What are algebraic identities?

Algebraic identities are equations that remain true for every value of the variables. They help simplify algebraic expressions, perform quick calculations, and solve factorisation problems efficiently.

What is the difference between an algebraic expression and an algebraic identity?

An algebraic expression is a combination of variables, numbers and mathematical operations, whereas an algebraic identity is an equation that is true for every permissible value of the variables.

Which algebraic identities are most important for Class 9 Maths Chapter 4?

The most important identities include (a+b)², (a−b)², a²−b², (x+a)(x+b), x³−y³ and the identity x³+y³+z³−3xyz. These identities are frequently used in NCERT exercises and CBSE examinations.

How do I choose the correct factorisation method?

First check for a common factor. If there is no common factor, try grouping, middle-term splitting or applying an algebraic identity. The correct method depends on the structure of the algebraic expression.

What are Algebra Tiles and why are they useful?

Algebra Tiles are visual models used to represent algebraic expressions. They help students understand multiplication, factorisation and algebraic identities by arranging tiles into rectangles.

Can I cancel terms in a rational expression?

No. You should cancel only common factors, not individual terms. Always factorise the numerator and denominator completely before cancelling.

Are these Class 9 Maths Chapter 4 solutions based on the latest NCERT Ganita Manjari (2026)?

Yes. All Chapter 4 solutions on Maths Gurukulam are prepared according to the latest NCERT Ganita Manjari (2026) textbook and follow the current CBSE guidelines with simple step-by-step explanations.

📘 NCERT Ganita Manjari (2026) • CBSE Class 9 Mathematics • Chapter 4

Class 9 Maths
Chapter 4Exploring Algebraic Identities

Stop Memorising Formulas.
Start Understanding the Patterns Behind Them.

Master algebra through visual learning, Algebra Tiles, factorisation techniques, rational expressions, and complete Class 9 Maths Chapter 4 Solutions designed to make every concept simple, logical, and easy to remember.

📖 Learn Concepts 📝 Solve NCERT Exercises

Class 9 Maths Chapter 4 Exploring Algebraic Identities

Your Learning Dashboard

Everything you need to master Exploring Algebraic Identities is organised here. Learn the concepts first, practise them through NCERT exercises, and finish with a quick revision before your exam.

⏱

Study Time

6–8 Hours

⭐

Difficulty

Moderate

📘

Exercises

4.1 – 4.5 + End

🧠

Think & Reflect

Included

📄

Revision

30–45 Min

🎯

Exam Ready

CBSE Focus

By the End of this Chapter You Will Be Able To

Use IdentitiesRead Visual ProofsFactoriseSimplify Rational ExpressionsMaster NCERT

YOUR LEARNING JOURNEY

Learn Chapter 4 Like a Classroom

Follow this guided learning path to build your understanding step by step—from recognising patterns to confidently solving every NCERT question.

👀

Observe

Recognise algebraic patterns.

🎨

Visualise

Understand through visual proofs and Algebra Tiles.

🧠

Understand

Know why every identity always works.

✍️

Apply

Expand and transform expressions correctly.

✂️

Factorise

Use identities and Split the Middle Term method.

🏆

Master

Solve every NCERT exercise with confidence.

CHAPTER NAVIGATION

Explore every part of Chapter 4 in one simple, guided learning path.

📖 CHOOSE YOUR EXERCISE

Exercise Navigation Class 9 Maths Chapter 4 Solutions

Follow the chapter in the same sequence as NCERT and build your understanding step by step.

EXERCISE 4.1⭐ Beginner

Algebraic Identities

Learn standard identities and expand expressions quickly and accurately.

You’ll Learn

Basic identitiesPattern recognitionExpansion

Start hereStart Solving →

EXERCISE 4.2⭐⭐ Easy

Applications of Identities

Use identities for smart calculations and simpler expression-based problems.

You’ll Learn

Smart calculationsApplicationsSimplification

Build confidenceStart Solving →

EXERCISE 4.3⭐⭐ Moderate

Factorisation

Factorise expressions using common factors and algebraic identities.

You’ll Learn

Common factorUsing identitiesReverse expansion

Core skillStart Solving →

EXERCISE 4.4⭐⭐⭐ Moderate

Split the Middle Term

Factorise quadratic expressions step by step using the grouping method.

You’ll Learn

Number pairsGroupingQuadratics

Step-by-stepStart Solving →

EXERCISE 4.5⭐⭐⭐ Challenging

Rational Expressions

Factorise first, then cancel common factors correctly.

