Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड

Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Text Book Questions and Answers.

BSEB Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड

Bihar Board Class 8 Maths गुणनखंड Ex 14.1

Given a polynomial function, identify the degree and leading coefficient calculator…. the following exercises, graph the polynomial functions using a calculator.

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)

Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड

प्रश्न 2.
दिए गए उदाहरण के आधार पर खाली जगह को भरिए-
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.1 Q2
उत्तर
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.1 Q2.1

Prime Factorization of 84 with a Factor Tree.

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 Solutions Chapter 14 गुणनखंड

Bihar Board Class 8 Maths गुणनखंड Ex 14.2

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 – \(\frac{n^{2}}{4}\)
(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 – \(\frac{n^{2}}{4}\)
= (3x)2 – \(\left(\frac{n}{2}\right)^{2}\)
= \(\left(3 x-\frac{n}{2}\right)\left(3 x+\frac{n}{2}\right)\)

(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)

Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड

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)

Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड

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

प्रश्न 1.
निम्नलिखित का भाग कीजिए
(a) -2x2yz का 4xyz से
(b) \(-\frac{1}{2}\) का \(\frac{x}{2}\) से
(c) (3x2)5 का (9x2)3 से
(d) (7x5)2 × (3y5)5 का 27y3 से
(e) 8x6y6 का -4x4y6 से
उत्तर
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q1
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q1.1

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
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q2
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q2.1

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 + \(\frac{1}{4}\)) ÷ (p – \(\frac{1}{2}\))
(h) (x2 – 5xy + 6y2) ÷ (x – 2y)
(i) (27x3 + 3x2 – 2x + 8) ÷ (3x – 2)
उत्तर
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q3
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q3.1
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q3.2
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q3.3
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q3.4
Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Ex 14.3 Q3.5

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