CBSE Notes Class 9 Maths Chapter 2 Polynomial

Polynomials are one of the key sections in the new CBSE Class 9 Maths syllabus. It helps develop an understanding of algebraic expressions and their operations. The chapter introduces the student to the definition of various types of polynomials as well as the operations related to polynomials. By properly defining the terms, recording formulae, offering solved examples, and providing revision advice, these notes truly seek to help students understand the ideas and provide them with a clear roadmap for solving maths problems. By using these notes, students may improve their problem-solving abilities and achieve high test scores.

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

Master the concepts of Polynomials with ease! Our free CBSE Class 9 Maths Chapter 2 notes 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 Notes Chapter 2 Polynomial

• There are practice problems, from simple identification of the degrees to challenging factorisation and applications of the theorems in the body of the notes.
• The content is written in simple language so that the student can easily access it and retain it in his or her memory.
• The notes include a polynomial graph with pictures to help students comprehend how the degree influences the curve and, consequently, the function of zeros.
• Important formulas and theorems at the end of each section allow them to revise for the exams.