Class 8 Maths Notes for Factorisation

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CLASS 8
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FACTORISATION
IMPORTANT CONCEPTS:

1. Prime factor form :A number written as a product of prime factors is said to be in the prime factor form.
For example: The prime factor form of 30 = 2 x 3 x 5.

2. Factors of algebraic expressions :Like any natural number, an algebraic expression is also the product of
its factors. In the case of algebraic expression, it is said to be an irreducible form instead of prime factor form.
For example : 8ab = 8 x a x b ( irreducible form)

3. Factorisation : Factorisation of an algebraic expression is writing it as a product of factors.These factors
may be numbers, algebraic variables or algebraic expressions.

4. Methods of Factorisation :
a) Methods of common factors.
b) Factorisation by regrouping terms
c) Factorisation using identities .

Methods of common factors:
Example : Factorise 12xy  24 y .
2

Solution : Factors of 12xy  2  2  3  x  y  y
2

Factors of 24 y  2  2  2  3  y
common factors of two terms are 2 , 2 , y.
12xy 2  24 y = 2  2  3  x  y  y   2  2  2  3  y 
= 2 × 2 × 3× y [(x × y ) + (2)]
= 12y × (xy + 2)
= 12y (xy + 2)
Factorisation by regrouping terms :
Example : Factorise 8p – q + pq – 8 by regrouping terms.
8p + q – pq – 8 = 8p – 8 – pq + q
= 8(p –1) – q(p – 1)
= (p – 1)(8 – q )

Factorisation using Identities :
Algebraic Identities : ( a + b) 2 = a2 + 2ab + b2
( a – b) 2 = a2 – 2ab + b2
( a + b)(a – b) = a2 – b2
( x+ a )( x + b) = x2 + (a + b)x + ab
4.Division of Algebraic Expressions
Consider 7 x  14 x
3

7 x3
7 x 3  14x 
14x

68

7 x x x
= 27 x

x2

2

SOME ILLUSTRATIONS:

1.The factorisation of 18a2 b + 15ab2 is

a) 3ab (6a + 5b) b) 3a 2b (6a + 5b) c) 3ab2 (6a + 5b) d) 3a 2b2 (6a + 5b)

Ans: a) 3ab (6a + 5b)

2. The common factor of x²y²z and x³y³z2 is

a) x²y²z b) x³y³z c) x²y³z d) x³y²z

Ans: a) x²y²z

3. The factorisation of 6xy – 4y + 6 – 9x is

a) (3x – 2)(2y – 3) b) (3x + 2)(2y – 3) c) (3x – 2)(2y + 3) d) (3x + 2)(2y + 3)

Ans: a) (3x – 2)(2y – 3)

4. The factors of 4y2 – 12y + 9 is:

a) (2y+3)2 b) (2y-3)2 c) (2y-3)(2y+3) d) None of the above

Ans: b) (2y-3)2

Case Study Based Questions:
1. Riddhi wants to make a garden in her house in the shape of a parallelogram ABCD, the area of which is 4x²-9
and side BC is of length 3m.
(i) If the height corresponding to side AB is 2x+3, then find the length of side CD.
(ii) she fenced the garden with wire which cost a total of rupees x. Find the rate of wire per unit length.
(iii) find the area of parallelogram shaped garden for x=4 units.
Ans: (i) 2 x  3
m
(ii) Rs4
(iii) 55 m2

Short Answer Questions:
69

1. Factorise p2 – 2p + 1.

Ans: p2 – 2p + 1 = p2 –p –p +1

= p(p-1) -1(p-1)

= (p-1)(p-1)

= (p-1)2

2. Factorise: 6ab + 12bc

Ans: : 6ab + 12bc = 2  3  a  b  2  2  3  b  c 

=
2  3  ba  2c 

=
6b(a  2c)

3. Expand using suitable identities :
7 x  52

Ans:

(7 x  5) 2  (7 x) 2  2  7 x  5  52 (a  b) 2  a 2  2ab  b 2 
= 49x  70x  25
2

Long Answer Questions:

1. Factorise the following using the identity ( a + b)(a – b) = a2 – b2

4x2 9 y2

a) 4x2 – 25y2 b) 9 16

Ans:

4 x2  25y 2  2 x  5 y 
2 2
a)

=
2x  5 y 2x  5 y 

4 x 2 9 y 2 2 x  3 y 
2 2
  2  2
b) 9 16 3 4

2 2
 2x   3y 
   
= 
3   4 

 2 x 3 y  2 x 3 y 
    
=
3 4  3 4 

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2
2. Factorize (x + y) – 4xy
Ans: (x + y) = x + 2xy + y2
2 2

(x + y)2 – 4xy = x2 + 2xy + y2 – 4xy
= x2 + y2 – 2xy
(x – y)2 = x2 + y2 – 2xy
(x + y)2 – 4xy = (x – y)2

PRACTICE

A) MCQ

1. Which of the following is the common factor of 25a²b and 55ab²?
(a) 5ab² (b) 5a²b (c) 5ab (d) 5a²b²

2. The common factor of 6a²b4c², 21a²b and 15a³ is
(a) 3a³ (b) 6a³ (c) 6a² (d) 3a²

3. The factorisation of x²yz + xy²z + xyz² is
(a) xyz(x + y + z) (b) x²yz (x + y + z)
(c) xy²z (x + y + z) (d) xyz² (x + y + z)

4. If x² – x – 42 = (x + k) (x + 6) then k =
(a) 6 (b) -6 (c) -7 (d) 7
8 8 6 6
5. The quotient of 12a b ÷ (- 4a b ) is
(a) 3a²b² (b) 3a²b (c) 3ab² (d) -3a²b²

