## Book: NCERT - Mathematics

### Chapter: 14. Factorisation

#### Subject: Maths - Class 8th

##### Q. No. 1 of Exercises 14.2

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##### Factorise the following expressions.(i) (ii) (iii) (iv) (v) (vi) (vii) (Hint: Expand first)(viii)

(i) a2 + 8a + 16

= a2 + 2× a × 4 + 42

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

Here a =a; b = 4

a2 + 2× a × 4 + 42

= (a + 4)2

= (a + 4)(a + 4)

(ii) p2 -10p + 25

= p2 - 2× 5 × p + 52

Using identity (a - b)2 = a2 - 2ab + b2

Here a =p; b = 5

p2 - 2× 5 × p + 52

= (p - 5)2

= (p-5)(p-5)

(iii) 25m2 + 30m + 9

= (5m)2 + 2 × 5 × 3 × m + 32

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

Here a =5m; b = 3

(5m)2 + 2 × 5 × 3 × m + 32

= (5m + 5)2

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

(iv) 49y2 + 84yz + 36z2

= (7y)2 + 2 × 7 × 6 × y × z + (6z)2

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

Here a =7y; b = 6z

(7y)2 + 2 × 7 × 6 × y × z + (6z)2

= (7y + 6z)2

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

(v) 4x2 - 8x + 4

= (2x)2 - 2 × 2 × 2× x + 22

Using identity (a - b)2 = a2 - 2ab + b2

Here a =2x; b = 2

(2x)2 - 2 × 2 × 2× x + 2

= (2x - 2)2

= (2x - 2)(2x - 2)

(vi) 121b2 - 88bc + 16c2

= (11b)2 - 2 × 11b × 4c + (4c)2

Using identity (a - b)2 = a2 - 2ab + b2

Here a =11b; b = 4c

(11b)2 - 2 × 11b × 4c + (4c)2

= (11b – 4c)2

(11b – 4c)(11b – 4c)

(vii) (l + m)- 4lm

Expand (l + m)2 = l2 + 2lm + m2

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

l2 + 2lm + m2 - 4lm

l2 - 2lm + m2

= l2 - 2 × l × m + m2

Using identity (a - b)2 = a2 - 2ab + b2

Here a =l; b = m

l2 - 2 × l × m + m2

= (l – m)2

= ( l - m)(l - m)

(viii) a⁴ + 2a2b2 + b⁴

= (a2)2 + 2 × a2 × b2 + (b2)2

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

Here a = a2; b = b2

(a2)2 + 2 × a2 × b2 + (b2)2

= (a2 + b2)2

(a2 + b2)(a2 + b2)

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