RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.3 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 3.3.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomial |

Exercise |
Ex 3.3 |

Number of Questions Solved |
4 |

Category |
RBSE Solutions |

## RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Ex 3.3

Question 1.

Find the remainder when dividing (RBSESolutions.com) the polynomial x^{4} + x^{3} – 3x^{2} + 3x + 1 by each of the following one degree expression.

(i) x – 1

(ii) \(x-\frac { 1 }{ 2 }\)

(iii) x + π

(iv) 3 + 2x

(v) x

Solution.

(i) Let p(x) = x^{4} + x^{3} – 3x^{2} + 3x + 1

zero of x – 1 is 1

∴p(1) = (1)^{4} + (1)^{3} – 3(1) + 1

= 1 + 1 – 3 + 3 + 1 = 3

Question 2.

Find the remainder when 2x^{3} + 20x^{2} – 5x + a is (RBSESolutions.com) divided by x + a.

Solution.

Let p(x) = 2x^{3} + 2ax^{2} – 5x + a

Zero of x + a is x = -a

∴ p(- a) = 2(- a)^{3} + 2a(- a)^{2} – 5(- a) + a

= – 2a^{3} + 2a^{3} + 5a + a = 6a

Question 3.

Verify whether x + 1 is (RBSESolutions.com) a factor of x^{3} + 3x^{2} + 3x + 1 or not.

Solution.

Let p(x) = x^{3} + 3x^{2} + 3x + 1

If x + 1 is a factor of p(x), then p(- 1) = 0

∴ p(- 1) = (- 1)^{3} + 3(- 1)^{2} + 3(- 1) + 1

= – 1 + 3 – 3 + 1 = 0

∵ p(- 1) = 0

∴ x + 1 is a factor of p(x).

Question 4.

Polynomial x^{3} + x^{2} – 4x + a and 2x^{3} + ax^{2} + 3x – 3 when divided by x – 2 gives (RBSESolutions.com) same remainder. Find the value of a.

Solution.

Let p(x) = x^{3} + x^{2} – 4x + a

Here, p(x) is divided by x – 2, so remainder is p(2).

i. e. p(2) = (2)^{3} + (2)^{2} – 4(2) + a

⇒ p( 2) = 8 + 4 – 8 + a = a + 4

Again let f(x) = 2x^{3} + ax^{2} + 3x – 3 is (RBSESolutions.com) divided by x – 2.

So remainder = f(2)

∴ f(2) = 2(2)^{3} + a(2)^{2} + 3(2) – 3

= 16 + 4a + 6 – 3 = 4a + 19

∵ p( 2) = f(2)

⇒ a + 4 = 4a + 19

⇒ 3a = – 15

⇒ a = – 5

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