RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.5 is part of RBSE Solutions for Class 9 Maths. Here we have given RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.5.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomial |

Exercise |
Ex 3.5 |

Number of Questions Solved |
12 |

Category |
RBSE Solutions |

## RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.5

Question 1.

Use suitable identities to find (RBSESolutions.com) the following products:

(i) (x + 3)(x + 7)

(ii) (x – 5)(x + 8)

(iii) (2x + 7)(3x – 5)

(iv) (5 – 3x)(3 + 2x)

(v) [latex]\left( { x }^{ 2 }+\frac { 3 }{ 5 } \right) \left( { x }^{ 2 }-\frac { 3 }{ 5 } \right) [/latex]

(vi) (x + 2)(x – 5)

Solution.

(i) ∵ (x + a)(x + b) = x^{2} + (a + b)x + ab

∴(x + 3)(x + 7) = x^{2} + (3 + 7)x + 3 x 7

= x^{2} + 10x + 21

(ii) ∵ (x + a)(x + b) = x^{2} + (a + b)x + ab

∴(x + (- 5))(x + 8)

= x^{2} + (- 5 + 8)x + (- 5) x (8)

= x^{3} + 3x – 40

(iii) We have, (2x + 7)(3x + 5)

Question 2.

Evaluate the following (RBSESolutions.com) products without multiplying directly.

(i) 104 x 109

(ii) 94 x 97

(iii) 103 x 97

Solution.

(i) 104 x 109 = (104 + 4)(100 + 9)

Using identity

(x + a)(x + b) = x² + (a + b)x + ab, we get

104 x 104

= (100)² + (4 + 9) x 100 + 4 x 9

= 10000 + 1300 + 36 = 11336

(ii) 94 x 97 = (100 – 6)(100 – 3)

= (100)² + [(- 6) + (- 3)] x 100 + (- 6) x (- 3)

= 10000 – 900 + 18 = 9118

(iii) 103 x 97 = (100 + 3)(100 – 3)

Using identity

(a + b)(a – b) = a² – b², we get

103 x 97 = (100)² – (3)²

= 10000 – 9 = 9991

Question 3.

Factorise the following (RBSESolutions.com) by using suitable identity

(i) x² + 6xy + 9y²

(ii) x² – 4x + 4

(iii) [latex]\frac { { x }^{ 2 } }{ 100 } [/latex] – y²

Solution.

(i) x² + 6xy + 9y²

= (x)² + 2.x.3y + (3y)²

= (x + 3y)² = (x + 3y)(x + 3y)

(ii) x² – 4x + 4

= (x)² – 2.x.2 + (2)²

= (x – 2)²

= (x – 2)(x – 2)

Question 4.

Expand each of the following, using (RBSESolutions.com) suitable identity.

(i) (2a – 3b – c)²

(ii) (2 + x – 2y)²

(iii) (a + 2b + 4c)²

(iv) (m + 2n – 5p)²

(v) (3a – 7b – c)²

(vi) [latex]{ \left( \frac { x }{ y } +\frac { y }{ z } +\frac { z }{ x } \right) }^{ 2 }[/latex]

Solution.

(i) We have, (2a – 3b – c)²

Using identity (a + b + c)²

= a² + b² + c² + 2ab + 2bc + 2ca, we get

∴(2a – 3b – c)²

= (2a)² + (- 3b)² + (-c)² + 2(2a)(- 3b) + 2(- 3b)(- c) + 2(- c)2a

= 4a² + 9b² + c² – 12ab + 6bc – 4ac

(ii) We have, (2 + x – 2y)²

= (2)² + (x)² + (- 2y)² + 2(2)(x) + 2(x)(- 2y) + 2(2)(- 2y)

= 4 + x² + 4y² + 4x – 4xy – 8y

Question 5.

Factorise:

(i) 9x² + 4y² + 16z² – 12xy – 16yz + 24xz

(ii) x² + 2y² + 8z² + 2√2xy – 8yz – 4√2xz

Solution.

(i) We have,

9x² + 4y² + 16z² – 12xy – 16yz + 24xz

= (3x)² + (- 2y)² + (4z)² + 2(3x)(- 2y) + 2(- 2y)(4z) + 2(3x)(4z)

[∵ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

= (3x – 2y + 4z)²

= (3x – 2y + 4z)(3x – 2y + 4z)

(ii) We have,

x² + 2y² + 8z² + 2√2xy – 8yz – 4√2xz = (x)² + (√2y)² + (- 2√2z)² + 2(x)(√2y) + 2(√2y)(- √2z) + 2(x)(- 2√2z)

[∵ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

= (x + √2y – 2√2z)²

= (x + √2y – 2√2z)(x + √2y – 2√2z)

Question 6.

Expand the following:

Solution.

(i) (3a – 2b)^{3}

= (3a)^{3} – (2b)^{3} – 3(3a)(2b) [3a – 2b]

= 27a^{3} – 8b^{3} – 18ab (3a – 2b)

= 27a^{3} – 8b^{3} – 54a^{2}b + 36ab^{2}

= 27a^{3} – 54a^{2}b + 36ab^{2} – 8b^{3}

(ii) (1 + 2x)^{3}

= (1)^{3} + (2x)^{3} + 3(1)(2x)[1 + 2x]

= 1 + 8x^{3} + 6x (1 + 2x)

= 1 + 8x^{3} + 6x + 12x^{2}

= 1 + 6x + 12x^{2} + 8x^{3}

Question 7.

