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NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th | Class 8 Mathematics by VP Classes PDF Download

Q: 4. What are the methods of factorisation?

Ans. There are several methods of factorisation, including:- Common factor method: Finding the common factors of a given expression and factoring them out.- Difference of squares method: Factoring an expression that can be written as the difference of two perfect squares.- Quadratic method: Factoring quadratic expressions by splitting the middle term or using the quadratic formula.- Grouping method: Grouping the terms of an expression in pairs and factoring out common factors.- Perfect square trinomial method: Factoring expressions that are perfect squares of binomials.

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 Page 1


  
Question: Divide:
(i) 24xy
2
z
3
 by 6yz
2
(ii) 63a
2
b
4
c
6
 by 7a
2
b
2
c
3
Solution: (i) We have 24xy
2
z
3
 ? 6yz
2
= 
24
6
23
2
xy z
yz
= 
222 3
23
¥ ¥ ¥ ¥¥ ¥¥ ¥ ¥
¥¥ ¥¥¥
x yyz zz
yz z
= 
22
1
¥¥ ¥ ¥ xyz
\ 24xy
2
x
3
 ? 6yz
2
= 4xyz
(ii) 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 
63a b c
7a b c
24 6
22 3
= 
33 7
7
24 6
22 3
¥¥ ¥ ¥ ¥
¥¥ ¥
ab c
ab c
= 3 ¥ 3 ¥ 
a
a
2
2
 ¥ 
b
b
4
2
 ¥ 
c
c
6
3
= 9 ¥ a
2–2
 ¥ b
4– 2
  c
6–3
Page 2


  
Question: Divide:
(i) 24xy
2
z
3
 by 6yz
2
(ii) 63a
2
b
4
c
6
 by 7a
2
b
2
c
3
Solution: (i) We have 24xy
2
z
3
 ? 6yz
2
= 
24
6
23
2
xy z
yz
= 
222 3
23
¥ ¥ ¥ ¥¥ ¥¥ ¥ ¥
¥¥ ¥¥¥
x yyz zz
yz z
= 
22
1
¥¥ ¥ ¥ xyz
\ 24xy
2
x
3
 ? 6yz
2
= 4xyz
(ii) 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 
63a b c
7a b c
24 6
22 3
= 
33 7
7
24 6
22 3
¥¥ ¥ ¥ ¥
¥¥ ¥
ab c
ab c
= 3 ¥ 3 ¥ 
a
a
2
2
 ¥ 
b
b
4
2
 ¥ 
c
c
6
3
= 9 ¥ a
2–2
 ¥ b
4– 2
  c
6–3
  
= 9 ¥ a
0
 + b
2
 ¥ c
3
= 9 ¥ 1 ¥ b
2
 ¥ c
3
 = 9b
2
c
3
\ 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 9b
3
c
3
EXERCISE 14.3
Question 1. Carry out the following divisions.
(i) 28x
4
 ? 56x (ii) –36y
3
 ? 9y
2
(iii) 66pq
2
r
3
 ? 11qr
2
(iv) 34x
3
y
3
z
3
 ? 51xy
2
z
3
(v) 12a
8
b
8
 ? (–6a
6
b
4
)
Solution: (i) We have 28x
4
 ? 56x = 
28
56
4
x
x
= 
227
22 2 7
4
1
¥¥ ¥
¥ ¥¥¥
x
x
= 
1
2
 ¥ x
4– 1
 = 
1
2
x
3
\ 28x
4
 ? 56x = 
1
2
x
3
(ii) We have –36y
3
= (–1) ¥ 2 ¥ 2 ¥ 3 ¥ 3 ¥ y ¥ y ¥ y
and 9y
2
= 3 ¥ 3 ¥ y ¥ y
\ –36y
3
 ? 9y
2
= 
-36
9
3
2
y
y
= 
() - ¥¥¥ ¥ ¥ ¥ ¥
¥¥ ¥
12 2 3 3
33
yy y
yy
= 
() - ¥¥¥ 12 2
1
y
 = –4y
(iii) We have 66pq
2
r
3
= 2 ¥ 3 ¥ 11 ¥ p ¥ q ¥ q ¥ r ¥ r ¥ r
and 11qr
2
= 11 ¥ q ¥ r ¥ r
\ 66pq
2
r
3
 ? 11qr
2
= 
66
11
23
2
pq r
qr
= 
23 11
11
¥ ¥ ¥¥¥ ¥ ¥ ¥
¥¥ ¥
p q qrrr
qr r
= 
23
1
¥ ¥¥¥ pq r
 = 6pqr
(iv) We have 34x
3
y
3
z
3
 ? 51xy
2
z
3
= 
34
51
33 3
23
xy z
xy z
= 
217
317
¥
¥
 ¥ 
x
x
3
 ¥ 
y
y
3
2
 ¥ 
z
z
3
3
Page 3


