NCERT Class 9 Maths Chapter 2 Polynomials Notes

Chapter 2 · CBSE · Class IX

🧮

Polynomials

NCERT Class 9 Mathematics Chapter 2 Polynomials Class 9 Algebra Polynomial Expressions Types of Polynomials Linear Polynomial Quadratic Polynomial Cubic Polynomial Zeroes of Polynomial Factorisation Remainder Theorem Algebraic Identities NCERT Solutions CBSE Class 9 Maths Mathematics Notes Math MCQs Math True False Polynomial Graphs Polynomial Questions Class 9 Maths Chapter 2

📘 Definition

A polynomial is an algebraic expression formed using variables, constants, coefficients, and non-negative integral powers of variables connected by mathematical operations such as addition, subtraction, and multiplication.

General form of a polynomial in one variable:

\[ a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_1x+a_0 \]

where:

\(a_n,a_{n-1},a_{n-2},\dots,a_0\) are constants called coefficients
\(x\) is the variable
\(n\) is a non-negative integer
\(a_n \neq 0\)

Important: Powers of variables in a polynomial can only be whole numbers such as \(0,1,2,3,\dots\). Negative powers, fractional powers, and variables in denominators are not allowed in polynomials.

🗒️ Parts Of A Polynomial

First Term Second Term Constant Term

Consider the polynomial:

\[ ax^2+bx+c \]

\(ax^2\)

Quadratic term (first term)

\(a\)

Coefficient of \(x^2\) (leading coefficient)

\(bx\)

Linear term (second term)

\(b\)

Coefficient of \(x\)

\(c\)

Constant term

\(x\)

Independent variable

🗒️ Terms In A Polynomial

Every separated part of a polynomial joined using \(+\) or \(-\) sign is called a term.

Example:

\[ 5x^3-2x^2+7x-9 \]

Terms are:

\(5x^3\)
\(-2x^2\)
\(7x\)
\(-9\)

🗒️ Degree Of A Polynomial

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

Polynomial	Highest Power	Degree
\(5x+7\)	\(1\)	1
\(3x^2+2x+1\)	\(2\)	2
\(7x^4+5x^2+1\)	\(4\)	4
\(8\)	\(0\)	0

Remember: A non-zero constant polynomial always has degree \(0\).

🗂️ Types / Category

Monomial Binomial Trinomial

Classification of Polynomials

Based on Number of Terms

Monomial

A polynomial with exactly one term. Consists of a single term with variables and coefficients. No addition/subtraction between terms.
Examples:
\(5x^2\), \(7\), \(-3y^4\), \(2abc\).

Binomial

A polynomial with exactly two terms. Contains two distinct terms separated by addition or subtraction.
Examples:
\(x + 3\), \(4x^2 - 7y\), \(2a + 5b^2\), \(p - 3q^4\).

Trinomial

A polynomial with exactly three terms. Contains three distinct terms connected by addition or subtraction.
Examples:
\(x^2 + 2x + 1\), \(3a^2 - 5ab + 2b^2\), \(y + 4 - 2y^2\), \(5x - 3 + 7x^3\).

🗂️ Types / Category

Constant Polynomial Linear Polynomial Quadratic Polynomial Cubic Polynomial

Classification of Polynomials

Based on Degree

Degree 0 - Constant Polynomial

Polynomial where highest power of variable is 0 (constant polynomial). No variable or variable power is zero.
Examples:
\(7\), \(-12\), \(5.2\), \(0\).

Degree 1 - Linear Polynomial

Polynomial where highest power of variable is 1 (linear polynomial). Forms: \(ax + b\).
Examples:
\(3x + 2\), \(x - 5\), \(-2x + 7\), \(4x\).

Degree 2 - Quadratic Polynomial

Polynomial where highest power of variable is 2 (quadratic polynomial). Forms: \(ax^2 + bx + c\).
Examples:
\(x^2 + 3x - 2\), \(4x^2 - 5\), \(2x^2 + x + 1\), \(y^2\).

Degree 3 - Cubic Polynomial

Polynomial where highest power of variable is 3 (cubic polynomial). Forms: \(ax^3 + bx^2 + cx + d\).
Examples:
\(x^3 - 2x^2 + x - 1\), \(3x^3 + 4\), \(x^3 + 2x^2 - 5x\), \(y^3\).

📌 Note

Expressions Which Are Not Polynomials

The following expressions are not polynomials:

Expression	Reason
\(\dfrac{1}{x}\)	Variable in denominator
\(\sqrt{x}\)	Fractional power
\(x^{-2}+3\)	Negative exponent
\(2^x\)	Variable as exponent

🎨 SVG Diagram

Polynomilas

As the degree of a polynomial increases, its graph becomes more curved and complex. Higher degree terms dominate the behaviour of the polynomial for large values of \(x\).

🗒️ Definiton

Standard Form of Polynomial

A polynomial is said to be in standard form when its terms are arranged in descending order of powers.

Example:

\[ 7x^4+3x^3-5x+2 \]

Here powers are arranged as:

\[ 4,3,1,0 \]

🤔 Did You Know?

Evaluation of Polynomial

Finding the value of a polynomial for a given value of variable is called evaluation.

Example

Evaluate:

\[ p(x)=2x^2+3x-1 \]

at \(x=2\)

Solution

Substitute \(x=2\):

\[ p(2)=2(2)^2+3(2)-1 \]

\[ =2(4)+6-1 \]

\[ =8+6-1 \]

\[ =13 \]

Zero Polynomial

A polynomial whose all coefficients are zero is called a zero polynomial.

\[ 0x^3+0x^2+0x+0 \]

\[ =0 \]

Degree of zero polynomial is not defined.

Real Life Applications of Polynomials

Used in physics for motion equations
Used in economics for profit and cost analysis
Used in engineering designs
Used in computer graphics and animations
Used in architecture and bridge construction
Used in statistics and data modelling

🌟 Importance

Importnat Concepts

❌ Common Mistakes

Wrong Concept	Correct Concept
Taking \(\sqrt{x}\) as polynomial	Fractional powers are not allowed
Considering \(\dfrac{1}{x}\) as polynomial	Negative exponents are not allowed
Degree of \(5\) taken as 5	Degree of non-zero constant is 0
Ignoring missing powers	Polynomial may skip powers

✏️ Example

Identify the degree, coefficients, and constant term of: \[ 5x^3-7x^2+4x-9 \]

Degree = highest power of variable
Numbers multiplying variables are coefficients
Term without variable is constant term

Find highest power
Identify coefficients
Locate constant term

Highest power of \(x\) is \(3\)

Therefore degree \(=3\)

Coefficients are:

\[ 5,-7,4 \]

Constant term:

\[ -9 \]

📋 Case Study

A school designs a rectangular garden where the length is represented by:

\[ (x+5)\text{ m} \]

and breadth is:

\[ (x+2)\text{ m} \]

The area of the garden is represented by a polynomial.

Questions

Find the polynomial representing area
Find degree of polynomial
Identify coefficient of \(x\)

Solution

Area:

\[ (x+5)(x+2) \]

\[ =x^2+2x+5x+10 \]

\[ =x^2+7x+10 \]

Degree:

\[ 2 \]

Coefficient of \(x\):

\[ 7 \]

⚡ Quick Revision

Polynomial powers must be whole numbers
Highest exponent gives degree
Constant polynomial has degree \(0\)
Zero polynomial has undefined degree
Terms are separated by \(+\) or \(-\)
Expressions with denominator variables are not polynomials

🧮

Coefficient

📘 Definition

The numerical factor attached to a variable term in an algebraic expression or polynomial is called its coefficient.

In simple words, the coefficient tells us:

How many times the variable is multiplied
The numerical value associated with the variable term

Consider the polynomial:

\[ ax^2+bx+c \]

Here:

\(a\) is the coefficient of \(x^2\)
\(b\) is the coefficient of \(x\)
\(c\) is the constant term

Important: Coefficients may be positive, negative, integers, fractions, or decimals.

✏️ Example

Understanding Coefficients with Examples

Polynomial Term	Coefficient	Variable Part
\(7x^2\)	\(7\)	\(x^2\)
\(-5x\)	\(-5\)	\(x\)
\(\dfrac{3}{2}y^3\)	\(\dfrac{3}{2}\)	\(y^3\)
\(-0.8a\)	\(-0.8\)	\(a\)

📌 Note

Invisible Coefficients

Sometimes coefficients are not written explicitly. In such cases, we assume their value carefully.

Expression	Actual Coefficient
\(x\)	\(1\)
\(-x\)	\(-1\)
\(x^2\)	\(1\)
\(-y^3\)	\(-1\)

Students often forget that the coefficient of \(x\) is \(1\) and not \(0\).

🗂️ Types / Category

Positive Coefficient Negative Coefficient

Types of Coefficients

➕

Positive Coefficient

Coefficient value is **greater than zero** (\(> 0\)). Graph rises to right, falls to left. Term contributes upward direction.
Examples:
\(+3x^2\), \(+5x\), \(+7\), \(+2.5y\).

➖

Negative Coefficient

Coefficient value is **less than zero** (\(< 0\)). Graph falls to right, rises to left. Term contributes downward direction.
Examples:
\(-4x^2\), \(-2x\), \(-9\), \(-1.8y\).

📊 Comparison Table

Difference Between Coefficient and Constant Term

Coefficient	Constant Term
Attached with variable	No variable attached
Example: \(5\) in \(5x\)	Example: \(7\) in \(x^2+3x+7\)
Can change with variable term	Independent term

✏️ Example

How to Find Coefficients

Find coefficients in:\[4x^3-7x^2+2x-9\]

Coefficient of \(x^3 = 4\)
Coefficient of \(x^2 = -7\)
Coefficient of \(x = 2\)
Constant term \(= -9\)

Identify the coefficient of \(a^2\) in:\[-11a^2+5a-3\]

The number multiplying \(a^2\) is:

\[ -11 \]

📌 Note

Coefficient Form in General Polynomial

In a general polynomial:

\[ a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \]

Each \(a\)-term represents a coefficient.

\(a_n\) → coefficient of \(x^n\)
\(a_{n-1}\) → coefficient of \(x^{n-1}\)
\(a_1\) → coefficient of \(x\)
\(a_0\) → constant term

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
Coefficient of \(-x\) is \(1\)	Coefficient is \(-1\)
Constant term is coefficient	Constant has no variable
Ignoring negative signs	Signs are part of coefficient
Exponent taken as coefficient	Exponent and coefficient are different

📋 Case Study

A student writes the polynomial:

\[ p(x)=6x^3-4x^2+x-8 \]

The teacher asks the following questions.

What is the coefficient of \(x^3\)?
What is the coefficient of \(x\)?
What is the constant term?

Solution

Coefficient of \(x^3 = 6\)
Coefficient of \(x = 1\)
Constant term \(= -8\)

⚡ Quick Revision

Coefficient is the numerical factor of variable term
Sign always belongs to coefficient
Coefficient of \(x\) is \(1\)
Coefficient of \(-x\) is \(-1\)
Constant term has no variable

🧮

Exponents

📘 Definition

The power raised to a variable in an algebraic expression or polynomial is called an exponent.

Exponents tell us how many times a variable is multiplied by itself.

Consider the polynomial:

\[ ax^2+bx+c \]

Here:

\(2\) is the exponent of \(x\) in \(ax^2\)
\(1\) is the exponent of \(x\) in \(bx\)
\(0\) is the exponent in constant term \(c\) because: \[ c=cx^0 \]

Important: In polynomials, exponents of variables must always be non-negative integers such as \(0,1,2,3,\dots\)

📌 Note

Meaning of Exponents

Exponents represent repeated multiplication.

Expression	Expanded Form	Exponent
\(x^2\)	\(x \times x\)	2
\(x^3\)	\(x \times x \times x\)	3
\(a^4\)	\(a \times a \times a \times a\)	4
\(y^1\)	\(y\)	1

📌 Note

Parts of an Exponential Term

Consider the term:

\[ 5x^3 \]

Part	Name
\(5\)	Coefficient
\(x\)	Base / Variable
\(3\)	Exponent

🧠 Remember

Allowed Exponents in Polynomials

Only non-negative whole number exponents are allowed in polynomials.

Expression	Polynomial or Not	Reason
\(x^3+2x+1\)	Polynomial	Exponent is whole number
\(\sqrt{x}\)	Not Polynomial	Exponent is \(\dfrac{1}{2}\)
\(x^{-2}+5\)	Not Polynomial	Negative exponent
\(3x^0+7\)	Polynomial	Exponent is \(0\)

🤔 Did You Know?

Zero Exponent Rule

Any non-zero number raised to the power \(0\) is always equal to \(1\).

\[ a^0=1 \quad \text{where } a \neq 0 \]

Example

\[ 5^0=1 \]

\[ x^0=1 \]

\[ 9x^0=9 \]

⚖️ Laws

Basic Laws of Exponents

Law	Formula
Product Law	\[ a^m \times a^n=a^{m+n} \]
Division Law	\[ \frac{a^m}{a^n}=a^{m-n} \]
Power of Power	\[ (a^m)^n=a^{mn} \]
Zero Power Law	\[ a^0=1 \]

🔗 Relations

Relation Between Degree and Exponents

The degree of a polynomial is determined by the highest exponent of the variable.

Example

Polynomial	Highest Exponent	Degree
\(2x+5\)	\(1\)	1
\(x^2+4x+1\)	\(2\)	2
\(5x^5-3x+7\)	\(5\)	5

✏️ Example

Identify the exponent of \(x\) in:\[7x^4+2x^2+9\]

Exponent of \(x\) in \(7x^4\) is \(4\)
Exponent of \(x\) in \(2x^2\) is \(2\)

Determine whether the expression is polynomial:\[x^{\frac{1}{2}}+3\]

Check exponent of variable
Verify whether exponent is whole number

Exponent of \(x\) is:

\[ \frac{1}{2} \]

Since it is not a whole number, the expression is not a polynomial.

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
\(x\) has no exponent	\(x=x^1\)
\(\sqrt{x}\) is polynomial	Exponent is fractional
\(\dfrac{1}{x}\) is polynomial	Exponent becomes negative
Constant term has no exponent	Exponent is \(0\)

📋 Case Study

A student writes the following expressions:

\[ x^3+2x+1 \]

\[ \sqrt{x}+5 \]

\[ \frac{1}{x}+7 \]

The teacher asks which of these are polynomials.

