NCERT Solutions Class 8 Maths Chapter 14 Factorisation

Question

Our NCERT Solutions Class 8 Math Chapter 14 Factorisation will assist you in understanding and resolving all NCERT question-answer concerns. We've covered every topic and answered every in-text and exercise question. For one-stop solutions for all concepts of Factorisation, please see our solution. Our NCERT solutions are created by a subject matter specialist who understands all of the concepts in this field. Experts have used various methods to make these solutions easier to understand for users, such as diagrams, different equations, graphs, shortcuts, tips, and tricks.

We have provided NCERT Solutions Class 8 with all the related topics that deal with the students' queries in this chapter of Factorisation. The following are some of the chapter's key points:

• Division of any polynomial
• Factorization
• Factorization of expressions by the method of common factors
• Factorize 12p2q + 8pq2 + 18pq
• Factorize 2a2 – b + 2a – ab
• Factorization by using the identity, x2 + (a + b)x + ab = (x + a) (x + b)
• Factorize a2 – 2a– 8

Our NCERT Solutions Class 8 Maths Chapter 7 Factorisation will assist students in easily comprehending the concepts. Our NCERT solution is ideal for last-minute examination preparation. To get good grades, read through all of the solutions provided here.

Answer 1

NCERT Solutions Class 8 Maths Chapter 14 Factorisation

1. Find the common factors of the given terms.

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14pq, 28p²q²

(iv) 2x, 3x², 4

(v) 6abc, 24ab², 12a²b

(vi) 16x³, -4x², 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x²y³, 10x³y², 6x²y²z

Solution:

(i) 12x, 36

(2 × 2 × 3 × x) and (2 × 2 × 3 × 3)

Common factors are 2 × 2 × 3 = 12

Hence, the common factor = 12

(ii) 2y, 22xy

= (2 × y) and (2 × 11 × x × y)

Common factors are 2 × y = 2y

Hence, the common factor = 2y

(iii) 14pq, 28p²q²

= (2 × 7 × p × q) and (2 × 2 × 7 × p × p × q × q)

Common factors are 2 × 7 × p × q = 14pq

Hence, the common factor = 14pq

(iv) 2x, 3x², 4

= (2 × x), (3 × x × x) and (2 × 2)

Common factor is 1

Hence, the common factor = 1 [∵ 1 is a factor of every number]

(v) 6abc, 24ab², 12a²b

= (2 × 3 × a × b × c), (2 × 2 × 2 × 3 × a × b × b) and (2 × 2 × 3 × a × a × b)

Common factors are 2 × 3 × a × b = 6ab

Hence, the common factor = 6ab

(vi) 16x³, -4x², 32x

= (2 × 2 × 2 × 2 × x × x × x), -(2 × 2 × x × x), (2 × 2 × 2 × 2 × 2 × x)

Common factors are 2 × 2 × x = 4x

Hence, the common factor = 4x

(vii) 10pq, 20qr, 30rp

= (2 × 5 × p × q), (2 × 2 × 5 × q × r), (2 × 3 × 5 × r × p)

Common factors are 2 × 5 = 10

Hence, the common factor = 10

(viii) 3x²y², 10x³y², 6x²y²z

= (3 × x × x × y × y), (2 × 5 × x × x × x × y × y), (2 × 3 × x × x × y × y × z)

Common factors are x × x × y × y = x²y²

Hence, the common factor = x²y².

2. Factorise the following expressions.

(i) 7x – 42

(ii) 6p – 12q

(iii) 7a² + 14a

(iv) -16z + 20z³

(v) 20l²m + 30alm

(vi) 5x²y – 15xy²

(vii) 10a² – 15b² + 20c²

(viii) -4a² + 4ab – 4ca

(ix) x²yz + xy²z + xyz²

(x) ax²y + bxy² + cxyz

Solution:

(i) 7x – 42 = 7(x – 6)

(ii) 6p – 12q = 6(p – 2q)

(iii) 7a² + 14a = 7a(a + 2)

(iv) -16z + 20z³ = 4z(-4 + 5z²)

(v) 20l²m + 30alm = 10lm(2l + 3a)

