NCERT Solutions for Class 9 Maths Chapter 2 Polynomials in Hindi and English Medium for download in PDF format updated for new academic session 202223 at new Curriculum. UP Board Students can also use the same solutions for their help.
NCERT Solutions for Class 9 Maths Chapter 2
As per the Uttar Pradesh Board, Prayagraj – the NCERT Books are now implemented for class 9 Maths for 20222023. 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 20222023 and Offline Solutions based on latest Curriculum for 20222023.
9th Maths Chapter 2 Sols in English & Hindi Medium
Class: 9  Maths (English and Hindi Medium) 
Chapter 2:  Polynomials 
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 (xa), 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 nonzero term, is called a monomial.
 A polynomial containing two nonzero terms, is called a binomial.
 A polynomial containing three nonzero 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
Determine whether polynomials x³ + x² + x + 1 has x + 1 a factor?
Let p(x) = x³ + x² + x + 1 Putting x + 1 = 0, we get, x = – 1 Using remainder theorem, when p(x)= x³ + x² + 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 x³ + x² + x + 1.
Use the Factor Theorem to determine whether g(x) is a factor of p(x): p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1.
p(x) = 2x³ + x² – 2x – 1 and g(x) = x + 1 Putting x + 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = 2x³ + x² – 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).
Find the value of k, if x – 1 is a factor of p(x) = x² + x + k.
p(x) = x² + x + k Putting x – 1 = 0, we get, x = 1 Using remainder theorem, When p(x) = x² + 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
Find the zero of the polynomial p(x) = x + 5.
p(x) = x + 5 Putting p(x) = 0, we get x + 5 = 0 ⇒ x = – 5 Hence, x = – 5 is a zero of the polynomial p(x).
What is the core motive of the chapter 2 Polynomials of class 9 Maths?
The core motive of chapter 2 Polynomials of class 9 Maths is to make the meaning of the following things clear to students.
 1. Meaning of polynomials in one variable.
 2. Terms of the polynomials.
 3. Meaning of coefficients.
 4. Meaning of zero polynomial.
 5. Degree of the polynomial.
 6. Types of the polynomial (Linear, Quadratic, Cubic).
 7. Zeroes of a polynomial.
 8. Remainder theorem.
 9. Factorization of polynomial, Factor theorem.
 10. Algebraic identities.
Is the chapter 2 Polynomials of class 9th Maths complicated?
No, chapter 2 Polynomials of class 9th Maths is not that complicated. Basically, in exercises 2.4 and 2.5, most of the students face a little problem. Exercises 2.1, 2.2, and 2.3 are quite easy compared to exercise 2.4 and 2.5.
How many exercises are there in 9th Maths Chapter 2 with least number of difficult questions?
In chapter 2 of class 9th Maths, there are five exercises. The first exercise (Ex 2.1) contains five questions, the second exercise (Ex 2.2) contains four questions, the third exercise contains only three questions, the fourth exercise contains five questions, and the fifth exercise (Ex 2.5) contains sixteen questions. So the third exercise (Ex 2.3) has the least number of questions.
How many theorems are there in the chapter 2 Polynomials of class 9 Maths?
There are two theorems (Remainder theorem and Factor theorem) in chapter 2 Polynomials of class 9 Maths. Both the theorems are important for the exams. The remainder theorem is used in exercise 2.3, and the Factor theorem is used in exercise 2.4.