Solution

\(x^3+2x+1\) is polynomial because exponents are whole numbers
\(\sqrt{x}+5\) is not polynomial because exponent is \(\dfrac{1}{2}\)
\(\dfrac{1}{x}+7\) is not polynomial because exponent becomes \(-1\)

⚡ Quick Revision

Exponent shows repeated multiplication
\(x=x^1\)
Constant term has exponent \(0\)
Only whole number exponents are allowed in polynomials
Highest exponent determines degree

🧮

Zero Polynomial

📘 Definition

A zero polynomial is a polynomial in which all coefficients are zero. Therefore, the value of the polynomial becomes zero for every value of the variable.

It is generally represented as:

\[ p(x)=0 \]

Examples:

\[ 0x^3+0x^2+0x+0 \]

\[ 0y^5+0y^2+0 \]

Both simplify to:

\[ 0 \]

Important: A zero polynomial gives output zero for every value of the variable.

💡 Concept

Understanding the Concept

Consider:

\[ p(x)=0x^2+0x+0 \]

For any value of \(x\):

\[ p(1)=0 \]

\[ p(5)=0 \]

\[ p(-100)=0 \]

Thus, the polynomial always remains zero.

🏷️ Properties

Properties

Characteristics of Zero Polynomial

Property	Description
All coefficients	Equal to zero
Value of polynomial	Always zero
Variable terms	May appear with coefficient zero
Degree	Not defined
Type	Special polynomial

🔎 Key Fact

Degree of Zero Polynomial

The degree of a polynomial is the highest power of the variable having a non-zero coefficient.

In a zero polynomial, every coefficient is zero.

Therefore, there is no highest power with non-zero coefficient.

Result: Degree of zero polynomial is not defined.

Why Degree is Undefined?

Consider:

\[ 0x^5+0x^3+0x+0 \]

Since every term becomes zero, we cannot identify any meaningful highest power.

📊 Comparison Table

Difference Between Zero Polynomial and Constant Polynomial

Zero Polynomial	Constant Polynomial
0	5, -2, 100
All coefficients are zero	Constant value is non-zero
Degree not defined	Degree is 0
Always equal to zero	Has fixed non-zero value

🎨 SVG Diagram

Graph of Zero Polynomial

✏️ Example

Examples of Zero Polynomial

Expression	Zero Polynomial?	Reason
\(0x^2+0x+0\)	Yes	All coefficients are zero
\(0a^5\)	Yes	Entire expression becomes zero
\(5\)	No	Non-zero constant polynomial
\(x-x\)	Yes	Simplifies to zero

✏️ Example

Determine whether the following is a zero polynomial:\[0x^4+0x^2+0\]

All coefficients are zero.

Therefore, it is a zero polynomial.

Find the degree of: 0

Degree of zero polynomial is not defined.

🌟 Importance

Importance of Zero Polynomial

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
Degree of zero polynomial is \(0\)	Degree is undefined
Zero polynomial is same as constant polynomial	Zero polynomial is special case
\(5\) is zero polynomial	\(5\) is non-zero constant polynomial
Ignoring simplified forms	\(x-x=0\) is zero polynomial

📋 Case Study

A student writes the expression:

\[ p(x)=3x^2-3x^2+5-5 \]

The teacher asks:

Is it a zero polynomial?
What is its value for \(x=10\)?
What is its degree?

Solution

Simplifying:

\[ p(x)=0 \]

Yes, it is a zero polynomial
Value at \(x=10\) is \(0\)
Degree is not defined

⚡ Quick Revision

All coefficients are zero
Value is always zero
Represented as \(p(x)=0\)
Degree is undefined
Graph lies on x-axis

🧮

Monomials

📘 Definition

A monomial is an algebraic expression consisting of only one term.

It may contain:

A constant
A variable
A product of constants and variables
Variables raised to non-negative integer exponents

General form of a monomial:

\[ ax^n \]

where:

\(a\) is the coefficient
\(x\) is the variable
\(n\) is a non-negative integer

Important: A monomial contains only one term and no addition or subtraction sign separating terms.

✏️ Example

Examples of Monomials

Expression	Why It Is a Monomial
\(5x^2\)	Single algebraic term
\(x\)	Single variable term
\(3\)	Constant monomial
\(-7a^3b^2\)	Product of variables with whole number exponents
\(\dfrac{4}{5}x^4\)	Fractional coefficient is allowed

🚨 Caution

Expressions Which Are Not Monomials

🚧 Caution

Expressions having more than one term or invalid exponents are not monomials.

Expression	Reason
\(x+2\)	Contains two terms
\(a-b\)	Contains subtraction of terms
\(\sqrt{x}\)	Exponent is fractional
\(\dfrac{1}{x}\)	Negative exponent involved
\(x^2+y^2\)	Contains multiple terms

📌 Note

Parts of a Monomial

Consider the monomial:

\[ 7x^3 \]

Part	Name
\(7\)	Coefficient
\(x\)	Variable
\(3\)	Exponent

🗂️ Types / Category

Types of Monomials

Constant Monomial

Monomial with **only constant** (no variables). Degree = 0.
Examples:
\(9,\quad-15,\quad.2,\quad0\).

Variable Monomial

Monomial with **only variable(s)** (no numerical coefficient except 1).
Examples:
\(x,\quad y^3,\quad ab\,\quad pqr\).

Product Monomial

Monomial as **product of coefficient and variables**. Numerical coefficient ≠ 0,1,-1.
Examples:
\(4x^2y,\quad-3ab^2,\quad5xy^2z,\quad2.5p^3q\).

📍 Key Point

Degree of a Monomial

The degree of a monomial is the sum of exponents of all variables.

Example

Monomial	Calculation	Degree
\(5x^2\)	\(2\)	2
\(3xy^2\)	\(1+2\)	3
\(7a^2b^3\)	\(2+3\)	5
\(8\)	\(0\)	0

📌

Note Degree of a constant monomial is always \(0\).

✖️ ➗ Algebraic Operations

Multiplication Division

Multiplication of Monomials

While multiplying monomials:

Multiply coefficients
Add exponents of same variables

\[ a^m \times a^n=a^{m+n} \]

Example

\[ 2x^2 \times 3x^3 \]

\[ =6x^{2+3} \]

\[ =6x^5 \]

Division of Monomials

During division:

Divide coefficients
Subtract exponents of same variables

\[ \dfrac{a^m}{a^n}=a^{m-n} \]

Example

\[ \frac{12x^5}{3x^2} \]

\[ =4x^{5-2} \]

\[ =4x^3 \]

🛠️ Application

Real Life Applications of Monomials

Area formulas such as \(s^2\)
Volume formulas like \(a^3\)
Physics equations involving speed and distance
Computer graphics and scaling
Engineering calculations

✏️ Example

Determine whether the following expression is a monomial:\[7x^2y^3\]

The expression contains only one term and exponents are whole numbers.

Therefore, it is a monomial.

Find the degree of:\[4a^2b^4\]

Identify exponents
Add exponents

\[ 2+4=6 \]

Degree of the monomial is \(6\).

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
\(x+y\) is monomial	It has two terms
\(\sqrt{x}\) is monomial	Exponent is fractional
Degree of \(3xy^2\) is \(2\)	Degree is \(1+2=3\)
Constants are not monomials	Constants are monomials of degree \(0\)

📋 Case Study

A teacher writes the following expressions on the board:

\[ 5x^2 \]

\[ x+y \]

\[ 7a^2b \]

Students are asked to identify monomials and find their degrees.

Solution

\(5x^2\) is a monomial of degree \(2\)
\(x+y\) is not a monomial because it has two terms
\(7a^2b\) is a monomial of degree \(2+1=3\)

⚡ Quick Revision

Monomial has only one term
Exponents must be whole numbers
Constants are monomials
Degree is sum of exponents
Fractional and negative exponents are not allowed

🧮

Binomials

📘 Definition

A binomial is an algebraic expression consisting of exactly two unlike non-zero terms joined by addition or subtraction.

General form of a binomial:

\[ a+b \]

where:

\(a\) and \(b\) are algebraic terms
The two terms are separated by \(+\) or \(-\)
Both terms must be non-zero

Important: A binomial always contains exactly two terms.

✏️ Example

Examples of Binomials

Expression	Reason
\(3x^2+x\)	Contains two non-zero terms
\(x+2\)	Variable term and constant term
\(y^3-5\)	Two unlike terms connected by subtraction
\(4a-7b\)	Two variable terms
\(5x^2-9\)	Quadratic variable term and constant

🚨 Caution

Expressions Which Are Not Binomials

🚧 Caution

Expressions having one term or more than two terms are not binomials.

Expression	Reason
\(5x\)	Only one term, so monomial
\(x^2+x+1\)	Contains three terms
\(0+x\)	Only one non-zero term remains
\(\sqrt{x}+2\)	Fractional exponent not allowed in polynomial
\(a+b+c+d\)	Contains four terms

📌 Note

Parts of a Binomial

Consider the binomial:

\[ 5x^2-3 \]

Part	Name
\(5x^2\)	First term
\(5\)	Coefficient of \(x^2\)
\(2\)	Exponent of \(x\)
\(-3\)	Constant term

🗂️ Types / Category

Types of Binomials

Variable Binomial

Binomial where both terms contain variables. No constant terms present.
Examples:
\(x + y,\quad 3x^2 + 2y,\quad ab + c^2d,\quad p^3 + 2q^4\).

Mixed Binomial

Binomial with one variable term + one constant term. Contains exactly one numerical constant.
Examples:
\(x + 5,\quad 3x^2 - 7,\quad 2y - 4,\quad p^3 + 9\).

📍 Key Point

Degree of a Binomial

The degree of a binomial is the highest exponent of the variable present in it.

Example

Binomial	Highest Exponent	Degree
\(x+2\)	\(1\)	1
\(5x^2-7\)	\(2\)	2
\(3a^4+a\)	\(4\)	4
\(7y^5-9y^2\)	\(5\)	5

➕ ➖ ✖️ ➗ Algebraic Operation

Addition and Subtraction of Binomials

Addition Example

Example

\[ (x+2)+(3x+5) \]

\[ =x+3x+2+5 \]

\[ =4x+7 \]

Subtraction Example

Example

\[ (5x+3)-(2x+1) \]

\[ =5x+3-2x-1 \]

\[ =3x+2 \]

Multiplication of Binomials

While multiplying binomials, each term of the first binomial is multiplied by each term of the second binomial.

Example

\[ (x+2)(x+3) \]

\[ =x(x+3)+2(x+3) \]

\[ =x^2+3x+2x+6 \]

\[ =x^2+5x+6 \]

🔢 Formula

Important Algebraic Identities Involving Binomials

Identity	Expansion
\[ (a+b)^2 \]	\[ a^2+2ab+b^2 \]
\[ (a-b)^2 \]	\[ a^2-2ab+b^2 \]
\[ (a+b)(a-b) \]	\[ a^2-b^2 \]

These identities are extremely important for CBSE Board examinations.

🛠️ Application

Real Life Applications of Binomials

Used in area and perimeter calculations
Important in algebraic modelling
Used in physics equations
Applied in engineering calculations
Used in computer algorithms and graphics

✏️ Example

Determine whether the following expression is a binomial:\[7x^2-5\]

The expression contains exactly two non-zero terms.

Therefore, it is a binomial.

Find the degree of:\[4a^3+a\]

Identify exponents of all terms
Choose the highest exponent

Highest exponent is:

\[ 3 \]

Therefore, degree of the binomial is \(3\).

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
\(x+x\) is binomial	Simplifies to \(2x\), which is monomial
\(0+x\) is binomial	Only one non-zero term remains
Degree is number of terms	Degree is highest exponent
Ignoring signs during multiplication	Signs affect every term

📋 Case Study

A student is asked to classify the following expressions:

\[ x+5 \]

\[ 2x^2+3x+1 \]

\[ 7y-9 \]

The student also has to find their degrees.

Solution

\(x+5\) is a binomial of degree \(1\)
\(2x^2+3x+1\) is not a binomial because it has three terms
\(7y-9\) is a binomial of degree \(1\)

⚡ Quick Revision

Binomial contains exactly two non-zero terms
Terms are connected using \(+\) or \(-\)
Highest exponent gives degree
Important identities are based on binomials
Combining like terms may change type of expression

🧮

Trinomials

📘 Definition

A trinomial is an algebraic expression consisting of exactly three non-zero unlike terms separated by addition or subtraction signs.

General form of a trinomial:

\[ a+b+c \]

where:

\(a\), \(b\), and \(c\) are algebraic terms
All three terms are non-zero
Terms are joined using \(+\) or \(-\)

Important: A trinomial always contains exactly three terms after simplification.

✏️ Example

Examples of Trinomials

Expression	Reason
\(5x^2+6x+5\)	Contains three terms
\(3y+2z-5\)	Three unlike terms
\(x^2+x+1\)	Quadratic trinomial
\(a+b-c\)	Three variable terms
\(7m^3+2m^2-9\)	Three non-zero terms

🚨 Caution

Expressions Which Are Not Trinomials

🚧 Caution

Expressions with fewer or more than three terms are not trinomials.

Expression	Reason
\(5x\)	Only one term, so monomial
\(x+2\)	Contains two terms, so binomial
\(x+y+z+1\)	Contains four terms
\(x+x+2\)	Simplifies to \(2x+2\), a binomial
\(\sqrt{x}+x+1\)	Fractional exponent not allowed in polynomial

📌 Note

Parts of a Trinomial

Consider the trinomial:

\[ 4x^2+3x-7 \]

Part	Name
\(4x^2\)	First term
\(3x\)	Second term
\(-7\)	Constant term
\(4\)	Coefficient of \(x^2\)
\(2\)	Exponent of \(x\)

🗂️ Types / Category

Types of Trinomials

Quadratic Trinomial

Trinomial where highest degree is 2. Standard form: \(ax^2 + bx + c\).
Examples:
\[\begin{aligned}x^2 + 5x + 6,\\ 3x^2 - 2x + 1,\\ 2y^2 + 4y - 7,\\ p^2 - 3p + 5\end{aligned}\]

Cubic Trinomial

Trinomial where highest degree is 3. Standard form: \(ax^3 + bx^2 + cx + d\) (leading \(x^3\) term).
Examples:
\[\begin{aligned}x^3 + x^2 + 1,\\ 2x^3 - 3x + 5,\\ y^3 + 2y^2 - 4y,\\ p^3 - x^2 + 7\end{aligned}\]

🗒️ Key Point

Degree of a Trinomial

The degree of a trinomial is the highest exponent of the variable present in it.

