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

## Class 8 Mathematics by VP Classes

Created by: Full Circle

## Class 8 : NCERT Solutions (Part - 3) - Factorisation, Mathematics Class 8th Class 8 Notes | EduRev

``` 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
2â€“2
¥ b
4â€“ 2
c
6â€“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
2â€“2
¥ b
4â€“ 2
c
6â€“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) â€“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
? (â€“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
4â€“ 1
=
1
2
x
3
\ 28x
4
? 56x =
1
2
x
3
(ii) We have â€“36y
3
= (â€“1) ¥ 2 ¥ 2 ¥ 3 ¥ 3 ¥ y ¥ y ¥ y
and 9y
2
= 3 ¥ 3 ¥ y ¥ y
\ â€“36y
3
? 9y
2
=
-36
9
3
2
y
y
=
() - ¥¥¥ ¥ ¥ ¥ ¥
¥¥ ¥
12 2 3 3
33
yy y
yy
=
() - ¥¥¥ 12 2
1
y
= â€“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
2â€“2
¥ b
4â€“ 2
c
6â€“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) â€“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
? (â€“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
4â€“ 1
=
1
2
x
3
\ 28x
4
? 56x =
1
2
x
3
(ii) We have â€“36y
3
= (â€“1) ¥ 2 ¥ 2 ¥ 3 ¥ 3 ¥ y ¥ y ¥ y
and 9y
2
= 3 ¥ 3 ¥ y ¥ y
\ â€“36y
3
? 9y
2
=
-36
9
3
2
y
y
=
() - ¥¥¥ ¥ ¥ ¥ ¥
¥¥ ¥
12 2 3 3
33
yy y
yy
=
() - ¥¥¥ 12 2
1
y
= â€“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
3â€“1
¥ y
3â€“ 2
¥ 3z
3â€“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
? (â€“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
= â€“2 ¥ a
8â€“ 6
¥ b
8â€“4
= â€“2 ¥ a
2
¥ b
4
= â€“2a
2
b
4
Question 2. Divide the given polynomial by the given monomial.
(i) (5x
2
â€“ 6x) ? 3x (ii) (3y
8
â€“ 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
â€“ p
6
q
3
) ? p
3
q
3
Solution: (i) ? (5x
2
â€“ 6x) ? 3x =
5x 6
3
2
- x
x
=
x
x
() 5x 6
3
-
=
5x 6
3
-
\ (5x
2
â€“ 6x) ? 3x =
1
3
(5x â€“ 6)
(ii) ? (3y
8
â€“ 4y
6
+ 5y
4
) ? y
4
=
34 5y
864
4
yy
y
-+
=
yy y
y
44 2
4
34 5 () -+
\ (3y
8
â€“ 4y
6
+ 5y
4
) ? y
4
= 3y
4
â€“ 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
2â€“2
¥ b
4â€“ 2
c
6â€“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) â€“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
? (â€“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
4â€“ 1
=
1
2
x
3
\ 28x
4
? 56x =
1
2
x
3
(ii) We have â€“36y
3
= (â€“1) ¥ 2 ¥ 2 ¥ 3 ¥ 3 ¥ y ¥ y ¥ y
and 9y
2
= 3 ¥ 3 ¥ y ¥ y
\ â€“36y
3
? 9y
2
=
-36
9
3
2
y
y
=
() - ¥¥¥ ¥ ¥ ¥ ¥
¥¥ ¥
12 2 3 3
33
yy y
yy
=
() - ¥¥¥ 12 2
1
y
= â€“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
3â€“1
¥ y
3â€“ 2
¥ 3z
3â€“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
? (â€“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
= â€“2 ¥ a
8â€“ 6
¥ b
8â€“4
= â€“2 ¥ a
2
¥ b
4
= â€“2a
2
b
4
Question 2. Divide the given polynomial by the given monomial.
(i) (5x
2
â€“ 6x) ? 3x (ii) (3y
8
â€“ 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
â€“ p
6
q
3
) ? p
3
q
3
Solution: (i) ? (5x
2
â€“ 6x) ? 3x =
5x 6
3
2
- x
x
=
x
x
() 5x 6
3
-
=
5x 6
3
-
\ (5x
2
â€“ 6x) ? 3x =
1
3
(5x â€“ 6)
(ii) ? (3y
8
â€“ 4y
6
+ 5y
4
) ? y
4
=
34 5y
864
4
yy
y
-+
=
yy y
y
44 2
4
34 5 () -+
\ (3y
8
â€“ 4y
6
+ 5y
4
) ? y
4
= 3y
4
â€“ 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