6. The factors of 6xy – 4y + 6 – 9x are:
a. (3x + 2) (2y + 3) b. (3x – 2) (2y – 3)
c. (3x – 2) (2y + 3) d. (3x –+2) (2y – 3)
2
7. The factors of 49p – 36 are:
2
a. (7p+6) b. (7p-6)2 c. (7p – 6 ) ( 7p + 6) d. none of the above

3 4 4 4 3 2
8. The common factor of 24a b , 36a c and 48a b c is:
3 3 3
a. 12a b. 24a c. 36a d. 48a3

9. The factors of xyz are:
a. x b. y c. z d. all of the above

10. The value of 3.5 × 3.5 – 2.5 × 2.5 is
(a) -6 (b) 6 (c) 60 (d) 1

B) SHORT ANSWER QUESTIONS

71

1. Subtract : (i) 5a2b2c2 from – 7a2b2c2
2. Simplify (ab – c)2 + 2abc
2
 4a 5b 
  
3. Expand using suitable identities :  5 4
3 3 2 2
4. Carry out the following divisions: (ii) 76x yz 19x y
5. Factorise the following : x2 – 8x + 16
6. Factorise the expressions. (lm + l) + m + 1
7. Factorise (a 2 – 2ab + b2 – c2 )
8. Factorise m4 – 256
9. Obtain the factors of z2 – 4z – 12.
10. Carry out the following divisions. 12a8b8 ÷ (– 6a6b4)

LONG ANSWER TYPE :
1. Find the common factors of the following:
(i) 6 xyz, 24 xy2 and 12 x2y
(ii) 3x2 y3, 10x3 y2 and 6x2 y2 z
2. Solve for (4x2 – 100) ÷ 6(x + 5)
3.Factorize the expressions and divide them as directed. (m2 – 14m – 32) ÷ (m + 2)
4. Factorize 9a (6a – 5b) – 12a2 (6a – 5b)
5. The area of a square is given by 4x2 + 12xy + 9y2. Find the side of the square.
6. The area of a circle is given by the expression 𝜋x2 + 6𝜋x + 9𝜋. Find the radius of the circle.
7. Factorise the expressions and divide them as directed:
(x4 – 16)  x3 + 2x2 + 4x + 8
8. Factorize (q2 – 10q + 21)

1 3
x 2  2  2  3x 
9. Factorize x x
2 2
10. If m – n = 16 and m + n = 400, then find mn.

CASE BASED QUESTIONS:-
1. Ramesh has a cuboidal box, the volume of which is x³+3x²+2x.
(i) write an expression for it's length, breadth and height in terms of x.
(ii) find the dimensions of box if its height 6m is the largest dimension.
(iii) how many rectangular sheets will be used to cover the base of box if each sheet is of dimension ( 1 x ) ?

2. The total amount paid by Neha for purchasing some books at the rate of Rs. x is x³-x . The number of books
is 1 more than the cost of each book.
(i) How many books she can purchase with the amount she has?
(ii) How many sets of books she can purchase from x³-x rupees if number
of books in each set is x-1?
(iii) If the cost of each book is Rs. 10, total how much money was spent in
purchasing 4 sets of books?

72

ANSWERS:

Solution of MCQs
1. c 2. d 3. a 4. c 5. d 6. b
7. c 8. a 9. d 10. b

Solution Short Answer Questions:
16a 2 25b 2 4 xz3
1. -2a2 b2 c2 . 2. a2b2 + c2 3.   2ab 4.
25 16 y
5. x  42 6. (l  1)(m  1) 7.(a-b-c)(a-b+c)
2
8. (m-2)(m+2)(m +4)
2 4
9. (z-6)(z+12) 10. -2a b

Solutions of Long Answers Questions:
1.(i) 6xy (ii) x2y2 2. ( x  5) 3. (m-16) 4. (2x +3y)2
3
5. (2x +3y) units 6. (x+3) units 7. (x-2) 8. (x-7)(x-3)

9.  x    x   3 
1 1
10. mn = 72
 x  x 

Solutions Of Case Studt Based Questions :

1.(i)x units , (x+2) units ,(x+1)units (ii) 5 units ,4 units (iii) (x+1)
2.(i) (x2 -1) (ii)x(x+1) (iii) Rs 440

TEST-1( 20 marks)

1. Factorise
(i) (16x5 – 144x3 )
(ii) x2 + 9x + 20 .
2. Find the common factors of the given terms.
(i) 20 x3 , – 4x2 , 32x
(ii) 12 pq, 20pqr, 30rp
3.Factorise (z – 7 + 7 x y – x y z)
4. The area of a rectangle is (x2 + 7x + 12). If its breadth is (x + 3), then
find its length
5. Perform the following divisions: (x3y3 + x2y3 – xy4 + xy) ÷ xy

TEST-2 ( 30 marks)

1. Factorise the following expressions.
(i) 3𝑝𝑞𝑟 – 6𝑝2 𝑞 2 𝑟 2 – 15𝑟 2
(ii) –xy – ay
2. Expand the following, using suitable identities

73

(x2y – xy2)2
3. Simplify (1.5p + 1.2q)2 – (1.5p – 1.2q)2
4. Factorise the expressions.
(i) 5y2 – 20y – 8z + 2yz
(ii) 10ab + 4a + 5b + 2
5. Factorise the expressions and divide them as directed.
39y3 (50y2 – 98) ÷26y2 (5y + 7)
6. The area of a square is given by 4x2 + 12xy + 9y2. Find the side of
the square.
7. Find the factors of 3m2 + 9m + 6.
8. Factorise
(ii) p4 – 81
(iii) x4 – (y + z)4

74

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