Example

Trinomial	Highest Exponent	Degree
\(x^2+5x+6\)	\(2\)	2
\(a^3+a^2+a\)	\(3\)	3
\(7y^5+y^2-1\)	\(5\)	5
\(m+n+5\)	\(1\)	1

➕ ➖ ✖️ ➗ Algebraic Operations

Addition and Subtraction of Trinomials Factorisation of Trinomials

Addition Example

Example

\[ (x^2+2x+1)+(2x^2+3x+5) \]

\[ =3x^2+5x+6 \]

Subtraction Example

Example

\[ (5x^2+4x+3)-(2x^2+x+1) \]

\[ =3x^2+3x+2 \]

Factorisation of Trinomials

Example

Many quadratic trinomials can be factorised into binomials.

\[ x^2+5x+6 \]

Find two numbers whose:

Product is \(6\)
Sum is \(5\)

The numbers are \(2\) and \(3\).

\[ x^2+5x+6=(x+2)(x+3) \]

🔢 Formula

Important Identities Related to Trinomials

Identity	Expansion
\[ (a+b)^2 \]	\[ a^2+2ab+b^2 \]
\[ (a-b)^2 \]	\[ a^2-2ab+b^2 \]
\[ (a+b+c)^2 \]	\[ a^2+b^2+c^2+2ab+2bc+2ca \]

🛠️ Application

Real Life Applications of Trinomials

Projectile motion equations in physics
Area and geometry calculations
Business profit modelling
Engineering design curves
Computer graphics and animation paths

✏️ Example

Determine whether the following expression is a trinomial:\[3x^2+2x+7\]

The expression contains exactly three non-zero terms.

Therefore, it is a trinomial.

Find the degree of:\[5a^4+a^2+1\]

Identify all exponents
Choose the highest exponent

Highest exponent is:

\[ 4 \]

Therefore, degree of the trinomial is \(4\).

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
\(x+x+2\) is trinomial	Simplifies to \(2x+2\), a binomial
Degree equals number of terms	Degree equals highest exponent
Ignoring signs while combining terms	Signs affect simplification
All three-term expressions are polynomials	Exponents must be non-negative integers

📋 Case Study

A teacher asks students to classify the following expressions:

\[ x^2+4x+4 \]

\[ x+x+2 \]

\[ 3a^2+5a-1 \]

Students also need to find the degree of each polynomial.

Solution

\(x^2+4x+4\) is a trinomial of degree \(2\)
\(x+x+2\) simplifies to \(2x+2\), so it is a binomial
\(3a^2+5a-1\) is a trinomial of degree \(2\)

⚡ Quick Revision

Trinomial contains exactly three non-zero terms
Terms are joined using \(+\) or \(-\)
Highest exponent gives degree
Quadratic trinomials are very important
Always simplify before classification

🧮

Degree of Polynomials

📘 Definition

The degree of a polynomial is the highest exponent of the variable present in the polynomial having a non-zero coefficient.

In simple words, the degree tells us the highest power appearing in the polynomial.

Examples:

\[ x^5+y^4+z-3 \]

\[ \Rightarrow \text{Degree }=5 \]

\[ z^3-z^2+4z-5 \]

\[ \Rightarrow \text{Degree }=3 \]

Important: Only terms with non-zero coefficients are considered while finding degree.

💡 Concept

Understanding the Concept of Degree

Consider the polynomial:

\[ 7x^4+3x^2+x+5 \]

Exponents of variable \(x\) are:

\[ 4,\;2,\;1,\;0 \]

The highest exponent is:

\[ 4 \]

Therefore:

\[ \text{Degree}=4 \]

✏️ Example

Examples of Degree of Polynomials

Polynomial	Highest Exponent	Degree
\(5x+7\)	\(1\)	1
\(x^2+4x+9\)	\(2\)	2
\(3x^5+2x^2+1\)	\(5\)	5
\(9\)	\(0\)	0
\(7a^3b^2\)	\(3+2\)	5

✏️ Example

Degree of Polynomial in More Than One Variable

For polynomials having more than one variable, the degree of a term is obtained by adding the exponents of all variables in that term.

Example

\[ 5x^2y^3 \]

Degree:

\[ 2+3=5 \]

Example

\[ x^2y^3+xy+5 \]

Degrees of terms:

\(x^2y^3 \Rightarrow 2+3=5\)
\(xy \Rightarrow 1+1=2\)
\(5 \Rightarrow 0\)

Highest degree:

\[ 5 \]

Therefore, polynomial degree is \(5\).

📌 Note

Classification of Polynomials Based on Degree

Degree	Name of Polynomial	Example
0	Constant Polynomial	\(7\)
1	Linear Polynomial	\(2x+5\)
2	Quadratic Polynomial	\(x^2+3x+1\)
3	Cubic Polynomial	\(x^3-2x+7\)
4	Quartic Polynomial	\(x^4+5x+1\)
5	Quintic Polynomial	\(x^5+x+2\)

Degree of Constant Polynomial

A non-zero constant polynomial has degree \(0\)

Example

\[ 5 \]

can be written as:

\[ 5x^0 \]

Since exponent is \(0\):

\[ \text{Degree}=0 \]

Board Important Fact: Degree of every non-zero constant polynomial is zero.

Degree of Zero Polynomial

Degree of the zero polynomial is not defined.

Reason:

Degree is determined by the highest power having non-zero coefficient. But in zero polynomial all coefficients are zero.

\[ 0x^4+0x^3+0x+0 \]

Students often confuse degree of zero polynomial with \(0\). Its degree is undefined.

Steps to Find Degree of a Polynomial

Simplify the polynomial if needed
Identify all variable terms
Find exponents of variables
For multivariable terms, add exponents
Select the highest exponent value

Always simplify like terms before finding degree.

✏️ Example

Find the degree of:\[4x^3+7x^2+2x+9\]

Highest exponent of \(x\) is:

\[ 3 \]

Therefore:

\[ \text{Degree}=3 \]

Find the degree of:\[5a^2b^3+2ab+1\]

Add exponents in each term
Choose highest value

Degree of:

\[ 5a^2b^3=2+3=5 \]

\[ 2ab=1+1=2 \]

Highest value:

\[ 5 \]

Therefore:

\[ \text{Degree}=5 \]

🌟 Importance

Importance of Degree of Polynomial

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
Degree is coefficient value	Degree is highest exponent
Degree of \(5\) is \(5\)	Degree is \(0\)
Degree of zero polynomial is \(0\)	Degree is undefined
Ignoring simplification before degree	Always simplify first
Degree of \(a^2b^3\) is \(3\)	Degree is \(2+3=5\)

🗒️ Case Atudy

A student writes the following polynomials:

\[ x^4+2x^2+7 \]

\[ 5a^2b^3+a \]

\[ 9 \]

The teacher asks students to determine their degrees.

Solution

\(x^4+2x^2+7\) has degree \(4\)
\(5a^2b^3+a\) has degree \(2+3=5\)
\(9\) is a constant polynomial of degree \(0\)

⚡ Quick Revision

Degree is highest exponent of variable
For multivariable terms, add exponents
Degree of non-zero constant polynomial is \(0\)
Degree of zero polynomial is undefined
Degree helps classify polynomials

🧮

Quadratic Polynomial

📘 Definition

A quadratic polynomial is a polynomial whose highest degree is \(2\).

General form of a quadratic polynomial:

\[ ax^2+bx+c \]

where:

\(a\), \(b\), and \(c\) are constants or coefficients
\(a \neq 0\)
\(x\) is the variable

Important: If \(a=0\), then the polynomial is no longer quadratic because the \(x^2\) term disappears.

💡 Concept

Understanding the General Form

In:

\[ ax^2+bx+c \]

Part	Meaning
\(ax^2\)	Quadratic term
\(bx\)	Linear term
\(c\)	Constant term
\(a,b,c\)	Coefficients
\(2\)	Highest exponent or degree

✏️ Example

Examples of Quadratic Polynomials

Polynomial	Degree	Quadratic or Not
\(x^2+5x+6\)	2	Yes
\(5x^2-3x+1\)	2	Yes
\(7y^2-9\)	2	Yes
\(x^3+2x+1\)	3	No
\(4x+5\)	1	No

📌 Note

Standard Form of Quadratic Polynomial

A quadratic polynomial is usually written in descending order of powers:

\[ ax^2+bx+c \]

Example

Rearrange:

\[ 5+2x+x^2 \]

Standard form:

\[ x^2+2x+5 \]

📌 Note

Degree of a Quadratic Polynomial

The degree of every quadratic polynomial is always:

\[ 2 \]

because the highest exponent of the variable is \(2\).

Degree depends on exponent and not on coefficient values.

🎨 SVG Diagram

Graph of a Quadratic Polynomial

The graph of a quadratic polynomial forms a special curve called a parabola.

The parabola may open upward or downward depending on the sign of coefficient \(a\).

🗂️ Types / Category

Types of Quadratic Polynomials

Complete Quadratic Polynomial

Quadratic with all three terms present: \(ax^2 + bx + c\).
All coefficients
\(a ≠ 0, b ≠ 0, c ≠ 0\).
Examples:
\[x^2 + 5x + 6\] \[3x^2 - 2x + 1\] \[2y^2 + 4y - 7\] \[p^2 - 3p + 5\]

Missing Linear Term

Quadratic where linear term \(bx\) is absent: \(ax^2 + c\).
Coefficient
\(b = 0\).
Examples:
\[x^2 + 9\] \[x^2 - 5\] \[3y^2 + 4\] \[p^2 - 7\]

Missing Constant Term

Quadratic where constant term \(c\) is absent: \(ax^2 + bx\).
Coefficient
\(c = 0\).
Examples:
\[x^2 + 4x\] \[3x^2 - 2x\] \[2y^2 + 5y\] \[p^2 - 3p\]

🔄 Process

Factorisation of Quadratic Polynomials

1

Many quadratic polynomials can be expressed as a product of two linear factors.
2

Example
3

\[x^2+5x+6\]
4

Find two numbers whose:
5

Product is 6 and Sum is 5
6

Numbers are \(2\) and \(3\).
7

\[x^2+5x+6=(x+2)(x+3)\]

🔄 Process

Zeros of Quadratic Polynomial

The values of the variable for which the polynomial becomes zero are called zeros or roots.

Example

1

\[x^2-5x+6\]
2

Factorising
3

\[(x-2)(x-3)\]
4

Therefore
5

\[x=2,\;3\]
6

are zeros of the polynomial.

🔢 Formula

Important Algebraic Identities Related to Quadratic Polynomials

Identity	Expansion
\[ (a+b)^2 \]	\[ a^2+2ab+b^2 \]
\[ (a-b)^2 \]	\[ a^2-2ab+b^2 \]
\[ (a+b)(a-b) \]	\[ a^2-b^2 \]

🛠️ Application

Real Life Applications of Quadratic Polynomials

Projectile motion in physics
Designing bridges and arches
Business profit optimisation
Engineering calculations
Computer graphics and animations
Area and geometry problems

✏️ Example

Determine whether the following is a quadratic polynomial:\[5x^2+3x-7\]

Yes, this is a quadratic polynomial because it is of the form \(ax^2 + bx + c\) where \(a = 5\), \(b = 3\), and \(c = -7\).

Find the degree of:\[7y^2-4y+9\]

Identify exponents
Choose highest exponent

The degree of the polynomial \(7y^2-4y+9\) is 2, as the highest exponent of \(y\) is 2.

🌟 Importance

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
\(a=0\) still gives quadratic polynomial	\(a\neq0\) is necessary
Degree depends on coefficient	Degree depends on exponent
Ignoring missing terms	Quadratic polynomial may have missing terms
Incorrect factorisation signs	Signs affect factors carefully
Every quadratic has two different roots	Roots may be equal or imaginary

📋 Case Study

A rectangular garden has length:

\[ (x+2)\text{ m} \]

and breadth:

\[ (x+3)\text{ m} \]

The area of the garden forms a polynomial.

Questions

Find the polynomial representing area
Identify its degree
State whether it is quadratic or not

Solution

Area:

\[ (x+2)(x+3) \]

\[ =x^2+3x+2x+6 \]

\[ =x^2+5x+6 \]

Polynomial is \(x^2+5x+6\)
Degree is \(2\)
Therefore, it is a quadratic polynomial

⚡ Quick Revision

Quadratic polynomial has degree \(2\)
General form is \(ax^2+bx+c\)
\(a\neq0\)
Graph forms parabola
Can often be factorised into linear factors

🧮

Cubic Polynomial

📘 Definition

A cubic polynomial is a polynomial whose highest degree is \(3\).

General form of a cubic polynomial:

\[ ax^3+bx^2+cx+d \]

where:

\(a\), \(b\), \(c\), and \(d\) are constants or coefficients
\(a \neq 0\)
\(x\) is the variable

Important: If the coefficient of \(x^3\) becomes zero, then the polynomial is no longer cubic.

💡 Concept

Understanding the General Form

In:

\[ ax^3+bx^2+cx+d \]

Part	Meaning
\(ax^3\)	Cubic term
\(bx^2\)	Quadratic term
\(cx\)	Linear term
\(d\)	Constant term
\(3\)	Degree of polynomial

✏️ Example

Examples of Cubic Polynomials

Polynomial	Degree	Cubic or Not
\(x^3+2x^2+5x+1\)	3	Yes
\(5x^3-7x+9\)	3	Yes
\(2a^3+a^2+1\)	3	Yes
\(x^4+3x^2+1\)	4	No
\(5x^2+3x+1\)	2	No

📌 Note

Standard Form of Cubic Polynomial

A cubic polynomial is usually written in descending order of powers:

\[ ax^3+bx^2+cx+d \]

Example

Rearrange:

\[ 1+2x+x^3+3x^2 \]

Standard form:

\[ x^3+3x^2+2x+1 \]

📌 Note

Degree of a Cubic Polynomial

Every cubic polynomial has degree:

\[ 3 \]

because the highest exponent of the variable is \(3\).

