RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 8 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 8. Students can also read RBSE Class 8 Maths Important Questions for exam preparation. Students can also go through RBSE Class 8 Maths Notes to understand and remember the concepts easily. Practicing the class 8 maths chapter 6 try these solutions will help students analyse their level of preparation.

RBSE Class 8 Maths Solutions Chapter 14 गुणनखंडन Ex 14.2

प्रश्न 1.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए
(i) a2 + 8a + 16
हल:
a2 + 8a + 16
= a2 + 2 × a × 4 + 42.
= (a + 4)2
[a2 + 2ab + b2 = (a + b) का प्रयोग करने पर]
= (a + 4) (a + 4)

RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2

(ii) P2 - 10p + 25
हल:
p2 - 10p + 25 = p2 - 2 × p × 5 + (5)2
= (p - 5)2
[a2 - 2ab + b2 = (a - b)2 का प्रयोग करने पर]
= (p - 5) (p - 5)

(iii) 25m2 + 30m + 9
हल:
25m2 + 30m + 9 = (5m)2 + 2 × 5m × 3 + (3)2
= (5m + 3)
[a2 + 2ab + b2 = (a + b)2 का प्रयोग करने पर
= (5m + 3) (5m + 3)

(iv) 49y2 + 84yz + 36z2
हल:
49y2 + 84yz + 36z2 = (7y)2 + 2 × 7y × 6z + (6z)2
= (7y + 6z)2
[a2 + 2ab + b2 = (a + b)2 का प्रयोग करने पर]
= (7y + 6z)(7y + 6z) 

RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2

(v) 4x2 - 8x + 4
हल:
4x2 - 8x + 4 = 4(x2 - 2x + 1)
= 4(x2 - 2 × x × 1 + 12)
=4(x - 1)2
[a - 2ab + b* = (a - b) का प्रयोग करने पर]
= 4(x - 1) (x - 1) उत्तर

(vi) 121b2 - 88bc + 16c2
हल:
121b2 - 88bc + 16c2 = (11b)2 - 2 × 11b × 4c + (4c)2
= (11b - 4c)
[a2 - 2ab + b2 = (a - b)2 का प्रयोग करने पर]
= (11b - 4c)(11b - 4c) 

(vii) (l + m)2 - 4lm
हल:
(l + m)2 - 4lm = l2 + 2lm + m2 - 4lm
= l2 - 2lm + m2
= (l - m)2 = (l - m) (l - m) 

(viii) a4 + 2a2b2 + b4
हल:
a4 + 2a2b2 + b4 = (a2)2 + 2 × a2 × b2 + (b2)2
= (a2 + b2)2
= (a2 + b2)(a2 + b2

प्रश्न 2.
गुणनखंड कीजिए
(i) 4p2 - 9q2
हल:
4p2 - 9q2 = (2p)2 - (3q)2
= (2p + 3q) (2p - 3q)
[a2 - b2 = (a + b) (a - b) का प्रयोग करने पर]

RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2

(ii) 63a2 - 112b2
हल:
63a2 - 112b2 = 7(9a2 - 16b2)
= 7[(3a)2 - (4b)2]
= 7(3a + 4b)(3a - 4b)
[a2 - b2 = (a + b) (a - b) का प्रयोग करने पर]

(iii) 49x2 - 36
हल:
49x2 - 36 = (7x)2 - (6)2
= (7x + 6) (7x - 6)
[a2 - b2 = (a + b) (a – b) का प्रयोग करने पर]

(iv) 16x5 - 144x3
हल:
16x5 - 144x3 = 16x3(x2 - 9)
= 16x3 (x2 - 32)
= 16x3(x + 3)(x - 3)
[a2 - b2 = (a + b) (a - b) का प्रयोग करने पर]

(v) (l + m)2 - (l - m)2
हल:
(l + m)2 - (l - m)2
= [(l + m) + (l - m)] [(l + m) - (l - m)]
[a2 - b2 = (a - b) (a + b) का प्रयोग करने पर]
= (2l) (2m) = 4lm

(vi) 9x2y2 - 16
हल:
9x2y2 - 16 = (3xy)2 - (4)2
= (3xy + 4)(3xy - 4)
[a2 - b2 = (a - b) (a + b) का प्रयोग करने पर]

RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2

(vii) (x2 - 2xy + y2) - z2
हल:
(x2 - 2xy + y2) - z2 = (x - y)2 - z2
= [(x - y) + z] [(x -y) - z]
= (x - y + z)(x - y - z)

