 → IRREDUCIBLE FACTORS
• Similarly for an algebraic expression
2𝑥2 + 4𝑥 = 2 𝑥2 + 2𝑥 → REDUCIBLE FACTOR
= 2𝑥( 𝑥 + 2 ) → IRREDUCIBLE FACTORS
In certain given algebraic expressions it is not obvious what
their factors are. So, we need to develop systematic methods to
factorise them, i.e. find their ( irreducible ) factors.
The methods are :
Method of Common Factors
Factorisation by Grouping or Regrouping Terms
Factorisation by using Identities ( by middle term split )
[ Checking the given expression for factorisation necessarily in the above sequence
will always help ]
Let us first learn how to find common factors of given expressions
• Ex 14.1 Q1 ( v) Find the common factors of the given terms
6𝑎𝑏𝑐, 24𝑎𝑏2, 12𝑎2𝑏
Soln. : 6𝑎𝑏𝑐 = 2 × 3 × 𝑎 × 𝑏 × 𝑐
24𝑎𝑏2 = 2 × 2 × 2 × 3 × 𝑎 × 𝑏 × 𝑏
12𝑎2𝑏 = 2 × 2 × 3 × 𝑎 × 𝑎 × 𝑏
Common factors = 2 × 3 × 𝑎 × 𝑏
= 6𝑎𝑏
[ Can you do it by mere observation, now ? ]
Q1 Find the common factors of the given terms
(iv) 2𝑥, 3𝑥2, 4
• Soln. : 2𝑥 = 2 × 𝑥
3𝑥2 = 3 × 𝑥 × 𝑥
4 =2×2
Common factor = 1
• Q7. Find the common factors of the given terms
c) 3𝑎3𝑏3, 10𝑎3𝑏3, 6𝑎²𝑏²𝑐
Soln : 3𝑎3𝑏3 = 3 × 𝑎 × 𝑎 × 𝑎 × 𝑏 × 𝑏 × 𝑏
10𝑎3𝑏2 = 2 × 5 × 𝑎 × 𝑎 × 𝑎 × 𝑏 × 𝑏
6𝑎2𝑏2𝑐 = 2 × 3 × 𝑎 × 𝑎 × 𝑏 × 𝑏 × 𝑐
Common factors = 𝑎 × 𝑎 × 𝑏 × 𝑏 = 𝑎2𝑏2
NCERT Ex. 14.1
Q2. Factorise the following expression
(iv) −16𝑧 + 20𝑧3
Soln. : −16𝑧 + 20𝑧3
Factorizing each term
= −2 × 2 × 2 × 2 × 𝑧 + 2 × 2 × 5 × 𝑧 × 𝑧 × 𝑧
Using method of common factors
= 2 × 2 × 𝑧 ( −2 × 2 + 5 × 𝑧 × 𝑧 )
= 4𝑧 ( −4 + 5𝑧2)
Q2. Factorise the following expression
(vi) 5𝑥2𝑦 − 15𝑥𝑦2
Soln. : 5𝑥2𝑦 − 15𝑥𝑦2
= 5×𝑥×𝑥×𝑦−3×5×𝑥×𝑦×𝑦
= 5 × 𝑥 × 𝑦 (𝑥 − 3𝑦)
= 5𝑥𝑦 ( 𝑥 − 3𝑦)
The factors of the given algebraic expression are 5, 𝑥, 𝑦 and (𝑥 − 3𝑦).
