# Class 8 NCERT Solutions- Chapter 14 Factorisation – Exercise 14.3

• Last Updated : 09 Mar, 2021

### (i)28x4 ÷ 56x

Solution:

28x4 = 2 × 2 ×7 × x × x × x × x

56x = 2 × 2 × 2 × 7 × x

28x4÷ 56x =                  (grouping 28x to cancel)

= ½ × x × x × x

= ½ x3

### (ii) -36y3 ÷ 9y2

Solution:

-36y3 = -2 × 2 × 3 × 3 × y × y × y

9y2= 3 × 3 × y × y

-36y3 ÷ 9y2            (grouping 9y2 to cancel)

= -(2 × 2 × y)

= -4y

### (iii) 66pq2r3  ÷ 11qr2

Solution:

66pq2r = 2 × 3 × 11 × p × q × q × r × r × r

11qr2  = 11 × q × r × r

66pq2r3 ÷ 11qr2       (grouping 11qr2 to cancel)

= (2 × 3 × p × q × r)

= 6pqr

### (iv) 34x3y3z3 ÷ 51xy2z3

Solution:

34x3y3z3 = 2 × 17 × x × x × x × y × y × y × z × z × z

51xy2z3  = 3 × 17 × x × y × y × z × z × z

34x3y3z3 ÷ 51xy2z3

(grouping 17xy2z3  to cancel)

x2y

### (v) 12a8b8 ÷ (-6a6b4)

Solution:

12a8b8 = 2 × 2 × 3 × a × a × a × a × a × a × a × a × b × b × b × b × b × b × b × b

-6a6b4 = -2 × 3 × a × a × a × a × a × a × b × b × b × b

12a8b8 ÷ (-6a6b4) =

= – (2 × a × a × b × b × b × b)             (grouping 6a6b to cancel)

= -2a2b4

### (i) (5x2  – 6x) ÷ 3x

Solution:

5x2  – 6x = (5 × x × x) – (2 × 3 × x)

= 5x × (x) – 6 × (x)

= x(5x – 6)

3x = 3 × (x)

(5x2 – 6x) ÷ 3x =     (grouping x to cancel)

### (ii)(3y8 – 4y6 + 5y4) ÷ y4

Solution:

3y8-4y6+5y4  = y [(3 × y × y × y × y) – (2× 2 × y × y) + (5)]

y4  = (y × y × y × y)

(3y8-4y6+5y4) ÷ y4       (grouping y4  to cancel)

= (3x4-4y2+5 )

### (iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2

Solution:

8 (x3y2z2 + x2y3z2 + x2y2z3 ) = 2 × 2 × 2 × x2y2z2  (x + y + z)

4 x x2y2z2  = 2 × 2 × x2y2z2

8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2          (grouping x2y2z2  to cancel)

= 2(x+y+z)

### (iv)(x3+2x2+3x) ÷ 2x

Solution:

x3+2x2+3x  = x ×  (x2+2x+3 )

(x3+2x2+3x) ÷ 2x =                (grouping x  to cancel)

### (v) (p3q6-p6q3) ÷ p3q3

Solution:

p3q6-p6q3  = p3q3(q3-p3)

(p3q6-p6q3) ÷ p3q3         (grouping p3q3  to cancel)

= q3– p3

### (i) (10x – 25) ÷ 5

Solution:

10x-25 = (5 × 2 × x) – (5 × 5)

= 5(2x-5)

(10x-25) ÷ 5 =     (grouping 5 to cancel)

= (2x – 5)

### (ii) (10x – 25) ÷ (2x – 5)

Solution:

10x-25 = 5(2x-5)

(10x-25)÷(2x-5) =              (grouping (2x-5)  to cancel)

= 5

### (iii)10y(6y+21)  ÷ 5(2y+7)

Solution:

10y(6y+21)  = 5 × 2 × y × 3 × (2y+7)

10y(6y+21) ÷ 5(2y+7) =            (grouping 5(2y+7)  to cancel)

= 2 × 3 × y

= 6y

### (iv) 9x2y2(3z-24) ÷ 27xy(z-8)

Solution:

9x2y2(3z-24) = 3 × 3 × x2 × y2 × 3 × (z-8)

27xy(z-8) = 3 × 3 × 3 × x × y × (z-8)

9x2y2(3z-24)÷27xy(z-8)=         (grouping (27xy(z-8))  to cancel)

= xy

### (v) 96abc(3a-12)(5b-30) ÷ 144 (a-4)(b-6)

Solution:

96abc(3a-12)(5b-30) = 2 × 2 × 2 × 2 × 2 × 3 × a × b × c × 3 × (a-4) × 5 × (b-6)

144(a-4)(b-6) = 2 × 2 × 2 × 2 × 3 × 3 × (a-4) × (b-6)

96abc(3a-12)(5b-30) ÷ 144(a-4)(b-6) =

= (2 × 5 × a × b × c)                (grouping (144(a-4)(b-6)) to cancel)