(viii) 25a2 - 4b2 + 28bc - 49c2
हल:
25a2 - 4b2 + 28bc - 49c2
= 25a2 - (4b2 - 28bc + 49c2)
= 25a2 - [(2b)2 - 2 × 2b × 7c + (7c)2]
= (5a)2 - (2b - 7c)2
= [5a + (2b - 7c)][5a - (2b - 7c)]
= (5a + 2b - 7c) (5a - 2b + 7c)

प्रश्न 3.
निम्नलिखित व्यंजकों के गुणनखंड कीजिए
(i) ax2 + bx
हल:
ax2 + bx = x(ax + b)

(ii) 7p2 + 21q2
हल:
7p2 + 21q2 = 7(p2 + 30q2)

(iii) 2x3 + 2xy2 + 2xz2
हल:
2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)

(iv) am2 + bm2 + bn2 + an2
हल:
am2 + bm2 + bn2 + an2
= (am2 + bm2) + (bn2 + an2)
= (a + b)m2 + (b + a)n2
= (a + b) (m2 + n2)

(v) (lm + 1) + m + 1
हल:
(lm + 1) + m + 1 = l(m + 1)+ 1(m + 1)
= (m + 1) (1 + 1)

(vi) y (y + z) + 9(y + z)
हल:
y (y + z) + 9(y + z) = (y + z) (y + 9)

RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2

(vii) 5y2- 20y - 8z +2yz
हल:
5y2 - 20y - 8z + 2yz = (5y2 - 20y) + 2yz - 8z.
= 5y(y - 4)+ 2z(y - 4)
= (y - 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2
हल:
10ab + 4a + 5b + 2 = (10ab + 5b) + (4a + 2)
= 5b(2a + 1) + 2(2a + 1)
= (2a + 1) (5b + 2)

(ix) 6xy - 4y + 6 - 9x
हल:
6xy - 4y + 6 - 9x = (6xy - 4y) - (9x - 6)
= 2y(3x - 2) - 3(3x - 2)
= (3x - 2)(2y - 3)

प्रश्न 4.
गुणनखंड कीजिए
(i) a4 - b4
हल:
a4 - b4 = (a2)2 - (b2)2
= (a2 + b2) (a2 - b2)
= (a2 + b2) (a + b) (a - b)

(ii) p4 - 81
हल:
p4 - 81 = (p2)2 - (9)2
= (p2 + 9) (p2 - 9)
= (p2 + 9) (p + 3) (p - 3)

(iii) x4 + - (y + z)4
हल:
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)]
= [x2 + (y + z)2](x + y + z) (x - y - z)

RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2

(iv) x4 - (x - z)4
हल:
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)
= z(2x - z) [x2 + (x - z)2]
= z(2x - z) (2x2 - 2xz + z2)

(v) a4 - 2a2b2 + b4
हल:
a4 - 2a2b2 + b4 = (a2)2 - 2 × a × b + (b2)2.
= (a2 - b2)2 = [(a + b) (a - b)]2
= (a + b)2 × (a - b)2
= (a + b) (a + b) (a - b) (a - b)

प्रश्न 5.
निम्नलिखित व्यंजकों के गुणनखंड कीजिए
(i) p2 + 6p + 8
हल:
p2 + 6p + 8 = (p2 + 6p + 9) - 1.
[8 = 9 - 1 का प्रयोग करने पर]
= (p2 + 2 × p × 3 + 3) - 1
= (p + 3)2 - 12
= (p + 3 + 1)(p + 3 - 1)
= (p + 4)(p + 2)

(ii) q2 - 10q + 21
हल:
q2 - 10q + 21 = (q2 - 10q + 25) - 4
[21 = 25 - 4 का प्रयोग करने पर]
= (q2 - 2 × q × 5 + 52) - 4
= (q - 5)2 - 22
= (q - 5 + 2) (q - 5 - 2)
= (q - 3) (q - 7)

RBSE Solutions for Class 8 Maths Chapter 14 गुणनखंडन Ex 14.2

(iii) p2 + 6p - 16
हल:
p2 + 6p - 16 = (p2 + 6p + 9) - 25
[- 16 = 9 - 25 का प्रयोग करने पर]
= (p + 3)2 - 52
= (p + 3 + 5) (p + 3 - 5)
= (p + 8) (p - 2)

Bhagya
Last Updated on May 26, 2022, 3:05 p.m.
Published May 26, 2022