Q2. Factorise the following expression
(ix) 𝑥2𝑦𝑧 + 𝑥𝑦2 𝑧 + 𝑥𝑦𝑧2
Soln. : Let us see an alternative / shorter method
HCF of (𝑥2, 𝑥 ) = 𝑥 [ 1st variable ]
HCF of (𝑦, 𝑦2) = 𝑦 [ 2nd variable ]
HCF of (z, 𝑧 2) = 𝑧 [ 3rd variable ]
𝑥2𝑦𝑧 + 𝑥𝑦2𝑧 + 𝑥𝑦𝑧2 = 𝑥𝑦𝑧( 𝑥 + 𝑦 + 𝑧)
Q2(x) 𝑎𝑥 2𝑦 + 𝑏𝑥𝑦2 + 𝑐𝑥𝑦𝑧
Soln. : HCF of (𝑎, 𝑏, 𝑐 ) =1
HCF of (𝑥2, 𝑥 ) =𝑥
HCF of (𝑦, 𝑦2) =𝑦
𝑎𝑥 2𝑦 + 𝑏𝑥𝑦2 + 𝑐𝑥𝑦𝑧 = 𝑥𝑦( 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧)
Common Identities
( ) 2 2 2
1. 𝑎 + 𝑏 = 𝑎 + 2𝑎𝑏 + 𝑏
e.g. 𝑥 2 + 8𝑥 + 16
𝑥 2 + 2(𝑥 )(4) + (4)2
Comparing all the three terms with the identity, used above,
and we get:
a = x;b = 4
𝑥 2 + 2(𝑥 )(4) + (4)2 = (𝑥 + 4)2
2. (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2
e.g. 𝑥 2 − 2𝑥𝑦𝑧 + 𝑦 2 𝑧 2
(𝑥 )2 − 2(𝑥 )(𝑦𝑧) + (𝑦𝑧)2
Comparing all the three terms with the identity, used above,
and we get:
𝑎 = 𝑥; 𝑏 = 𝑦𝑧
𝑥 2 − 2𝑥𝑦𝑧 + 𝑦 2 𝑧 2 = (𝑥 − 𝑦𝑧)
3. (𝑎2 − 𝑏 2 ) = 𝑎2 – 𝑏 2
9
e.g. 81 −
𝑎2
( 3 )2
(9)2 −
(𝑎 ) 2
2
3
(9)2 − ( )
𝑎
Using above identity:
3 3
(9 − ) (9 + )
𝑎 𝑎
4. (𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 𝑏 3
e.g. 𝑥 3 + 9𝑥 2 + 27𝑥 + 27
(𝑥 )3 + 3(𝑥 )2 (3) + 3(𝑥 )(3)2 + (3)3
Comparing from above identity:
𝑎 = 𝑥; 𝑏 = 3
3 2 ( ) 2
𝑥 + 9𝑥 + 27𝑥 + 27 = 𝑥 + 3
5. (𝑎 − 𝑏)3 = 𝑎3 − 3𝑎2 𝑏 + 3𝑎𝑏 2 − 𝑏 3
e.g. 𝑠 3 + 3𝑠 2 𝑎2 + 3𝑠𝑎4 + 𝑎6
(𝑠)3 + 3(𝑠)2 (𝑎2 ) + 3(𝑠)(𝑎2 )2 + (𝑎2 )3
Comparing from above Identity:
𝑎 = 𝑠; 𝑏 = 𝑎2
3 2 2 4 6 ( 2 )3
𝑠 + 3𝑠 𝑎 + 3𝑠𝑎 + 𝑎 = 𝑠 + 𝑎
𝑎3 + 𝑏 3 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 )
e.g. 64 + 8𝑧 3
(4)3 + (2𝑧)3
Comparing from above identity:
𝑎 = 4; 𝑏 = 2𝑧
64 + 8𝑧 3 = (4 + 2𝑧)((4)2 − (4)(2𝑧) + (2𝑧)2 )
(4 + 2𝑧)(16 − 8𝑧 + 4𝑧 2 )
8(2 + 𝑧)(4 − 2𝑧 + 𝑧 2 )
𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 )
e.g. 𝑎−3 − 𝑏 −3
((𝑎−1 )3 − (𝑏 −1 )3 )
Comparing form above identity:
𝑎 = (𝑎−1 ); 𝑏 = (𝑏 −1 )
(𝑎−1 − 𝑏 −1 )((𝑎−1 )2 + (𝑎−1 )(𝑏 −1 ) + (𝑏 −1 )2 )
Practice Questions
Factorise: 12a^2b + 15ab^2
Factorise: 6xy−4y+6−9x
Factorise: 49p^2−36
Factorise: 2x^2yz+2xy^2z+4xyz
Factorise: 16x^4−y^4
