NCERT Solutions for Class 8 Maths Chapter 14 – Factorisation

NCERT Solutions for Class 8 Maths Chapter 14 – Factorisation. Furthermore, here we’ve provided you with the latest solution for Class 8 Maths Chapter 14 – Factorisation. As a result here you’ll find solutions to all the exercises. This NCERT Class 8 solution will help you to score good marks in your exam.

Students can refer to our solution for NCERT Class 8 Maths Chapter 14 – Factorisation. The Chapter 14 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise-wise latest solution.

NCERT Solutions for Class 8 Maths Chapter 14 – Factorisation Exercise Wise Solution

Exercise 14.1 – Page 220 of NCERT
Exercise 14.2 – Page 223 of NCERT
Exercise 14.3 – Page 227 of NCERT
Exercise 14.4 – Page 228 of NCERT

NCERT Solutions for Class 8 Maths Chapter 14 – Factorisation Exercise 14.1 Solution

Here you’ll find NCERT Chapter 14 – Factorisation Exercise 14.1 Solution
Exercise 14.1: Solutions of Questions on Page Number: 220

Q1: Find the common factors of the terms

1. 12x, 36

2. 2y, 22xy

3. 14pq, 28p²q²

4. 2x, 3x², 4

5. 6abc, 24ab², 12a²b
   
6. 16x³, -4x², 32x
   
7. 10pq, 20qr, 30rp

8. 3x²y³, 10x³y², 6x²y²z

Answer:

1. 12x = 2 x 2 x 3 x x

36 = 2 x 2 x 3 x 3

The common factors are 2, 2, 3.

And, 2 x 2 x 3 = 12

2. 2y = 2 x y

22xy = 2 x 11 x x x y

The common factors are 2, y.

And, 2 x y = 2y

3. 14pq = 2 x 7 x p x q

28p²q² = 2 x 2 x 7 x p x p x q x q

The common factors are 2, 7, p, q.

And, 2 x 7 x p x q = 14pq

4. 2x = 2 x x
   3x² = 3 x x x x
   4 = 2 x 2

The common factor is 1.

5. 6abc = 2 x 3 x a x b x c
   24ab² = 2 x 2 x 2 x 3 x a x b x b
   12a²b = 2 x 2 x 3 x a x a x b

The common factors are 2, 3, a, b.

And, 2 x 3 x a x b = 6ab

6. 16x³ = 2 x 2 x 2 x 2 x x x x x x
   -4x² = -1 x 2 x 2 x x x x
   32x = 2 x 2 x 2 x 2 x 2 x x

The common factors are 2, 2, x.

And, 2 x 2 x x = 4x

7. 10pq = 2 x 5 x p x q
   20qr = 2 x 2 x 5 x q x r
   30rp = 2 x 3 x 5 x r x p

The common factors are 2, 5.

And, 2 x 5 = 10

8. 3x²y³ = 3 x x x x x y x y x y
   10x³y² = 2 x 5 x x x x x x x y x y
   6x²y²z = 2 x 3 x x x x x y x y x z
   The common factors are x, x, y, y. And, x x x x y x y = x²y²

Q2: Factorise the following expressions

1. 7x – 42

2. 6p – 12q

3. 7a² + 14a
4. -16z + 20z³

5. 20l²m + 30 alm

6. 5x²y – 15xy²
7. 10a² – 15b² + 20c²

8. -4a² + 4ab – 4 ca

9. x²yz + xy²z + xyz²

10. ax²y + bxy² + cxyz

Answer:

1. 7x = 7 x x

42 = 2 x 3 x 7

The common factor is 7.

∴ 7x – 42 = (7 x x) – (2 x 3 x 7) = 7 (x – 6)

2. 6p = 2 x 3 x p

12q = 2 x 2 x 3 x q

The common factors are 2 and 3.

∴ 6p – 12q = (2 x 3 x p) – (2 x 2 x 3 x q)

= 2 x 3 [p – (2 x q)]

= 6 (p – 2q)

3. 7a² = 7 x a x a

14a = 2 x 7 x a

The common factors are 7 and a.

∴ 7a² + 14a = (7 x a x a) + (2 x 7 x a)

= 7 x a [a + 2] = 7a (a + 2)

4. 16z = 2 x 2 x 2 x 2 x z

20z³ = 2 x 2 x 5 x z x z x z

The common factors are 2, 2, and z.

