Class 8 NCERT Solutions- Chapter 14 Factorisation – Exercise 14.3

Question 1. Carry out the following divisions.

(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: 

66pq²r³  = 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 6a⁶b⁴  to cancel)

= -2a²b⁴

Question 2. Divide the given polynomial by the given monomial.

(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  = y⁴  [(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

Question 3. Work out the following divisions.

(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  

Question 4. Divide as directed.

(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)

Question 5. Factorise the expressions and divide them as directed. 

(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)