Degree depends on exponent and not on coefficient values.

The graph of a cubic polynomial generally forms an S-shaped curve.

Depending on the coefficient of \(x^3\), the graph may rise or fall.

🗂️ Types / Category

Types of Cubic Polynomials

Complete Cubic Polynomial

Cubic with all four terms present: \(ax^3 + bx^2 + cx + d\).
All coefficients
\(a ≠ 0, b ≠ 0, c ≠ 0, d ≠ 0\).
Examples:
\[x^3 + 2x^2 + 5x + 1\] \[3x^3 - 7x + 9\] \[2a^3 + a^2 + 1\]

Missing Quadratic Term

Cubic where quadratic term \(bx^2\) is absent: \(ax^3 + cx + d\).
Coefficient
\(b = 0\).
Examples:
\[x^3 + 5x + 6\] \[3x^3 - 2x + 1\] \[2a^3 + 1\]

Missing Linear Term

Cubic where linear term \(cx\) is absent: \(ax^3 + bx^2 + d\).
Coefficient
\(c = 0\).
Examples:
\[x^3 + 2x^2 + 6\] \[3x^3 - 7x^2 + 9\] \[2a^3 + a^2 + 1\]

Constant term is absent.

Cubic where constant term \(d\) is absent: \(ax^3 + bx^2 + cx\).
Coefficient
\(d = 0\).
Examples:
\[x^3 + 2x^2 + 5x\] \[3x^3 - 7x + 9\] \[2a^3 + a^2\]

🔄 Process

Factorisation of Cubic Polynomials

Cubic polynomials can often be factorised into linear and quadratic factors.

Example

1

Step 1: Given polynomial
2

\[x^3 - 6x^2 + 11x - 6\]
3

Step 2: Try possible rational roots using Rational Root Theorem
4

\[±1, ±2, ±3, ±6\]
5

Step 3: Test x=1: \[1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 ✅\]
6

Step 4: Synthetic Division with root x=1:
7

\begin{array}{r|r} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array}
8

Quotient: \[x^2 - 5x + 6\]
9

Step 5: Factorize quadratic: \[x^2 - 5x + 6 = (x-2)(x-3)\]
10

Step 6: Complete factorisation: \[(x-1)(x-2)(x-3)\]
11

Step 7: Zeros are: \[1, 2, 3\]

📘 Definition

Zeros of Cubic Polynomial

The values of variable for which the polynomial becomes zero are called zeros or roots.

A cubic polynomial may have:

Three real zeros
One real and two imaginary zeros
Repeated zeros

Example

\[ x^3-4x \]

Factorising:

\[ x(x^2-4) \]

\[ =x(x-2)(x+2) \]

Zeros are:

\[ 0,\;2,\;-2 \]

🔢 Formula

Important Algebraic Identities Related to Cubic Polynomials

Identity	Expansion
\[ (a+b)^3 \]	\[ a^3+3a^2b+3ab^2+b^3 \]
\[ (a-b)^3 \]	\[ a^3-3a^2b+3ab^2-b^3 \]
\[ a^3+b^3 \]	\[ (a+b)(a^2-ab+b^2) \]
\[ a^3-b^3 \]	\[ (a-b)(a^2+ab+b^2) \]

Cube identities are very important for factorisation and board examinations.

🛠️ Application

Real Life Applications of Cubic Polynomials

Volume calculations such as cubes and containers
Engineering and architecture designs
Computer graphics and animations
Business and economics modelling
Physics motion equations

✏️ Example

Determine whether the following is a cubic polynomial:\[4x^3+2x^2-5\]

Highest exponent of variable is:

\[ 3 \]

Therefore, it is a cubic polynomial.

Find the degree of:\[7y^3+5y^2+y+1\]

Identify exponents
Select highest exponent

Highest exponent:

\[ 3 \]

Therefore:

\[ \text{Degree}=3 \]

🌟 Importance

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
\(a=0\) still gives cubic polynomial	\(a\neq0\) is necessary
Degree depends on number of terms	Degree depends on highest exponent
Ignoring missing terms	Cubic polynomial may have missing terms
Using wrong cube identities	Signs in cube formulas are very important
Not simplifying before factorisation	Simplification should be done first

📋 Case Study

A cube-shaped water tank has side length:

\[ (x+2)\text{ m} \]

Volume of cube:

\[ (x+2)^3 \]

Expand the polynomial and identify its degree.

Solution

\[ (x+2)^3=x^3+3x^2(2)+3x(2^2)+2^3 \]

\[ =x^3+6x^2+12x+8 \]

The polynomial is cubic
Degree of polynomial is \(3\)

⚡ Quick Revision

Cubic polynomial has degree \(3\)
General form is \(ax^3+bx^2+cx+d\)
\(a\neq0\)
Graph generally forms S-shaped curve
Cube identities are important for factorisation

🧮

Polynomial in One Variable of Degree n

📘 Definition

A polynomial in one variable is an algebraic expression involving only one variable raised to non-negative integer powers.

General form of a polynomial of degree \(n\):

\[ a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_1x+a_0 \]

where:

\(x\) is the variable
\(a_0,a_1,a_2,\dots,a_n\) are constants or coefficients
\(a_n \neq 0\)
\(n\) is a non-negative integer

Important: Since only one variable is involved, such polynomials are called polynomials in one variable.

💡 Concept

Consider the polynomial:

\[ a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \]

Symbol	Meaning
\(x\)	Variable
\(a_n,a_{n-1},\dots,a_0\)	Coefficients
\(a_0\)	Constant term
\(n\)	Degree of polynomial
\(a_nx^n\)	Leading term
\(a_n\)	Leading coefficient

✏️ Example

Examples of Polynomial in One Variable

Polynomial	Variable	Degree
\(5x^2+3x+1\)	\(x\)	2
\(7y^3-2y+5\)	\(y\)	3
\(4a^5+9\)	\(a\)	5
\(8m-1\)	\(m\)	1
\(9\)	No variable	0

🔎 Key Fact

Conditions for a Valid Polynomial

For an algebraic expression to be a polynomial:

Exponents of variables must be non-negative integers
Variables cannot appear in denominators
Variables cannot have fractional exponents
Highest degree coefficient \(a_n\) must not be zero

Valid Examples

\[ x^3+2x+1 \]

\[ 5y^2-7 \]

Invalid Examples

\[ \frac{1}{x}+2 \]

\[ \sqrt{x}+1 \]

📌 Note

Leading Term and Leading Coefficient

The term having the highest exponent is called the leading term.

The coefficient of the leading term is called the leading coefficient.

Example

🗒️ \[6x^4 3x^2+x 5\]

Leading term: \[ 6x^4 \]
Leading coefficient: \[ 6 \]
Degree: \[ 4 \]

🎨 SVG Diagram

As the degree of the polynomial increases, the graph becomes more curved and complex.

⭐ Special Case

Zero Polynomial as a Special Case

If all coefficients become zero:

\[ a_0=a_1=a_2=\cdots=a_n=0 \]

then the polynomial becomes:

\[ p(x)=0 \]

This is called the zero polynomial.

Very Important: Degree of the zero polynomial is not defined.

🗒️ Classifications

Classification Based on Degree

Degree	Name	Example
0	Constant Polynomial	\(5\)
1	Linear Polynomial	\(2x+3\)
2	Quadratic Polynomial	\(x^2+5x+1\)
3	Cubic Polynomial	\(x^3+2x+1\)
4	Quartic Polynomial	\(x^4+7\)

✏️ Example

Identify the degree and leading coefficient of:\[8x^5+3x^2-x+7\]

Highest exponent:

\[ 5 \]

Therefore:

Degree: \[ 5 \]
Leading coefficient: \[ 8 \]

Determine whether the following is a polynomial:\[\sqrt{x}+2\]

Check exponent of variable
Verify whether exponent is non-negative integer

Exponent of \(x\) is:

\[ \frac{1}{2} \]

Since exponent is fractional, it is not a polynomial.

🌟 Importance

Importance of Polynomials in One Variable

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
Degree depends on coefficient	Degree depends on exponent
\(\sqrt{x}\) is polynomial	Fractional exponent not allowed
Degree of zero polynomial is \(0\)	Degree is undefined
Leading coefficient is highest number	Leading coefficient belongs to highest degree term
\(\dfrac{1}{x}\) is polynomial	Negative exponent involved

📋 Case Study

A student writes the polynomial:

\[ p(x)=5x^4-2x^2+x-9 \]

The teacher asks:

Identify the degree
Find the leading coefficient
State whether it is polynomial in one variable

Solution

Highest exponent is \(4\)
Degree: \[ 4 \]
Leading coefficient: \[ 5 \]
Only one variable \(x\) is present, so it is polynomial in one variable

⚡ Quick Revision

Polynomial in one variable contains only one variable
General form: \[ a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \]
\(a_n\neq0\)
Degree is highest exponent
Zero polynomial has undefined degree

🧮

Zeroes of Polynomials

📘 Definition

The zeroes of a polynomial are the values of the variable that make the polynomial equal to zero.

For a polynomial \(p(x)\), if:

\[ p(c)=0 \]

then \(c\) is called a zero of the polynomial.

Zeroes are also known as:

Roots
Solutions
Factors' values

Important: Zeroes are the x-values where the graph of the polynomial touches or cuts the x-axis.

💡 Concept

Understanding the Concept of Zeroes

Consider the polynomial:

\[ p(x)=x-3 \]

Put:

\[ x=3 \]

Then:

\[ p(3)=3-3 \]

\[ =0 \]

Therefore:

\[ 3 \]

is a zero of the polynomial.

✏️ Example

Examples of Zeroes of Polynomials

Polynomial	Zeroes	Verification
\(x-5\)	\(5\)	\(5-5=0\)
\(x^2-4\)	\(2,-2\)	\(2^2-4=0\)
\(x^2-5x+6\)	\(2,3\)	\((2)^2-5(2)+6=0\)
\(x^3-x\)	\(0,1,-1\)	\(0^3-0=0\)

🔄 Process

How to Find Zeroes of a Polynomial

To find the zeroes of a polynomial:

Set the polynomial equal to zero
Solve the resulting equation for the variable

Example

1

\[x^3 - 6x^2 + 11x - 6 = 0\]
2

Apply Rational Root Theorem: possible roots \(±1, ±2, ±3, ±6\)
3

Test \(x = 1\): \(1^3 - 6(1)^2 + 11(1) - 6 = 0\) ✅
4

Synthetic division with root \(x = 1\):
5

\[ \begin{array}{r|r} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array} \]
6

Quotient: \(x^2 - 5x + 6 = 0\)
7

Middle-term splitting: \(x^2 - 2x - 3x + 6 = 0\)
8

\[x(x - 2) - 3(x - 2) = 0\]
9

\[(x - 2)(x - 3) = 0\]
10

Zero-product property: \(x - 2 = 0\) or \(x - 3 = 0\)
11

\[x = 1, 2, 3\]

🔗 Relations

Relationship Between Zeroes and Factors

If\[ p(a)=0 \] then \[ (x-a) \]

is a factor of the polynomial.

Example

\[p(x)=x^2-9\]

Factorising:

\[ x^2-9=(x-3)(x+3) \]

Hence zeroes are:

\[ 3,-3 \]

📌 Note

Number of Zeroes of a Polynomial

The degree of a polynomial gives the maximum possible number of zeroes.

Degree	Polynomial Type	Maximum Number of Zeroes
1	Linear Polynomial	1
2	Quadratic Polynomial	2
3	Cubic Polynomial	3
4	Quartic Polynomial	4

A polynomial may have fewer real zeroes than its degree.

🎨 SVG Diagram

Graphical Meaning of Zeroes

Zeroes are the points where the graph intersects the x-axis.

The marked points on the x-axis represent the zeroes of the polynomial.

⭐ Special Case

Zeroes of Special Polynomials

Non-Zero Constant Polynomial

A non-zero constant polynomial has no zeroes.

\[ p(x)=5 \]

Since:

\[ 5\neq0 \]

for every value of \(x\), it has no zeroes.

Important: Non-zero constant polynomials have no zeroes.

Zero Polynomial

For the zero polynomial:

\[ p(x)=0 \]

Every value of \(x\) satisfies:

\[ p(x)=0 \]

Therefore:

Every real number is a zero of the zero polynomial.

📖 Theory

Factor Theorem

Introduction to Factor Theorem

If:

\[ p(a)=0 \]

then:

\[ (x-a) \]

is a factor of the polynomial.

This concept is known as the Factor Theorem.

Example

\[ p(x)=x^2-9 \]

Since:

\[ p(3)=3^2-9=0 \]

therefore:

\[ (x-3) \]

is a factor.

✏️ Example

Find the zeroes of:\[x^2-9\]

Factorising:

\[ x^2-9=(x-3)(x+3) \]

Therefore:

\[ x=3,\;-3 \]

are the zeroes.

Verify whether \(2\) is a zero of:\[p(x)=x^2-5x+6\]

Substitute \(x=2\)
Check whether result becomes zero

\[ p(2)=2^2-5(2)+6 \]

\[ =4-10+6 \]

\[ =0 \]

Therefore, \(2\) is a zero of the polynomial.

🌟 Importance

Importance of Zeroes of Polynomials

⚡ Exam Tip

❌ Common Mistakes

Incorrect Understanding	Correct Understanding
Zeroes are coefficients	Zeroes are values making polynomial zero
Non-zero constant polynomial has one zero	It has no zeroes
Only positive numbers can be zeroes	Zeroes may be negative, positive, or zero
Degree always equals number of real zeroes	Degree gives maximum possible zeroes
Ignoring sign errors during substitution	Signs must be handled carefully

📋 Case Study

A ball is thrown upward and its height is represented by:

\[ h(x)=x^2-5x+6 \]

The ball touches the ground when height becomes zero.