You’ll Learn

Rational expressionsFactorise firstRestrictions

Challenge yourselfStart Solving →

END EXERCISE⭐⭐⭐ Exam Ready

Complete Chapter Revision

Test the whole chapter with mixed NCERT questions.

You’ll Revise

IdentitiesFactorisationRational expressions

Final revisionStart Solving →

🎯 Suggested Order

Each exercise builds on the previous one, so following this sequence will make the chapter much easier to understand.

4.1 → 4.2 → 4.3 → 4.4 → 4.5 → End Exercise

📖 LEARN BEFORE YOU SOLVE

Before You Learn Algebraic Identities

Don’t memorise formulas first. Understand the ideas behind them and every identity will become easier to remember and apply.

💡

One Idea. Many Uses.

Algebraic Identities are mathematical patterns that always work. Once you understand them, you can expand expressions, factorise them, perform quick calculations and simplify rational expressions with confidence.

What Every Student Should Know

Before solving Exercise 4.1, understand these six important ideas.

🔍

Patterns Create Identities

Every identity begins with a repeating mathematical pattern. If the pattern is always true, it becomes an identity.

💡 Every identity starts with observation.

⚖️

Identity ≠ Equation

An identity is true for every value of the variables, while an equation is true only for specific value(s).

💡 An identity always works.

📐

Why NCERT Uses Diagrams

Squares and rectangles show exactly where every term comes from. Understanding the picture makes the identity easy.

💡 First understand, then memorise.

🔄

Expansion & Factorisation

Expansion opens brackets. Factorisation closes them again. They are simply opposite directions of the same algebraic identity.

💡 Factorisation is reverse expansion.

🚀

Why Learn Identities?

One identity helps you expand expressions, factorise quickly, perform smart calculations and simplify rational expressions.

💡 One concept. Many applications.

🎯

Where Will You Use Them?

Exercise 4.1 introduces identities. Later exercises use the same identities for factorisation, Algebra Tiles and rational expressions.

💡 Learn once. Apply everywhere.

🌟 CHAPTER SUCCESS FORMULA

Understand the Pattern,
Not the Formula

Every algebraic identity is built on a simple mathematical pattern. If you understand the pattern, you can expand, factorise and simplify expressions without memorising every formula separately.

👨‍🏫 Study This Chapter Like a Topper

🔍 Observe 📐 Visualise ✍️ Understand 🧮 Apply 🧩 Factorise 🏆 Master

Don’t ask “Which formula should I memorise?”
Ask “Which pattern do I recognise?”

⚡ QUICK FORMULA REFERENCE

Chapter 4 Formula Wall

All 11 NCERT Algebraic Identities in one place. Perfect for quick revision before solving questions or revising for exams.

📚

Your One-Minute Formula Revision Sheet

This Formula Wall is designed for fast revision. Learn the detailed concepts inside the exercise pages, then return here whenever you need a quick formula recap.

✅ 11 NCERT Identities 🔍 Pattern Recognition ⚡ One-Minute Revision 🎯 Exam Ready

💡 Maths Gurukulam Tip
Don’t memorise every formula separately. Recognise the pattern first— choosing the correct identity then becomes easy.

Identity 1 ⭐ MOST USED

Square of a Sum

(x + y)² = x² + 2xy + y²

📌 WHEN TO USE

When the entire expression with a plus sign is squared.

🧠 MEMORY TRICK

First² + Twice Product + Last²

👀 RECOGNISE THIS PATTERN

(a+b)² (3x+5)² (m+n)²

📖 Detailed explanation in Exercise 4.1

Learn More →

Identity 2 ⭐ VERY IMPORTANT

Square of a Difference

(x − y)² = x² − 2xy + y²

📌 WHEN TO USE

When the entire expression with a minus sign is squared.

🧠 MEMORY TRICK

First² − Twice Product + Last²

👀 RECOGNISE THIS PATTERN

(a−b)² (5x−2)² (m−n)²

⚠️ COMMON MISTAKE

Students often make the last term negative.

❌ x² − 2xy − y²
✅ x² − 2xy + y²

📖 Detailed explanation in Exercise 4.2

Learn More →

Identity 3 ⭐ EXAM FAVOURITE

Difference of Squares

(x + y)(x − y) = x² − y²

📌 WHEN TO USE

When one bracket has a plus (+) sign and the other has a minus (−) sign.

🧠 MEMORY TRICK

Same Terms • Opposite Signs → Squares Only

👀 RECOGNISE THIS PATTERN

(a+b)(a−b) (5x+3)(5x−3) (m+n)(m−n)

💡 QUICK FACT

The middle terms cancel automatically, leaving only the difference of the two squares.

