Open App

Class 8 Exam > Class 8 Notes > Class 8 Mathematics by VP Classes > NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th

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.

Download, print and study this document offline

Please wait while the PDF view is loading

 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) +

	Class 8 Mathematics by VP Classes 90 docs\|16 tests

Class 8 Mathematics by VP Classes

90 docs|16 tests

Join Course for Free

Top Courses for Class 8

View all

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.

Related Exams

Class 8

About this Document

	4.85/5 Rating
	Nov 21, 2024 Last updated

Document Description: NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th for Class 8 2024 is part of Class 8 Mathematics by VP Classes preparation. The notes and questions for NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th have been prepared according to the Class 8 exam syllabus. Information about NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th covers topics like and NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th.

Introduction of NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th in English is available as part of our Class 8 Mathematics by VP Classes for Class 8 & NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th in Hindi for Class 8 Mathematics by VP Classes course. Download more important topics related with notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th | Class 8 Mathematics by VP Classes

Description

Full syllabus notes, lecture & questions for NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th | Class 8 Mathematics by VP Classes - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Class 8 Mathematics by VP Classes | Best notes, free PDF download

Information about NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th

In this doc you can find the meaning of NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th defined & explained in the simplest way possible. Besides explaining types of NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th theory, EduRev gives you an ample number of questions to practice NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th tests, examples and also practice Class 8 tests

	Class 8 Mathematics by VP Classes 90 docs\|16 tests

Class 8 Mathematics by VP Classes

90 docs|16 tests

Join Course for Free

Download as PDF

Explore Courses for Class 8 exam

Top Courses for Class 8

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Semester Notes

shortcuts and tricks

study material

NCERT Solutions (Part - 3) - Factorisation

practice quizzes

past year papers

MCQs

Objective type Questions

mock tests for examination

Extra Questions

Summary

ppt

Important questions

Sample Paper

Mathematics Class 8th | Class 8 Mathematics by VP Classes

NCERT Solutions (Part - 3) - Factorisation

Free

NCERT Solutions (Part - 3) - Factorisation

Mathematics Class 8th | Class 8 Mathematics by VP Classes

Viva Questions

video lectures

Previous Year Questions with Solutions

pdf

Exam

;

Additional Information about NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th for Class 8 Preparation

NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th Free PDF Download

The NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th is an invaluable resource that delves deep into the core of the Class 8 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th now and kickstart your journey towards success in the Class 8 exam.

Importance of NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th

The importance of NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th cannot be overstated, especially for Class 8 aspirants. This document holds the key to success in the Class 8 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th Notes

NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th Notes on EduRev are your ultimate resource for success.

NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th Class 8 Questions

The "NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th on the App

Students of Class 8 can study NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th is prepared as per the latest Class 8 syllabus.

Education Revolution

Signup to see your scores go up
within 7 days!

Continue with Google

Takes less than 10 seconds to signup