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BSEB Class 8 Maths Chapter 14 गुणनखंड Textbook Solutions PDF: Download Bihar Board STD 8th Maths Chapter 14 गुणनखंड Book Answers |
Bihar Board Class 8th Maths Chapter 14 गुणनखंड Textbooks Solutions PDF
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Board | BSEB |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 8th |
Subject | Maths Chapter 14 गुणनखंड |
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Provider | Hsslive |
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BSEB Class 8th Maths Chapter 14 गुणनखंड Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 8th Maths Chapter 14 गुणनखंड Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:Bihar Board Class 8 Maths गुणनखंड Ex 14.1
गुणनखंड ज्ञात कीजिए Class 8 Bihar Board Chapter 14 प्रश्न 1.
दिए गए पदों में सार्व (उभयनिष्ठ) गुणनखंड ज्ञात कीजिए-
(a) 9y, 27
(b) 5x, 25x
(c) 7ab, -14ab
(d) -16x2y2, -x2y2z2
(e) 17x, 102y
(f) 11xyz, 100z
(g) a2bc, ab2c, abc2
(h) 2x, 3y, 5z
(i) 20x2y2, 30y2z2, 40z2x2
(j) 2x (a + b)(b + c), x (a + b)
उत्तर
(a) 9y = 3 × 3 × y
27 = 3 × 3 × 3
सार्व उभयनिष्ठ गुणनखण्ड = 3 × 3 = 9
(b) 5x = 5 × x
25x = 5 × 5 × x
सार्व गुणनखण्ड = 5x
(c) 7ab = 7 × a × b
-14ab = -2 × 7 × a × b
सार्व गुणनखण्ड = 7ab
(d) -16x2y2 = -2 × 2 × 2 × 2 × x × x × y × y
x2y2z2 = -x × x × y × y × z × z
सार्व गुणनखण्ड = x2y2
(e) 17x = 17 × x
102y = 6 × 17 × y
सार्व गुणनखण्ड = 17
(f) 11xyz = 11 × x × y × z
100z = 2 × 2 × 5 × 5 × z
सार्व गुणनखण्ड = z
(g) a2bc = a × a × b × c
ab2c = a × b × b × c
abc2 = a × b × c × c
सार्व गुणनखंड = a × b × c = abc
(h) 2x = 2 × x × 1
3y = 3 × y × 1
5z = 5 × z × 1
सार्व गुणनखंड = 1
(i) 20x2y2 = 2 × 2 × 5 × x × x × y × y
30y2z2 = 2 × 3 × 5 × y × y × z × z
40z2x2 = 2 × 2 × 2 × 5 × z × z × x × x
सार्व गुणनखण्ड = 2 × 5 = 10
(j) 2x (a + b) (b + c) = 2 × x × (a + b) × (b + c)
x (a + b) = x × (a + b)
सार्व गुणनखण्ड = x(a + b)
Gunankhand Class 8 Bihar Board Chapter 14 प्रश्न 2.
दिए गए उदाहरण के आधार पर खाली जगह को भरिए-
उत्तर
गुणनखंड कक्षा 8 Bihar Board Chapter 14 प्रश्न 3.
निम्नलिखित का गुणनखण्ड ज्ञात कीजिए-
(a) 12x2 – 15y2 – 24x2z2
(b) -6a2 + 36a – 24ab
(c) 3a2 + ab + 9a + 3b
(d) 6ab – 4b + 6 – 9a
(e) ab2 + a2b + ac + bc
(f) a2bc + b2ca + c2ab + a + b + c
(g) a(b – c) + d(c – b)
(h) 3y(y + 3) + 6y(3y + 9)
(i) a3 – 3a2 + a – 3
(j) ab2 – bc2 – ab + c2
(k) xy(a2 + b2) + ab (x2 + y2)
उत्तर
(a) 12x2 – 15y2 – 24x2z2
= 3 × 2 × 2 × x × x – 3 × 5 × y × y – 2 × 2 × 2 × 3 × x × x × z × z
= 3(4x2 – 5y2 – 8x2z2)
(b) -6a2 + 36a – 24ab
= -2 × 3 × a × a + 2 × 2 × 3 × 3 × a – 2 × 2 × 2 × 3 × a × b
= -6a (a – 6 + 4b)
(c) 3a2 + ab + 9a + 3b
= a (3a + b) + 3(3a + b)
= (3a + b) (a + b)
(d) 6ab – 4b + 6 – 9a
= 2b (3a – 2) – 3 (3a – 2)
= (3a – 2) (2b – 3)
(e) ab2 + a2b + ac + bc
= ab (b + a) + c (b + a)
= (b + a) (ab + c)
(f) a2bc + b2ca + c2ab + a + b + c
= abc (a + b + c) + 1 (a + b + c)
= (a + b + c) (abc + 1)
(g) a(b – c) + d (c – b)
= ab – ac + dc – db
= a(b – c) – b(b – c)
= (a – b) (b – c)
(h) 3y (y + 3) + 6y (3y + 9)
= 3y2 + 9y + 18y2 + 54y
= 21y2 + 63y
= 21y (y + 3)
(i) a3 – 3a2 + a – 3
= a2(a – 3) + 1 (a – 3)
= (a – 3) (a2 + 1)
(j) ab2 – bc2 – ab + c2
= b(ab – c2) – 1(ab – c2)
= (ab – c2) (b – 1)
(h) xy(a2 + b2) + ab (x2 + y2)
= xya2 + xyb2 + abx2 + aby2
= ay(ax + by) + bx(ax + by)
= (ay + bx) (ax + by)
Bihar Board Class 8 Maths गुणनखंड Ex 14.2
Gunankhand Math Class 8 Bihar Board Chapter 14 प्रश्न 1.