📖 Detailed explanation in Exercise 4.2

Learn More →

Identity 4 ⭐ VERY USEFUL

Product of Two Binomials

(x + a)(x + b) = x² + (a + b)x + ab

📌 WHEN TO USE

When both brackets begin with the same variable but have different constants.

🧠 MEMORY TRICK

Square → Sum → Product

👀 RECOGNISE THIS PATTERN

(x+2)(x+5) (m+3)(m+7) (p+a)(p+b)

💡 QUICK FACT

The coefficient of x is the sum of the constants, while the constant term is their product.

📖 Detailed explanation in Exercise 4.3

Learn More →

Identity 5 ⭐ NEW CONCEPT

Square of Three Terms

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

📌 WHEN TO USE

When the sum of three terms is squared.

🧠 MEMORY TRICK

Three Squares + Twice Every Pair

👀 RECOGNISE THIS PATTERN

(a+b+c)² (x+y+1)² (m+n+p)²

💡 QUICK FACT

Square each term once, then add twice the product of every possible pair.

📖 Detailed explanation in Exercise 4.3

Learn More →

Identity 6 ⭐ CUBE IDENTITY

Cube of a Sum

(x + y)³ = x³ + 3x²y + 3xy² + y³

📌 WHEN TO USE

When the sum of two terms is raised to the power 3.

🧠 MEMORY TRICK

Cube • 3 • 3 • Cube

👀 RECOGNISE THIS PATTERN

(a+b)³ (2x+1)³ (m+n)³

💡 QUICK FACT

The coefficients 1, 3, 3, 1 always appear in the expansion of a cube of a sum.

📖 Detailed explanation in Exercise 4.4

Learn More →

Identity 7 ⭐ CUBE IDENTITY

Cube of a Difference

(x − y)³ = x³ − 3x²y + 3xy² − y³

📌 WHEN TO USE

When the difference of two terms is raised to the power 3.

🧠 MEMORY TRICK

Same 1 • 3 • 3 • 1 Pattern,
But the Signs Go + − + −

👀 RECOGNISE THIS PATTERN

(a−b)³ (2x−1)³ (m−n)³

💡 QUICK FACT

The coefficients remain 1, 3, 3, 1; only the signs alternate because of subtraction.

📖 Detailed explanation in Exercise 4.4

Learn More →

Identity 8 ⭐ FACTORISATION

Sum of Two Cubes

x³ + y³ = (x + y)(x² − xy + y²)

📌 WHEN TO USE

To factorise the sum of two perfect cubes.

🧠 MEMORY TRICK

Same Sign Outside • Opposite Sign Inside

👀 RECOGNISE THIS PATTERN

a³+b³ 8x³+27 64+y³

💡 QUICK FACT

The sign between x² and xy is always minus.

📖 Detailed explanation in Exercise 4.5

Learn More →

Identity 9 ⭐ FACTORISATION

Difference of Two Cubes

x³ − y³ = (x − y)(x² + xy + y²)

📌 WHEN TO USE

To factorise the difference of two perfect cubes.

🧠 MEMORY TRICK

Same Sign Outside • Same Sign Inside

👀 RECOGNISE THIS PATTERN

a³−b³ 27x³−8 125−y³

💡 QUICK FACT

The sign between x² and xy is always plus.

📖 Detailed explanation in Exercise 4.5

Learn More →

Identity 10 ⭐ ADVANCED

Three Cubes Identity

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

📌 WHEN TO USE

To factorise expressions involving the sum of three cubes and −3xyz.

🧠 MEMORY TRICK

Sum of Variables × Difference of Squares

👀 RECOGNISE THIS PATTERN

a³+b³+c³−3abc x³+y³+1−3xy

💡 QUICK FACT

This is the most advanced identity in this chapter and is mainly used for factorisation.

📖 Detailed explanation in Exercise 4.5

Learn More →

Identity 11 ⭐ PRACTICE SPECIAL

Product of Two Binomials

(x + a)(x + b) = x² + (a + b)x + ab

📌 WHEN TO USE

When both brackets begin with the same variable but have different constants.

🧠 MEMORY TRICK

Square • Sum • Product

👀 RECOGNISE THIS PATTERN

(x+2)(x+5) (m+3)(m+7) (p+a)(p+b)

💡 QUICK FACT

The middle term is obtained by adding the constants first, then multiplying by x.

📖 Detailed explanation in Exercise 4.5

Learn More →

Identity 12 ⭐ ADVANCED

General Product of Two Binomials

(ax + b)(cx + d) = acx² + (ad + bc)x + bd

📌 WHEN TO USE

When both brackets have different coefficients of the variable.

