CBSE Notes Class 9 Maths Chapter 2 Polynomial

Polynomials are algebraic expressions made using variables, constants, and powers, and they play a major role in solving mathematical equations. They help us describe patterns, relationships, and real-life quantities in a simple symbolic form, making algebra easier to understand and apply.

In CBSE Maths Chapter 2 Polynomials Class 9 Notes, the chapter introduces types of polynomials, degree of a polynomial, zeroes of a polynomial, and the relationship between zeroes and coefficients of linear and quadratic polynomials. These CBSE Class 9 Maths notes explain each concept in a clear and step-wise manner, along with easy examples that strengthen conceptual understanding and support accurate problem solving in exams.

1.0Download CBSE Class 9 Maths Notes Chapter 2 Polynomial - Free PDF

Master the concepts of Polynomials with ease! Our free CBSE Class 9 Maths Chapter 2 Notes PDF are carefully prepared to help students grasp the fundamentals quickly and effectively.

2.0CBSE Class 9 Maths Notes Chapter 2 Polynomial - Revision Notes

Important Concepts in Polynomials

Polynomial Definition: A polynomial is an algebraic expression consisting of variables, coefficients, and non-negative integer exponents.

Types of Polynomials:

• Monomial: A polynomial with one term (e.g., 5x).
• Binomial: A polynomial with two terms (e.g., x²+ 3).
• Trinomial: A polynomial with three terms (e.g., x³ + x²+ x ).

Degree and Zeros of a Polynomial

• Degree of a Polynomial: The degree of a polynomial is often defined as the highest power of any variable in the expression. For example, in 4x³ + 2x + 1, the degree is 3.
• Zeros of a Polynomial: The zeros of any polynomial are the variable values that make the polynomial equal to zero. For example, if p (x) = x−3, then x = 3 is a zero of the polynomial.

Definitions

• Constant Polynomial: A polynomial with degree 0 (e.g., 5).
• Linear Polynomial: A polynomial of degree 1 (e.g., 3x + 4).
• Quadratic Polynomial: A polynomial of degree 2 (e.g., x² − 5x + 6).
• Cubic Polynomial: A polynomial of degree 3 (e.g., x³+ 2x²− x).
• Remainder Theorem: States that if a polynomial p (x) is divided by x−a the remainder is p(a).
• Factor Theorem: If p(a) = 0, then x−a is a factor of p (x).

Formulas

• Remainder Theorem:$\text{ Remainder }=p(a)\text{ when }p(x)\text{ is divided by }x-a$
• Factor Theorem: $p(a)=0⟹(x-a)\text{ is a factor of }p(x)$
• Polynomials in Division:

Dividend = (Divisor × Quotient) + Remainder

Or, 

p (x) = (x−a) q (x)+ r

Here, q (x) is the quotient, and r is the remainder.

For the division of polynomials, follow the following steps 

Step 1: Write the terms of both polynomials in descending order of degrees (if they are not already in that order).

Step 2: Divide the first term of the dividend (the polynomial you are dividing) by the first term of the divisor (the polynomial you are dividing by).

Step 3: Multiply the whole divisor by the term you just found (from step 2).

Step 4: Subtract the result obtained in Step 3 from the dividend.

Step 5: Now repeat the division with a new polynomial.

Step 6: Subtract again

Step 7: Check whether the degree of the remainder is less than the degree of the divisor.

Tips and Tricks

• Identifying the Degree Quickly: Focus on the term with the highest power to determine the degree of the polynomial.
• Applying Remainder Theorem: Substitute x = a directly into the polynomial to quickly find the remainder.
• Factorization Made Easy: Use the Factor Theorem to check for simple factors like x−1 or x+2.
• Understanding Graphs: Visualise the zeros of polynomials as points where the graph intersects the x-axis.
• A table summarising the types of polynomials and their degrees for quick revision:

3.0Solved Problems

Example 1: Divide the polynomial P(x) = 2x³−3x²+4x−5 by (x−2). Find the remainder using the Remainder Theorem.

Solution: give that x-2 is divisor 

x - 2 = 0;  x = 2 

P(x) = 2x³ − 3x² + 4x − 5 put x = 2 

P(2) = 2 x 2³−3 x 2²+ 4 x 2 − 5 

= 16 - 12 + 8 - 5 = 7 

Example 2: Determine if (x−2) is a factor of the polynomial P(x) = x³ − 4x² + 3x + 2.

Solution: x-2 = 0; x = 2 

Put x = 2 in P(x) = x³ − 4x² + 3x + 2 to check whether (x-2) is factor or not. 

P(2) = 2³ − 42² + 3 x 2 + 2

P(2) = 8 – 16 + 6 + 2 

= 16 – 16 = 0 

Since p(2) = 0, hence x-2 is a factor of the given equation.  

Example 3: Find the remainder when $\mathrm{p}(x)=x^3-4x+6$ is divided by 𝑥−2.

Solution:

Using the Remainder Theorem, substitute x=2 into p (x):

$P(2)=2^3-4(2)+6=8-8+6=6.$

The remainder is 6.

Example 4: Divide 3x⁴ + x³ - 17x² + 19x - 6 by 3x² + 7x - 6. 

Solution: 

Here, in this example, the remainder is 0. 

Example 5: Find the value of ‘a’ if x – a is a factor of x³ – ax² + 2x + a – 1. 

Solution: 

Let p(x) = x³ – ax² + 2x + a – 1 

Since (x – a) is a factor of p(x), p(a) = 0.

x - a = 0; x = a 

p(a) = a³ – a(a)² + 2a + a – 1 = 0 

a³ – a³ + 2a + a – 1 = 0 

3a = 1 

Hence, a = 1/3

Example 6: If x + y = 12 and xy = 27, find the value of x³ + y³.

Solution: 

Using identity 

= (x+y)² = x² + 2xy + y²

= 12² = x² + y² + 227 

= x² + y² = 144 - 54 = 90

Now, x³ + y³ = (x + y) (x² + y²– xy) 

= 12 [90 - 27] 

= 12 × 63 = 756

Example 7: Factorise: 9x² + 4y² + 16z² + 12xy – 16yz – 24xz

Solution: 

Using the identity 

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

We can also write this equation 9x² + 4y² + 16z² + 12xy – 16yz – 24xz as

$=(3x{)}^2+(2y{)}^2+(-4z{)}^2+2\times 3x\times 2y-2\times 2y\times -4z+2\times 3x\times -4z$

= (3x+2y-4z)² = (3x+2y-4z)(3x+2y-4z)

4.0Key Features of CBSE Class 9 Maths Chapter 2 Polynomial Notes

• Concept-Focused Explanation – The notes clearly explain the definition of a polynomial, terms, coefficients, variables, and degree with simple language so students can build a strong algebraic foundation.
• Types of Polynomials – Covers linear, quadratic, cubic, and higher-degree polynomials along with classification based on number of terms (monomial, binomial, trinomial) for better conceptual clarity.
• Standard Form & Identification – Helps students identify standard form, leading coefficient, constant term, and degree — important for solving NCERT and exam-oriented questions.
• Solved Examples (NCERT Based) – Important textbook problems solved in a step-wise method to match CBSE exam pattern.
• CBSE Pattern Coverage – Prepared strictly according to the latest CBSE syllabus to help students score better in school exams and competitive exams.