NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

Here students can find Free NCERT Class 9 Maths Chapter 2 Notes. These notes are created by an expert team at careers360 consdering the latest need of students. These notes includes sort tricks, formulae, concepts and step by step explanation of every topic. The NCERT Class 9 Maths Chapter 2 Notes contains systematic explanations of topics using examples and exercises. Polynomials class 9 notes gives examples and link to NCERT exercises. Polynomials class 9 notes pdf download includes FAQ’s or frequently asked questions about class 9 maths chapter 2 notes.

Also, students can refer,

• NCERT Class 9 Maths Notes
  
• NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
• NCERT Exemplar Class 9 Maths Chapter 2 Solutions Polynomials

Polynomials class 9 notes:

Polynomial is an algebraic expression that consists of constants, variables, and exponents. Polynomial expressions are such that the variables have only positive integral powers.

P(x) = a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + ............... + a₂x² + a₁x + a₀

where a₀, a₁, a₂,.......a_n are constants, a_n ≠ 0

Example:

• polynomials are represented by f(x), g(x) etc.

• In the above polynomial 5x², 2y, and 7 are the terms of the polynomial.

• 5 is the coefficient of the x².

• X and y are the variables.

• 7 is the constant term that has no variable

Polynomials In One Variable

When there is only one variable in the expression then it is called the polynomial in one variable.

Example: X+2=4

Types of Polynomials On The Basis of The Number of Terms

Types of Polynomials On The Basis of The Number of Degrees

The highest value of the power of a variable in the polynomial is the degree of the polynomial.

Zeroes of A Polynomial

If f(x) is a polynomial then the number ‘a’ will be the zero of the polynomial with f(a) = 0. Thus, we can find the zero of the polynomial by equating the equation to zero.

The root of the polynomial is essentially the x-intercept of the polynomial.

If the polynomial has one root, then it will intersect the x-axis at one point only and if it has two roots then it will intersect at two points.

Remainder Theorem

The property of division which follows in the basic division is :

Dividend = (Divisor × Quotient) + Remainder

The same thing follows the division of polynomials.

f(x) and g(x) are two polynomials in which the degree of f(x) ≥ degree of g(x) and g(x) ≠ 0 are given then we can get the h(x) and n(x) so that:

F(x) = g(x) h(x) + n(x),

where n(x) = 0 or degree of n(x) < degree of g(x).

It says that f(x) divided by g(x), gives h(x) as quotient and n(x) as a remainder.

Division of A Polynomial With A Monomial

(3x³+x²+x)/x = 3x³ / x + x² / x + x / x=3x²+x+1

‘x’ is common in the above polynomial, therefore

3x³ + x² + x = x (3x²+x+1)

3x²+x+1 and x the factors of 3x³ + x² + x

Steps of Division of A Polynomial With A Non – Zero Polynomial

Divide x² - 3x -10 by 2 + x

Step 1: Write the dividend and divisor in descending order in the standard form. x² - 3x -10 and x + 2

Divide the first term of dividend with the first term divisor.

x²/x = x this will be the first term of quotient.

Step 2: Now, we will multiply the divisor with this term of the quotient and subtract it from the dividend.

Step 3: Now repeat the process again by dividing the dividend with the divisor.

Step 4: – (5x/x) = – 5

Step 5:

The remainder = 0

Hence x²- 3x – 10 = (x + 2)(x - 5) + 0

Dividend = (Divisor × Quotient) + Remainder

Factor Theorem

Factor theorem says that if f(y) is a polynomial with degree n≥1 and ‘s’ is a real number, then

1. (y - s) is a factor of f(y), if f(t) = 0

2. F (t) = 0 if (y – s) is a factor of f(y).

Factorization of Polynomials

Three methods of the factorization are :

1. By taking out the common factor

To factorize x² –x, we can do it by taking x common.

x(x – 1) such that x and x-1 are factors of x² – x.

2. By grouping

ab + bc + ax + cx = (ab + bc) + (ax + cx)

= b(a + c) + x(a + c)

= (a + c)(b + x)

3. By splitting the middle term

x² + (p + q)x + pq

(x + p)(x + q)

The middle term is split in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

Algebraic Identities

(a+b)² = a²+2ab+b²

a²+b² = (a+b)²-2ab

(a-b)² = a²-2ab+b²

a²+b² = (a-b)²+2ab

(a+b+c)² = a² + b² + c² + 2(ab + bc + ca)

(a+b)³ = a³ + b³ + 3ab(a+b)

a³ + b³ = (a+b)³ - 3ab(a+b)

(a-b)³ = a³ - b³ - 3ab(a-b)

a³ - b³ = (a-b)³ + 3ab(a-b)

a² - b² = (a+b)(a-b)

a³ - b³ = (a-b)(a²+ab+b²)

a³ + b³ = (a+b)(a²-ab+b²)

Significance of NCERT class 9 maths chapter 2 notes

• Learn about polynomials, which are math expressions with variables and numbers.
• Understand the degree of a polynomial, which helps in solving math problems.
• Explore how to add and subtract polynomials, building a strong foundation in algebra.
• Discover algebraic identities, useful in solving equations and simplifying expressions.
• Explore visual representations of polynomials, making it easier to understand math concepts.
• Master factorization, a key skill for breaking down and solving polynomial expressions.
• See how polynomials are applied in real-life situations, connecting math to the world around you.
• Improve problem-solving skills through practical examples and scenarios.
• Build a solid math foundation for higher classes by mastering the content in Chapter 2.

NCERT Class 9 Notes Chapter wise

NCERT solutions of class 9 subject wise

• NCERT solutions for class 9 Maths
• NCERT solutions for class 9 Science

NCERT Class 9 Exemplar Solutions for Other Subjects:

• NCERT Exemplar Class 9 Maths solutions
• NCERT Exemplar Class 9 Science solutions