(vi) 5x²y – 15xy² = 5xy(x – 3y)

(vii) 10a² – 15b² + 20c² = 5(2a² – 3b² + 4c²)

(viii) -4a² + 4ab – 4ca = 4a(-a + b – c)

(ix) x²yz + xy²z + xyz² = xyz(x + y + z)

(x) ax²y + bxy² + cxyz = xy(ax + by + cz)

3. Factorise:

(i) x² + xy + 8x + 8y

(ii) 15xy – 6x + 5y – 2

(iii) ax + bx – ay – by

(iv) 15pq + 15 + 9q + 25p

(v) z – 7 + 7xy – xyz

Solution:

(i) x² + xy + 8x + 8y

Grouping the terms, we have

x² + xy + 8x + 8y

= x(x + y) + 8(x + y)

= (x + y)(x + 8)

Hence, the required factors = (x + y)(x + 8)

(ii) 15xy – 6x + 5y – 2

Grouping the terms, we have

(15xy – 6x) + (5y – 2)

= 3x(5y – 2) + (5y – 2)

= (5y – 2)(3x + 1)

(iii) ax + bx – ay – by

Grouping the terms, we have

= (ax – ay) + (bx – by)

= a(x – y) + b(x – y)

= (x – y)(a + b)

Hence, the required factors = (x – y)(a + b)

(iv) 15pq + 15 + 9q + 25p

Grouping the terms, we have

= (15pq + 25p) + (9q + 15)

= 5p(3q + 5) + 3(3q + 5)

= (3q + 5) (5p + 3)

Hence, the required factors = (3q + 5) (5p + 3)

(v) z – 7 + 7xy – xyz

Grouping the terms, we have

= (-xyz + 7xy) + (z – 7)

= -xy(z – 7) + 1 (z – 7)

= (-xy + 1) (z – 1)

Hence the required factor = -(1 – xy) (z – 7)

Answer 2

4. Factorise the following expressions.

(i) a² + 8a +16

(ii) p² – 10p + 25

(iii) 25m² + 30m + 9

(iv) 49y² + 84yz + 36z²

(v) 4x² – 8x + 4

(vi) 121b² – 88bc + 16c²

(vii) (l + m)² – 4lm. (Hint: Expand (l + m)² first)

(viii) a⁴ + 2a²b² + b⁴

Solution:

(i) a² + 8a + 16

Here, 4 + 4 = 8 and 4 × 4 = 16

a² + 8a +16

= a² + 4a + 4a + 4 × 4

= (a² + 4a) + (4a + 16)

= a(a + 4) + 4(a + 4)

= (a + 4) (a + 4)

= (a + 4)²

(ii) p² – 10p + 25

Here, 5 + 5 = 10 and 5 × 5 = 25

p² – 10p + 25

= p² – 5p – 5p + 5 × 5

= (p² – 5p) + (-5p + 25)

= p(p – 5) – 5(p – 5)

= (p – 5) (p – 5)

= (p – 5)²

(iii) 25m² + 30m + 9

Here, 15 + 15 = 30 and 15 × 15 = 25 × 9 = 225

25m² + 30m + 9

= 25m² + 15m + 15m + 9

= (25m² + 15m) + (15m + 9)

= 5m(5m + 3) + 3(5m + 3)

= (5m + 3) (5m + 3)

= (5m + 3)²

(iv) 49y² + 84yz + 36z²

Here, 42 + 42 = 84 and 42 × 42 = 49 × 36 = 1764

49y² + 84yz + 36z²

= 49y² + 42yz + 42yz + 36z²

= 7y(7y + 6z) +6z(7y + 6z)

= (7y + 6z) (7y + 6z)

= (7y + 6z)²

(v) 4x² – 8x + 4

= 4(x² – 2x + 1) [Taking 4 common]

= 4(x² – x – x + 1)

= 4[x(x – 1) -1(x – 1)]

= 4(x – 1)(x – 1)

