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

गुणनखंड ज्ञात कीजिए Class 8 Bihar Board 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

(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 प्रश्न 2.
दिए गए उदाहरण के आधार पर खाली जगह को भरिए-

उत्तर

गुणनखंड कक्षा 8 Bihar Board 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)

(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

Bihar Board Class 8 Math Solution 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)

(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 8th Class Math Solution 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)

(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)

Bseb Class 8 Math Solution 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)

(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

Class 8 Math Bihar Board 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

Gunankhand Math Class 8 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

Class 8 Maths 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 + \(\frac{1}{4}\)) ÷ (p – \(\frac{1}{2}\))
(h) (x² – 5xy + 6y²) ÷ (x – 2y)
(i) (27x³ + 3x² – 2x + 8) ÷ (3x – 2)
उत्तर