NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.2 Factorisation

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.2 Factorisation, has been designed by the NCERT to test the knowledge. Class 8 maths chapter 14 exercise 14.2 contains 5 questions and each question has explained stepwise. If you are class 8th student you must be looking for the maths class 8 chapter 14 exercise 14.2 solution for your exams preparation. Here we are providing complete ncert maths class 8 chapter 14 exercise 14.2.

NCERT solutions for class 8 maths chapter 14 exercise 14.2

• Factorisation using identities
• Factors of the form ( x + a) ( x + b)

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.2 Factorisation

1. Factorise the following expressions.

i) a² + 8a + 16

Sol.: –
= a² + 2 × 4 × a + 4²
= (a + 4)²  {∴ (a + b)² = a² + 2ab + b² }

ii) p² – 10p + 25

Sol.: –
= p² – 2 × 5 × p + 5²
= (p – 5)²  {∴ (a – b)² = a² – 2ab + b² }

iii) 25m² + 30m + 9

Sol.: –
= (5m)² + 2 × 5m × 3 + 3²
= (5m + 3)²   {∴ (a + b)² = a² + 2ab + b² }

iv) 49y² + 84yz + 36z²

Sol.: –
= (7y)² + 2 × 7y × 6z + (6z)²
= (7y + 6z)²    {∴ (a + b)² = a² + 2ab + b² }

v) 4x² -8x + 4

Sol.: –
= (2x)² – 2 × 2x × 2 + 2²
= (2x – 2)²    {∴ (a – b)² = a² – 2ab + b² }
= [2(x – 1)]²
= 4(x – 1)²

vi) 121b² – 88bc +16c²

Sol.: –
= (11b)² – 2 × 11b × 4c + (4c)²
= (11b – 4c)²  {∴ (a – b)² = a² – 2ab + b² }

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

Sol.: –
= l² + 2lm + m² – 4lm
= l² – 2lm + m²
= (l – m)²    {∴ (a – b)² = a² – 2ab + b² }

viii) a⁴ + 2a²b² + b⁴

Sol.: –
= (a²)² + 2 × a² × b² + (b²)²
= (a² + b²)²  {∴ (a + b)² = a² + 2ab + b² }

2. Factorise. 

i) 4p² – 9q²

Sol.: –
= (2p)² – (3q)²
= (2p + 3q) (2p – 3q)   {∴ (a² – b²) = (a + b) (a – b) }

ii) 63a² – 112b²

Sol.: –
= 7(9a² – 16b²)
= 7((3a)² – (4b)²)
= 7 (3a + 4b) (3a – 4b)   {∴ (a² – b²) = (a + b) (a – b) }

iii) 49x² – 36

Sol.: – (7x)² – 6²
= (7x + 6) (7x – 6)       {∴ (a² – b²) = (a + b) (a – b) }

iv) 16x⁵ – 144x³

Sol.: –
= 16x³(x² – 9)
= 16x³(x² – 3²)
= 16x³ (x -3) (x + 3)    {∴ (a² – b²) = (a + b) (a – b) }

v) (l + m)² – (l – m)²

Sol.: –
= (l + m + l – m) (l + m – l +m)     {∴ (a² – b²) = (a + b) (a – b) }
= (2l) (2m)
= 4lm

vi) 9x²y² – 16

Sol.: –
= (3xy)² – 4²
= (3xy + 4) (3xy – 4)   {∴ (a² – b²) = (a + b) (a – b) }

vii) (x² – 2xy + y²) – z²

Sol.: –
= (x – y)² – z²
(x – y – z) (x – y + z)    {∴ (a² – b²) = (a + b) (a – b) }

viii) 25a² – 4b² + 28bc – 49c²

Sol.: – 25a² – (4b² – 28bc + 49c²)
= (5a)² – {(2b)² – 2 × 2b × 7c + (7c)²}
= (5a)² – (2b – 7c)²     {∴ (a – b)² = a² – 2ab + b² }
= (5a + 2b – 7c) (5a – 2b +7c)    {∴ (a² – b²) = (a + b) (a – b) }

3. Factorise the expressions.

i) ax² + bx

Sol.: – ax² + bx
= x(ax + b)

ii) 7p² + 21q²

Sol.: – 7p² + 21q²
= 7(p² + 3q²)

iii) 2x³ + 2xy² + 2xz²

Sol.: – 2x³ + 2xy² + 2xz²
= 2x(x² + y² + z²)

iv) am² + bm² + bn² + an²

Sol.: – am² + bm² + bn² + an²
= m²(a + b) + n²(a + b)
= (a + b) (m² + n²)

v) (lm + l) + m +1

Sol.: – (lm + l) + m +1
= l(m +1) +1 (m + 1)
= (m + 1) (l + 1)

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

Sol.: – y(y + z) + 9(y + z)
= (y + z) (y +9)

vii) 5y² – 20y – 8z + 2yz

Sol.: – 5y² – 20y – 8z + 2yz
= 5y(y – 4) + 2z(y – 4)
= (y – 4) (5y + 2z)

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

Sol.: – 10ab + 4a +5b +2
= 2a(5b + 2) + (5b + 2)
= (5b +2) (2a + 1)

ix) 6xy – 4y + 6 – 9x

Sol.: – 6xy – 4y + 6 – 9x
= 2y(3x -2) – 2(3x – 2)
= (3x – 2) (2y – 2)

4. Factorise:

i) a⁴ – b⁴

Sol.: – a⁴ – b⁴
= (a²)² – (b²)²
= (a² + b²) (a² – b²)     { ∴ (a² – b²) = (a + b) (a – b) }
= (a² + b²) (a + b) (a – b)

ii) p⁴ – 81

Sol.: – p⁴ – 81
= (p²)² – (3²)²
= (p² + 3²) (p² – 3²)   { ∴ (a² – b²) = (a + b) (a – b) }
= (p² + 9) (p + 3) (p – 3)

iii) x⁴ – (y + z)⁴

Sol.: – x⁴ – (y + z)⁴
= (x²)² – ((y + z)²)²
= [x² + (y + z)²] [(x² – (y + z)²]   { ∴ (a² – b²) = (a + b) (a – b) }
= [x² + (y + z)²] (x + y + z) (x – y – z)

iv) x⁴ – (x – z)⁴

Sol.: – x⁴ – (x – z)⁴
= (x²)² – ((x – z)²)²
= [x² + (x – z)²] [(x² – (x – z)²]      {∴ (a² – b²) = (a + b) (a – b) }
= [x² + (x – z)²] (x + x – z) (x – x + z)
= z (2x – z) [x² + (x – z)²]

v) a⁴ – 2a²b² + b⁴

Sol.: – a⁴ – 2a²b² + b⁴
= (a²)² – 2×a²×b² + (b²)²
= (a² – b²)²    {∴ (a – b)² = a² – 2ab + b² }
= ((a + b) (a – b))²
= (a + b)² (a – b)²   { ∴ (a² – b²) = (a + b) (a – b) }

5. Factorise the following expressions.

i) p² + 6p + 8

Sol.: – p² + 6p + 8
= p² + 4p + 2p + 8
= p(p +4) + 2(p + 4)
= (p + 4) (p + 2)

ii) q² -10q + 21

Sol.: – q² -10q + 21
= q² – 7q – 3q + 21
= q(q – 7) – 3(q – 7)
= (q – 3) (q – 7)

iii) p² + 6p – 16

Sol.: – 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

• Exercise 14.1 – Factorisation
• Exercise 14.2 – Factorisation
• Exercise 14.3 – Factorisation
• Exercise 14.4 – Factorisation

Download NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.2 Factorisation