∴ -16z + 20z³ = – (2 x 2 x 2 x 2 x z) + (2 x 2 x 5 x z x z x z)

= (2 x 2 x z) [- (2 x 2) + (5 x z x z)]

= 4z (- 4 + 5z²)

5. 20l²m = 2 x 2 x 5 x l x l x m

30alm = 2 x 3 x 5 x a x l x m

The common factors are 2, 5, l, and m.

∴ 20l²m + 30alm = (2 x 2 x 5 x l x l x m) + (2 x 3 x 5 x a x l x m)

= (2 x 5 x l x m) [(2 x l) + (3 x a)]

= 10lm (2l + 3a)

6. 5x²y = 5 x x x x x y

15xy² = 3 x 5 x x x y x y

The common factors are 5, x, and y.

∴ 5x²y – 15xy² = (5 x x x x x y) – (3 x 5 x x x y x y)

= 5 x x x y [x – (3 x y)]

= 5xy (x – 3y)

7. 10a² = 2 x 5 x a x a
   15b² = 3 x 5 x b x b
   20c² = 2 x 2 x 5 x c x c

The common factor is 5.

10a² – 15b² + 20c² = (2 x 5 x a x a) – (3 x 5 x b x b) + (2 x 2 x 5 x c x c)

= 5 [(2 x a x a) – (3 x b x b) + (2 x 2 x c x c)]

= 5 (2a² – 3b² + 4c²)

8. 4a² = 2 x 2 x a x a

4ab = 2 x 2 x a x b

4ca = 2 x 2 x c x a

The common factors are 2, 2, and a.

∴ -4a² + 4ab – 4ca = – (2 x 2 x a x a) + (2 x 2 x a x b) – (2 x 2 x c x a)

= 2 x 2 x a [- (a) + b – c]

= 4a (-a + b – c)

9. x²yz = x x x x y x z
   xy²z = x x y x y x z
   xyz² = x x y x z x z

The common factors are x, y, and z.

∴ x²yz + xy²z + xyz² = (x x x x y x z) + (x x y x y x z) + (x x y x z x z)

= x x y x z [x + y + z]

= xyz (x + y + z)

10. ax²y = a x x x x x y
    bxy² = b x x x y x y
    cxyz = c x x x y x z

The common factors are x and y. a

Q3: Factorise

1. x² + xy + 8x + 8y

2. 15xy – 6x + 5y – 2

3. ax + bx – ay – by
4. 15pq + 15 + 9q + 25p
5. z – 7 + 7xy – xyz

Answer:

1. x² + xy + 8x + 8y = x x x + x x y + 8 x x + 8 x y

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

= (x + y) (x + 8)

2. 15xy – 6x + 5y – 2 = 3 x 5 x x x y – 3 x 2 x x + 5 x y – 2

= 3x (5y – 2) + 1 (5y – 2)

= (5y – 2) (3x + 1)

3. ax + bx – ay – by = a x x + b x x – a x y – b x y

= x (a + b) – y (a + b)

= (a + b) (x – y)

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

= 3 x 5 x p x q + 3 x 3 x q + 5 x 5 x p + 3 x 5

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

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

(v) z – 7 + 7xy – xyz = z – x x y x z – 7 + 7 x x x y

= z (1 – xy) – 7 (1 – xy)

= (1 – xy) (z – 7)

NCERT Solutions for Class 8 Maths Chapter 14 – Factorisation Exercise 14.2 Solution

Here you’ll find NCERT Chapter 14 – Factorisation Exercise 14.2 Solution
Exercise 14.2: Solutions of Questions on Page Number: 223

Q1: 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

Answer:

(i) a² + 8a + 16 = (a)² + 2 x a x 4 + (4)²

= (a + 4)² [(x + y)² = x² + 2xy + y²]

(ii) p² – 10p + 25 = (p)² – 2 x p x 5 + (5)²

= (p – 5)² [(a – b)² = a² – 2ab + b²]

(iii) 25m² + 30m + 9 = (5m)² + 2 x 5m x 3 + (3)²

= (5m + 3)² [(a + b)² = a² + 2ab + b²]

(iv) 49y² + 84yz + 36z² = (7y)² + 2 x (7y) x (6z) + (6z)²

= (7y + 6z)² [(a + b)² = a² + 2ab + b²] (v) 4x² – 8x + 4 = (2x)² – 2 (2x) (2) + (2)²

= (2x – 2)² [(a – b)² = a² – 2ab + b²]

= [(2) (x – 1)]² = 4(x – 1)²

(vi) 121b² – 88bc + 16c² = (11b)² – 2 (11b) (4c) + (4c)²

= (11b – 4c)² [(a – b)² = a² – 2ab + b²]

