RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.2 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 3.2.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomial |

Exercise |
Ex 3.2 |

Number of Questions Solved |
4 |

Category |
RBSE Solutions |

## RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.2

Question 1

Find the value of the (RBSESolutions.com) polynomial 2x^{3} – 13x^{2} + 17x + 12 at

(i) x = 2

(ii) x = – 3

(iii) x = 0

(iv) x = – 1

Solution.

Let p(x) = 2x^{3} – 13x^{2} + 17x + 12

(i) p(2) = 2(2)^{3} – 13(2)^{2} + 17(2) + 12

= 16 – 52 + 34 + 12

= 62 – 52 = 10

(ii) p(-3) = 2(-3)^{3} – 13(-3)^{2} + 17(-3) + 12

= – 54 – 117 – 51 + 12

= – 222 + 12

= – 210

(iii) p(0) = 2(0)^{3} – 13(0)^{2} + 17(0) + 12

= 0 – 0 + 0 + 12

= 12

(iv) p(-1) = 2(- 1)^{3} – 13(-1)^{2} + 17(- 1) + 12

= – 2 – 13 – 17 + 12

= – 32 + 12

= – 20

Question 2.

Find p(2), p(1) and p(0) for each (RBSESolutions.com) of the following polynomials.

(i) p(x) = x^{2} – x + 1

(ii) p(y) = (y + 1)(y – 1)

(iii) p(x) = x^{3}

(iv) p(t) = 2 + t + t^{2} – t^{3}

Solution.

(i) The given polynomial is

p(x) = x^{2} – x + 1

∴ p(2) = 2^{2} – 2 + 1 = 3

[replacing x by 2]

p(1) = 1^{2} -1 + 1 = 1

[replacing x by 1]

p(0) = (0)^{2} – (0) + 1 = 1

[replacing x by 0]

(ii) The given (RBSESolutions.com) polynomial is

p(y) = (y + 1)(y – 1)

p(2) = (2 + 1)(2 – 1) = 3

p(1) = (1 + 1)(1 – 1) = 0

p(0) = (0 + 1)(0 – 1) = – 1 .

(iii) The given polynomial is p(x) = x^{3}

p( 2) = 2^{3} = 8

p(1) = 1^{3} = 1

p(0) = 0^{3} = 0

(iv) The given (RBSESolutions.com) polynomial is

p(t) = 2 + t + t^{2} – t^{3}

p(2) = 2 + 2 + (2)^{2} – (2)^{3}

= 8 – 8 = 0

p(1) = 2 + 1 + (1)^{2} – (1)^{3}

= 4 – 1 = 3

p(0) = 2 + 0 + (0)^{2} – (0)^{3}

= 2 – 0

= 2

Question 3.

Verify whether the following are (RBSESolutions.com) zeroes of the polynomial, indicated against them.

(i) p(x) = x^{2} – 1; x = 1,-1

(ii) p(x) = 2x + 1; x = \(x=\frac { -1 }{ 2 }\)

(iii) p(x) = 4x + 5; \(x=\frac { -5 }{ 4 }\)

(iv) p(x) = 3x^{2}; x = 0

(v) p(x) = (x – 3)(x + 5); x = 3, – 5

(vi) p(x) = ax + b; \(x=\frac { -b }{ a }\)

(vii) p(x) = 3x^{2} – 1; \(x=\frac { -1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } \)

(viii) p(x) = 3x + 2; \(x=\frac { -2 }{ 3 }\)

Solution.

p(x) = x^{2} – 1; x = 1, – 1

∴ p(1) = (1)^{2} – 1

= 1 – 1 = 0

p(- 1) = (- 1)^{2} – 1

= 1 – 1 = 0

Yes, both 1,-1 are zeroes (RBSESolutions.com) of the polynomials p(x) = x^{2} – 1.

Question 4.

Find the zeroes of the (RBSESolutions.com) polynomial in each case.

(i) p(x) = x – 4

(ii) p(x) = 4x

(iii) p(x) = bx, b ≠ 0

(iv) p(x) = x + 3

(v) p(x) = 2x – 1

(vi) p(x) = 3x + 7

(vii) p(x) = cx + d, c ≠ 0, c and d are real numbers.

Solution.

For zeroes of the (RBSESolutions.com) polynomial, p(x) = 0

(i) x – 4 = 0 ⇒ x = 4

Hence, 4 is the zero of the polynomial p(x).

(ii) 4x = 0 ⇒ x = 0

Hence, 0 is the zero of the polynomial p(x).

(iii) bx = 0 ⇒ x = 0

Hence, 0 is the zero of the polynomial p(x).

(iv) x + 3 = 0 ⇒ x = – 3

Hence, – 3 is the zero (RBSESolutions.com) of the polynomial p(x).

(v) 2x – 1 = 0 ⇒ 2x = 1 ⇒ x = \(\frac { 1 }{ 2 }\)

Hence, \(\frac { 1 }{ 2 }\) is the zero of the polynomial p(x).

(vi) 3x + 7 = 0 ⇒ 3x = -7 ⇒ x = \(\frac { -7 }{ 3 }\)

Hence, \(\frac { -7 }{ 3 }\) is the zero of the polynomial p(x).

(vii) p(x) = cx + d, c ≠ 0, c and d are real numbers.

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