Questions

Find the zeroes of the polynomial
Interpret the zeroes physically

Solution

Factorising:

\[ x^2-5x+6=(x-2)(x-3) \]

Therefore:

\[ x=2,\;3 \]

These values represent the times when the ball reaches ground level.

⚡ Quick Revision

Zeroes make polynomial equal to zero
If \(p(a)=0\), then \(a\) is a zero
Zeroes are also called roots
Zeroes are x-intercepts of graph
Every real number is a zero of zero polynomial

🧮

Important Points to Remember

💡 Concept

Core Concepts Every Student Must Remember

A zero of a polynomial need not be zero

The value which makes the polynomial equal to zero is called its zero. That value itself may or may not be \(0\).

Example

\[ p(x)=x-5 \]

Since:

\[ p(5)=5-5=0 \]

Therefore, \(5\) is a zero of the polynomial, even though it is not \(0\).

\(0\) may be a zero of a polynomial

If substituting \(x=0\) makes the polynomial equal to zero, then \(0\) is a zero of the polynomial.

Example

\[ p(x)=x^2+5x \]

Put:

\[ x=0 \]

\[ p(0)=0^2+5(0)=0 \]

Hence, \(0\) is a zero of the polynomial.

Every linear polynomial has one and only one zero

A linear polynomial has degree \(1\) and is generally written as:

\[ ax+b \]

where:

\[ a\neq0 \]

Solving:

\[ ax+b=0 \]

\[ x=-\frac{b}{a} \]

Therefore, every linear polynomial always has exactly one zero.

Graphically, a linear polynomial intersects the x-axis at exactly one point.

A polynomial can have more than one zero

Higher degree polynomials may have multiple zeroes.

Example

\[ p(x)=x^2-9 \]

Factorising:

\[ x^2-9=(x-3)(x+3) \]

Therefore:

\[ x=3,\;-3 \]

are two zeroes of the polynomial.

🔎 Key Fact

Additional Important Points for CBSE Board Examinations

Concept	Important Fact
Degree of Polynomial	Highest exponent of variable gives degree
Degree of Non-Zero Constant Polynomial	Always \(0\)
Degree of Zero Polynomial	Not defined
Zeroes of Polynomial	Values making polynomial equal to zero
Linear Polynomial	Can have only one zero
Quadratic Polynomial	Can have at most two zeroes
Cubic Polynomial	Can have at most three zeroes
Factor Theorem	If \(p(a)=0\), then \((x-a)\) is a factor
Polynomial Condition	Exponents must be non-negative integers

📝 Summary

Summary of Key Points

Zeroes are values making polynomial zero
Linear polynomials have exactly one zero
Higher degree polynomials may have multiple zeroes
Non-zero constant polynomials have no zeroes
Zero polynomial has every real number as zero
Factor Theorem relates zeroes to factors

🧮

Factorisation of Polynomials

📘 Definition

Factorisation of polynomials is the process of expressing a polynomial as a product of two or more simpler polynomials called factors.

These factors cannot be simplified further and are called irreducible factors.

Polynomial factorisation is similar to finding prime factors of integers.

Example

Step-by-step factorization:

Start with: \[ x^2 - 5x + 6 \]

Find two numbers that multiply to 6 and add to -5: -2 and -3.
Rewrite: \[ x^2 - 2x - 3x + 6 \]
Group: \[ x(x - 2) - 3(x - 2) \]
Factor: \[ (x - 2)(x - 3) \]

Verification: \[ (x - 2)(x - 3) = x^2 - 5x + 6 \]

Important: Multiplying all factors again gives the original polynomial.

🌟 Importance

Why Factorisation is Important

Factorisation helps in:

Finding zeroes of polynomials
Simplifying algebraic expressions
Solving equations easily
Reducing complicated calculations
Understanding graphs of polynomials

Factorisation is one of the most important topics for CBSE board examinations.

💡 Concept

Common Factorisation Techniques

Some common techniques for factorising polynomials include:

Common Factor Method

\[ 3x^2+6x \]

\[ =3x(x+2) \]
Grouping Method

\[ ax+ay+bx+by \]

\[ =a(x+y)+b(x+y) \]

\[ =(a+b)(x+y) \]
Using Algebraic Identities

\[ a^2-b^2 \]

\[ =(a-b)(a+b) \]
Quadratic Factorisation

\[ x^2+5x+6 \]

\[ =(x+2)(x+3) \]

🔄 Process

Factorisation Process Explained

Factorisation by Taking Common Factor

1

If all terms contain a common factor, it is taken outside the bracket.
2

\[4x^2+8x\]
3

Common factor: 4x
4

\[4x(x+2)\]

Factorisation by Grouping

1

Group terms in pairs.
2

\[x^2+3x+2x+6\]
3

Group: \[(x^2+3x)+(2x+6)\]
4

Factor each group: \[x(x+3)+2(x+3)\]
5

Take common factor: \[(x+2)(x+3)\]

Factorisation Using Identities

1

Identify the identity: \[a^2-b^2\]
2

Apply: \[x^2-9\]
3

\[=x^2-3^2\]
4

\[=(x-3)(x+3)\]

Factorisation of Quadratic Polynomials

1

Find two numbers that multiply to constant term and add to coefficient of x.
2

\[x^2+5x+6\]
3

Numbers: 2 and 3
4

\[=x^2+2x+3x+6\]
5

\[=x(x+2)+3(x+2)\]
6

\[=(x+3)(x+2)\]

✏️ Example

Factorise:\[2x^2+4x\]

Factorising by taking common factor:

\[ 2x^2+4x \]

\[ =2x(x+2) \]

Factorise:\[9x^2-25\]

Identify the identity: \(a^2-b^2\)
Apply the identity to factorise

Factorising using difference of squares:

Recognize: \[ 9x^2 - 25 = (3x)^2 - 5^2 \]

Apply identity \( a^2 - b^2 = (a - b)(a + b) \):

\[ (3x - 5)(3x + 5) \]

Verification: \[ (3x - 5)(3x + 5) = 9x^2 - 25 \]

🌟 Importance

⚡ Exam Tip

❌ Common Mistakes

Forgetting to check for a common factor before applying other methods
Mistakenly applying the wrong algebraic identity
Incorrectly handling signs in the factorisation process
Not verifying the factorisation by multiplying the factors back

📋 Case Study

A rectangular playground has dimensions:

\[ (x+2)\text{ m and }(x+5)\text{ m} \]

The area polynomial is:

\[ x^2+7x+10 \]

Questions

Factorise the polynomial
Find the dimensions from factors

Solution

Find two numbers whose:

Product is \(10\)
Sum is \(7\)

Numbers are:

\[ 2,\;5 \]

Therefore:

\[ x^2+7x+10=(x+2)(x+5) \]

Hence, dimensions are:

\[ (x+2)\text{ m and }(x+5)\text{ m} \]

⚡ Quick Revision

Factorisation expresses polynomial as product of factors
Always check common factor first
Algebraic identities simplify factorisation
Factors help find zeroes
Verification is done by multiplication

🧮

Factor Theorem

📘 Definition

The Factor Theorem establishes a direct relationship between the factors and zeroes of a polynomial.

If \(p(x)\) is a polynomial of degree:

\[ n\geq1 \]

and \(a\) is a real number, then:

\[ (x-a) \] is a factor of \(p(x)\), if: \[ p(a)=0 \]
If: \[ (x-a) \] is a factor of \(p(x)\), then: \[ p(a)=0 \]

Important: A number \(a\) is a zero of the polynomial if and only if \((x-a)\) is its factor.

💡 Concept

Understanding the Concept

Consider the polynomial:

\[ p(x)=x^2-5x+6 \]

Put:

\[ x=2 \]

Then:

\[ p(2)=2^2-5(2)+6 \]

\[ =4-10+6 \]

\[ =0 \]

Since:

\[ p(2)=0 \]

therefore:

\[ (x-2) \]

is a factor of the polynomial.

📌 Note

Statement of Factor Theorem

🔬 Proof

Proof of Factor Theorem

By the Remainder Theorem, when a polynomial \(p(x)\) is divided by:

\[ (x-a) \]

then:

\[ p(x)=(x-a)q(x)+p(a) \]

where:

\(q(x)\) is the quotient polynomial
\(p(a)\) is the remainder

Part (i)

If:

\[ p(a)=0 \]

then:

\[ p(x)=(x-a)q(x) \]

Hence:

\[ (x-a) \]

is a factor of \(p(x)\).

Part (ii)

Suppose:

\[ (x-a) \]

is a factor of \(p(x)\).

Then:

\[ p(x)=(x-a)g(x) \]

for some polynomial \(g(x)\).

Putting:

\[ x=a \]

we get:

\[ p(a)=(a-a)g(a) \]

\[ =0 \]

Therefore:

\[ p(a)=0 \]

Hence proved.

🔗 Relations

Relation Between Factor and Zero

Zero of Polynomial	Corresponding Factor
\(2\)	\((x-2)\)
\(-3\)	\((x+3)\)
\(5\)	\((x-5)\)
\(-a\)	\((x+a)\)

🛠️ Application

Applications of Factor Theorem

Finding factors of polynomials
Finding zeroes or roots
Simplifying factorisation
Checking divisibility of polynomials
Solving polynomial equations

Factor theorem is widely used in higher algebra and calculus.

✏️ Example

Verify whether \((x-1)\) is a factor of \(p(x)=x^2-4x+3\).

Put:

\[ x=1 \]

Then:

\[ p(1)=1^2-4(1)+3 \]

\[ =1-4+3 \]

\[ =0 \]

Therefore:

\[ (x-1) \]

is a factor of the polynomial.

Check whether: \((x+2)\) is a factor of \(p(x)=x^3+3x^2-4x-12\).

Find corresponding zero
Substitute value in polynomial
Check whether result becomes zero

Corresponding zero for \((x+2)\) is:

\[ -2 \]

Put:

\[ x=-2 \]

Then:

\[ p(-2)=(-2)^3+3(-2)^2-4(-2)-12 \]

\[ =-8+12+8-12 \]

\[ =0 \]

Therefore:

\[ (x+2) \]

is a factor of the polynomial.

⚡ Exam Tip

❌ Common Mistakes

Confusing signs when substituting values
Forgetting to change the sign when substituting \((x+a)\)
Mistakenly using the wrong value for substitution

📋 Case Study

A company models its daily profit using:

\[ p(x)=x^3-6x^2+11x-6 \]

The manager wants to check whether:

\[ (x-2) \]

is a factor of the polynomial.

Solution

Substitute:

\[ x=2 \]

Then:

\[ p(2)=2^3-6(2)^2+11(2)-6 \]

\[ =8-24+22-6 \]

\[ =0 \]

Therefore:

\[ (x-2) \]

is a factor of the polynomial.

⚡ Quick Revision

If \(p(a)=0\), then \((x-a)\) is a factor
If \((x-a)\) is factor, then \(p(a)=0\)
Factor theorem is based on remainder theorem
Zeroes and factors are directly related
Graph intersects x-axis at zeroes

🧮

Example 1: Verification Using Factor Theorem

❓ Question

Examine whether \[x+2 \] is a factor of \[x^3+3x^2+5x+6\]

💡 Concept

Concept Used

We use the Factor Theorem.

According to Factor Theorem:

\[ (x-a)\text{ is a factor of }p(x) \]

if and only if

\[ p(a)=0 \]

Since the factor is:

\[ x+2 \]

corresponding zero is:

\[ -2 \]

🗺️ Roadmap

Roadmap to Solve

Identify the corresponding zero of the factor.
Substitute the zero into the polynomial.
Simplify carefully.
If the result becomes zero, then the factor exists.

🧩 Solution

Let:

\[ p(x)=x^3+3x^2+5x+6 \]

Since:

\[ x+2=0 \]

therefore:

\[ x=-2 \]

Now substitute:

\[ x=-2 \]

into the polynomial.

\[ \begin{aligned} p(-2) &=(-2)^3+3(-2)^2+5(-2)+6 \\ &=-8+3(4)-10+6 \\ &=-8+12-10+6 \\ &=0 \end{aligned} \]

Since: \[ p(-2)=0 \] therefore, \[ x+2 \] is a factor of \[ x^3+3x^2+5x+6 \]

Verification by Actual Factorisation

Let us divide the polynomial by:

\[ x+2 \]

The factorisation becomes:

\[ x^3+3x^2+5x+6 \]

\[ =(x+2)(x^2+x+3) \]

Hence the factor theorem is verified.

Alternative Short Method

For quick board examination calculations:

Replace: \[ x \] with: \[ -2 \]
Simplify directly.
If answer is: \[ 0 \] then factor exists.

⚡ Exam Tip

For factor: \[ (x+a) \] substitute: \[ x=-a \]
Sign mistakes are very common in cubic polynomials.
Simplify powers carefully: \[ (-2)^2=4 \] but \[ (-2)^3=-8 \]

❌ Common Mistakes

One common mistake is to confuse the signs when substituting values into the polynomial.

🧮

Example 2: Finding Unknown Constant Using Factor Theorem

❓ Question

Find the value of \[k\] if \[x-1\] is a factor of \[4x^3+3x^2-4x+k\]

💡 Concept

Concept Used

This question is based on the Factor Theorem.

According to Factor Theorem:

\[(x-a)\text{ is a factor of }p(x)\]

if and only if

\[p(a)=0\]

Since:

\[x-1\]

is a factor, corresponding zero is:

\[1\]

🗺️ Roadmap

Solution Roadmap

Identify the corresponding zero of the factor.
Substitute the value into the polynomial.
Use: \[ p(1)=0 \]
Solve for unknown constant \(k\).

🧩 Solution

Let:

\[ p(x)=4x^3+3x^2-4x+k \]

Since:

\[ x-1 \]

is a factor of \(p(x)\), therefore:

\[ p(1)=0 \]

Substitute:

\[ x=1 \]

\[ \begin{aligned} p(1) &=4(1)^3+3(1)^2-4(1)+k \\ &=4+3-4+k \\ &=3+k \end{aligned} \]

\[ \Rightarrow3+k=0 \]

\[ k=-3 \]

Verification

Substitute:

\[ k=-3 \]

into the polynomial:

\[ p(x)=4x^3+3x^2-4x-3 \]

Now calculate:

\[ p(1)=4(1)^3+3(1)^2-4(1)-3 \]

\[ =4+3-4-3 \]

\[ =0 \]

Hence verified.

