# NCERT Solutions for Class 9 Maths Chapter 2

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials in Hindi and English Medium, PDF format as well as online digital contents updated for new academic session 2020-21 following the new CBSE Curriculum. UP Board Students can also use the same solutions for their help. As per the Uttar Pradesh Board, Prayagraj – the NCERT Books are now implemented for class 9 Maths for 2020-2021. So, the students of class 9 (High School) can download UP Board Solutions for Class 9 Maths Chapter 2 from the links given below. 9th Maths Chapter 2 Solutions are available in Hindi and English Medium. Videos related to each exercise are also given below, which help the students to know the steps of solutions.

Download NCERT Solutions Apps and UP Board Solutions app 2020-2021 and Offline Solutions based on latest Curriculum for 2020-2021.Page Contents

## NCERT Solutions for Class 9 Maths Chapter 2

Class: 9 | Maths (English and Hindi Medium) |

Chapter 2: | Polynomials |

### 9th Maths Chapter 2 Sols in English & Hindi Medium

### Class 9th Maths Chapter 2 Polynomials Solutions

- NCERT SolutionsClass 9 Maths Chapter 2 Exercise 2.1Read more
- NCERT SolutionsClass 9 Maths Chapter 2 Exercise 2.2Read more
- NCERT SolutionsClass 9 Maths Chapter 2 Exercise 2.3Read more
- NCERT SolutionsClass 9 Maths Chapter 2 Exercise 2.4Read more
- NCERT SolutionsClass 9 Maths Chapter 2 Exercise 2.5Read more

#### Class 9 Maths Chapter 2 Exercise 2.1 Solutions in Videos

#### Class 9 Maths Chapter 2 Exercise 2.2 Solutions in Videos

#### Class 9 Maths Chapter 2 Exercise 2.3 Solutions in Videos

#### Class 9 Maths Chapter 2 Exercise 2.4 Solutions in Videos

#### Class 9 Maths Chapter 2 Exercise 2.5 Solutions in Videos

#### Important Terms on 9th Maths Chapter 2

- A combination of constants and variables, connected by four fundamental arithmetical operations +, -, x and / is called an algebraic expression. e.g. 6x² – 5y² + 2xy
- An algebraic expression which have only whole numbers as the exponent of one variable, is called polynomial in one variable. e.g. 3x³ + 2x² – 7x + 5 etc.
- The part of a polynomial separated from each other by + or – sign is called a term and each term of a polynomial has a coefficient.
- Highest power of the variable in a polynomial, is known as degree of that polynomial.
- The value obtained on putting a particular value of the variable in polynomial is called value of the polynomial at the value of variable.
- Zero of a polynomial p(x) is a number alpha, such that p(alpha) = 0. It is also called root pf polynomial equation p(x) = 0.
- Let f(x) be any polynomial of degree n,(n ≥ 1) and a be any real number. If f(x) is divided by the linear polynomial (x-a), then the remainder is f(a).
- Let f(x) be a polynomial of degree n,(n ≥ 1) and a be any real number. Then,

i). If f(a) = 0, then (x – a) is a factor of f(x).

ii). If (x – a) is a factor of f(x), then f(a) = 0.

##### Polynomial on the Basis of Number of Terms

Polynomial on the Basis of Number of Terms

- A polynomial containing one non-zero term, is called a monomial.
- A polynomial containing two non-zero terms, is called a binomial.
- A polynomial containing three non-zero terms, is called a trinomial.

###### Polynomial on the Basis of Degree of Variables

- A polynomial of degree 0, is called a constant polynomial.
- A polynomial of degree 1, is called a linear polynomial.
- A polynomial of degree 2, is called a quadratic polynomial.
- A polynomial of degree 3, is called a cubic polynomial.
- A polynomial of degree 4, is called a biquadratic polynomial.

### Important Questions on 9th Maths Chapter 2

=x^2 + (4 + 10)x + 4 × 10

[∵〖(x + a)(x + b) = x〗^2 + (a + b) x + ab]

= x^2 + 14x + 40

Putting x + 1 = 0,

we get, x = – 1

Using remainder theorem,

when p(x)=x^3+x^2+x+1 is divided by x + 1,

remainder is given by p(-1)

=〖(-1)^3 + (-1)〗^2 + (-1) + 1

= – 1 + 1 – 1 + 1

= 0

Since, remainder p(-1) = 0,

Hence x + 1 is a factor of x^3 + x^2 + x + 1.

g(x) = x + 1

Putting x + 1 = 0,

we get, x = -1

Using remainder theorem,

when p(x) = 2x^3 + x^2 – 2x – 1 is divided by g(x) = x + 1, remainder is given by p(-1)

=〖(-1)^3 + (-1)〗^2 + (-1) + 1

= – 1 + 1 – 1 + 1

= 0

Since, remainder p(-1) = 0,

hence g(x) is a factor of p(x).

= (100 + 3)(100 + 7)

= (100)^2 + (3 + 7)100 + 3 × 7

[∵〖(x + a)(x + b) = x〗^2 + (a + b)x + ab]

= 10000 + 1000 + 21

= 11021

=(3x)^2 + 2 × 3x × y + y^2

=(3x + y)^2

[∵a^2 + 2ab + b^2 = (a + b)^2]

= (100)^3 + (-1)^3 + 3(100)^2 (-1) + 3(100) (-1)^2

[〖∵(a + b)〗^3 = a^3 + b^3 + 3a^2 b + 3ab^2 ]

= 1000000 – 1 -30000 + 300

= 970299

Putting x – 1 = 0, we get, x = 1

Using remainder theorem,

When p(x) = x^2 + x + k is divided by x – 1, remainder is given by p(1)

= (1)^2 + (1) + k

= 2 + k

Since x – 1 is a factor of p(x), hence remainder p(1) = 0

⇒ 2 + k = 0

⇒ k = -2

x – a = 0 रखने पर, x = a

शेषफल प्रमेय के अनुसार

p(x) =〖x^3 – ax〗^2 + 6x – a को x – a से भाग देने पर शेषफल

p(a) =〖(a)^3 – a(a)〗^2 + 6(a) – a

= a^3 – a^3 + 6a – a

= 5a

Putting p(x) = 0, we get

x + 5 = 0

⇒ x = – 5

Hence, x = – 5 is a zero of the polynomial p(x).