निम्नलिखित व्यंजकों का गुणनखंड ज्ञात कीजिए-
(a) 1 + 2x + x2
(b) a2b2 – 6abc + 9c2
(c) 1 – (a – b)2
(d) 16 (a – b)2 – 9 (a + b)
(e) (x + y)2 – 10 (x + y) + 25
(f) (a + b)2 – 4ab
(g) 4x2 – y2+ 4y – 4
(h) 9x2 – 𝑛24
(i) a2 + a + 4 + 3a
(j) x2 + 6x + 8
(k) y2 – 13y + 30
(l) x2 + 9x – 22
उत्तर
(a) 1 + 2x + x2
= 1 + x + x + x2
= 1(1 + x) + x(1 + x)
= (1 + x)(1 + x)
(b) a2b2 – 6abc + 9c2
= a2b2 – 3abc – 3abc + 9c2
= ab(ab – 3c) – 3c(ab – 3c)
= (ab – 3c) (ab – 3c)
(c) 1 – (a – b)2
= 1 – (a2 – 2ab – b2)
= 1 – a2 + 2ab + b2
= (1 – a – b)(1 + a – b)
(d) 16(a – b)2 – 9 (a + b)2
= 16(a2 – 2ab + b2) – 9 (a2 + 2ab + b2)
= 16a2 – 32ab + 16b2 – 9a2 – 18ab – 9b2
= 7a2 + 7b2 – 50ab
= 7a2 – 50ab + 7b2
= 7a2 – ab – 49ab + 7b2
= a(7a – b) – 7b (7a – b)
= (7a – b)(a – 7b)
(e) (x + y)2 – 10(x + y) + 25
= (x + y)2 – 5(x + y) – 5(x + y) + 25
= (x + y) [(x + y) – 5] – 5 [(x + y) – 5]
= (x + y – 5) (x + y – 5)
(f) (a + b)2 – 4ab
= a2 + 2ab + b2 – 4ab
= a2 – 2ab + b2
= a2 – ab – ab + b2
= a(a – b) – b(a – b)
= (a – b)(a – b)
(g) 4x2 – y2 + 4y – 4
= 4x2 – (y + 2)2
= (2x – y + 2) (2x + y – 2)
(h) 9x2 – 𝑛24
= (3x)2 – (𝑛2)2
= (3𝑥−𝑛2)(3𝑥+𝑛2)
(i) a2 + a + 4 + 3a
= a2 + 4a + 4
= a2 + 2a + 2a + 4
= a(a + 2) + 2(a + 2)
= (a + 2) (a + 2)
(j) x2 + 6x + 8
= x2 + 4x + 2x + 8
= x(x + 4) + 2(x + 4)
= (x + 4) (x + 2)
(k) y2 – 13y + 30
= y2 – 10y – 3y + 30
= y(y – 10) – 3 (y – 10)
= (y – 10) (y – 3)
(l) x2 + 9x – 22
= x2 + 11x – 2x – 22
= x(x + 11) – 2(x + 11)
= (x + 11)(x – 2)
