# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

## BSEB Class 8 Maths Chapter 14 गुणनखंड Textbook Solutions PDF: Download Bihar Board STD 8th Maths Chapter 14 गुणनखंड Book Answers

 BSEB Class 8 Maths Chapter 14 गुणनखंड Textbook Solutions PDF: Download Bihar Board STD 8th Maths Chapter 14 गुणनखंड Book Answers

BSEB Class 8th Maths Chapter 14 गुणनखंड Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 8th Maths Chapter 14 गुणनखंड Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 8th Maths Chapter 14 गुणनखंड books are published by the Bihar Board Publishers. These Bihar Board Class 8th Maths Chapter 14 गुणनखंड textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 8th Maths Chapter 14 गुणनखंड book solutions pdf online from this page.

## Bihar Board Class 8th Maths Chapter 14 गुणनखंड Books Solutions

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### 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
= (4x+ 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

## Bihar Board Class 8th Maths Chapter 14 गुणनखंड Textbooks for Exam Preparations

Bihar Board Class 8th Maths Chapter 14 गुणनखंड Textbook Solutions can be of great help in your Bihar Board Class 8th Maths Chapter 14 गुणनखंड exam preparation. The BSEB STD 8th Maths Chapter 14 गुणनखंड Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 8th Maths Chapter 14 गुणनखंड Books State Board syllabus with maximum efficiency.

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