Alternative Short Method

For quick board examination calculations:

Shortcut Method for Board Exams

Replace: \[ x \] with: \[ 1 \]
Simplify expression.
Equate result to zero.
Solve for unknown constant.

👁️ Observation

Important Observation

The question does not ask to fully factorise the polynomial.

It only asks to find the value of \[ k \] for which: \[ x-1 \] becomes a factor.

⚡ Exam Tip

For factor: \[ (x-a) \] substitute: \[ x=a \]
Simplify powers before addition or subtraction.
Carefully combine constants.
Always write: \[ p(a)=0 \] before substitution.

❌ Common Mistakes

A common mistake is to forget to set:

\[ p(a)=0 \]

before substituting the value into the polynomial.

🧮

Example 3: Factorisation by Splitting the Middle Term

❓ Question

Factorise \[x^2+17x+5\] by:

Splitting the middle term
Using Factor Theorem

💡 Concept

Concept Used

In quadratic factorisation of the form:

\[ ax^2+bx+c \]

we find two numbers such that:

Their product equals: \[ a\times c \]
Their sum equals: \[ b \]

This method is called splitting the middle term.

🗺️ Roadmap

Roadmap to Solve

Multiply coefficient of \(x^2\) and constant term. /li>
Find two numbers whose product equals that value. /li>
Their sum should equal coefficient of \(x\). /li>
Split the middle term. /li>
Factorise by grouping. /li>

🧩 Solution

Method 1: Factorisation by Splitting the Middle Term

Given polynomial:

\[ 6x^2+17x+5 \]

Compare with:

\[ ax^2+bx+c \]

Here:

Coefficient	Value
\(a\)	\(6\)
\(b\)	\(17\)
\(c\)	\(5\)

Calculate:

\[ a\times c=6\times5=30 \]

We now find two numbers whose:

Product is: \[ 30 \]
Sum is: \[ 17 \]

First Number	Second Number	Sum
\(1\)	\(30\)	\(31\)
\(2\)	\(15\)	\(17\)
\(3\)	\(10\)	\(13\)
\(5\)	\(6\)	\(11\)

Therefore:

\[ 17x=2x+15x \]

Split the middle term:

\[ \begin{aligned} 6x^2+17x+5 &= 6x^2+2x+15x+5 \\ &= 2x(3x+1)+5(3x+1) \\ &= (3x+1)(2x+5) \end{aligned} \]

Method 2: Using Factor Theorem

Let:

\[ p(x)=6x^2+17x+5 \]

We test possible rational zeroes.

Checking \(x=-\frac{1}{3}\)

\[ \begin{aligned} p\left(-\frac{1}{3}\right) &= 6\left(-\frac{1}{3}\right)^2 +17\left(-\frac{1}{3}\right) +5 \\ &= 6\left(\frac{1}{9}\right) -\frac{17}{3} +5 \\ &= \frac{2}{3} -\frac{17}{3} +\frac{15}{3} \\ &=0 \end{aligned} \]

Therefore:

\[ \left(x+\frac{1}{3}\right) \]

is a factor.

Multiplying by \(3\):

\[ (3x+1) \]

is a factor.

Dividing the polynomial by:

\[ (3x+1) \]

gives:

\[ (2x+5) \]

Hence:

\[ 6x^2+17x+5=(3x+1)(2x+5) \]

Verification

Multiply the factors:

\[ (3x+1)(2x+5) \]

\[ \begin{aligned} (3x+1)(2x+5) &= 6x^2+15x+2x+5 \\ &= 6x^2+17x+5 \end{aligned} \]

Hence verified.

👁️ Observation

Important Observation

While splitting middle term:

Product uses: \[ a\times c \]
Sum uses coefficient of: \[ x \]
After splitting, factorisation by grouping is performed.

⚡ Exam Tip

Always compare with: \[ ax^2+bx+c \]
Find product: \[ a\times c \] carefully.
Keep sign of middle term in mind.
Verify final answer by multiplication.
Factor theorem helps confirm factors quickly.

❌ Common Mistakes

Some common mistakes to avoid:

Forgetting to check the sign of the middle term.
Incorrectly calculating the product of coefficients.
Not verifying the final answer by multiplication.

🧮

Example 4: Factorisation by Splitting Middle Term and Factor Theorem

❓ Question

Factorise: \[6x^2+17x+5\] by:

Splitting the middle term
Using Factor Theorem

💡 Concept

Concept Used

A quadratic polynomial:

\[ ax^2+bx+c \]

can often be factorised by:

Splitting the middle term
Using possible rational zeroes and Factor Theorem

Both methods must give the same final factorisation.

🧩 Solution

Method 1: Factorisation by Splitting the Middle Term

Given polynomial:

\[ 6x^2+17x+5 \]

Compare with:

\[ ax^2+bx+c \]

Coefficient	Value
\(a\)	\(6\)
\(b\)	\(17\)
\(c\)	\(5\)

Calculate:

\[ a\times c=6\times5=30 \]

We now find two numbers whose:

Product is: \[ 30 \]
Sum is: \[ 17 \]

First Number	Second Number	Product	Sum
\(1\)	\(30\)	\(30\)	\(31\)
\(2\)	\(15\)	\(30\)	\(17\)
\(3\)	\(10\)	\(30\)	\(13\)
\(5\)	\(6\)	\(30\)	\(11\)

Therefore:

\[ 17x=2x+15x \]

Split the middle term:

\[ \begin{aligned} 6x^2+17x+5 &= 6x^2+2x+15x+5 \\ &= 2x(3x+1)+5(3x+1) \\ &= (3x+1)(2x+5) \end{aligned} \]

Method 2: Factorisation Using Factor Theorem

Let:

\[ p(x)=6x^2+17x+5 \]

Divide entire polynomial by \(6\):

\[ p(x)=6\left(x^2+\frac{17}{6}x+\frac{5}{6}\right) \]

Let:

\[ q(x)=x^2+\frac{17}{6}x+\frac{5}{6} \]

Possible rational zeroes are obtained from factors of:

\[ \frac{5}{6} \]

Possible Values of \(a\)	Possible Values of \(b\)
\(1\)	\(\frac{5}{6}\)
\(\frac{1}{2}\)	\(\frac{5}{3}\)
\(\frac{1}{3}\)	\(\frac{5}{2}\)

Since all coefficients are positive, positive values cannot make the polynomial zero.

Therefore, we only test negative values.

Checking Possible Root \(x=-\dfrac{1}{3}\)

\[ \begin{aligned} q\left(-\frac{1}{3}\right) &= \left(-\frac{1}{3}\right)^2 +\frac{17}{6}\left(-\frac{1}{3}\right) +\frac{5}{6} \\ &= \frac{1}{9} -\frac{17}{18} +\frac{5}{6} \\ &= \frac{2-17+15}{18} \\ &=0 \end{aligned} \]

Therefore:

\[ \left(x+\frac{1}{3}\right) \]

is a factor.

Multiplying by \(3\):

\[ (3x+1) \]

is a factor of the original polynomial.

Checking Possible Root \(x=-\dfrac{5}{2}\)

\[ \begin{aligned} q\left(-\frac{5}{2}\right) &= \left(-\frac{5}{2}\right)^2 +\frac{17}{6}\left(-\frac{5}{2}\right) +\frac{5}{6} \\ &= \frac{25}{4} -\frac{85}{12} +\frac{5}{6} \\ &= \frac{75-85+10}{12} \\ &=0 \end{aligned} \]

Therefore:

\[ \left(x+\frac{5}{2}\right) \]

is also a factor.

Multiplying by \(2\):

\[ (2x+5) \]

becomes a factor of original polynomial.

Final Factorisation

\[ \begin{aligned} 6x^2+17x+5 &= (3x+1)(2x+5) \end{aligned} \]

Verification

Multiply the factors:

\[ \begin{aligned} (3x+1)(2x+5) &= 6x^2+15x+2x+5 \\ &= 6x^2+17x+5 \end{aligned} \]

Hence verified.

👁️ Observation

Splitting method is faster for most quadratic polynomials.
Factor theorem confirms factors mathematically.
Rational roots are usually tested using factors of constant term.
Always verify by multiplication.

⚡ Exam Tip

First compare polynomial with: \[ ax^2+bx+c \]
Use: \[ a\times c \] carefully while splitting middle term.
In Factor Theorem, substitute possible rational roots systematically.
Negative values are important when coefficients are positive.

❌ Common Mistakes

Forgetting to check all possible rational roots.
Incorrectly applying the factor theorem.
Misapplying the splitting method.
Not verifying the factors by multiplication.

🧮

Algebraic Identities

📘 Definition

Algebraic identities are algebraic equations that remain true for every value of the variables involved.

These identities help in:

Fast calculations
Factorisation of polynomials
Expansion of expressions
Solving equations
Competitive and board examination problems

Algebraic identities are extremely important in NCERT Mathematics Class IX and X.

🔢 Formula

Standard Algebraic Identities

Identity	Expanded Form
\[ (x+y)^2 \]	\[ x^2+2xy+y^2 \]
\[ (x-y)^2 \]	\[ x^2-2xy+y^2 \]
\[ x^2-y^2 \]	\[ (x+y)(x-y) \]
\[ (x+a)(x+b) \]	\[ x^2+(a+b)x+ab \]
\[ (x+y+z)^2 \]	\[ x^2+y^2+z^2+2xy+2yz+2zx \]
\[ (x+y)^3 \]	\[ x^3+y^3+3xy(x+y) \]
\[ (x-y)^3 \]	\[ x^3-y^3-3xy(x-y) \]
\[ x^3+y^3+z^3-3xyz \]	\[ (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \]

📐 Derivation

Derivations of Important Identities

Derivation of \((x+y)^2\)

\[ \begin{aligned} (x+y)^2 &= (x+y)(x+y) \\ &= x(x+y)+y(x+y) \\ &= x^2+xy+xy+y^2 \\ &= x^2+2xy+y^2 \end{aligned} \]

Derivation of \((x-y)^2\)

\[ \begin{aligned} (x-y)^2 &= (x-y)(x-y) \\ &= x^2-xy-xy+y^2 \\ &= x^2-2xy+y^2 \end{aligned} \]

Derivation of \(x^2-y^2\)

\[ \begin{aligned} (x+y)(x-y) &= x^2-xy+xy-y^2 \\ &= x^2-y^2 \end{aligned} \]

🧮

Example 5: Fast Multiplication Using Algebraic Identity

❓ Question

Evaluate: \[ 105\times106 \] without multiplying directly.

💡 Concept

Concept Used

This question uses the algebraic identity:

This identity helps in performing large multiplications mentally and quickly.

Such techniques are widely used in mental mathematics, competitive examinations, and board exams.

Understanding the Shortcut

Both numbers:

\[ 105 \quad \text{and} \quad 106 \]

are close to:

\[ 100 \]

So we rewrite them as:

\[ 105=100+5 \]

\[ 106=100+6 \]

Now the identity becomes easy to apply.

🗺️ Roadmap

Roadmap to Solve

Rewrite numbers near a common base.
Apply the identity: \[ (x+a)(x+b) \]
Simplify step-by-step.
Obtain answer without long multiplication.

🧩 Solution

Rewrite the numbers:

\[ 105=100+5 \]

\[ 106=100+6 \]

Therefore:

\[ \begin{aligned} 105\times106 &= (100+5)(100+6) \\ &= 100^2+(5+6)(100)+(5)(6) \\ &= 10000+11(100)+30 \\ &= 10000+1100+30 \\ &=11130 \end{aligned} \]

Verification by Direct Multiplication

Using ordinary multiplication:

\[ \begin{aligned} 105\times106 &= 105(100+6) \\ &= 10500+630 \\ &=11130 \end{aligned} \]

Hence the answer is verified.

🌟 Importance

⚡ Exam Tip

Always identify numbers close to a common base.
Learn algebraic identities thoroughly.
Keep calculations systematic.
Identities save time in board examinations.
Verify answers whenever possible.

❌ Common Mistakes

Forgetting to account for the product of the differences from the base.
Incorrectly applying the identity to numbers that are not close to a common base.
Making arithmetic errors during the verification step.

🧮

ample 6: Factorisation Using Difference of Squares Identity

❓ Question

Factorise: \[\frac{25x^2}{4}-\frac{y^2}{9}\]

💡 Concept

This problem uses the identity:

This identity is called the difference of squares identity.

Whenever an expression is written as the subtraction of two perfect squares, this identity can be applied directly.

Understanding the Expression

We first rewrite each term as a perfect square.

Expression	Perfect Square Form
\[ \frac{25x^2}{4} \]	\[ \left(\frac{5x}{2}\right)^2 \]
\[ \frac{y^2}{9} \]	\[ \left(\frac{y}{3}\right)^2 \]

Therefore:

\[ \frac{25x^2}{4}-\frac{y^2}{9} = \left(\frac{5x}{2}\right)^2 - \left(\frac{y}{3}\right)^2 \]

🗺️ Roadmap

Solution Roadmap

Identify two perfect squares.
Rewrite the expression in square form.
Apply: \[ a^2-b^2=(a+b)(a-b) \]
Simplify the factors.

🧩 Solution

Given:

\[ \frac{25x^2}{4}-\frac{y^2}{9} \]

Rewrite each term:

\[ \begin{aligned} \frac{25x^2}{4}-\frac{y^2}{9} &= \left(\frac{5x}{2}\right)^2 - \left(\frac{y}{3}\right)^2 \\ &= \left(\frac{5x}{2}+\frac{y}{3}\right) \left(\frac{5x}{2}-\frac{y}{3}\right) \end{aligned} \]

Verification

Multiply the factors:

\[ \begin{aligned} &\left(\frac{5x}{2}+\frac{y}{3}\right) \left(\frac{5x}{2}-\frac{y}{3}\right) \\ &= \left(\frac{5x}{2}\right)^2 - \left(\frac{y}{3}\right)^2 \\ &= \frac{25x^2}{4} - \frac{y^2}{9} \end{aligned} \]

Hence verified.

