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

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

## Haryana Board 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.2

प्रश्न 1.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए
(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
(viii) a4 + 2a2b2 + b4
हल:
हम जानते हैं कि सर्वसमिका
(a + b)2 = a2 + 2ab + b2
तथा, (a – b)2 = a2 – 2ab + b2

(i) a2 + 8a + 16
= a2 + 2 × a × 4 + (4)2.
= a2 + 8a + 16
= (a + 4)2

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

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

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

(v) 4x2 – 8x + 4
= (2x)2 – 2 × 2x × 2 + (2)2
= (2x – 2)2
= 22(x – 1)
= 4(x – 1)2
∴ 4x2 – 8x + 4 = 4(x – 1)2

(vi) 121b2 – 88bc + 16c2
= (11b)2 – 2 × 11b × 4c + (4c)2
= (11b – 4c)2
∴ 121b2 – 44bc – 44bc +16c2

(vii) (l + m)2 – 4lm
(l + m)2 का प्रयोग करने पर-
(l + m)2 = l2 + 2lm + m2

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

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

प्रश्न 2.
गुणनखंड कीजिए
(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
हल:
(i) 4p2 – 9q2
सर्वसमिका a2 – b2 = (a + b)(a – b) से
∴ 4p2 – 9q2 = (2p)2 – (3q)2
= (2p + 3q) (2p – 3q)
∴ 4p2 – 9q2 = (2p + 3q) (2p – 3q)

(ii) 63a2 – 112b2
दोनों पदों में 7 उभयनिष्ठ है
∴ 63a2 – 112b2 = 7(9a2 – 16b2)
= 7[(3a)2 – (4b)2]
= 7 (3a + 4b) (3a – 4b)
∴ 63a2 – 112b2 = 7 (3a + 4b) (3a – 4b)

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

(iv) 16x5 – 144x3
= 16x3(x2 – 9)[16 x2 उभ्यनिष्ठ लेने पर]
= 16x3 (x2 – 32)
= 16x3[(x + 3)(x – 3)]
∴ 6x5 – 144x3 = 16x3[(x + 3)(x – 3)]

(v) (l + m)2 – (l – m)2
सर्वसमिका
(a + b)2 = a2 + 2ab + b2
तथा, (a – b)2 = a2 – 2ab + b2
∴ (l + m)2 = l2 + 2lm + m2
(l – m)2 = l2 – 2lm + m2

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

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

(vii) (x2 – 2xy + y2) – z2
पहला पद, x2 – 2xy + y2
= (x)2 – 2 × x × y + (y)2
= (x – y)2
∴ (x2 – 2xy + y2) – z2 = (x – y)2 – (z)2
= (x – y + z) (x – y – z)
अतः (x2 – 2xy + y2) – z2 = (x – y + z) (x – y – z)

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

प्रश्न 3.
निम्नलिखित व्यंजकों के गुणनखंड कीजिए
(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
हल:
(i) ax2 + bx
दोनों पदों में x उभयनिष्टि है।
ax2 + bx = x (ax + b)
∴ ax2 + bx= x (ax + b)

(ii) 7p2 + 21q2
दोनों पदों में 7 उभयनिष्टि है।
7p2 + 21q2 = 7(p2 + 3q2)
∴ 7p2 + 21q2 = 7(p2 + 3q2)

(iii) 2x3 + 2xy2 + 2xz2
इस योजक में 2x उभयनिष्टि है।
2x3 + 2xy2 + 2xz2
= 2x(x2 + y2 + z2)
∴ 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)

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

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

(vi) y(y + z) + 9(y + z)
= (y + z)(y + 9)
∴ 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)
∴ 5y2 – 20y – 8z + 2yz = (y – 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2
= (10ab + 4a) + (5b + 2)
= 2a (5b + 2) + 1(5b + 2)
= (5b + 2) (2a + 1)
∴ 10ab + 4a + 5b + 2 = (5b + 2) (2a + 1)

(ix) 6xy – 4y + 6 – 9x
= (6xy – 9x) – (4y – 6)
= 3x(2y – 3) – 2(2y – 3)
= (2y – 3) (3x – 2).
∴ 6xy – 4y + 6 – 9x = (2y – 3) (3x – 2).

प्रश्न 4.
गुणनखंड कीजिए
(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) (सर्वसमिका a2 – b2 = (a + b) (a- b) से)
= (a2 + b2) [a2 – b2]
= (a2 + b2) [(a + b) (a- b)]
अतः a4 – b4 = (a2 + b2) (a + b) (a- b)

(ii) p4 – 81
= p4 – 81
= (p2)2 – (9)2
= (p2 + 9) (p2 – 9)
= (p2 + 9) (p2 – 32)
= (p2 + 9) (p + 3) (p – 3)
∴ p4 – 81 = (p2 + 9) (p + 3) (p – 3)

(iii) x4 – (y + z)4
= (x2)2 – [(y + z)2]2
= [x2 + (y + z)2] [x2 – (y + z)2]
= [(x2 + (y + z)2)] [(x + y + z) (x – y – z)]
∴ x4 – (y + z)4 = (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]
= [x2 + (x – z)2] [(x + x – z) (x – x + z)]
= [x2 + (x – z)2] (2x – z) z
∴ x4 – (x – z)4 = z (2x – 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
∴ a4 – 2a2b2 + b4 = (a + b)2 (a – b)2

प्रश्न 5.
निम्नलिखित व्यंजकों के गुणनखंड कीजिए
(i) p2 + 6p + 8
(ii) q2 – 10q + 21
(iii) p2 + 6p – 16
हल:
(i) p2 + 6p + 8
अंक 8 को दो संख्याओं में इस प्रकार विभाजित करते हैं कि उनका योग व्यंजक के बीच वाली संख्या 6 के बराबर तथा उनका गुणनफल 8 हो-
अर: 8 = 4 × 2 (∴ 4 + 2 = 6)
अतः व्यंजक के मध्य पद 6p को 4p तथा 2p में विभाजित करेंगे ।
∴ p2 + 6p + 8 ⇒ p2 + 4p + 2p + 8
दो-दो संख्या लेकर याग्म बनाने पर-
= (p2 + 4p) + (2p + 8)
= p(p + 4) + 2(p + 4)
= (p + 4) (p + 2)

(ii) q2 – 10q + 21
21 = 7 × 3 (∴ 7 + 3 = 10)
अतः q2 – 7q – 3q + 21
युग्म बनाने पर,
= (q2 – 7q) – (3q – 21)
= q(q – 7) – 3(q – 7)
= (q – 3)(q – 7)
अतः q2 – 10q + 21 = (q – 3)(q – 7)

(iii) p2 + 6p – 16
व्यंजक के आखिरी पद 16 को दो संख्याओं में इस प्रकार विभाजित करेंगे कि उनका गुणनफल 16 तथा उनका अन्तर 6 हो जाय, क्योंकि पद 16 से पूर्व ॠणात्कक (-) चित्र है ।
16 = 8 × 2 (8 – 2 = 6)
अतः p2 + 6p – 16 = p2 + 8p – 2p – 16
युग्म बनाने पर = (p2 + 8p) – (2p + 16)
= p (p + 8) – 2 (p + 8)
= (p + 8)(p – 2)
अतः p2 + 6p – 16 = (p + 8)(p – 2)