  
Question: Divide:
(i) 24xy
2
z
3
 by 6yz
2
(ii) 63a
2
b
4
c
6
 by 7a
2
b
2
c
3
Solution: (i) We have 24xy
2
z
3
 ? 6yz
2
= 
24
6
23
2
xy z
yz
= 
222 3
23
¥ ¥ ¥ ¥¥ ¥¥ ¥ ¥
¥¥ ¥¥¥
x yyz zz
yz z
= 
22
1
¥¥ ¥ ¥ xyz
\ 24xy
2
x
3
 ? 6yz
2
= 4xyz
(ii) 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 
63a b c
7a b c
24 6
22 3
= 
33 7
7
24 6
22 3
¥¥ ¥ ¥ ¥
¥¥ ¥
ab c
ab c
= 3 ¥ 3 ¥ 
a
a
2
2
 ¥ 
b
b
4
2
 ¥ 
c
c
6
3
= 9 ¥ a
2–2
 ¥ b
4– 2
  c
6–3
  
= 9 ¥ a
0
 + b
2
 ¥ c
3
= 9 ¥ 1 ¥ b
2
 ¥ c
3
 = 9b
2
c
3
\ 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 9b
3
c
3
EXERCISE 14.3
Question 1. Carry out the following divisions.
(i) 28x
4
 ? 56x (ii) –36y
3
 ? 9y
2
(iii) 66pq
2
r
3
 ? 11qr
2
(iv) 34x
3
y
3
z
3
 ? 51xy
2
z
3
(v) 12a
8
b
8
 ? (–6a
6
b
4
)
Solution: (i) We have 28x
4
 ? 56x = 
28
56
4
x
x
= 
227
22 2 7
4
1
¥¥ ¥
¥ ¥¥¥
x
x
= 
1
2
 ¥ x
4– 1
 = 
1
2
x
3
\ 28x
4
 ? 56x = 
1
2
x
3
(ii) We have –36y
3
= (–1) ¥ 2 ¥ 2 ¥ 3 ¥ 3 ¥ y ¥ y ¥ y
and 9y
2
= 3 ¥ 3 ¥ y ¥ y
\ –36y
3
 ? 9y
2
= 
-36
9
3
2
y
y
= 
() - ¥¥¥ ¥ ¥ ¥ ¥
¥¥ ¥
12 2 3 3
33
yy y
yy
= 
() - ¥¥¥ 12 2
1
y
 = –4y
(iii) We have 66pq
2
r
3
= 2 ¥ 3 ¥ 11 ¥ p ¥ q ¥ q ¥ r ¥ r ¥ r
and 11qr
2
= 11 ¥ q ¥ r ¥ r
\ 66pq
2
r
3
 ? 11qr
2
= 
66
11
23
2
pq r
qr
= 
23 11
11
¥ ¥ ¥¥¥ ¥ ¥ ¥
¥¥ ¥
p q qrrr
qr r
= 
23
1
¥ ¥¥¥ pq r
 = 6pqr
(iv) We have 34x
3
y
3
z
3
 ? 51xy
2
z
3
= 
34
51
33 3
23
xy z
xy z
= 
217
317
¥
¥
 ¥ 
x
x
3
 ¥ 
y
y
3
2
 ¥ 
z
z
3
3
  