🌟 Importance

Importance of Difference of Squares Identity

⚡ Exam Tip

Always check whether terms are perfect squares.
Difference of squares works only with subtraction.
Carefully simplify fractional squares.
Verify by multiplying factors again.

❌ Common Mistakes

Confusing difference of squares with sum of squares.
Applying the identity to addition instead of subtraction.
Forgetting to check if terms are perfect squares.
Incorrectly simplifying fractional parts.

🧮

Example 6: Factorisation Using Difference of Squares Identity

❓ Question

Factorise: \[ \frac{25x^2}{4}-\frac{y^2}{9} \]

💡 Concept

Concept Used

This problem uses the identity:

\[ x^2-y^2=(x+y)(x-y) \]

This identity is called the difference of squares identity.

Whenever an expression is written as the subtraction of two perfect squares, this identity can be applied directly.

Understanding the Expression

We first rewrite each term as a perfect square.

Expression	Perfect Square Form
\[ \frac{25x^2}{4} \]	\[ \left(\frac{5x}{2}\right)^2 \]
\[ \frac{y^2}{9} \]	\[ \left(\frac{y}{3}\right)^2 \]

Therefore:

\[ \frac{25x^2}{4}-\frac{y^2}{9} = \left(\frac{5x}{2}\right)^2 - \left(\frac{y}{3}\right)^2 \]

🗺️ Roadmap

Solution Roadmap

Identify two perfect squares.
Rewrite the expression in square form.
Apply: \[ a^2-b^2=(a+b)(a-b) \]
Simplify the factors.

🧩 Solution

Given:

\[ \frac{25x^2}{4}-\frac{y^2}{9} \]

Rewrite each term:

Verification

Multiply the factors:

Hence verified.

🌟 Importance

⚡ Exam Tip

Always check whether terms are perfect squares.
Difference of squares works only with subtraction.
Carefully simplify fractional squares.
Verify by multiplying factors again.

❌ Common Mistakes

Confusing difference of squares with sum of squares.
Forgetting to check if terms are perfect squares.
Incorrectly applying the formula to addition instead of subtraction.
Making arithmetic errors while simplifying fractional squares.

🧮

Example 7: Evaluating Using Algebraic Identity

❓ Question

Evaluate: \[ (109)^3 \] without direct multiplication.

💡 Concept

Algebraic Identity for Cube of a Binomial

🗺️ Roadmap

Solution Roadmap

Rewrite the number using a convenient base.
Apply: \[ (x+y)^3=x^3+y^3+3xy(x+y) \]
Calculate each term separately.
Add all values carefully.

🧩 Solution

Using:

\[ (109)^3=(100+9)^3 \]

Apply the identity:

\[ \begin{aligned} (100+9)^3 &= 100^3+9^3+3(100)(9)(100+9) \\ &= 1000000+729+2700(109) \end{aligned} \]

Now calculate:

\[ \begin{aligned} 2700(109) &= 2700(100+9) \\ &= 270000+24300 \\ &=294300 \end{aligned} \]

Therefore:

\[ \begin{aligned} (109)^3 &= 1000000+729+294300 \\ &=1295029 \end{aligned} \]

Verification

Using standard multiplication:

\[ 109\times109\times109 \]

also gives:

\[ 1295029 \]

Hence verified.

🛠️ Application

Real Life Applications

Fast numerical computation
Competitive examination shortcuts
Engineering approximations
Computer algorithm optimisation
Mental mathematics techniques

⚡ Exam Tip

Learn cube identities thoroughly.
Keep powers and products separate while solving.
Simplify calculations step-by-step.
Carefully handle large numbers.

❌ Common Mistakes

Confusing cube identities with square identities.
Forgetting to include the middle terms in the expansion.
Making arithmetic errors while computing the final result.

🧮

Example 8: Evaluating Using Algebraic Identity

❓ Question

Evaluate: \[ (91)^3 \] without direct multiplication.

💡 Concept

Algebraic Identity

🗺️ Roadmap

Solution Roadmap

Rewrite the number using a convenient base.
Apply: \[ (x-y)^3=x^3-y^3-3xy(x-y) \]
Calculate each part separately.
Simplify carefully.

🧩 Solution

Using:

\[ (91)^3=(100-9)^3 \]

Apply the identity:

\[ \begin{aligned} (100-9)^3 &= 100^3-9^3-3(100)(9)(100-9) \\ &= 1000000-729-2700(91) \end{aligned} \]

Now calculate:

\[ \begin{aligned} 2700(91) &= 2700(100-9) \\ &= 270000-24300 \\ &=245700 \end{aligned} \]

Therefore:

\[ \begin{aligned} (91)^3 &= 1000000-729-245700 \\ &= 999271-245700 \\ &=753571 \end{aligned} \]

Verification

Using direct multiplication:

\[ 91\times91\times91 \]

also gives:

\[ 753571 \]

Hence verified.

🛠️ Application

Real Life Applications

Mental mathematics techniques
Competitive examination shortcuts
Engineering computations
Computer science algorithms
Scientific approximations

⚡ Exam Tip

Use correct cube identity carefully.
Negative signs are extremely important.
Simplify large products step-by-step.
Verify calculations whenever possible.

❌ Common Mistakes

Forgetting to apply the correct cube identity.
Ignoring the sign of the terms in the expansion.
Making arithmetic errors while simplifying the expression.

🧮

Example 9: Factorisation Using Perfect Cube Identity

❓ Question

Factorise: \[ 8x^3+27y^3+36x^2y+54xy^2 \]

💡 Concept

Concept Used

This question uses the identity:

\[ x^3+y^3+3x^2y+3xy^2=(x+y)^3 \]

Whenever an expression contains: cubes of two terms along with \(3x^2y\) and \(3xy^2\), it can often be written as a perfect cube.

🗺️ Roadmap

Solution Roadmap

Rewrite terms as cubes.
Compare with cube identity.
Identify variables \(a\) and \(b\).
Write final answer as: \[ (a+b)^3 \]

🧩 Solution

Given:

\[ 8x^3+27y^3+36x^2y+54xy^2 \]

Rewrite cube terms:

\[ \begin{aligned} 8x^3 &= (2x)^3 \\ 27y^3 &= (3y)^3 \end{aligned} \]

Rewrite remaining terms:

\[ \begin{aligned} 36x^2y &= 3(2x)^2(3y) \\ 54xy^2 &= 3(2x)(3y)^2 \end{aligned} \]

Therefore:

\[ \begin{aligned} 8x^3+27y^3+36x^2y+54xy^2 &= (2x)^3+(3y)^3 \\ &\quad+ 3(2x)^2(3y) + 3(2x)(3y)^2 \\ &= (2x+3y)^3 \end{aligned} \]

Verification

Expand:

\[ (2x+3y)^3 \]

\[ \begin{aligned} (2x+3y)^3 &= (2x)^3 +(3y)^3 \\ &\quad+ 3(2x)^2(3y) + 3(2x)(3y)^2 \\ &= 8x^3 +27y^3 +36x^2y +54xy^2 \end{aligned} \]

Hence verified.

🌟 Importance

Importance for CBSE Board Aspirants

⚡ Exam Tip

Memorise cube identities properly.
Rewrite terms carefully before factorising.
Check coefficients systematically.
Verify answer by expansion.

❌ Common Mistakes

Confusing cube identities with square identities.
Incorrectly identifying the variables \(a\) and \(b\).
Making arithmetic errors while rewriting terms.
Forgetting to verify the final answer.

🧮

Important Points to Remember

🗒️ Revision

Important Definitions and Concepts

Monomial
A polynomial consisting of exactly one term is called a monomial.

\[ 5x^2,\quad 7y,\quad -3 \]
Binomial
A polynomial consisting of exactly two terms is called a binomial.

\[ x+2,\quad 3x^2-5 \]
Trinomial
A polynomial consisting of exactly three terms is called a trinomial.

\[ x^2+5x+6 \]
Linear Polynomial
A polynomial whose degree is one is called a linear polynomial.

\[ 2x+3 \]
Quadratic Polynomial
A polynomial whose degree is two is called a quadratic polynomial.

\[ ax^2+bx+c \]

where:

\[ a\ne0 \]
Cubic Polynomial
A polynomial whose degree is three is called a cubic polynomial.

\[ ax^3+bx^2+cx+d \]

where:

\[ a\ne0 \]
Zeroes of a Polynomial
A real number:

\[ a \]

is called a zero of polynomial:

\[ p(x) \]

if:

\[ p(a)=0 \]

In this case, \(a\) is also called a root of the equation:

\[ p(x)=0 \]
Special Cases of Zeroes
- Every linear polynomial has exactly one zero.
- A non-zero constant polynomial has no zero.
- Every real number is a zero of the zero polynomial.
Factor Theorem
If:

\[ p(a)=0 \]

then:

\[ (x-a) \]

is a factor of:

\[ p(x) \]

Conversely, if:

\[ (x-a) \]

is a factor of:

\[ p(x) \]

then:

\[ p(a)=0 \]

🗒️ Classifications

Classification Chart of Polynomials

Type	Number of Terms	Degree	Example
Monomial	1	Any	\(5x^2\)
Binomial	2	Any	\(x+3\)
Trinomial	3	Any	\(x^2+2x+1\)
Linear Polynomial	Any	1	\(2x+5\)
Quadratic Polynomial	Any	2	\(x^2+3x+2\)
Cubic Polynomial	Any	3	\(x^3+2x^2+1\)

NCERT · Class IX · Chapter 2

Polynomials
Complete AI Learning Engine

Master every concept with structured theory, interactive tools, AI-powered solving, and concept-building problems — all in one place.

8 Modules All Formulas Step-by-Step Solver Practice Quiz Flashcards

📖Core Concepts

Every foundational idea — from defining a polynomial to Remainder and Factor Theorems.

1 · What is a Polynomial? ▼

A polynomial in one variable x is an algebraic expression of the form:

p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀

where aₙ, aₙ₋₁ ... a₀ are real numbers (called coefficients), x is a variable, n is a non-negative integer, and aₙ ≠ 0.

Key Requirement: The exponent of the variable must be a whole number (0, 1, 2, 3, …). Expressions with fractional, negative, or variable exponents are NOT polynomials.

✅ Polynomials

3x² + 5x − 7
√2 y³ − 4y + 1
5 (constant polynomial)
x (linear)

❌ Not Polynomials

x⁻² + 4 (negative exponent)
√x + 3 = x^(1/2) + 3
1/x + 5 = x⁻¹ + 5
2^x + 1 (variable exponent)

2 · Degree of a Polynomial ▼

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

Constant

7, −√3, 0

Linear

2x + 5, 3y − 1

Quadratic

x² − 4x + 4

Cubic

2x³ − x + 9

Biquadratic

x⁴ − 16

nth Degree

aₙxⁿ + …

Special case: The zero polynomial (p(x) = 0) has no degree — it is undefined (or sometimes said to have degree −∞).

3 · Zeroes of a Polynomial ▼

A real number c is called a zero of polynomial p(x) if p(c) = 0.

Geometric meaning: On a graph, the zeroes are the x-coordinates where the curve crosses or touches the x-axis.

Worked Example

Find the zero of p(x) = 2x − 6.

Set p(x) = 0

2x − 6 = 0

Solve for x

2x = 6 → x = 3

Verify: p(3) = 2(3) − 6 = 0 ✓

A polynomial of degree n can have at most n real zeroes.

4 · Remainder Theorem ▼

Remainder Theorem

If p(x) is divided by (x − a), the remainder = p(a)

No long division needed — just substitute x = a into p(x).

Worked Example

Find the remainder when p(x) = x³ + 3x² + 3x + 1 is divided by (x + 1).

Divisor is (x − (−1)), so a = −1

Substitute x = −1

p(−1) = (−1)³ + 3(−1)² + 3(−1) + 1
= −1 + 3 − 3 + 1 = 0

Remainder = 0 (so (x+1) is actually a factor!)

5 · Factor Theorem ▼

Factor Theorem

(x − a) is a factor of p(x) ⟺ p(a) = 0

The Factor Theorem is a special case of the Remainder Theorem (remainder = 0).

Forward direction: If p(a) = 0, conclude (x − a) is a factor.

Reverse direction: If (x − a) is a factor, then p(a) must equal 0.

Worked Example

Show (x − 2) is a factor of p(x) = x³ − 4x² + x + 6.

p(2) = (2)³ − 4(2)² + (2) + 6
= 8 − 16 + 2 + 6 = 0 ✓

Since p(2) = 0, by Factor Theorem, (x − 2) is a factor.

6 · Algebraic Identities ▼

These are universally true equations — valid for all values of the variables. They are used for fast expansion and factorisation.

Identity I

(x + y)² = x² + 2xy + y²

Identity II

(x − y)² = x² − 2xy + y²

Identity III

x² − y² = (x + y)(x − y)

Identity IV

(x + a)(x + b) = x² + (a+b)x + ab

Identity V

(x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Identity VI

(x+y)³ = x³+y³+3xy(x+y)

Identity VII

(x−y)³ = x³−y³−3xy(x−y)

Identity VIII

x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²−xy−yz−zx)

Special case of VIII: If x + y + z = 0, then x³ + y³ + z³ = 3xyz.

7 · Methods of Factorisation ▼

Four key methods to factorise polynomials:

Method 1 — Common Factor

Take out the HCF from all terms.

3x² + 6x = 3x(x + 2)

Method 2 — Regrouping

Group terms and factor each group.

ax + ay + bx + by
= a(x+y) + b(x+y) = (a+b)(x+y)

Method 3 — Using Identities

Recognise the pattern of an identity.

x² − 9 = x² − 3² = (x+3)(x−3)

Method 4 — Factor Theorem + Splitting

Find a zero using trial, confirm via Factor Theorem, then divide to get remaining factor.

p(x) = x³ − 2x² − x + 2
p(1) = 1−2−1+2 = 0 → (x−1) is a factor
÷(x−1) → x² − x − 2 = (x−2)(x+1)
∴ p(x) = (x−1)(x−2)(x+1)

📐Complete Formula Sheet

Every formula from the chapter — grouped, labelled, and explained.