(vii) (l + m)² – 4lm = l² + 2lm + m² – 4lm

= l² – 2lm + m²

= (l – m)² [(a – b)² = a² – 2ab + b²]

(viii) a⁴ + 2a²b² + b⁴ = (a²)² + 2 (a²) (b²) + (b²)²

= (a² + b²)² [(a + b)² = a² + 2ab + b²]

Q2: 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

Answer:

(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)

= 16 x³ [(x)² – (3)²]

= 16 x³(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)] [Using identity a² – b² = (a – b) (a + b)]

= (l + m – l + m) (l + m + l – m)

= 2m x 2l

= 4ml

= 4lm

(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 – (z)² [(a – b)² = a² – 2ab + b²]

= (x – y – z) (x – y + z) [a² – b² = (a – b) (a + b)]

(viii) 25a² – 4b² + 28bc – 49c²

= 25a² – (4b² – 28bc + 49c²)

= (5a)² – [(2b)² – 2 x 2b x 7c + (7c)²]

= (5a)² – [(2b – 7c)²]

[Using identity (a – b)² = a² – 2ab + b²]

= [5a + (2b – 7c)] [5a – (2b – 7c)]

[Using identity a² – b² = (a – b) (a + b)]

= (5a + 2b – 7c) (5a – 2b + 7c)

Q3: Factorise the expressions

1. ax² + bx

2. 7p² + 21q²

3. 2x³ + 2xy² + 2xz²

4. am² + bm² + bn² + an²

5. (lm + l) + m + 1
6. y(y + z) + 9(y + z)

7. 5y² – 20y – 8z + 2yz

8. 10ab + 4a + 5b + 2

9. 6xy – 4y + 6 – 9x

Answer:

(i) ax² + bx = a x x x x + b x x = x(ax + b)

(ii) 7p² + 21q² = 7 x p x p + 3 x 7 x q x q = 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz² = 2x(x² + y² + z²)

(iv) am² + bm² + bn² + an² = am² + bm² + an² + bn²

= m²(a + b) + n²(a + b)

= (a + b) (m² + n²)

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

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

= (l + l) (m + 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 = 10ab + 5b + 4a + 2

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

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

(ix) 6xy – 4y + 6 – 9x = 6xy – 9x – 4y + 6

= 3x(2y – 3) – 2(2y – 3)

= (2y – 3) (3x – 2)

Q4: Factorise

1. a⁴ – b⁴

2. p⁴ – 81

3. x⁴ – (y + z)⁴

4. x⁴ – (x – z)⁴
5. a⁴ – 2a²b² + b⁴

Answer:

(i) a⁴ – b⁴ = (a²)² – (b²)²

= (a² – b²) (a² + b²)

= (a – b) (a + b) (a² + b²)

(ii) p⁴ – 81 = (p²)² – (9)²

= (p² – 9) (p² + 9)

= [(p)² – (3)²] (p² + 9)

= (p – 3) (p + 3) (p² + 9)

(iii) x⁴ – (y + 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]

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

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

= [x² – (x – z)²] [x² + (x – z)²]

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

= z(2x – z) [x² + x² – 2xz + z²]

= z(2x – z) (2x² – 2xz + z²)

(v) a⁴ – 2a²b² + b⁴ = (a²)² – 2 (a²) (b²) + (b²)²

= (a² – b²)²

= [(a – b) (a + b)]²

= (a – b)² (a + b)²

Q5: Factorise the following expressions

(i) p² + 6p + 8

(ii) q² – 10q + 21

(iii) p² + 6p – 16

Answer:

(i) p² + 6p + 8

It can be observed that, 8 = 4 x 2 and 4 + 2 = 6

∴ p² + 6p + 8 = p² + 2p + 4p + 8

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

= (p + 2) (p + 4)

(ii) q² – 10q + 21

It can be observed that, 21 = (-7) x (-3) and (-7) + (-3) = – 10

∴ q² – 10q + 21 = q² – 7q – 3q + 21

= q(q – 7) – 3(q – 7)

= (q – 7) (q – 3)

(iii) p² + 6p – 16

It can be observed that, 16 = (-2) x 8 and 8 + (-2) = 6

p² + 6p – 16 = p² + 8p – 2p – 16

= p(p + 8) – 2(p + 8)

= (p + 8) (p – 2)

NCERT Solutions for Class 8 Maths Chapter 14 – Factorisation Exercise 14.3 Solution

Here you’ll find NCERT Chapter 14 – Factorisation Exercise 14.3 Solution

Exercise 14.3: Solutions of Questions on Page Number: 227

Q1: Carry out the following divisions.