= 
2
3
 ¥ x
3–1
 ¥ y
3– 2
 ¥ 3z
3–3
= 
2
3
 ¥ x
2
 ¥ y
1
 ¥ z
0
 = 
2
3
¥ x
2
y ¥ 1
= 
2
3
x
2
y
(v) We have 12a
8
b
8
 ? (–6a
6
b
4
)= 
12
6
88
64
ab
ab -
= 
22 3
12 3
88
64
¥¥ ¥ ¥
-¥ ¥ ¥ ¥
ab
ab
= 
2
1 -
 ¥ 
a
a
8
6
 ¥ 
b
b
8
4
= –2 ¥ a
8– 6
 ¥ b
8–4
 = –2 ¥ a
2
 ¥ b
4
= –2a
2
b
4
Question 2. Divide the given polynomial by the given monomial.
(i) (5x
2
 – 6x) ? 3x (ii) (3y
8
 – 4y
6
 + 5y
4
) ? y
4
(iii) 8(x
3
y
2
z
2
 + x
2
y
3
z
2
 + x
2
y
2
z
3
) ? 4x
2
y
2
z
2
(iv) (x
3
 + 2x
2
 + 3x) ? 2x
(v) (p
3
q
6
 – p
6
q
3
) ? p
3
q
3
Solution: (i) ? (5x
2
 – 6x) ? 3x = 
5x 6
3
2
- x
x
= 
x
x
() 5x 6
3
-
 = 
5x 6
3
-
\ (5x
2
 – 6x) ? 3x = 
1
3
(5x – 6)
(ii) ? (3y
8
 – 4y
6
 + 5y
4
) ? y
4
= 
34 5y
864
4
yy
y
-+
= 
yy y
y
44 2
4
34 5 () -+
\ (3y
8
 – 4y
6
 + 5y
4
) ? y
4
= 3y
4
 – 4y
2
 + 5
(iii) ? 8(x
3
y
2
z
2
 + x
2
y
3
z
2
 + x
2
y
2
z
3
) ? 4x
2
y
2
z
2
= 
32 2 23 2 2 2 3
22 2
8(xyz xyz xyz)
4xyz
++
= 
24
4
22 2
22 2
¥¥ + +
¥
xy z x y z
xy z
[]
= 
2(x y z)
1
¥+ +
 = 2(x + y + z)
Page 4


  
Question: Divide:
(i) 24xy
2
z
3
 by 6yz
2
(ii) 63a
2
b
4
c
6
 by 7a
2
b
2
c
3
Solution: (i) We have 24xy
2
z
3
 ? 6yz
2
= 
24
6
23
2
xy z
yz
= 
222 3
23
¥ ¥ ¥ ¥¥ ¥¥ ¥ ¥
¥¥ ¥¥¥
x yyz zz
yz z
= 
22
1
¥¥ ¥ ¥ xyz
\ 24xy
2
x
3
 ? 6yz
2
= 4xyz
(ii) 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 
63a b c
7a b c
24 6
22 3
= 
33 7
7
24 6
22 3
¥¥ ¥ ¥ ¥
¥¥ ¥
ab c
ab c
= 3 ¥ 3 ¥ 
a
a
2
2
 ¥ 
b
b
4
2
 ¥ 
c
c
6
3
= 9 ¥ a
2–2
 ¥ b
4– 2
  c
6–3
  
= 9 ¥ a
0
 + b
2
 ¥ c
3
= 9 ¥ 1 ¥ b
2
 ¥ c
3
 = 9b
2
c
3
\ 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 9b
3
c
3
EXERCISE 14.3
Question 1. Carry out the following divisions.
(i) 28x
4
 ? 56x (ii) –36y
3
 ? 9y
2
(iii) 66pq
2
r
3
 ? 11qr
2
(iv) 34x
3
y
3
z
3
 ? 51xy
2
z
3
(v) 12a
8
b
8
 ? (–6a
6
b
4
)
Solution: (i) We have 28x
4
 ? 56x = 
28
56
4
x
x
= 
227
22 2 7
4
1
¥¥ ¥
¥ ¥¥¥
x
x
= 
1
2
 ¥ x
4– 1
 = 
1
2
x
3
\ 28x
4
 ? 56x = 
1
2
x
3
(ii) We have –36y
3
= (–1) ¥ 2 ¥ 2 ¥ 3 ¥ 3 ¥ y ¥ y ¥ y
and 9y
2
= 3 ¥ 3 ¥ y ¥ y
\ –36y
3
 ? 9y
2
= 
-36
9
3
2
y
y
= 
() - ¥¥¥ ¥ ¥ ¥ ¥
¥¥ ¥
12 2 3 3
33
yy y
yy
= 
() - ¥¥¥ 12 2
1
y
 = –4y
(iii) We have 66pq
2
r
3
= 2 ¥ 3 ¥ 11 ¥ p ¥ q ¥ q ¥ r ¥ r ¥ r
and 11qr
2
= 11 ¥ q ¥ r ¥ r
\ 66pq
2
r
3
 ? 11qr
2
= 
66
11
23
2
pq r
qr
= 
23 11
11
¥ ¥ ¥¥¥ ¥ ¥ ¥
¥¥ ¥
p q qrrr
qr r
= 
23
1
¥ ¥¥¥ pq r
 = 6pqr
(iv) We have 34x
3
y
3
z
3
 ? 51xy
2
z
3
= 
34
51
33 3
23
xy z
xy z
= 
217
317
¥
¥
 ¥ 
x
x
3
 ¥ 
y
y
3
2
 ¥ 
z
z
3
3
  
= 
2
3
 ¥ x
3–1
 ¥ y
3– 2
 ¥ 3z
3–3
= 
2
3
 ¥ x
2
 ¥ y
1
 ¥ z
0
 = 
2
3
¥ x
2
y ¥ 1
= 
2
3
x
2
y
(v) We have 12a
8
b
8
 ? (–6a
6
b
4
)= 
12
6
88
64
ab
ab -
= 
22 3
12 3
88
64
¥¥ ¥ ¥
-¥ ¥ ¥ ¥
ab
ab
= 
2
1 -
 ¥ 
a
a
8
6
 ¥ 
b
b
8
4
= –2 ¥ a
8– 6
 ¥ b
8–4
 = –2 ¥ a
2
 ¥ b
4
= –2a
2
b
4
Question 2. Divide the given polynomial by the given monomial.
(i) (5x
2
 – 6x) ? 3x (ii) (3y
8
 – 4y
6
 + 5y
4
) ? y
4
(iii) 8(x
3
y
2
z
2
 + x
2
y
3
z
2
 + x
2
y
2
z
3
) ? 4x
2
y
2
z
2
(iv) (x
3
 + 2x
2
 + 3x) ? 2x
(v) (p
3
q
6
 – p
6
q
3
) ? p
3
q
3
Solution: (i) ? (5x
2
 – 6x) ? 3x = 
5x 6
3
2
- x
x
= 
x
x
() 5x 6
3
-
 = 
5x 6
3
-
\ (5x
2
 – 6x) ? 3x = 
1
3
(5x – 6)
(ii) ? (3y
8
 – 4y
6
 + 5y
4
) ? y
4
= 
34 5y
864
4
yy
y
-+
= 
yy y
y
44 2
4
34 5 () -+
\ (3y
8
 – 4y
6
 + 5y
4
) ? y
4
= 3y
4
 – 4y
2
 + 5
(iii) ? 8(x
3
y
2
z
2
 + x
2
y
3
z
2
 + x
2
y
2
z
3
) ? 4x
2
y
2
z
2
= 
32 2 23 2 2 2 3
22 2
8(xyz xyz xyz)
4xyz
++
= 
24
4
22 2
22 2
¥¥ + +
¥
xy z x y z
xy z
[]
= 
2(x y z)
1
¥+ +
 = 2(x + y + z)
  
\
8
4
32 2 2 3 2 2 2 3
22 2
() xy x x y z x y z
xy z
++
= 2(x + y + z)
(iv) ? (x
3
 + 2x
2
 + 3x) ? 2x = 
xx x
x
32
23
2
++
= 
xx x
x
()
2
23
2
++
= 
1
2
(x
2
 + 2x + 3)
\ (x
3
 + 2x
2
 + 3x) ? 2x = 
1
2
(x
2
 + 2x + 3)
(v) ? (P
3
q
6
 – p
6
q
3
) ? p
3
q
3
= 
pq p q
pq
36 6 3
33
-
= 
pq q p
pq
33 3 3
33
[] -
 = 
qp
1
33
-
\ (p
3
q
6
 – p
6
q
3
) ? p
3
q
3
= q
3
 – p
3
Question 3. Work out the following divisions:
(i) (10x – 25) ? 5 (ii) (10x – 25) ? (2x – 5)
(iii) 10y(6y + 21) ? 5(2y + 7) (iv) 9x
2
y
2
(3z – 24) ? 27xy(z – 8)
(v) 96abc(3a – 12)(5b – 30) ? 144(a – 4)(b – 6)
Solution: (i) ? 10x – 25 = 5(2x – 5)
\
10 25
5
x -
= 
52 5
5
() x -
= (2x – 5)
Thus, (10x – 25) ? 5 = 2x – 5
(ii) ? 10x – 25 = 5(2x – 5)
\
10 25
25
x
x
-
-
= 
52 5
25
()
()
x
x
-
-
 = 5
Thus, (10x – 25) ? (2x – 5) = 5
(iii) ? 6y + 21 = 3(2y + 7)
\
10 6 21
52 7
yy
y
()
()
+
+
= 
10 3 2 7
52 7
yy
y
¥¥ +
+
()
()
= 2 ¥ y ¥ 3 = 6y
\ 10y(6y + 21) ? 5(2y + 7) = 6y
(iv) ? 3z – 24 = 3(z – 8)
\
93 24
27 8
22
xy z
xy z
()
()
-
-
= 
22
33 x y 3 (z 8)
33 3 x y (z 8)
´´ ´ ´´ -
´´ ´ ´ ´ -
Page 5