= 4(x – 1)²

(vi) 121b² – 88bc + 16c²

Here, 44 + 44 = 88 and 44 × 44 = 121 × 16 = 1936

121b² – 88bc + 16c²

= 121b² – 44bc – 44bc + 16c²

= 11b(11b – 4c) – 4c(11b – 4c)

= (11b – 4c) (11b – 4c)

= (11b – 4c)²

(vii) (l + m)² – 4lm

Expanding (l + m)², we get

l² + 2lm + m² – 4lm

= l² – 2lm + m²

= l² – Im – lm + m²

= l(l – m) – m(l – m)

= (l – m) (l – m)

= (l – m)²

(viii) a⁴ + 2a²b² + b⁴

= a⁴ + a²b² + a²b² + b⁴

= a²(a² + b²) + b²(a² + b²)

= (a² + b²)(a² + b²)

= (a² + b²)²

5. Factorise.

(i) 4p² – 9q²

(ii) 63a² – 112b²

(iii) 49x² – 36

(iv) 16x⁵ – 144x³

(v) (l + m)² – (l – m)²

(vi) 9x²y² – 16

(vii) (x² – 2xy + y²) – z²

(viii) 25a² – 4b² + 28bc – 49c²

Solution:

(i) 4p² – 9q²

= (2p)² – (3q)²

= (2p – 3q) (2p + 3q)

[∵ a² – b² = (a + b)(a – b)]

(ii) 63a² – 112b²

= 7(9a² – 16b²)

= 7 [(3a)² – (4b)²]

= 7(3a – 4b)(3a + 4b)

[∵ a² – b² = (a + b)(a – b)]

(iii) 49x² – 36 = (7x)² – (6)²

= (7x – 6) (7x + 6)

[∵ a² – b² = (a + b)(a – b)]

(iv) 16x⁵ – 144x³ = 16x³ (x² – 9)

= 16x³ [(x)² – (3)²]

= 16x³(x – 3)(x + 3)
[∵ a² – b² = (a + b)(a – b)]

(v) (l + m)² – (l – m)²

= (l + m) – (l – m)] [(l + m) + (l – m)]

[∵ a² – b² = (a + b)(a – b)]

= (l + m – l + m)(l + m + l – m)

= (2m) (2l)

= 4ml

(vi) 9x²y² – 16 = (3xy)² – (4)²

= (3xy – 4)(3xy + 4)

[∵ a² – b² = (a + b)(a – b)]

(vii) (x² – 2xy + y²) – z²

= (x – y)² – z²

= (x – y – z) (x – y + z)

[∵ a² – b² = (a + b)(a – b)]

(viii) 25a² – 4b² + 28bc – 49c²

= 25a² – (4b² – 28bc + 49c²)

= (5a)² – (2b – 7c)²

= [5a – (2b – 7c)] [5a + (2b – 7c)]

= (5a – 2b + 7c)(5a + 2b – 7c)

Answer 3

6. Factorise the expressions.

(i) ax² + bx

(ii) 7p² + 21q²

(iii) 2x³ + 2xy² + 2xz²

(iv) am² + bm² + bn² + an²

(v) (lm + l) + m + 1

(vi) y(y + z) + 9(y + z)

(vii) 5y² – 20y – 8z + 2yz

(viii) 10ab + 4a + 5b + 2

(ix) 6xy – 4y + 6 – 9x

Solution:

(i) ax² + bx

= x(ax + 5)

(ii) 7p² + 21q²

= 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz²

= 2x(x² + y² + z²)

(iv) am² + bm² + bn² + an²

= m² (a + b) + n²(a + b)

= (a + b)(m² + n²)

(v) (lm + l) + m + 1

= l(m + 1) + (m + 1)

= (m + 1) (l + 1)

(vi) y(y + z) + 9(y + z)

= (y + z)(y + 9)

(vii) 5y² – 20y – 8z + 2yz

= 5y² – 20y + 2yz – 8z

= 5y(y – 4) + 2z(y – 4)

= (y – 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2

= 2a(5b + 2) + 1(5b + 2)

= (5b + 2)(2a + 1)

(ix) 6xy – 4y + 6 – 9x

= 6xy – 4y – 9x + 6

= 2y(3x – 2) – 3(3x – 2)

