# RBSE Solutions for Class 8 Maths Chapter 10 गुणनखण्ड Ex 10.2

RBSE Solutions for Class 8 Maths Chapter 10 गुणनखण्ड Ex 10.2 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 10 गुणनखण्ड Exercise 10.2.

 Board RBSE Textbook SIERT, Rajasthan Class Class 8 Subject Maths Chapter Chapter 10 Chapter Name गुणनखण्ड Exercise Exercise 10.2 Number of Questions 3 Category RBSE Solutions

## Rajasthan Board RBSE Class 8 Maths Chapter 10 गुणनखण्ड Ex 10.2

प्रश्न 1
निम्नलिखित के गुणनखण्ड कीजिए
(i) a2 – 4
(ii) a2 – 49b2
(iii) p3 – 121p
(iv) (a – b)2 – c2
(v) a4 – b4
(vi) 5x3 – 125x
(vii) 63a2 – 112b2
(viii) 9x2y2 – 16
(ix) (l+ m) – (l – m)2
हल:
(i) a2 – 4
a2 – 4 = (a)2 – (2)2
= (a – 2) (a + 2)

(ii) a2 – 49b2
a2 – 49b2 = (a)2(7b)2
= (a – 7b) (a + 7b)

(iii) p3 – 121p
p3 – 121p =p (p2 – 121)
=p {(p2– (11)2}
= p (p – 11) (p + 11)

(iv) (a – b)2 – c2
(a – b)2 – c2
= (a – b – c) (a – b + c)

(v) a4 – b4
a4 – b4
= (a2)2 – (b2)2
= (a2 – b2) (a2 + b2)
= (a – b) (a + b) (a2 + b2)

(vi) 5x3 – 125x
5x3 – 125x
= 5 (x2  – 25)
= 5x (x2 – 52)
= 5x (x – 5) (x + 5)

(vii) 63a2 – 112b2
63a2 – 112b2
=7 (9a2 – 16b2)
= 7 {(3a)2 – (4b)2}
= 7(3a – 4b) (3a + 4b)

(viii) 9x2y2 – 16
9x2y2 – 16 = (3xy)2 – (4)2
= (3xy – 4) (3xy + 4)

(ix) (l + m)2 – (l – m)2
(l + m)2 – (l – m)
= {(l + m) + (l – m)} {(l + m) + (l – m)}
= (l + m – l + m) (l + m + l – m)
= (2m) (2l)
= 4lm

प्रश्न 2
निम्नलिखित के गुणनखण्ड कीजिए
(i) lx2 + mx
(ii) 2x3 + 2xy2 + 2xz2
(iii) a(a + b) + 4 (a + b)
(iv) (xy + y) + x +1
(v) 5a2 – 15a – 6c + 2ac
(vi) am2 + bm2 + bn2+ an2
हल:
(i) lx2 + mx
lx2 + mx
= x (lx + m)

(ii) 2x3 + 2xy2 + 2xz2
2x3 + 2xy2 +2xz2
= 2x (x2 + y2 + z2)

(iii) a(a + b) + 4 (a + b)
a(a + b) + 4 (a + b)
= (a + b)(a + 4)

(iv) (xy + y) + x + 1
(xy + y) + (x + 1)
= y (x + 1) + 1 (x + 1)
= (x + 1) (y + 1)

(v) 5a2 – 15a – 6c + 2ac
5a – 15a – 6c + 2ac
= 5a2 – 15a + 2ac – 6c
= 5a (a – 3) + 2c (a – 3)
= (a – 3) (5a + 2c)

(vi) am2 + bm2 + bn2 + an2
am2 + bm2 + bn2 + an2
= am2 + bm2 + an2 + bn2
= m2 (a + b) + n2 (a + n)
= (a + b) (m2 + n2)

प्रश्न 3
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए
(i) x2 + 5x + 6
(ii) q2 + 11q + 24
(iii) m2 – 10m + 21
(iv) x2 + 6x – 16
(v) x2 – 7x – 18
(vi) k2 – 11k – 102
(vii) y2 + 2y – 48.
(viii) d2 – 4d – 45
(ix) m2 + 16m + 63
(x) n2 – 19n – 92
(xi) p2 – 10p + 16
(xii) x2 + 4x – 45
हल:
(i) x2 + 5x + 6
x2 + 5x + 6
= x2 + 2x + 3x + 6
= x (x + 2) + 3(x + 2)
= (x + 2) (x + 3)

(ii) q2 + 11q + 24
q2 + 11q + 24
= q2 + 3q + 8q + 24
= q (q + 3) + 8 (q + 3)
= (q + 3) (q + 8)

(iii) m2 – 10m + 21
m2 – 10m + 21
= m2 – 3m – 7m + 21
= m (m – n) – 7 (m – 3)
= (m – 3) (m – 7)

(iv) x2 + 6x – 16
x2 + 6x – 16
= x2 + 8x – 2x – 16
= x (x + 8) – 2 (x + 8)
= (x + 8) (x – 2)

(v) x2 – 7x – 18
x2 – 7x – 18
= x2 – 9x + 2x – 18
= x (x – 9) + 2 (x – 9)
= (x – 9) (x + 2)

(vi) k2 – 11k – 102
k2 – 11k – 102
=k2 – 17k + 6k – 102
=k (k – 17) + 6 (k – 17)
= (k – 17) (k + 6)

(vii) y2 + 2y – 48
y2 + 2y – 48
= y2 + 8y – 6y – 48
= y (y + 8) – 6 (y + 8)
= (y + 8) (y – 6)

(viii) d2 – 4d – 45
= d2 – 4d – 45
= d2 – 9d + 5d – 45
= d (d – 9) + 5 (d – 9)
= (d – 9) (d + 5)

(ix) m2 + 16m + 63
m2 + 16m + 63
= m2 + 9m + 7m + 63
= m (m + 9) + 7 (m + 9)
= (m + 9) (m + 7)

(x) n2 – 19n – 92
n2 – 19n – 92
= n2 – 23n + 4n – 92
= n (n – 23) + 4 (n – 23)
= (n – 23) (n + 4)

(xi) p2 – 10p + 16
p2 – 10p + 16
= p2 – 8p – 2p + 16
= p (p – 8) – 2(p – 8)
= (p – 8) (p – 2)

(xii) x2 + 4x – 45
x2 + 4x – 45
= x2 + 9x – 5x – 45
= x (x + 9) – 5(x + 9)
= (x + 9) (x – 5)

We hope the RBSE Solutions for Class 8 Maths Chapter 10 गुणनखण्ड Ex 10.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 10 गुणनखण्ड Exercise 10.2, drop a comment below and we will get back to you at the earliest.