  
Question: Divide:
(i) 24xy
2
z
3
 by 6yz
2
(ii) 63a
2
b
4
c
6
 by 7a
2
b
2
c
3
Solution: (i) We have 24xy
2
z
3
 ? 6yz
2
= 
24
6
23
2
xy z
yz
= 
222 3
23
¥ ¥ ¥ ¥¥ ¥¥ ¥ ¥
¥¥ ¥¥¥
x yyz zz
yz z
= 
22
1
¥¥ ¥ ¥ xyz
\ 24xy
2
x
3
 ? 6yz
2
= 4xyz
(ii) 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 
63a b c
7a b c
24 6
22 3
= 
33 7
7
24 6
22 3
¥¥ ¥ ¥ ¥
¥¥ ¥
ab c
ab c
= 3 ¥ 3 ¥ 
a
a
2
2
 ¥ 
b
b
4
2
 ¥ 
c
c
6
3
= 9 ¥ a
2–2
 ¥ b
4– 2
  c
6–3
  
= 9 ¥ a
0
 + b
2
 ¥ c
3
= 9 ¥ 1 ¥ b
2
 ¥ c
3
 = 9b
2
c
3
\ 63a
2
b
4
c
6
 ? 7a
2
b
2
c
3
= 9b
3
c
3
EXERCISE 14.3
Question 1. Carry out the following divisions.
(i) 28x
4
 ? 56x (ii) –36y
3
 ? 9y
2
(iii) 66pq
2
r
3
 ? 11qr
2
(iv) 34x
3
y
3
z
3
 ? 51xy
2
z
3
(v) 12a
8
b
8
 ? (–6a
6
b
4
)
Solution: (i) We have 28x
4
 ? 56x = 
28
56
4
x
x
= 
227
22 2 7
4
1
¥¥ ¥
¥ ¥¥¥
x
x
= 
1
2
 ¥ x
4– 1
 = 
1
2
x
3
\ 28x
4
 ? 56x = 
1
2
x
3
(ii) We have –36y
3
= (–1) ¥ 2 ¥ 2 ¥ 3 ¥ 3 ¥ y ¥ y ¥ y
and 9y
2
= 3 ¥ 3 ¥ y ¥ y
\ –36y
3
 ? 9y
2
= 
-36
9
3
2
y
y
= 
() - ¥¥¥ ¥ ¥ ¥ ¥
¥¥ ¥
12 2 3 3
33
yy y
yy
= 
() - ¥¥¥ 12 2
1
y
 = –4y
(iii) We have 66pq
2
r
3
= 2 ¥ 3 ¥ 11 ¥ p ¥ q ¥ q ¥ r ¥ r ¥ r
and 11qr
2
= 11 ¥ q ¥ r ¥ r
\ 66pq
2
r
3
 ? 11qr
2
= 
66
11
23
2
pq r
qr
= 
23 11
11
¥ ¥ ¥¥¥ ¥ ¥ ¥
¥¥ ¥
p q qrrr
qr r
= 
23
1
¥ ¥¥¥ pq r
 = 6pqr
(iv) We have 34x
3
y
3
z
3
 ? 51xy
2
z
3
= 
34
51
33 3
23
xy z
xy z
= 
217
317
¥
¥
 ¥ 
x
x
3
 ¥ 
y
y
3
2
 ¥ 
z
z
3
3
  
