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Polynomials Class 9 Notes Maths Chapter 2


In mathematics, a variable is denoted by a symbol that can take any real value, often represented by letters such as x, y, z, etc. Expressions like 2x, 3x, -x, and -1/2x are examples of algebraic expressions, specifically in the form (a constant) × x. When the constant is unknown, it is denoted as a, b, c, etc. 

Polynomials Class 9 Notes Maths Chapter 2

A polynomial is an algebraic expression consisting of variables, coefficients, and exponents, combined using addition, subtraction, and multiplication operations.

  • Example: 2�3−5�2+3�−72x35x2+3x7 is a polynomial.

“The expression which contains one or more terms with non – zero coefficient is called polynomial”

Understanding Expressions with Examples

  • Consider a square of side 3 units, where the perimeter is given by the sum of the lengths of its four sides. Polynomials Class 9 Notes Maths Chapter 2
  • If each side is x units, the perimeter is expressed as 4x units, showcasing how the value of the variable influences the result. 
  • The area of the square, denoted as x² square units, is an example of an algebraic expression.

Polynomials in One Variable

Polynomials are algebraic expressions with variables, coefficients, and exponents. When the exponents are whole numbers, the expressions are termed polynomials in one variable.

 Example: x³ - x² + 4x + 7 and 3y² + 5y.

Terms and Coefficients

In a polynomial like x² + 2x, x² and 2x are referred to as terms. Each term has a coefficient—in -x³ + 4x² + 7x - 2, coefficients are -1, 4, 7, and -2. The term x⁰ (where x⁰ = 1) is also present.

Constant Polynomials and Zero Polynomials

Constants like 2, -5, and 7 are examples of constant polynomials. The constant polynomial 0 is termed the zero polynomial, a significant concept in polynomial theory.

Non-Polynomial Expressions

Expressions like x + 2/x, x⁻¹, and x + 3√(x) aren't polynomials due to non-whole number exponents.

Polynomials Class 9 Notes Maths Chapter 2

Notation for Polynomials

Polynomials can be denoted by symbols like p(x), q(x), or r(x), where the variable is x. Examples include:

  1. �(�)=2�2+5�−3p(x)=2x2+5x3
  2. �(�)=�3−1q(x)=x31
  3. �(�)=�3+�+1r(y)=y3+y+1
  4. �(�)=2−�−�2+6�5s(u)=2uu2+6u5

Degree of a Polynomial

The degree of a polynomial is the highest power of its variable. For example, in 3x⁷ - 4x⁶ + x + 9, the degree is 7. Constant polynomials have a degree of 0.


 Polynomials Class 9 Notes Maths Chapter 2�(�)=�5−�4+3

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Linear, Quadratic, and Cubic Polynomials

  • Linear Polynomial: Degree 1, written as ��+�ax+b where �≠0a is not equal to 0. Examples: 2�−12x1, 2�+12y+1, 2−�2u
  • Quadratic Polynomial: Degree 2, expressed as ��2+��+�ax2+bx+c where �≠0a is not equal to 0. Examples: 5−�25y2, 4�+5�24y+5y2, 6−�−�26yy2
  • Cubic Polynomial: Degree 3, in the form ��3+��2+��+�ax3+bx2+cx+d where �≠0a is not equal to zero 0. Examples: 4�34x3, 2�3+12x3+1, 5�3+�25x3+x2

General Form of a Polynomial

A polynomial in one variable of degree n is written as: ����+��−1��−1+…+�1�+�0anxn+an1xn-1++a1x+a0 where �0,�1,…,��a0,a1,an are constants and ��≠0an=0.

Zero Polynomial and Beyond

The zero polynomial, denoted as 0, has an undefined degree. Polynomials can extend to more than one variable, like �2+�2+���x2+y2+xyz in three variables.

Zeroes of a Polynomial

Consider the polynomial �(�)=5�3−2�2+3�−2p(x)=5x32x2+3x2. To find the value of �(�)p(x) at different points, substitute the given values for x

For Example, �(1)

Given Polynomial: p(x) = 5x3 - 2x2 + 3x - 2

Example Calculation:

1. For x = 1:

   p(1) = 5 - 2 + 3 - 2 = 4

2. For x = 0:

   p(0) = 0 - 0 + 0 - 2 = -2

3. For x = -1:

   p(-1) = -5 - 2 - 3 - 2 = -12

In summary, for the given polynomial �(�)=5�3−2�2+3�−2p(x)=5x32x2+3x2:

  • �(1)=4p(1)=4
  • �(0)=−2p(0)=2
  • �(−1)=−12p(1)=12

These values are found by substituting the respective values of x into the polynomial expression.

Example: Value of Polynomials at Given Points

(i) For �(�)=5�2−3�+7p(x)=5x23x+7 at �=1x=1: �(1)=5(1)2−3(1)+7=9p(1)=5(1)23(1)+7=9

(ii) For �(�)=3�3−4�+11q(y)=3y34y+11 at �=2y=2: �(2)=3(2)3−4(2)+11=27−8+11=30q(2)=3(2)34(2)+11=278+11=30

(iii) For �(�)=4�4+5�3−�2+6p(t)=4t4+5t3t2+6 at �=�t=a: �(�)=4�4+5�3−�2+6p(a)=4a4+5a3a2+6

Identifying Zeros of Polynomials

When evaluating �(�)=�−1p(x