1. 28x⁴  ÷ 56x

2. -36y³  ÷ 9y²

3. 66pq²r³  ÷11qr²

4. 34x³y³z³  ÷ 51xy²z³
5. 12a⁸b⁸  ÷ (-6a⁶b⁴)

Answer:

(i) 28x⁴ = 2 × 2 × 7 × x × x × x × x

56x = 2 × 2 × 2 × 7 × x

(ii) 36y³ = 2 × 2 × 3 × 3 × y × y × y

9y² = 3 × 3 × y × y

(iii) 66 pq² r³ = 2 × 3 × 11 × p × q × q × r × r × r

11qr² = 11 × q × r × r

(iv) 34 x³y³z³ = 2 × 17 × x × x × x × y × y × y × z × z × z

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

(v) 12a⁸b⁸ = 2 × 2 × 3 × a⁸ × b⁸

6a⁶b⁴ = 2 × 3 × a⁶ × b⁴

= – 2a²b⁴

Q2: Divide the given polynomial by the given monomial.

1. (5x² – 6x)  ÷ 3x
2. (3y⁸ – 4y⁶ + 5y⁴)  ÷ y⁴

3. 8(x³y²z² + x²y³z² + x²y²z³)  ÷ 4x²y²z²

4. (x³ + 2x² + 3x)  ÷ 2x
5. (p³q⁶ – p⁶q³)  ÷ p³q³

Answer :

(i) 5x² – 6x = x(5x – 6)

(ii) 3y⁸ – 4y⁶ + 5y⁴ = y⁴(3y⁴ – 4y² + 5)

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

(iv) x³ + 2x² + 3x = x(x² + 2x + 3)

(v) p³q⁶ – p⁶q³ = p³q³(q³ – p³)

Q3: 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)

Answer :

(i)

(ii)

(iii)

(iv)

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

Q4: 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)

Answer:

(i)

= 5(3x + 1)

(ii)
= 2y (x + 5)

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

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

(v)

= (x + 2) (x + 3)

Q5: 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)

Answer:

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

= y (y + 2) + 5 (y + 2)

= (y + 2) (y + 5)

(ii) m² – 14m – 32 = m² + 2m – 16m – 32

= m (m + 2) – 16 (m + 2)

= (m + 2) (m – 16)

(iii) 5p² – 25p + 20 = 5(p² – 5p + 4)

= 5[p² – p – 4p + 4]

= 5[p(p – 1) – 4(p – 1)]

= 5(p – 1) (p – 4)

(iv) 4yz(z² + 6z – 16) = 4yz [z² – 2z + 8z – 16]

= 4yz [z(z – 2) + 8(z – 2)]

= 4yz(z – 2) (z + 8)

(v) 5pq(p² – q²) = 5pq (p – q) (p + q)

(vi) 12xy(9x² – 16y²) = 12xy[(3x)² – (4y)²] = 12xy(3x – 4y) (3x + 4y)

(vii) 39y³(50y² – 98) = 3 × 13 × y × y × y × 2[(25y² – 49)]

= 3 × 13 × 2 × y × y × y × [(5y)² – (7)²]

= 3 × 13 × 2 × y × y × y (5y – 7) (5y + 7) 26y²(5y + 7) = 2 × 13 × y × y × (5y + 7)

39y³(50y² – 98) ÷ 26y² (5y + 7)

NCERT Solutions for Class 8 Maths Chapter 14 – Factorisation Exercise 14.4 Solution

Here you’ll find NCERT Chapter 14 – Factorisation Exercise 14.4 Solution
Exercise 14.4: Solutions of Questions on Page Number: 228

Q1: Find and correct the errors in the statement:

4(x – 5) = 4x – 5

Answer:

L.H.S. = 4(x – 5) = 4 x x – 4 x 5 = 4x – 20 ≠ R.H.S.
The correct statement is 4(x – 5) = 4x – 20

Q2: Find and correct the errors in the statement: x(3x + 2) = 3x² + 2

Answer:

L.H.S. = x(3x + 2) = x x 3x + x x 2 = 3x² + 2x ≠ R.H.S.
The correct statement is x(3x + 2) = 3x² + 2x

Q3: Find and correct the errors in the statement: 2x + 3y = 5xy

Answer:

L.H.S = 2x + 3y ≠ R.H.S.