= 
2
3
 ¥ x
3–1
 ¥ y
3– 2
 ¥ 3z
3–3
= 
2
3
 ¥ x
2
 ¥ y
1
 ¥ z
0
 = 
2
3
¥ x
2
y ¥ 1
= 
2
3
x
2
y
(v) We have 12a
8
b
8
 ? (–6a
6
b
4
)= 
12
6
88
64
ab
ab -
= 
22 3
12 3
88
64
¥¥ ¥ ¥
-¥ ¥ ¥ ¥
ab
ab
= 
2
1 -
 ¥ 
a
a
8
6
 ¥ 
b
b
8
4
= –2 ¥ a
8– 6
 ¥ b
8–4
 = –2 ¥ a
2
 ¥ b
4
= –2a
2
b
4
Question 2. Divide the given polynomial by the given monomial.
(i) (5x
2
 – 6x) ? 3x (ii) (3y
8
 – 4y
6
 + 5y
4
) ? y
4
(iii) 8(x
3
y
2
z
2
 + x
2
y
3
z
2
 + x
2
y
2
z
3
) ? 4x
2
y
2
z
2
(iv) (x
3
 + 2x
2
 + 3x) ? 2x
(v) (p
3
q
6
 – p
6
q
3
) ? p
3
q
3
Solution: (i) ? (5x
2
 – 6x) ? 3x = 
5x 6
3
2
- x
x
= 
x
x
() 5x 6
3
-
 = 
5x 6
3
-
\ (5x
2
 – 6x) ? 3x = 
1
3
(5x – 6)
(ii) ? (3y
8
 – 4y
6
 + 5y
4
) ? y
4
= 
34 5y
864
4
yy
y
-+
= 
yy y
y
44 2
4
34 5 () -+
\ (3y
8
 – 4y
6
 + 5y
4
) ? y
4
= 3y
4
 – 4y
2
 + 5
(iii) ? 8(x
3
y
2
z
2
 + x
2
y
3
z
2
 + x
2
y
2
z
3
) ? 4x
2
y
2
z
2
= 
32 2 23 2 2 2 3
22 2
8(xyz xyz xyz)
4xyz
++
= 
24
4
22 2
22 2
¥¥ + +
¥
xy z x y z
xy z
[]
= 
2(x y z)
1
¥+ +
 = 2(x + y + z)
  
\
8
4
32 2 2 3 2 2 2 3
22 2
() xy x x y z x y z
xy z
++
= 2(x + y + z)
(iv) ? (x
3
 + 2x
2
 + 3x) ? 2x = 
xx x
x
32
23
2
++
= 
xx x
x
()
2
23
2
++
= 
1
2
(x
2
 + 2x + 3)
\ (x
3
 + 2x
2
 + 3x) ? 2x = 
1
2
(x
2
 + 2x + 3)
(v) ? (P
3
q
6
 – p
6
q
3
) ? p
3
q
3
= 
pq p q
pq
36 6 3
33
-
= 
pq q p
pq
33 3 3
33
[] -
 = 
qp
1
33
-
\ (p
3
q
6
 – p
6
q
3
) ? p
3
q
3
= q
3
 – p
3
Question 3. Work out the following divisions:
(i) (10x – 25) ? 5 (ii) (10x – 25) ? (2x – 5)
(iii) 10y(6y + 21) ? 5(2y + 7) (iv) 9x
2
y
2
(3z – 24) ? 27xy(z – 8)
(v) 96abc(3a – 12)(5b – 30) ? 144(a – 4)(b – 6)
Solution: (i) ? 10x – 25 = 5(2x – 5)
\
10 25
5
x -
= 
52 5
5
() x -
= (2x – 5)
Thus, (10x – 25) ? 5 = 2x – 5
(ii) ? 10x – 25 = 5(2x – 5)
\
10 25
25
x
x
-
-
= 
52 5
25
()
()
x
x
-
-
 = 5
Thus, (10x – 25) ? (2x – 5) = 5
(iii) ? 6y + 21 = 3(2y + 7)
\
10 6 21
52 7
yy
y
()
()
+
+
= 
10 3 2 7
52 7
yy
y
¥¥ +
+
()
()
= 2 ¥ y ¥ 3 = 6y
\ 10y(6y + 21) ? 5(2y + 7) = 6y
(iv) ? 3z – 24 = 3(z – 8)
\
93 24
27 8
22
xy z
xy z
()
()
-
-
= 
22
33 x y 3 (z 8)
33 3 x y (z 8)
´´ ´ ´´ -
´´ ´ ´ ´ -
  
