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 + 3m^{2}) (RBSESolutions.com)और (3lm – 5m^{2})

(vi) ( + 3b^{2}) और (4a^{2} – b^{2})

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 x^{2} – 14 x x + 15x – 35

= 6x^{2} – 14x + 15x – 35

= 6x^{2} + 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 p^{2} + 0.75 x pq – 0.75 x qp – 0.25 x q^{2}

= 2.25p^{2} + 0.75pq – 0.75pq – 0.25q^{2}

= 2.25p^{2} – 0.25q^{2}

(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 + 3m^{2}) और (3lm – 5m^{2} )

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

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

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

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

= 6l^{2} m^{2} – lm^{3} – 15m^{4}

(vi) ( + 3b^{2}) और (4a^{2} – b^{2})

प्रश्न 2

गुणनफल ज्ञात कीजिए

(i) (3x + 8) (5 – 2x)

(ii) (x + 3y) (3x -y)

(iii) (a^{2} + b) (a + b^{2})

(iv) (p^{2} -q^{2}) (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 – 6x^{2} + 40 – 16x

= – 6x^{2} + 15x – 16x + 40

= – 6x^{2} – 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

= 3x^{2} – xy + 9yx – 3y^{2}

= 3x^{2} – xy + 9xy – 3y^{2}

= 3x^{2}+ 8xy – 3y^{2}.

(iii) (a^{2} + b) (a + b^{2})

(a^{2} + b)(a + b^{2}) = a^{2} (a + b^{2}) + b (a + b^{2})

= a^{2} x a + a^{2} x b^{2} + b x a + b x b^{2}

= a^{3} + a^{2}b^{2} + ba + b^{3}

= a^{3} + a^{2}b^{2} + ab + b^{3} .

(iv) (p^{2} – q^{2}) (2p+q)

(p^{2} – q^{2}) (2p + q) = p^{2} (2p + q) – q^{2} (2p + q)

= p^{2} x 2p + p^{2} x q – q^{2} x 2p – q^{2} x q

= 2p^{3} + p^{2}q – 2q^{2}p – q^{3}

प्रश्न 3

सरल कीजिए

(i) (x + 5) (x – 7) + 35

(ii) (a^{2} – 3) (b^{2} + 3) + 5

(iii) (t + s^{2}) (t^{2} – s)

(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)

(v) (a + b) (a^{2} – ab + b^{2})

(vi) (a + b + c)(a + b – c)

(vii) (a + b) (a – b) – a^{2} + b^{2}

हल:

(i) (x + 5) (x – 7) + 35

(x + 5) (x – 7) + 35 = x (x – 7) + 5 (x – 7) + 35

= x^{2} – 7x + 5x – 35 + 35

= x^{2} – 2x

(ii) (a^{2} – 3) (b^{2} + 3) + 5

(a^{2} – 3) (b^{2}+ 3) + 5 = a^{2} (b^{2} + 3) – 3 (b^{2} + 3) + 5

= a^{2}b^{2} + 3a^{2} – 3b^{2} – 9 + 5

= a^{2}b^{2} + 3a^{2} – 3b^{2} – 4

(iii) (t + s^{2}) (t^{2} – s)

(t + s^{2}) (t^{2} – s) = t (t^{2} – s) + s^{2} (t^{2} – s)

= t^{3} – ts + s^{2}t^{2} – s^{2}

= 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 + 2ac + 2bd

= ac + ac + 2ac – ad + ad + bc – bc – bd – bd – 2bd

= 4ac

(v) (a + b) (a^{2} – ab + b^{2})

= a (a^{2} – ab + b^{2}) + b (a^{2} – ab + b^{2})

= a^{3} – a^{2}b + ab^{2} + ba^{2} – ab^{2} + b^{3}

= a^{3} + b^{3} – a^{2}b + a^{2}b + ab^{2} – ab^{2}

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

(vi) (a + b + c)(a + b – c) = a(a + b – c) + b (a + b – c) + c (a + b – c )

= a^{2} + ab – ac + ab + b^{2} – bc + ac + bc – c^{2}

= a^{2} + b^{2} – c^{2} + 2ab

(vii) (a + b) (a – b) – a^{2} + b^{2}

(a + b) (a – b) – a^{2} + b^{2} = a (a – b) + b (a – b) – a^{2} + b^{2}

= a^{2}– ab + ab – b^{2} – a^{2} + b^{2}

= a^{2} – a^{2} – b^{2} + b^{2} – ab + ab

= 0 + 0 + 0

= 0

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

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