Standard Forms

Type	Standard Form	Example
Linear	ax + b, a ≠ 0	5x − 3
Quadratic	ax² + bx + c, a ≠ 0	2x² + 3x − 5
Cubic	ax³ + bx² + cx + d, a ≠ 0	x³ − 6x² + 11x − 6

Remainder & Factor Theorem

Remainder Theorem

Remainder of p(x) ÷ (x−a) = p(a)

Factor Theorem

(x−a) is a factor ⟺ p(a) = 0

All 8 Algebraic Identities — Quick Reference

#	Identity	Tip for use
I	(a+b)² = a² + 2ab + b²	Expansion of perfect square sum
II	(a−b)² = a² − 2ab + b²	Expansion of perfect square diff
III	a²−b² = (a+b)(a−b)	Difference of squares → two factors
IV	(x+a)(x+b) = x²+(a+b)x+ab	Product of two binomials
V	(a+b+c)² = a²+b²+c²+2ab+2bc+2ca	Square of trinomial
VI	(a+b)³ = a³+b³+3ab(a+b)	Cube of sum
VII	(a−b)³ = a³−b³−3ab(a−b)	Cube of difference
VIII	a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)	Sum of cubes identity

Derived result: If a + b + c = 0, then a³ + b³ + c³ = 3abc. This is very frequently tested!

Sum & Difference of Cubes

Sum of Cubes

a³ + b³ = (a+b)(a²−ab+b²)

Derived from Identity VIII with c = 0

Difference of Cubes

a³ − b³ = (a−b)(a²+ab+b²)

Very useful in factorisation problems

Useful Derived Identities

(a+b)² − (a−b)² = 4ab ← diff of squares identity
(a+b)² + (a−b)² = 2(a²+b²) ← add I and II
a² + b² = (a+b)² − 2ab ← rearrange I
a² + b² = (a−b)² + 2ab ← rearrange II
(a+b+c)³ use = a³+b³+c³+3(a+b)(b+c)(c+a)

🧮Step-by-Step Solver

Pick a problem type, enter values, and get a full worked solution instantly.

Select Problem Type

💡 Solver Pro Tips

→Remainder Theorem: Always write the divisor as (x − a) first. For (x + 3), a = −3, not +3.

→Factor Theorem: Test integer divisors of the constant term first when looking for zeroes by trial.

→Identities: Recognise the "shell" — (something)² or (something)³ — then map your expression to it.

📝Concept-Building Problems

Original, non-textbook questions with full step-by-step solutions — grouped by concept.

Concept A Identifying Polynomials

Q1 · Classify: Is p(x) = 3x^(4/3) − 2x + 5 a polynomial? Why or why not? ▼

Answer: No, it is NOT a polynomial.

Examine the exponent of x in the first term.

3x^(4/3) — exponent is 4/3, a fraction

Apply definition: exponents must be non-negative integers (whole numbers).

Conclusion

Since 4/3 is not a whole number, this expression is NOT a polynomial. It is an irrational algebraic expression.

Q2 · State the degree of p(x) = 7x⁵ − 3x³ + 2x² − x + 11. What type of polynomial is it? ▼

List the powers of x in each term.

x⁵, x³, x², x¹, x⁰

The highest power is 5.

Conclusion

Degree = 5 → a polynomial of degree 5

It has 5 terms, so it is also called a pentanomial (though degree-name is "degree-5 polynomial").

Concept B Zeroes of Polynomials

Q3 · Verify that x = −2 and x = 3 are zeroes of p(x) = x² − x − 6. ▼

Evaluate p(−2)

p(−2) = (−2)² − (−2) − 6 = 4 + 2 − 6 = 0 ✓

Evaluate p(3)

p(3) = (3)² − (3) − 6 = 9 − 3 − 6 = 0 ✓

Conclusion

Both values give p(x) = 0, confirming x = −2 and x = 3 are zeroes. Note: A quadratic has at most 2 zeroes — we've found both!

Bonus: Using these zeroes, we can factor: p(x) = (x+2)(x−3).

Q4 · Find k if x = 2 is a zero of p(x) = kx² − 3x − 2. ▼

Since x = 2 is a zero, p(2) = 0.

Substitute x = 2

k(2)² − 3(2) − 2 = 0
4k − 6 − 2 = 0
4k − 8 = 0

Solve for k

4k = 8 → k = 2

Verify: p(2) = 2(4) − 6 − 2 = 8 − 8 = 0 ✓

Concept C Remainder Theorem

Q5 · Without division, find the remainder when p(x) = 2x³ − 5x² + 4x − 3 is divided by (2x − 1). ▼

Write divisor in form (x − a)

2x − 1 = 0 → x = 1/2
So a = 1/2

Evaluate p(1/2)

p(1/2) = 2(1/2)³ − 5(1/2)² + 4(1/2) − 3
= 2(1/8) − 5(1/4) + 2 − 3
= 1/4 − 5/4 + 2 − 3
= (1 − 5)/4 − 1
= −4/4 − 1 = −1 − 1 = −2

∴ Remainder = −2

Q6 · If the remainder of p(x) = x³ + 2x² + kx − 3 divided by (x − 2) is 5, find k. ▼

By Remainder Theorem: p(2) = 5

Expand p(2)

(2)³ + 2(2)² + k(2) − 3 = 5
8 + 8 + 2k − 3 = 5
13 + 2k = 5

Solve

2k = 5 − 13 = −8 → k = −4

Concept D Factor Theorem & Factorisation

Q7 · Fully factorise p(x) = x³ − 6x² + 11x − 6. ▼

Try x = 1 (factor of constant term 6)

p(1) = 1 − 6 + 11 − 6 = 0 ✓ → (x−1) is a factor

Divide p(x) by (x−1) using synthetic/long division

p(x) ÷ (x−1) = x² − 5x + 6

Factorise the quotient x² − 5x + 6

x² − 5x + 6 = (x−2)(x−3) [product=6, sum=−5]

Complete factorisation

p(x) = (x−1)(x−2)(x−3)

Q8 · Factorise 8x³ + y³ + 27z³ − 18xyz. ▼

Rewrite using cubes

8x³ = (2x)³, 27z³ = (3z)³
So: (2x)³ + y³ + (3z)³ − 3·(2x)·y·(3z)

Apply Identity VIII: a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)

a = 2x, b = y, c = 3z

Result

= (2x+y+3z)(4x²+y²+9z²−2xy−3yz−6xz)

Concept E Identities — Evaluation Tricks

Q9 · Evaluate 99³ without a calculator using identities. ▼

Write 99 = (100 − 1)

Apply Identity VII: (a−b)³ = a³ − 3a²b + 3ab² − b³

(100−1)³ = 100³ − 3(100²)(1) + 3(100)(1²) − 1³
= 1,000,000 − 30,000 + 300 − 1

Calculate

= 970,299

Q10 · If a + b + c = 0, prove a³ + b³ + c³ = 3abc, and use it to evaluate 2³ + (−3)³ + 1³. ▼

Use Identity VIII

a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)

Since a+b+c = 0, RHS = 0

a³+b³+c³−3abc = 0 → a³+b³+c³ = 3abc

Check: 2 + (−3) + 1 = 0 ✓. Apply result.

2³ + (−3)³ + 1³ = 3 × 2 × (−3) × 1 = −18

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💡Tips & Tricks

Speed shortcuts, pattern recognition tricks, and exam-smart strategies.

🔍 Spotting the Identity — Pattern Recognition

⚡Perfect Square pattern: Three terms with the middle term being ±2 × (√first term)(√last term) → use Identity I or II.

⚡Difference of squares: Two terms, both perfect squares, subtracted → (a²−b²) = (a+b)(a−b).

⚡Sum/Difference of cubes: Two terms, both perfect cubes — find a, b, apply a³±b³ identity.

⚡Trinomial identity check: Check if the three terms can be written as a², b², c² and the cross terms match 2ab, 2bc, 2ca → Identity V.

🧮 Smart Calculation Tricks

💰Evaluate 103²: = (100+3)² = 10000 + 600 + 9 = 10609. Use (a+b)² with a=100, b=3.

💰Evaluate 997²: = (1000−3)² = 1000000 − 6000 + 9 = 994009. Use (a−b)².

💰Evaluate 98 × 102: = (100−2)(100+2) = 100² − 2² = 10000 − 4 = 9996. Use a²−b².

💰Evaluate 2.5³: = (5/2)³ = 125/8 = 15.625. Convert to fraction first.

🎯 Remainder Theorem Speed Tips

→Divisor (ax + b): Set ax + b = 0 → x = −b/a. Then find p(−b/a).

→Check factor first: If you suspect a factor, use zero-check first. If it's a factor (remainder=0), you get a bonus factorisation!

→Rational Root Theorem: For integer polynomials, possible rational zeroes = (factors of constant term) / (factors of leading coefficient).

📝 Exam Strategy Tips

📌State the theorem before applying it: Write "By Remainder Theorem…" or "By Factor Theorem…" to earn step marks in board exams.

📌Always verify zeroes: After finding a zero, substitute back to confirm p(c) = 0.

📌Factorisation check: Multiply your factors back. The product should equal the original polynomial.

📌Degree = number of zeroes (max): A degree-3 polynomial has at most 3 real zeroes — never more.

⚠️Common Mistakes

Every mistake students typically make — with corrections and memory anchors.

❌ Mistakes with Polynomials & Degree

✗Mistake: Thinking √x = x^(1/2) is a polynomial because "it's just x to some power."
Fix: The exponent 1/2 is not a whole number. Only exponents in {0, 1, 2, 3, …} are allowed in polynomials.

✗Mistake: Saying p(x) = 5 has no degree ("it's just a number").
Fix: p(x) = 5 = 5x⁰ — it's a degree-0 polynomial (constant polynomial), not zero polynomial.

✗Mistake: Confusing "zero polynomial" (p(x) = 0) with "zero of a polynomial" (value where p(c) = 0).
Fix: Zero polynomial = the polynomial itself equals zero always. Zero of a polynomial = specific x-value that makes p(x) = 0.

❌ Mistakes with Remainder Theorem

✗Mistake: For divisor (x + 3), substituting x = +3 instead of x = −3.
Fix: (x + 3) = (x − (−3)), so a = −3. Always set divisor = 0 and solve for x.

✗Mistake: For divisor (2x − 1), substituting x = 1 instead of x = 1/2.
Fix: Set 2x − 1 = 0 → x = 1/2. Do NOT just read off the number from the divisor.

✗Mistake: Forgetting that Remainder Theorem only applies when divisor is linear (degree 1).
Fix: For quadratic or higher divisors, Remainder Theorem doesn't directly apply.

❌ Mistakes with Algebraic Identities

✗Mistake: Expanding (a + b)² as a² + b² (forgetting the middle term).
Fix: (a + b)² = a² + 2ab + b². The middle term 2ab is always there.

✗Mistake: Writing (a + b)³ = a³ + b³.
Fix: (a+b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a+b). Never skip the middle terms.

✗Mistake: In a³+b³+c³−3abc identity, writing (a+b+c)(a²+b²+c²+ab+bc+ca) — wrong signs inside!
Fix: Correct: (a+b+c)(a²+b²+c²−ab−bc−ca). The three ab, bc, ca terms are NEGATIVE.

✗Mistake: In (x+y+z)², writing all cross terms as 2xy instead of 2xy + 2yz + 2zx.
Fix: Every pair of distinct terms gives a cross term: (2xy) + (2yz) + (2zx).

❌ Mistakes in Factorisation

✗Mistake: Stopping after finding one factor by Factor Theorem — not completing the full factorisation.
Fix: After finding (x − a) as a factor, divide the polynomial and factorise the quotient further.

✗Mistake: Applying x³+y³+z³ = 3xyz without checking that x+y+z = 0 first.
Fix: This result only holds if x + y + z = 0. Always verify the condition before applying.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025

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NCERT Class 9 Maths Chapter 2 Polynomials Notes

NCERT Class 9 Maths Chapter 2 Polynomials Notes — Complete Notes & Solutions · academia-aeternum.com

Introduction to Polynomials Polynomials form the foundation of algebra and higher mathematics. They are mathematical expressions consisting of variables, coefficients, and constants, combined using operations like addition, subtraction, multiplication, and non-negative integer exponents. In this post, polynomials are discussed in great detail. The different types of polynomials—such as monomials, binomials, trinomials, quadratic, and cubic expressions—are explained with suitable examples.…

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Polynomials

Chapter Snapshot

Why This Chapter Matters for Exams

Key Concept Highlights

Important Formula Capsules

What You Will Learn

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🏆 Exam Strategy & Preparation Tips

Classification of Polynomials

Classification of Polynomials

Expressions Which Are Not Polynomials

Standard Form of Polynomial

Evaluation of Polynomial

Example

Solution

Zero Polynomial

Real Life Applications of Polynomials

Questions

Solution

Invisible Coefficients

Types of Coefficients

Difference Between Coefficient and Constant Term

Coefficient Form in General Polynomial

Solution

Meaning of Exponents

Parts of an Exponential Term

Allowed Exponents in Polynomials

Zero Exponent Rule

Basic Laws of Exponents

Relation Between Degree and Exponents

Solution

Degree of Zero Polynomial

Why Degree is Undefined?

Solution

Parts of a Monomial

Multiplication of Monomials

Division of Monomials

Solution

Parts of a Binomial

Addition Example

Subtraction Example

Multiplication of Binomials

Solution

Addition Example

Subtraction Example

Factorisation of Trinomials

Solution

Degree of Constant Polynomial

Degree of Zero Polynomial

Steps to Find Degree of a Polynomial

Solution

Example

Questions

Solution

Degree of a Cubic Polynomial

Factorisation of Cubic Polynomials

Example

Zeros of Cubic Polynomial

Example

Important Algebraic Identities Related to Cubic Polynomials

Solution

Examples of Polynomial in One Variable

Conditions for a Valid Polynomial

Valid Examples

Invalid Examples

Leading Term and Leading Coefficient

Example

Zero Polynomial as a Special Case

Classification Based on Degree

Solution

Examples of Zeroes of Polynomials

How to Find Zeroes of a Polynomial

Example

Number of Zeroes of a Polynomial

Graphical Meaning of Zeroes

Zeroes of Special Polynomials

Non-Zero Constant Polynomial

Zero Polynomial

Introduction to Factor Theorem

Example