= 
x
x
2
 ¥ 
y
y
2
 = x
2–1
 y
2–1
= xy
\ 9x
2
y
2
(3z – 24) ? 27xy(z – 8) = xy
(v) ? 3a – 12 = 3(a – 4)
5b – 30 = 5(b – 6)
96 = 2 ¥ 2 ¥ 2 ¥ 2 ¥ 2 ¥ 3
and 144 = 2 ¥ 2 ¥ 2 ¥ 2 ¥ 3 ¥ 3
\
96 3 12 5 30
144( 4 6
abc a b
ab
()( )
)( )
--
--
= 
2222 2 3 3 4 5 6
2 2 223 3 4 6
¥ ¥¥¥¥ ¥ ¥ ¥ ¥ ¥ - ¥ ¥ -
¥¥¥¥ ¥ ¥ - ¥ -
ab c a b
ab
() ( )
()( )
= 2 ¥ 5 ¥ a ¥ b ¥ c = 10abc
Thus, 96abc (3a – 12)(5b – 36) ? 144(a – 4)(b – 6) = 10abc
Question 4. Divide as directed.
(i) 5(2x + 1)(3x + 5) ? (2x + 1) (ii) 26xy(x + 5)(y – 4) ? 13x(y – 4)
(iii) 52pqr(p + q)(q + r)(r + p) ? 104pq(q + r)(r + p)
(iv) 20(y + 4)(y
2
 + 5y + 3) ? 5(y + 4) (v) x(x + 1)(x + 2) (x + 3) ? x(x + 1)
Solution: (i) We have 5(2x + 1)(3x + 5) ? (2x + 1) = 
52 1 3 5
21
()( )
()
xx
x
++
+
= 
53 5
1
¥+ () x
 = 5(3x + 5)
\ 5(2x + 1)(3x + 5) ? (2x + 1) = 5(3x + 5)
(ii) We have 26xy(x + 5)(y – 4) ? 13x(y – 4)= 
26 5 4
13 4
xy x y
xy
()( )
()
+-
-
= 
2y(x 5)
1
¥+
\ 26xy(x + 5)(y – 4) ? 13x(y – 4) = 2y(x + 5)
(iii) We have
52pqr (p q) (q r) (r p)
104pq (q r) (r p)
++ +
++
= 
52
252
¥¥ ¥ ¥ + + +
¥¥ ¥ + +
pq r p q q r r p
pqq r r p
()( )( )
()( )
 = 
rq r ¥+ ()
2
\ 52pqr(p + q)(q + r)(r +p) ? 104pq(q + r)(r + p) = 
1
2
r(p q) +

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FAQs on NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th - Class 8 Mathematics by VP Classes

1. What is factorisation in mathematics?

Ans. Factorisation is a mathematical process of expressing a number or an algebraic expression as a product of its factors. It involves finding the factors that multiply together to give the original number or expression.

2. Why is factorisation important in mathematics?

Ans. Factorisation is important in mathematics because it helps in simplifying complex expressions and solving equations. It also helps in finding the common factors and simplifying fractions. Factorisation is used in various areas of mathematics such as algebra, number theory, and calculus.

3. How do you factorise a quadratic expression?

Ans. To factorise a quadratic expression, we need to find the two binomial factors that, when multiplied together, give the quadratic expression. We can use methods like splitting the middle term, completing the square, or using the quadratic formula to factorise quadratic expressions.

4. What are the methods of factorisation?

Ans. There are several methods of factorisation, including: - Common factor method: Finding the common factors of a given expression and factoring them out. - Difference of squares method: Factoring an expression that can be written as the difference of two perfect squares. - Quadratic method: Factoring quadratic expressions by splitting the middle term or using the quadratic formula. - Grouping method: Grouping the terms of an expression in pairs and factoring out common factors. - Perfect square trinomial method: Factoring expressions that are perfect squares of binomials.

5. How is factorisation used in real-life situations?

Ans. Factorisation is used in various real-life situations, including: - In finance and economics, factorisation is used to calculate interest rates, loan payments, and investment returns. - In computer science, factorisation is used in cryptography algorithms to ensure secure communication. - In physics and engineering, factorisation is used to simplify complex equations and models. - In statistics, factorisation is used in data analysis and regression analysis to identify the underlying factors affecting a set of data. - In chemistry, factorisation is used to balance chemical equations and calculate molecular weights.

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