The correct statement is 2x + 3y = 2x + 3y

Q4: Find and correct the errors in the statement: x + 2x + 3x = 5x
Answer :

L.H.S = x + 2x + 3x =1x + 2x + 3x = x(1 + 2 + 3) = 6x ≠ R.H.S.
The correct statement is x + 2x + 3x = 6x

Q5: Find and correct the errors in the statement: 5y + 2y + y – 7y = 0

Answer:

L.H.S. = 5y + 2y + y – 7y = 8y – 7y = y ≠ R.H.S.
The correct statement is 5y + 2y + y – 7y = y

Q6: Find and correct the errors in the statement: 3x + 2x = 5x²

Answer:

L.H.S. = 3x + 2x = 5x ≠ R.H.S

The correct statement is 3x + 2x = 5x

Q7: Find and correct the errors in the statement: (2x)² + 4(2x) + 7 = 2x² + 8x + 7

Answer:

L.H.S = (2x)² + 4(2x) + 7 = 4x² + 8x + 7 ≠ R.H.S

The correct statement is (2x)² + 4(2x) + 7 = 4x² + 8x + 7

Q8: Find and correct the errors in the statement: (2x)² + 5x = 4x + 5x = 9x

Answer:

L.H.S = (2x)² + 5x = 4x² + 5x ≠ R.H.S.
The correct statement is (2x)² + 5x = 4x² + 5x

Q9: Find and correct the errors in the statement: (3x + 2)² = 3x² + 6x + 4

Answer:

L.H.S. = (3x + 2)² = (3x)² + 2(3x)(2) + (2)² [(a + b)² = a² + 2ab + b²]

= 9x² + 12x + 4 ≠ R.H.S

The correct statement is (3x + 2)² = 9x² + 12x + 4

Q10: Find and correct the errors in the statement: (y – 3)² = y² – 9

Answer:

L.H.S = (y – 3)² = (y)² – 2(y)(3) + (3)² [(a – b)² = a² – 2ab + b²]

= y² – 6y + 9 ≠ R.H.S

The correct statement is (y – 3)² = y² – 6y + 9

Q11: Find and correct the errors in the statement: (z + 5)² = z² + 25

Answer:

L.H.S = (z + 5)² = (z)² + 2(z)(5) + (5)² [(a + b)² = a² + 2ab + b²]

= z² + 10z + 25 ≠ R.H.S

The correct statement is (z + 5)² = z² + 10z + 25

Q12: Find and correct the errors in the statement: (2a + 3b) (a – b) = 2a² – 3b²

Answer:

L.H.S. = (2a + 3b) (a – b) = 2a x a + 3b x a – 2a x b – 3b x b

= 2a² + 3ab – 2ab – 3b² = 2a² + ab – 3b² ≠ R.H.S.

The correct statement is (2a + 3b) (a – b) = 2a² + ab – 3b²

Q13: Find and correct the errors in the statement: (a + 4) (a + 2) = a² + 8

Answer:

L.H.S. = (a + 4) (a + 2) = (a)² + (4 + 2) (a) + 4 x 2

= a² + 6a + 8 ≠ R.H.S

The correct statement is (a + 4) (a + 2) = a² + 6a + 8

Q14: Find and correct the errors in the statement: (a – 4) (a – 2) = a² – 8

Answer:

L.H.S. = (a – 4) (a – 2) = (a)² + [(- 4) + (- 2)] (a) + (- 4) (- 2)

= a² – 6a + 8 ≠ R.H.S.

The correct statement is (a – 4) (a – 2) = a² – 6a + 8

Q15: Find and correct the errors in the statement:

Answer:

L.H.S =
The correct statement is

Q16: Find and correct the errors in the statement:
Answer:

The correct statement is

Q17: Find and correct the errors in the statement:

Answer:

L.H.S =

The correct statement is

Q18: Find and correct the errors in the statement:
Answer:

L.H.S. = ≠ R.H.S.

The correct statement is

Q19 : Find and correct the errors in the statement:

Answer:

L.H.S. =
The correct statement is

Q20: Find and correct the errors in the statement:
Answer:

L.H.S. =

The correct statement is

NCERT Class 8 Maths All Chapters Solution 

Chapter 1: Rational Numbers

Chapter 2: Linear Equations in One Variable

Chapter 3: Understanding Quadrilaterals

Chapter 4: Practical Geometry

Chapter 5: Data Handling

Chapter 6: Squares and Square root

Chapter 7: Cubes and Cube Roots

Chapter 8: Comparing Quantities

Chapter 9: Arithmetic Expressions

Chapter 10: Visualising Solid Shapes

Chapter 11: Mensuration

Chapter 12: Exponents and Powers

Chapter 13: Direct and Inverse Proportions

Chapter 14: Factorisation

Chapter 15: Introduction to Graphs

Chapter 16: Playing With Numbers