Find the value of the following (RBSESolutions.com) by using suitable identity.

(i) (98)^{3}

(ii) (103)^{3}

(iii) (999)^{3}

Solution.

(i) (98)^{3}

= (100 – 2)^{3}

= (100)^{3} – (2)^{3} – 3(100)(2) [100 – 2]

= 1000000 – 8 – 60000 + 1200

= 941192

(ii) (103)^{3}

= (100 + 3)^{3}

= (100)^{3} + (3)^{3} + 3(100)(3) [100 + 3]

= 1000000 + 27 + 90000 + 2700

= 1092727

(iii) (999)^{3}

= (1000 – 1)^{3}

= (1000)^{3} – (1)^{3} – 3(1000)(1) [1000 – 1]

= 100000000 – 1 – 3000000 + 3000

= 997002999

Question 8.

Factorise the following:

(i) x^{3} + 8y^{3} + 6x^{2}y + 12xy^{2}

(ii) 27a^{3} – 8b^{3} – 54a^{2}b + 36ab^{2}

(iii) 27 – 125x^{3} – 135x + 225x^{2}

(iv) 125x^{3} + 64y^{3} – 300x^{2}y + 240xy^{2}

Solution.

(i) x^{3} + 8y^{3} + 6x^{2}y + 12xy^{2}

= (x)^{3} + (2y)^{3} + 3(x)(2y) [x + 2y]

= (x + 2y)^{3}

[∵ (a + b)^{3} = a^{3} + b^{3} + 3 ab(a + b)]

(ii) 27a^{3} – 8b^{3} – 54a^{2}b + 36ab^{2}

= (3a)^{3} – (2b)^{3} – 3(3a)(2b) [3a – 2b]

= (3a – 2b)^{3}

[∵ (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)]

(iii) 27 – 125x^{3} – 135x + 225x^{2}

= (3)^{3} – (5x)^{3} – 3(3)(5x) [3 – 5x]

= (3 – 5x)^{3}

[∵ (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)]

(iv) 125x^{3} – 64y^{3} – 300x^{2}y + 240xy^{2}

= (5x)^{3} – (4y)^{3} – 3(5x)(4y) [5x – 4y]

= (5x – 4y)^{3}

[∵ (a – b) = a^{3} – b^{3} – 3ab(a – b)]

Question 9.

Factorise the following:

(i) 64a^{3} + 21b^{3}

(ii) 125x^{3} – 8y^{3}

Solution.

(i) 64a^{3} + 27b^{3}

= (4a)^{3} + (3b)^{3}

= (4a + 3b)[(4a)^{2} – 4a.3b + (3b)^{2}]

[∵ a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})]

= (4a + 3b)[ 16a^{2} – 12a6 + 9b^{2}]

(ii) 125x^{2} – 8y^{3}

= (5x)^{3} – (2y)^{3}

= (5x – 2y)[(5x)^{2} + 5x.2y + (2y)^{2}]

[ a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})]

= (5x – 2y)[25x^{2} + 10xy + 4y^{2}]

Question 10.

Verify that (i) x^{3} + y^{3} + z^{3} – 3xyz

Solution.

(i) RHS

Question 11.

If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3xyz.

Solution.

We are given that x + y + z = 0 and we know that

x^{3} + y^{3} + z^{3} – 3xyz

= (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)

Since x + y + z = 0, then

x^{3} + y^{3} + z^{3} – 3xyz

= 0 × (x^{2} + y^{2} + z^{2} – xy – yz – zx)

⇒ x^{3} + y^{3} + z^{3} – 3xyz = 0

⇒ x^{3} + y^{3} + z^{3} = 3xyz

Question 12.

Without actually calculating the cubes, find (RBSESolutions.com) the value of each of the following:

(i) (30)^{3} + (20)^{3} + (- 50)^{3}

(ii) (- 15)^{3} + (28)^{3} + (- 13)^{3}

Solution.

(i) (30)^{3} + (20)^{3} + (- 50)^{3}

We know that a^{3} + b^{3} + c^{3} = 3abc, iff a + b + c = 0

Here 30 + 20 – 50 = 0 i.e. a + b + c = 0

⇒ (30)^{3} + (20)^{3} + (- 50)^{3}

= 3 x (30) x (20) x (- 50)

= – 90000

(ii) (- 15)^{3} + (28)^{3} + (- 13)^{3}

We know (RBSESolutions.com) that a^{3} + b^{3} + c^{3} = 3abc, if a + b + c = 0

Here a + b + c = – 15 + 28 + (- 13) = 0

(- 15)^{3} + (28)^{3} + (- 13)^{3}

= 3 x (- 15) x 28 x (- 13)

= 16380

We hope the given RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.5 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.5, drop a comment below and we will get back to you at the earliest.

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