8 Ka Gunankhand Bihar Board Chapter 14 प्रश्न 2.
निम्नलिखित व्यंजकों का गुणनखण्ड कीजिए-
(a) x2 – 6x – 135
(b) 8(x + y)3 – 50(x + y)
(c) 4x2 + 9y2 + 12xy – 1
(d) 75 – x2 + 10x
(e) 12a2 – 27
(f) ax2 – bx2 + by2 – ay2
उत्तर
(a) x2 + 6x – 135
= x2 – 15x + 9x – 135
= x (x – 15) + 9 (x – 15)
= (x – 15) (x + 9)
(b) 8(x + y)3 – 50(x + y)
= 2 (x + y) (2x + 2y – 5) (2x + 2y + 5)
= 2 (x + y) (2x + 2y – 5) (2x + 2y + 5)
(c) 4x2 + 9y2 + 12xy – 1
= (2x)2 + (3y)2 + 2(2x) (3y) – 1
= (2x + 3y)2 – 12
= (2x + 3y + 1) (2x + 3y – 1)
(d) 75 – x2 + 10x
= -x2 + 10x + 75
= -x2 + 15x – 5x + 75
= x (x – 15) – 5 (x – 15)
= (x – 15) (x – 5)
(e) 12a2 – 27
= 3 (4a2 – 9)
= 3 [(2a)2 – 32]
= 3 (2a – 3) (2a + 3)
(f) ax2 – bx2 + by2 – ay2
= x2 (a – b) + y2 (b – a)
= (x2 – y2) (a – b)
= (x + y) (x – y) (a – b)
Class 8 Maths Chapter 14 Bihar Board Chapter 14 प्रश्न 3.
निम्नलिखित व्यंजकों का गुणनखंडन कीजिए-
(a) 16x4 – 81y4
(b) x4 – 1
(c) x4 – (x – y)4
(d) 9x2 – 4y2 – 3x + 2y
(e) (x + y) + 4 (x + y)2 + 4x + 4y
उत्तर
(a) 16x4 – 81y4
= ((2x)2)2 – ((3y)2)2
= (2x + 3y)2 – (2x – 3y)2
= (4x2 + 12xy + 9y2) (4x2 – 12xy + 9y2)
= (2x – 3y) (2x + 3y) (4×2 + 9y2)
(b) x4 – 1
= (x2)2 – (12)2
= (x2 + 1) (x2 – 1)
= (x – 1) (x + 1) (x2 + 1)
(c) x4(x – y)4
= (x2)2 – ((x – y)2)2
= (x2 – x – y) (x2 – 2xy + y2)2
= y(2x – y) (2x2 – 2xy + y2)
(d) 3x2 – 4y2 – 3x + 2y
= (3x)2 – (2y)2 – 3x + 2y
= (3x + 2y) (3x – 2y) – (3x + 2y)
= (3x + 2y) (3x + 2y – 1)
(e) (x + y)3 + 4(x + y) + 4x + 4y
= x3 + 3x2y + 3xy2 + y2 + 4 (x2 + 2xy + y2) + 4x + 4y
= x3 + 3x2y + 3xy2 + y2 + 4x2 + 8xy + 4y2 + 4x + 4y
= (x + y) (x + y + z) (x + y + z)
Bihar Board Class 8 Maths गुणनखंड Ex 14.3
गुणनखंड ज्ञात कीजिए Class 8 Bihar Board Chapter 14 प्रश्न 1.
निम्नलिखित का भाग कीजिए
(a) -2x2yz का 4xyz से
(b) −12 का 𝑥2 से
(c) (3x2)5 का (9x2)3 से
(d) (7x5)2 × (3y5)5 का 27y3 से
(e) 8x6y6 का -4x4y6 से
उत्तर
8 का गुणनखंड Bihar Board Chapter 14 प्रश्न 2.
दिए गए बहुपद को एकपदी से भाग कीजिए-
(a) (5m3 – 30m2) ÷ 5m
(b) (12x4 – 6x2) ÷ (-3x2)
(c) (5x2 – 15x) ÷ (x – 3)
(d) (6x4 + 9x3 – 12x2) ÷ 3x2
उत्तर
Chapter 14 Maths Class 8 Bihar Board प्रश्न 3.
(a) (a2 + 8a + 16) ÷ (a + 4)
(b) {(a + b)2 – 4ab} ÷ (a – b)2
(c) (a4 – b4) ÷ (a2 – ab)
(d) (x4 – 81) ÷ (x2 + 9)
(e) 121x2 + 16y2 – 88xy ÷ 4y – 11x
(f) (x2 – x – 30) ÷ (x – 6)
(g) (p2 – p + 14) ÷ (p – 12)
(h) (x2 – 5xy + 6y2) ÷ (x – 2y)
(i) (27x3 + 3x2 – 2x + 8) ÷ (3x – 2)
उत्तर
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