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CBSE Class 8 Mathematics Revision Notes Chapter 14 – Factorisation
In this Class 8 Chapter 14 Mathematics notes, students will learn about improving food and resources. In addition, in this Class 8 Chapter 14 math notes, students will get to know the significant details of the chapter that are important for their final examination. Along with Chapter 14 Mathematics Class 8 notes, Extramarks will provide students with essential questions that can be asked to prepare them quickly. Moreover, Class 8 Mathematics notes Chapter 14 will be a student’s lastminute revision guide providing all the necessary information. These notes are based on the CBSE syllabus.
Factorisation of a number or an algebraic expression means writing it as a product of numbers, variables, or expressions. Such numbers, variables, or expressions are known as factors.
 Factorisation of numbers – A number could be split as a product of two or more factors depending upon the number given. For example, we can write the number 15 as the product of 1 and 15 or the product of 3 and 5, or the product of 1,3, and 5. So, 1,15,3 and 5 are the factors of the number 30. Prime factors are the factors that are prime numbers.
 Factorisation of algebraic expressions – Any factor can be a number, variable, or expression. For example, factors of 5ab are 5, a, and b. Likewise, we can write 3ab+6a as 3ab+6a=3xax(b+2), where 3, a, and (b+2) are the factors.
Method Of Common Factors
Here, to find the factors of an algebraic expression, we have to first write each term of an expression as a product of irreducible factors. Then, We can take the common factors and thus determine the factors of that algebraic expression by using the distributive law. For example, let’s find the factors of the expression 4x+2xy. Firstly, find the factors of each term.
So, we can write 4x as 4x=22x and 2xy as 2xy. Thus, we can write 4x+2xy=(22x)+(2xy), where we can notice that 2 and x are common in both terms. So, 4x+2xy=2xx(2+y). So, the common factors are 2, x, and (2+y).
These notes contain all the topics in the NCERT books concisely. Download the Class 8 Mathematics Chapter 5 notes on this page to improve your exam results.
Factorisation By Regrouping Terms:
When the terms of any given expression do not have any single common factor, we rearrange the terms to group them, resulting in factorisation. This method of group forming is known as regrouping. There can be more than one way of regrouping for a given expression. Let’s take the example of the expression z−7+7xy−xyz. Here we can see the terms do not have any common factors, so we will rearrange the terms as z−xyz−7+7xy and take common z and −7 from the first two and last two terms, respectively. We get;
z−xyz−7+7xy=zx(1−xy)−7x(1−xy)
=(z−7)x(1−xy)
Hence, the factors are (z−7) and (1−xy).
Once you go through these notes, you can easily solve CBSE sample papers to test your understanding. These notes also contain CBSE extra questions that will help student test their understanding.
Factorisation using identities
Expression of the forms  Identity of factorisation 
a² + 2ab + b²  (a + b)² 
a² – 2ab + b²  (a – b)² 
a² – b²  (a + b)(a – b) 
a² + b² + c² +2ab + 2bc + 2ac  (a + b + c)² 
A³ + 3a²b + 3ab² + b³  (a + b)³ 
a² + (x + y) a + xy  (a + x)(a + y) 
Aside from the above identities, many other identities are used. Significantly, the signs of the terms should be taken care of when identities are used.
Division Of Algebraic Expressions:
In the case of algebraic expressions, unlike numbers, the division operations are different, and factorisation plays a key role. The division is carried out while keeping factorisation as the main player, as seen in the following cases. We can observe that the examples considered upon division give 00 as their remainder.
 Division of a monomial by another monomial – Monomials are written as products of their irreducible factors as they have only one term. In this case, the numerator and denominator are factored, and the common factors are cancelled.
 Let’s suppose we need to divide 14x²yz³ by 7xyz. We can write this as follows;
14x²yz³7xyz= 27xxyzzz 7xyz
Hence, the quotient is 2xz².
 Division of a polynomial by a monomial – In this case, each term of the polynomial is separately divided by the monomial. Hence, each term is written into its factor form and then divided separately. For example, let us divide the polynomial 4a³+2a²+6a by the monomial 2a. So, first, we can write the polynomial as;
4a³+2a²+6a= (22aaa)+(2aa)+(23a)
and the monomial as(2xa).
Hence,4a³+2a²+6a2a=(2x2xaxaxa + (2xaxa)+(3x2xa)(2xa)
 Division of polynomial by polynomial – In this case also, both numerator and denominator are factored, and the factors common in both are cancelled. For example, let us divide (7x² + 14x) by (x+2), then we can write;
(7x² + 14x)(x+2)=(7xx)+(27x)(x+2)
Take 7x common from the two terms of the numerator,
7xx(x+2)(x2)=7x
Hence, the quotient is 7x.
