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)