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 + 3m^{2}) and (3lm – 5m^{2})

= 2lm (3lm – 5m^{2}) + 3m^{2} (3lm – 5m^{2})

= 2lm × 3lm – 2lm × 5m^{2} + 3m^{2} × 3lm – 3m^{2} × 5m^{2}

= (2 × 3) l^{2}m^{2} – (2 × 5) lm^{3} + (3 × 3) lm^{3} – (3 × 5) m^{4}

= 6l2m^{2} – 10lm^{3} + 9lm^{3} – 15m^{4}

= 6l2m^{2} – lm^{3} – 15m^{4}

(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

= a^{3} + a²b² + ba + b^{3}

= a^{3} + a²b² + ab + b^{3}

(iv) (p² – q²) (2p + q)

= p² (2p + q) – q² (2p + q)

= p² x 2p + p² x q – q² x 2p – q² x q

= 2p^{3} + p²q – 2q²p – q^{3}

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)

= t^{3} – ts + s^{2}t^{2} – s^{3}

= t^{3} – s^{3} + s^{2}t^{2} – 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)

= a^{3} – a²b + ab² + ba² – ab² + b^{3}

= a^{3} + b^{3} – a²b + a²b + ab² – ab²

= a^{3} + b^{3}

(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

We hope the given RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Exercise 9.2, drop a comment below and we will get back to you at the earliest.

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