RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.2

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

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

Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.2

Question 1.
Multiply the given binomials
(i) (2x + 5) and (3x – 7)
(ii) (x – 8) and (3y + 5)
(iii) (1.5p – 0.5q) and (1.5p + 0.5q)
(iv) (a + 3b) and (x + 5)
(v) (21m + 3m²) and (3lm – 5m²)
(vi) (\(\frac { 3 }{ 4 }\)a² + 3b²) and (4a² – \(\frac { 5 }{ 3 }\)b²)
Solution
(i) (2x + 5) and (3x – 7)
= 2x(3x – 7) + 5(3x – 7)
= 2x × 3x – 2x × 7 + 5 × 3x – 5 × 7
= 2 × x × 3 × x – 2 × x × 7 + 5 × 3 × x – 35
= 2 × 3 × x × x – 2 × 7 × x + 15 × x – 35
= 6 × x² – 14 × x + 15x – 35
= 6x² – 14x + 15x – 35
= 6x² + x – 35

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

(iii) (1.5p – 0.5q) and (1.5p + 0.5q)
= 1.5p (1.5p + 0.5q) – 0.5q (1.5p + 0.5q)
= 1.5p × 1.5p + 1.5p × 0.5q – 0.5q × 1.5p – 0.5q × 0.5q
= 1.5 × p × 1.5 × p + 1.5 × p × 0.5 × q – 0.5 × q × 1.5 × p – 0.5 × q × 0.5 × q
= 1.5 × 1.5 × p × p + 1.5 × 0.5 × p × q – 0.5 × 1.5 × q × p – 0.5 × 0.5 × q × q
= 2.25 × p² + 0.75 × pq – 0.75 × qp – 0.25 × q²
= 2.25p² + 0.75pq – 0.75pq – 0.25q²
= 2.25p² – 0.25q²

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

(v) (2lm + 3m²) and (3lm – 5m²)
= 2lm (3lm – 5m²) + 3m² (3lm – 5m²)
= 2lm × 3lm – 2lm × 5m² + 3m² × 3lm – 3m² × 5m²
= (2 × 3) l²m² – (2 × 5) lm³ + (3 × 3) lm³ – (3 × 5) m⁴
= 6l2m² – 10lm³ + 9lm³ – 15m⁴
= 6l2m² – lm³ – 15m⁴

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

Question 2.
Find the product
(i) (3x + 8) (5 – 2x)
(ii) (x + 3y) (3x – y)
(iii) (a² + b) (a + b²)
(iv) (p² – q²) (2p + q)
Solution
(i) (3x + 8) (5 – 2x)
= 3x (5 – 2x) + 8(5 – 2x)
= 3x × 5 – 3x × 2x + 8 × 5 – 8 × 2x
= 15x – 6x² + 40 – 16x
= – 6x² + 15x – 16x + 40
= – 6x² – x + 40

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

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

(iv) (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

Question 3.
Simplify
(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
Solution
(i) (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 (b² + 3) – 3 (b² + 3) + 5
= a²b² + 3a² – 3b² – 9 + 5
= a²b² + 3a² – 3b² – 4

(iii) (t + s²) (t² – s)
= t (t² – s) + s² (t² – s)
= t³ – ts + s²t² – s³
= 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 + 2 ac + 2 bd
= ac + ac + 2ac – ad + ad + be – bc – bd – bd – 2 bd
= 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 (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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