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.