= (3x – 2) (2y – 3)

7. Factorise.

(i) a⁴ – b⁴

(ii) p⁴ – 81

(iii) x⁴ – (y + z)⁴

(iv) x⁴ – (x – z)⁴

(v) a⁴ – 2a²b² + b⁴

Solution:

(i) a⁴ – b⁴ – (a²)² – (b²)²

[∵ a² – b² = (a – b)(a + b)]

= (a² – b²) (a² + b²)

= (a – b) (a + b) (a² + b²)

(ii) p⁴ – 81 = (p²)² – (9)²

= (p² – 9) (p² + 9)

[∵ a² – b² = (a – b)(a + b)]

= (p – 3)(p + 3) (p² + 9)

(iii) x⁴ – (y + z)⁴ = (x²)² – [(y + z)²]²

[∵ a² – b² = (a – b)(a + b)]

= [x² – (y + z)²] [x² + (y + z)²]

= [x – (y + z)] [x + (y + z)] [x² + (y + z)²]

= (x – y – z) (x + y + z) [x² + (y + z)²]

(iv) x⁴ – (x – z)⁴ = (x²)² – [(y – z)²]²

= [x² – (y – z)²] [x² + (y – z)²]

= (x – y + z) (x + y – z) (x² + (y – z)²]

(v) a⁴ – 2a²b² + b⁴

= a⁴ – a²b² – a²b² + b⁴

= a²(a² – b²) – b²(a² – b²)

= (a² – b²)(a² – b²)

= (a² – b²)²

= [(a – b) (a + b)]²

= (a – b)² (a + b)²

8. Factorise the following expressions.

(i) p² + 6p + 8

(ii) q² – 10q + 21

(iii) p² + 6p – 16

Solution:

(i) p² + 6p + 8

Here, 2 + 4 = 6 and 2 × 4 = 8

p² + 6p + 8

= p² + 2p + 4p + 8

= p (p + 2) + 4(p + 2)

= (p + 2) (p + 4)

(ii) q² – 10q + 21

Here, 3 + 7 = 10 and 3 × 7 = 21

q² – 10q + 21

= q² – 3q – 7q + 21

= q(q – 3) – 7(q – 3)

= (q – 3) (q – 7)

(iii) p² + 6p – 16

Here, 8 – 2 = 6 and 8 × 2 = 16

p² + 6p – 16

= p² + 8p – 2p – 16

= p(p + 8) – 2(p + 8)

= (p + 8) (p – 2)

9. Carry out the following divisions.

(i) 28x⁴ ÷ 56x

(ii) -36y³ ÷ 9y²

(iii) 66pq²r³ ÷ 11qr²

(iv) 34x³y³z³ ÷ 51xy²z³

(v) 12a⁸b⁸ ÷ (-6a⁶b⁴)

Solution:

10. Divide the following polynomial by the given monomial.

(i) (5x² – 6x) ÷ 3x

(ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴

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

(iv) (x³ + 2x² + 3x) ÷ 2x

(v) (p³q⁶ – p⁶q³) ÷ p³q³

Solution:

Answer 4

11. Work out the following divisions.

(i) (10x – 25) ÷ 5

(ii) (10x – 25) ÷ (2x – 5)

(iii) 10y(6y + 21) ÷ 5(2y + 7)

(iv) 9x²y²(3z – 24) ÷ 27xy(z – 8)

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

Solution:

12. 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)(y² + 5y + 3) ÷ 5(y + 4)

(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)

Solution:

13. Factorise the expressions and divide them as directed.

(i) (y² + 7y + 10) ÷ (y + 5)

(ii) (m² – 14m – 32) ÷ (m + 2)

(iii) (5p² – 25p + 20) ÷ (p – 1)

(iv) 4yz(z² + 6z – 16) ÷ 2y(z + 8)

(v) 5pq(p² – q²) ÷ 2p(p + q)

(vi) 12xy(9x² – 16y²) ÷ 4xy(3x + 4y)

(vii) 39y³(50y² – 98) ÷ 26y²(5y + 7)

Solution: