# RBSE Solutions for Class 8 Maths Chapter 9 बीजीय व्यंजक Ex 9.2

RBSE Solutions for Class 8 Maths Chapter 9 बीजीय व्यंजक Ex 9.2 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 9 बीजीय व्यंजक Exercise 9.2.

 Board RBSE Textbook SIERT, Rajasthan Class Class 8 Subject Maths Chapter Chapter 9 Chapter Name बीजीय व्यंजक Exercise Exercise 9.2 Number of Questions 3 Category RBSE Solutions

## Rajasthan Board RBSE Class 8 Maths Chapter 9 बीजीय व्यंजक Ex 9.2

प्रश्न 1
नीचे दिए गए द्विपदों का गुणा कीजिए
(i) (2x + 5) और (3x – 7)
(ii) (x – 8) और (3y + 5)
(iii) (1.5p – 0.5q) और (1.5p + 0.5q)
(iv) (a + 3b) और (x + 5)
(v) (2lm + 3m2) (RBSESolutions.com)और (3lm – 5m2)
(vi) ( $$\frac { 3 }{ 4 }$$ + 3b2) और (4a2 –$$\frac { 5 }{ 3 }$$ b2)
हल:
(i) (2x + 5) और (3x – 7)
(2x + 5) (3x -7)
= 2 x (3x – 7) + 5(3x – 7)
= 2x x 3x – 2x x 7 + 5 x 3x – 5 x 7
= 2 x x x 3 x x – 2 x x x 7 + 5 x 3 x x – 35
= 2 x 3 x x x x -2 x 7 x x + 15 x x – 35
= 6 x x2 – 14 x x + 15x – 35
= 6x2 – 14x + 15x – 35
= 6x2 + x – 35

(ii) (x – 8) और (3y + 5)
(x – 8) (3y + 5) = x(3y + 5) – 8(3y + 5)
= x x 3y + x x 5 – 8 x 3y – 8 x 5
= x x 3 x y + 5x – 8 x 3 x y – 40
= 3 x x x y + 5x – 24 x y – 40
= 3x x y + 5x – 24y – 40
= 3xy + 5x – 24y – 40

(iii) (1.5p – 0.5q) और (1.5p + 0.5q)
(1.5p – 0.5q) (1.5p + 0.5q) = 1.5p (1.5p + 0.57q). – 0.5q (1.5p + 0.57)
= 1.5p x 1.5p + 1.5p x 0.5q – 0.5q x 1.5p – 0.5q x 0.5q
= 1.5 x P x 1.5 x p + 1.5 x P x 0.5 x q – 0.5 x q x 1.5 x P – 0.5 x q x 0.5 x q
= 1.5 x 1.5 x p x p + 1.5 x 0.5 x P x q – 0.5 x 1.5 x q x p – 0.5 x 0.5 x q x q
= 2.25 x p2 + 0.75 x pq – 0.75 x qp – 0.25 x q2
= 2.25p2 + 0.75pq – 0.75pq – 0.25q2
= 2.25p2 – 0.25q2

(iv) (a + 3b) और (x + 5)
(a + 3b) (x + 5) = a (x + 5) + 3b (x + 5)
= a x x + a x 5 + 3b x x + 3b x 5
= ax + 5a + 3bx + 3 x 5 x b
= ax + 5a + 3bx + 15b

(v) (2lm + 3m2) और (3lm – 5m2 )
(2lm + 3mऔर ) (3lm – 5mऔर ) = 2lm (3lm – 5m2) + 3m2 (3lm – 5m2)
= 2lm x 3lm – 2lm x 5m2 + 3m2 x 3lm – 3m2 x 5m2
= (2 x 3) l2m2 – (2 x 5) lm3 + (3 x 3) lm3 – (3 x 5) m4
= 6l2m2 – 10lm2 + 9lm3 – 15m4
= 6l2 m2 – lm3 – 15m4

(vi) ( $$\frac { 3 }{ 4 }$$ + 3b2) और (4a2 –$$\frac { 5 }{ 3 }$$ b2)

प्रश्न 2
गुणनफल ज्ञात कीजिए
(i) (3x + 8) (5 – 2x)
(ii) (x + 3y) (3x -y)
(iii) (a2 + b) (a + b2)
(iv) (p2 -q2) (2p + q)
हल:
(i) (3x + 8) (5 – 2x)
(3x + 8) (5 – 2x) = 3x (5 – 2x) + 8(5 – 2x)
= 3x x 5 – 3x x 2x + 8 x 5 – 8 x 2x
= 15x – 6x2 + 40 – 16x
= – 6x2 + 15x – 16x + 40
= – 6x2 – x + 40

(ii) (x + 3y) (3x – y)
(x + 3y) (3x – y) = x (3x – y) + 3y (3x – y)
= x x 3x – x x y + 3y x 3x – 3y x y
= 3x2 – xy + 9yx – 3y2
= 3x2 – xy + 9xy – 3y2
= 3x2+ 8xy – 3y2.

(iii) (a2 + b) (a + b2)
(a2 + b)(a + b2) = a2 (a + b2) + b (a + b2)
= a2 x a + a2 x b2 + b x a + b x b2
= a3 + a2b2 + ba + b3
= a3 + a2b2 + ab + b3 .

(iv) (p2 – q2) (2p+q)
(p2 – q2) (2p + q) = p2 (2p + q) – q2 (2p + q)
= p2 x 2p + p2 x q – q2 x 2p – q2 x q
= 2p3 + p2q – 2q2p – q3

प्रश्न 3
सरल कीजिए
(i) (x + 5) (x – 7) + 35
(ii) (a2 – 3) (b2 + 3) + 5
(iii) (t + s2) (t2 – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (a + b) (a2 – ab + b2)
(vi) (a + b + c)(a + b – c)
(vii) (a + b) (a – b) – a2 + b2
हल:
(i) (x + 5) (x – 7) + 35
(x + 5) (x – 7) + 35 = x (x – 7) + 5 (x – 7) + 35
= x2 – 7x + 5x – 35 + 35
= x2 – 2x

(ii) (a2 – 3) (b2 + 3) + 5
(a2 – 3) (b2+ 3) + 5 = a2 (b2 + 3) – 3 (b2 + 3) + 5
= a2b2 + 3a2 – 3b2 – 9 + 5
= a2b2 + 3a2 – 3b2 – 4

(iii) (t + s2) (t2 – s)
(t + s2) (t2 – s) = t (t2 – s) + s2 (t2 – s)
= t3 – ts + s2t2 – s2
= t3 – s3+ s2t2 – ts

(iv) (a + b)(c – d) + (a – b) (c + d) + 2(ac + bd)
(a + b) (c – d) + (a – b)(c + d) + 2(ac + bd) = a (c – d) + b (c – d) + a (c + d) – b(c + d) +2 (ac + bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac – ad + ad + bc – bc – bd – bd – 2bd
= 4ac

(v) (a + b) (a2 – ab + b2)
= a (a2 – ab + b2) + b (a2 – ab + b2)
= a3 – a2b + ab2 + ba2 – ab2 + b3
= a3 + b3 – a2b + a2b + ab2 – ab2
= a3 + b3

(vi) (a + b + c)(a + b – c) = a(a + b – c) + b (a + b – c) + c (a + b – c )
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + b2 – c2 + 2ab

(vii) (a + b) (a – b) – a2 + b2
(a + b) (a – b) – a2 + b2 = a (a – b) + b (a – b) – a2 + b2
= a2– ab + ab – b2 – a2 + b2
= a2 – a2 – b2 + b2 – ab + ab
= 0 + 0 + 0
= 0

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