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NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

Edited By Ramraj Saini | Updated on Mar 18, 2024 12:43 PM IST

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.

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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) = anxn + an-1xn-1 + an-2xn-2 + ............... + a2x2 + a1x + a0

where a0, a1, a2, are constants, an ≠ 0



  • 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.

The root 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

  1. 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 Class 9 Exemplar Solutions for Other Subjects:


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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg


An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)


Option 2)

\; K\;

Option 3)


Option 4)


In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)


Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)


Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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