NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3.

- Polynomials Class 9 Ex 2.1
- Polynomials Class 9 Ex 2.2
- Polynomials Class 9 Ex 2.3
- Polynomials Class 9 Ex 2.4
- Polynomials Class 9 Ex 2.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Polynomials |

Exercise |
Ex 2.3 |

Number of Questions Solved |
3 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

**Ex 2.3 Class 9 Maths Question 1.**

Find the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by

(i) x + 1

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

(iii) x

(iv) x+π

(v) 5+2x

**Solution:**

(1) By remainder theorem, the required remainder is equal to p(-1).

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

p(-1) = (-1)^{3} + 3(-1)^{2} + 3(-1) +1 =-1+3-3+1=0

Hence, the required remainder = p(-1) = 0

**Ex 2.3 Class 9 Maths Question 2.**

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

**Solution:**

Let p(x) = x^{3}-ax^{2}+6x-a

By remainder theorem, when p(x) is divided by x – a,

Then,remainder = p(a)

p(a)= a^{3}-a . a^{2}+6a-a

=a^{3} – a^{3} +6a-a = 5a

**Ex 2.3 Class 9 Maths Question 3.**

Check, whether 7 + 3x is a factor of 3x^{3} + 7x.

**Solution:**

Let f(x) = 3x^{3} + 7x and g(x) = 7 + 3x

On putting g(x) = 0, we get

7 + 3x = 0 3x = – 7

=> x = \(\frac { -7 }{ 3 }\)

Thus, zero of g(x) is x =\(\frac { -7 }{ 3 }\).

We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3, drop a comment below and we will get back to you at the earliest.