These CBSE revision notes are made according to the CBSE syllabus. The questions in these notes are based on CBSE’s previous year’s question papers. These notes contain important questions and formulas to help students improve their marks in their examinations.
FAQs (Frequently Asked Questions)
1. Find the common factors of the given terms (i) 12x, 36 (ii)2y, 22xy (iii)14pq, 28p2q2 (iv)2x, 3x2, 4 (v) 6abc, 24ab2, 12a2b (vi) 16x3,  4x2, 32x (vii)10 pq, 20qr, 30rp (viii)3x2 y3, 10x3y2, 6x2y2z Solution:
We will first find out the factors of each term and then check which factors are common in each term.
(i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
Here, 2, 2, and 3 are the common factors.
Thus, 2 × 2 × 3 = 12
(ii) 2y = 2 × y
22xy = 2 × 11 × x × y
The common factors are 2, y.
Thus, 2 × y = 2y
(iii) 14pq = 2 × 7 × p × q
28p2q2 = 2 × 2 × 7 × p × p × q × q
The common factors are 2, 7, p, q.
Thus, 2 × 7 × p × q = 14pq
(iv) 2x = 2 × x
3x2 = 3 × x × x
4 = 2 × 2
The common factor is 1.
(v) 6abc = 2 × 3 × a × b × c
24ab2 = 2 × 2 × 2 × 3 × a × b × b
12a2b = 2 × 2 × 3 × a × a × b
The common factors are 2, 3, a, b.
Thus, 2 × 3 × a × b = 6ab
(vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x
4x2 = 1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
The common factors are 2, 2, x.
Thus, 2 × 2 × x = 4x
(vii) 10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
The common factors are 2, 5.
Thus, 2 × 5 = 10
(viii) 3x2y3 = 3 × x × x × y × y × y
10x3y2 = 2 × 5 × x × x × x × y × y
6x2y2z = 2 × 3 × x × x × y × y × z
The common factors are x, x, y, and y.
Thus, x × x × y × y = x2 y2
2. Factorise the following expressions (i) 7x  42 (ii) 6p 12q (iii) 7a2+14a (iv) 16z + 20z3 (v) 20l2m + 30alm (vi) 5x2y 15xy2 (vii) 10a2 15b2 + 20c2 (viii) 4a2 + 4ab  4ca (ix) x2yz + xy2z + xyz2 (x) ax2y + bxy2 + cxyz
(i) 7x – 42
7x = 7 × x
42 = 2 × 3 × 7
The common factor is 7.
Therefore, 7x – 42 = (7 × x) – (2 × 3 × 7 )
= 7(x – 6)
(ii) 6p 12q
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
The common factors are 2 and 3.
Therefore, 6p 12q = (2 × 3 × p) – (2 × 2 × 3 × q)
= 2 × 3(p – 2 × q)
= 6(p – 2q)
(iii) 7a2+14a
7a2= 7 × a × a
14a = 2 × 7 × a
The common factors are 7 and a.
Therefore, 7a2 + 14a = (7 × a × a) + (2 × 7 × a)
= 7 × a(a + 2)
= 7a(a + 2)
(iv) 16z + 20z3
16z = 1 × 2 × 2 × 2 × 2 × z
20z3 = 2 × 2 × 5 × z × z × z
The common factors are 2, 2, and z.
Therefore, 16z + 20z3 = (2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)
= (2 × 2 × z)[(2 × 2) + (5 × z × z)]
= 4z(4 + 5z2)
(v) 20l2m + 30alm
20l2m= 2 × 2 × 5 × l × l × m
30alm = 2 × 3 × 5 × a × l × m
The common factors are 2, 5, l and m.
Therefore, 20l2m + 30alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m)
= (2 × 5 × l × m)[(2 × l) + (3 × a)]
= 10lm(2l + 3a)
(vi) 5x2y 15xy2
5x2y = 5 × x × x × y
15xy2 = 3 × 5 × x × y × y
The common factors are 5, x, and y.
Therefore, 5x2y – 15xy2 = (5 × x × x × y) – (3 × 5 × x × y × y)
= 5 × x × y[x – (3 × y)]
= 5xy(x – 3y)
(vii) 10a2 15b2 + 20c2
10a2 = 2 × 5 × a × a
15b2 = 3 × 5 × b × b
20c2 = 2 × 2 × 5 × c × c
The common factor is 5.
