# HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.2

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.2 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 14 Factorization Exercise 14.2

Question 1.
Factorise the following expressions :
(i) a2 + 8a + 16
(ii) p2 – 10p + 25
(iii) 25m2 + 30m + 9
(iv) 49 y2 + 84yz + 36z2
(v) 4x2 – 8x + 4
(vi) 121b2 – 88bc + 16c2
(vii) (l + m)2 – 4lm
[HintExpand (l + m)2 First]
(viii) a4 + 2a2b2 + b4
Solution:
(i) a2 + 8a + 16
= a2 + 4a + 4a + 16
= a (a + 4) + 4 (a + 4)
= (a + 4) (a + 4)

(ii) p2 – 10p + 25
= p2 – 5p – 5p + 25
= p(p – b) – 5(p – 5)
= (p – 5)(p – b)

(iii) 25m2 + 30m + 9
= 25m2 + 15m + 15m + 9
= 5m (5m + 3) + 3 (5m + 3)
= (5m + 3) (5m + 3)

(iv) 49y2 + 84yz + 36z2
= 49y2 + 42yz + 42yz + 36z2
= 7y (7y + 6z) + 6z (7y + 6z)
= (7y + 6z) (7y + 6z)

(v) 4x2 – 8x + 4
= 4x2 – 4x – 4x + 4
= 4x (x – 1) – 4(x – 1)
= (x – 1) (4x – 4)

(vi) 121b2 – 88bc + 16c2
= 121b2 – 44bc – 44bc +16c2
= 11b (11b – 4c) – 4c(11b – 4c)
= (11b – 4c) (11b – 4c)

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

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

Question 2 .
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)
(viii) 25a2 – 4b2 + 28bc – 49c2
Solution:
(i) ∵ a2 – b2 = (a + b)(a – b)
∴ 4p2 – 9q2 = (2p)2 – (3q)2
= (2p + 3q) (2p – 3q)2

(ii) 63a2 – 112b2 = 7(9a2 – 16b2)
= 7[(3a)2 – (4b)2]
= 7[(3a + 4b) (3a – 4b)]
= 7 (3a + 4b) (3a – 4b)

(iii) 49x2 – 36 = (7x)2 – (6)2
= (7x + 6) (7x – 6)

(iv) 16x5 – 144x3 = x (16x4 – 144x2)
= x [(4x2)2 – (12x)2]
= x [(4x2 + 12x) (4x2 – 12x)]
= x (4x2 + 12x) (4x2 – 12x)

(v) (l + m)2 -(l – m)2
= [{(l + m) + (l – m)} – {(l + m) – (l – m)}]
= [{l + m + l – m}{l + m – l + m}]
= 2l × 2m = 4lm

(vi) 9x2y2 – 16 = (3xy)2 – (4)2
= (3xy + 4) (3xy – 4)

(vii) (x2 – 2xy + y2) – z2
= (x – y)2 – z2
= (x – y + z) (x – y – z)

(viii) 25a2 – 4b2 + 28bc – 49c2
= 25a2 – (4b2 – 28bc + 49c2)
= (5a)2 – [(26)2– 2 x 2b x 7c + (7c)2]
= (5a)2 – (2b – 7c)2
= [(5a) (2b – 7c)] [(5a) – (2b – 7c)]
= (5a + 2b – 7c) (5a – 2b + 7c)

Question 3.
Factorise the expressions:
(i) ax2 + 6x
(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 + b)
(ii) 7p2 + 21q2 = 7(p2 + 3q2)
(iii) 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)

(iv) am2 + 6m2 + bn2 + an2
= m2(a + b) + n2(b + a)
= m2(a + b) + n2(a + b)
= (a + b) (m2 + n2)

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

Question 4.
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) (a2 – b2)

(ii) p4 – 81 = (p2)2 – (9)2
= (p2 + 9) (p2 – 9)

(iii) x4 – (y + z)4
= (x2)2 – [(y + z)2]2
= [x2 + (y + z)2] [x2 – (y + z)2]
= (x2 + y2 + 2yz + z2) [(x2 – (y2 + 2yz + z2)]
= (x2 + y2 + z2 + 2yz) (x2 – y2 – z2 – 2yz)

(iv) x4 – (x – z)4
= (x2)2 – [(x – z)2]2
= [x2 + (x – z)2] [x2 – (x – z)2]
= (x2 + x2 – 2xz + z2) [x2(x2 – 2xz + z)2]
= (2x2 + z2 – 2xz) (x2 – x2 + 2xz – z2)
= (2x2 + z2 – 2xz) (2xz – z2)

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

Question 5.
Factorise the following expressions:
(i) p2 + 6p + 8
(ii) q2 – 10q + 21
(iii) p2 + 6p – 16
Solution:
(i) p2 + 6p + 8
= p2 + 2 × p × 3 + 9 – 1
= p2 + 2 × p × 3 + (3)2 – 1
= (p + 3)2 – 1
[∵ a2 + 2ab + b2 = (a + b)2]
= (p + 3)2 – (1)2
[∵ a2 – b2 = (a + b) (a – b)]
= (p + 3 + 1)(p + 3 – 1)
= (p + 4) (p + 2)

(ii) q2 – 10q + 21
= q2 – 2 × q × 5+ (5)2 – 4
= (q – 5)2 – (2)2
[∵ a2 – 2ab + b2 = (a – b)2 ]
= (q – 5 + 2) × (q – 5 – 2)
[∵ a2 – b2 = (a + b) (a – b)
= (q – 3)(q – 7)

(iii) p2 + 6p – 16
= p2 + 2 × p × 3 + (3)2 – 25
= (p + 3)2 – (5)2
= (p + 3 + 5)(p + 3 – 5)
= (p + 8)(p – 2)