â€“ 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
â€“ p
6
q
3
) ? p
3
q
3
= q
3
â€“ p
3
Question 3. Work out the following divisions:
(i) (10x â€“ 25) ? 5 (ii) (10x â€“ 25) ? (2x â€“ 5)
(iii) 10y(6y + 21) ? 5(2y + 7) (iv) 9x
2
y
2
(3z â€“ 24) ? 27xy(z â€“ 8)
(v) 96abc(3a â€“ 12)(5b â€“ 30) ? 144(a â€“ 4)(b â€“ 6)
Solution: (i) ? 10x â€“ 25 = 5(2x â€“ 5)
\
10 25
5
x -
=
52 5
5
() x -
= (2x â€“ 5)
Thus, (10x â€“ 25) ? 5 = 2x â€“ 5
(ii) ? 10x â€“ 25 = 5(2x â€“ 5)
\
10 25
25
x
x
-
-
=
52 5
25
()
()
x
x
-
-
= 5
Thus, (10x â€“ 25) ? (2x â€“ 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 â€“ 24 = 3(z â€“ 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
2â€“2
¥ b
4â€“ 2
c
6â€“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) â€“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
? (â€“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
4â€“ 1
=
1
2
x
3
\ 28x
4
? 56x =
1
2
x
3
(ii) We have â€“36y
3
= (â€“1) ¥ 2 ¥ 2 ¥ 3 ¥ 3 ¥ y ¥ y ¥ y
and 9y
2
= 3 ¥ 3 ¥ y ¥ y
\ â€“36y
3
? 9y
2
=
-36
9
3
2
y
y
=
() - ¥¥¥ ¥ ¥ ¥ ¥
¥¥ ¥
12 2 3 3
33
yy y
yy
=
() - ¥¥¥ 12 2
1
y
= â€“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
3â€“1
¥ y
3â€“ 2
¥ 3z
3â€“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
? (â€“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
= â€“2 ¥ a
8â€“ 6
¥ b
8â€“4
= â€“2 ¥ a
2
¥ b
4
= â€“2a
2
b
4
Question 2. Divide the given polynomial by the given monomial.
(i) (5x
2
â€“ 6x) ? 3x (ii) (3y
8
â€“ 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
â€“ p
6
q
3
) ? p
3
q
3
Solution: (i) ? (5x
2
â€“ 6x) ? 3x =
5x 6
3
2
- x
x
=
x
x
() 5x 6
3
-
=
5x 6
3
-
\ (5x
2
â€“ 6x) ? 3x =
1
3
(5x â€“ 6)
(ii) ? (3y
8
â€“ 4y
6
+ 5y
4
) ? y
4
=
34 5y
864
4
yy
y
-+
=
yy y
y
44 2
4
34 5 () -+
\ (3y
8
â€“ 4y
6
+ 5y
4
) ? y
4
= 3y
4
â€“ 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
â€“ 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
â€“ p
6
q
3
) ? p
3
q
3
= q
3
â€“ p
3
Question 3. Work out the following divisions:
(i) (10x â€“ 25) ? 5 (ii) (10x â€“ 25) ? (2x â€“ 5)
(iii) 10y(6y + 21) ? 5(2y + 7) (iv) 9x
2
y
2
(3z â€“ 24) ? 27xy(z â€“ 8)
(v) 96abc(3a â€“ 12)(5b â€“ 30) ? 144(a â€“ 4)(b â€“ 6)
Solution: (i) ? 10x â€“ 25 = 5(2x â€“ 5)
\
10 25
5
x -
=
52 5
5
() x -
= (2x â€“ 5)
Thus, (10x â€“ 25) ? 5 = 2x â€“ 5
(ii) ? 10x â€“ 25 = 5(2x â€“ 5)
\
10 25
25
x
x
-
-
=
52 5
25
()
()
x
x
-
-
= 5
Thus, (10x â€“ 25) ? (2x â€“ 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 â€“ 24 = 3(z â€“ 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
2â€“1
y
2â€“1
= xy
\ 9x
2
y
2
(3z â€“ 24) ? 27xy(z â€“ 8) = xy
(v) ? 3a â€“ 12 = 3(a â€“ 4)
5b â€“ 30 = 5(b â€“ 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 â€“ 12)(5b â€“ 36) ? 144(a â€“ 4)(b â€“ 6) = 10abc
Question 4. Divide as directed.
(i) 5(2x + 1)(3x + 5) ? (2x + 1) (ii) 26xy(x + 5)(y â€“ 4) ? 13x(y â€“ 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 â€“ 4) ? 13x(y â€“ 4)=
26 5 4
13 4
xy x y
xy
()( )
()
+-
-
=
2y(x 5)
1
¥+
\ 26xy(x + 5)(y â€“ 4) ? 13x(y â€“ 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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