Therefore, 10a2 15b2 + 20c2 = (2 × 5 × a × a) – (3 × 5 × b × b) + (2 × 2 × 5 × c × c)
= 5[(2 × a × a) – (3 × b × b) + (2 × 2 × c × c)]
= 5(2a2 – 3b2 + 4c2 )
(viii) 4a2 + 4ab – 4ca
4a2 = 2 × 2 × a × a
4ab = 2 × 2 × a × b
4ca = 2 × 2 × c × a
The common factors are 2, 2, and a.
Therefore, – 4a2 + 4ab – 4ca = (2 × 2 × a × a) + (2 × 2 × a × b) – (2 × 2 × c × a)
= 2 × 2 × a (a + b – c)
= 4a(a + b – c)
(ix) x2yz + xy2z + xyz2
x2yz = x × x × y × z
xy2z= x × y × y × z
xyz2 = x × y × z × z
The common factors are x, y, and z.
Therefore, x2yz + xy2z + xyz2 = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)
= x × y × z (x + y + z)
= xyz(x + y + z)
(x) ax2y + bxy2 + cxyz
ax2y= a × x × x × y
bxy2 = b × x × y × y
cxyz= c × x × y × z
The common factors are x and y.
Therefore, ax2y + bxy2 + cxyz = (a × x × x × y) + (b × x × y × y) + (c × x × y × z)
= (x × y)[(a × x) + (b × y) + (c × z)]
= xy(ax + by + cz)
3. Factorise: (i) x2+ xy + 8x + 8 y (ii) 15xy 6x + 5 y  2 (iii) ax + bx  ay  by (iv) 15 pq +15 + 9q + 25 p (v) z  7 + 7xy  xyz
(i) x2 + xy + 8x + 8y
= x × x + x × y + 8 × x + 8 × y
= x(x + y) + 8(x + y)
= (x + y) (x + 8)
(ii) 15xy – 6x + 5y – 2
= 3 × 5 × x × y – 3 × 2 × x + 5 × y – 2
= 3x (5y – 2) + 1(5y – 2)
= (5y – 2) (3x + 1)
(iii) ax + bx – ay – by = a × x + b × x – a × y – b × y
= x(a + b) – y(a + b)
= (a + b)(x – y)
(iv) 15pq + 15 + 9q + 25 p
= 15pq + 9q + 25p + 15
= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5
= 3q(5p + 3) + 5(5p + 3)
= (5p + 3)(3q + 5)
(v) z – 7 + 7xy – xyz
= z – xyz – 7 + 7xy
= z – x × y × z – 7 + 7 × x × y
= z(1 xy) – 7(1 – xy)
= (1 – xy)(z – 7)
4. Find the common factors of the given terms (i) 12x, 36 (ii)2y, 22xy (iii)14pq, 28p2q2 (iv)2x, 3x2, 4 (v) 6abc, 24ab2, 12a2b (vi) 16x3,  4x2, 32x (vii)10 pq, 20qr, 30rp (viii)3x2 y3, 10x3y2, 6x2y2z
We will first find out the factors of each term and then check which factors are common in each term.
(i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
Here, 2, 2, and 3 are common factors.
Thus, 2 × 2 × 3 = 12
(ii) 2y = 2 × y
22xy = 2 × 11 × x × y
The common factors are 2, y.
Thus, 2 × y = 2y
(iii) 14pq = 2 × 7 × p × q
28p2q2 = 2 × 2 × 7 × p × p × q × q
The common factors are 2, 7, p, q.
Thus, 2 × 7 × p × q = 14pq
(iv) 2x = 2 × x
3x2 = 3 × x × x
4 = 2 × 2
The common factor is 1.
(v) 6abc = 2 × 3 × a × b × c
24ab2 = 2 × 2 × 2 × 3 × a × b × b
12a2b = 2 × 2 × 3 × a × a × b
The common factors are 2, 3, a, b.
Thus, 2 × 3 × a × b = 6ab
(vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x
4x2 = 1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
The common factors are 2, 2, x.
Thus, 2 × 2 × x = 4x
(vii) 10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
The common factors are 2, 5.
Thus, 2 × 5 = 10
(viii) 3x2y3 = 3 × x × x × y × y × y
10x3y2 = 2 × 5 × x × x × x × y × y
6x2y2z = 2 × 3 × x × x × y × y × z
The common factors are x, x, y, y.
Thus, x × x × y × y = x2 y2