# 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 of the student on the following topics:-

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

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

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)

The next Exercise for NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3 – Factorisation can be accessed by clicking here

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