= 10abc

### (i) 5(2x+1)(3x+5) ÷ (2x+1)

Solution:

= 5(3x+1)               (grouping (2x+1) to cancel)

### (ii) 26xy(x+5)(y-4)÷13x(y-4)

Solution:

26xy(x+5)(y-4) = 2 × 13 × x × y × (x+5) × (y-4)

26xy(x+5)(y-4)÷13x(y-4) =                 (grouping 13x(y-4)  to cancel)

= (2 × y × (x+5))

= 2y(x+5)

### (iii) 52pqr(p+q)(q+r)(r+p)÷104pq(q+r)(r+p)

Solution:

52pqr(p+q)(q+r)(r+p)  = 13 × 2 × 2 × pqr(p+q)(q+r)(r+p)

104pq(q+r)(r+p) = 13 × 2 × 2 × 2 × pq(q+r)(r+p)

52pqr(p+q)(q+r)(r+p)÷104pq(q+r)(r+p) =

=                      (grouping (52pq(q+r)(r+p))  to cancel)

### (iv) 20(y+4)(y2+5y+3)÷5(y+4)

Solution:

20(y+4)(y2+5y+3) = 2 × 2 × 5 × (y+4) × (y2+5y+3)

20(y+4)(y2+5y+3)÷5(y+4) =            (grouping (5(y+4))  to cancel)

= 2 × 2 × (y2+5y+3)

= 4(y2+5y+3)

### (v) x(x+1)(x+2)(x+3) ÷ x(x+1)

Solution:

= (x+2)(x+3)                           (grouping x(x+1)  to cancel)

### (i) (y2+7y+10) ÷ (y+5)

Solution:

(y2+7y+10) = (y2+5y+2y+10)

= (y(y+5) + 2(y+5))                                          (2 + 5 = 7  &  2 × 5 = 10)

= (y+5) (y+2)

(y2+7y+10) ÷ (y+5) =           (grouping (y+5)  to cancel)

= (y+2)

### (ii) (m2-14m-32)÷(m+2)

Solution:

(m2-14m-32) =  (m2-16m+2m-32 )

= (m(m-16) + 2(m-16))                                                     (-16 + 2 = -14  &  -16 × 2 = -32)

= (m+2) (m-16)

(m2-14m-32)÷(m+2) =               (grouping (m+2)  to cancel)

= (m-16)

### (iii) (5p2-25p+20) ÷ (p-1)

Solution:

(5p2-25p+20) = (5p2-20p-5p+20)

=(5p(p-4)-5(p-4))                                                                (-20 – 5 = -25 )

=(5p-5) (p-4)

=5 (p-1) (p-4)

(5p2-25p+20)÷(p-1) =                 (grouping (p-1)  to cancel)

= 5(p-4)

### (iv) 4yz(z2+6z-16)÷2y(z+8)

Solution:

4yz(z2+6z-16)  = 2 × 2 × y × z × (z2+8z-2z-16)

= 2 × 2 × y × z × (z(z+8)-2(z+8))                                                      (8 + (-2) = 6 & 8 × (-2) = -16)

= 2 × 2 × y × z × (z+8) (z-2))

4yz(z2+6z-16) ÷ 2y(z+8) =      (grouping 2y(z+8)  to cancel)

= 2 × z × (z-2)

= 2z(z-2)

### (v) 5pq(p2-q2)÷2p(p+q)

Solution:

(p2-q2) = (p+q) (p-q)                                                                         (IDENTITY a2-b2 = (a+b)(a-b) )

5pq(p2-q2)÷2p(p+q) =                               (grouping p(p+q)  to cancel)

=

### (vi) 12xy(9x2-16y2) ÷ 4xy(3x+4y)

Soln.

12xy(9x2-16y2) = 2 × 2 × 3 × ((3x)2-(4y)2)

12xy(9x2-16y2) = 2 × 2 × 3 × (3x+4y) (3x-4y)                                                                              (IDENTITY a2-b2 = (a+b)(a-b) )

12xy(9x2-16y2) ÷ 4xy(3x+4y) =                        (grouping 4xy(3x+4y)  to cancel)

= 3 (3x-4y)

### (vii) 39y3(50y2-98) ÷ 26y2(5y+7)

Solution:

39y3(50y2-98) = 3 × 13 × y3 × 2 × (25y2-49)

= 3 × 13 × y3 × 2 × ((5y)2-(7)2)                                                                                        (IDENTITY a2-b2 = (a+b)(a-b) )

= 3 × 13 × y3 × 2 × (5y+7) (5y-7)

26y2(5y+7) = 2 × 13 × y2 × (5y+7)

39y3(50y2-98)÷26y2(5y+7) =     (grouping 26y2(5y+7)  to cancel)

= (3 × y × (5y-7))

= 3y(5y-7)

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