Class 8 NCERT Solutions- Chapter 14 Factorisation - Exercise 14.3

Exercise 14.3 in Chapter 14 of the Class 8 NCERT Mathematics textbook focuses on the factorization of algebraic expressions. This exercise builds upon the concepts of factors and multiples introduced earlier and applies them to algebraic expressions. Students will learn various methods to factorize expressions, including the common factor method, regrouping method, and factorization of quadratic expressions.

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)

Summary

Chapter 14 of the Class 8 NCERT Mathematics textbook, titled "Factorisation," is a crucial component in the study of algebra. The chapter begins by revisiting the concepts of factors and multiples, then extends these ideas to algebraic expressions. Students learn to identify common factors in algebraic terms and use this skill to factorise expressions. The chapter introduces various methods of factorisation, including the common factor method, regrouping method, and specific techniques for quadratic expressions. It emphasizes the importance of recognising patterns in expressions, such as the difference of squares and perfect square trinomials. Through numerous examples and exercises, students develop the ability to break down complex expressions into their simplest factors. This skill is fundamental in solving equations, simplifying fractions, and understanding more advanced algebraic concepts. The chapter also touches on the applications of factorisation in solving real-world problems. By the end of the chapter, students should be proficient in factorising a wide range of algebraic expressions, laying a strong foundation for future mathematical studies in areas such as quadratic equations, polynomials, and calculus.