🧠 MEMORY TRICK

First × First • Cross Add • Last × Last

👀 RECOGNISE THIS PATTERN

(2x+3)(5x+4) (3m+1)(2m+7) (ax+b)(cx+d)

💡 QUICK FACT

Multiply the first terms, add the cross products for the middle term, and multiply the last terms.

📖 Detailed explanation in Exercise 4.3

Learn More →

👀 SEE THE PROOF

Visual Proof Gallery

Discover how squares, rectangles, and cubes transform into algebraic identities. See the picture. Understand the formula.

🟧 Geometry → 🟦 Area / Volume → 🟩 Algebraic Identity

🟧 VISUAL PROOF 1

Square of a Sum

Watch how one square creates an entire algebraic identity.

💡 Why does 2ab appear?

The figure contains two identical yellow rectangles. Each rectangle has area ab. Therefore, ab + ab = 2ab

a² + ab + ab + b²

↓

(a+b)² = a² + 2ab + b²

🟦 VISUAL PROOF 2

Square of a Difference

Start with one large square. Remove two regions. The remaining square reveals the identity.

Large Square

Area = a²

Rectangle Removed

Area = ab

Rectangle Removed

Area = b(a−b)

Remaining Square

Area = (a−b)²

💡 Why is the middle term −2ab ?

Two rectangular regions are removed from the large square. After simplifying, −ab − ab = −2ab

a² − ab − b(a−b)

↓

a² − ab − ab + b²

↓

(a−b)² = a² − 2ab + b²

🟩 VISUAL PROOF 3

Difference of Two Squares

NCERT proves this identity by rearranging one square into a rectangle.

Large Square

Area = a²

Vertical Strip

Width = b

Small Square

Area = b²

↷

Rearrange

Rectangle (a+b)(a−b)

💡 What’s Happening?

The large square has area a². NCERT separates one small square of area b². The remaining coloured part is rearranged to form a rectangle whose dimensions are (a+b) × (a−b)

a²

↓

(a+b)(a−b)+b²

↓

a²−b²=(a+b)(a−b)

🧠 Memory Trick

Whenever you see a² − b², think of two brackets, not two squares.

🟪 VISUAL PROOF 4

Square of Three Terms

One large square is divided into smaller squares and rectangles. Their combined areas build the identity naturally.

👀 What Can You See?

🟥

3 Squares

a², b² and c²

🟧

2ab

Two identical rectangles.

🟨

2bc

Another pair of rectangles.

🟩

2ca

The final rectangle pair.

💡 The Secret

The figure contains three squares and three pairs of identical rectangles. Each pair becomes a doubled product.

a² + b² + c²

2ab + 2bc + 2ca

↓

(a+b+c)² = a²+b²+c² + 2ab+2bc+2ca

🧠 Memory Trick

Remember it as 3 Squares + 3 Double Products. That’s the entire identity.

🧩 ALGEBRA TILES

Algebra Tiles
Learning Lab

Build Shapes. Discover Algebra.

Explore how NCERT Class 9 Maths Chapter 4 – Exploring Algebraic Identities uses Algebra Tiles to make multiplication, factorisation and algebraic identities easy to understand. Instead of memorising formulas, you’ll build expressions with simple shapes and discover the answer yourself.

👀 Observe

→

🧩 Build

→

📐 Visualise

→

✂️ Factorise

→

✅ Solve

🧩 STEP 1 • MEET THE TILES

Every Algebraic Expression Starts With Three Simple Tiles

Instead of memorising symbols, NCERT helps you see algebra as shapes. Every expression in this chapter is built using just three tiles.

🟧 Square Tile

x²

Represents a square with side x.

x × x = x²

🟩 Rectangle Tile

Represents one variable tile.

x × 1 = x

🟨 Unit Tile

Represents one unit square.

1 × 1 = 1

💡 Amazing Fact

Every algebraic identity, multiplication, factorisation and quadratic expression in Class 9 Maths Chapter 4 is created using different combinations of these three simple tiles.

🧩 STEP 2 • BUILD AN EXPRESSION

Build an Expression

Collect the Algebra Tiles for the expression before arranging them into a rectangle.

Expression

x² + 5x + 6

↓

Convert each term into its matching Algebra Tile.

Collect the Tiles

1 Square Tile = x²

5 Rectangle Tiles = 5x

6 Unit Tiles = 6

🤔 Think Before Moving On

You have collected 1 Square Tile, 5 Rectangle Tiles and 6 Unit Tiles.

Can you arrange them into one perfect rectangle?

📐 STEP 3 • BUILD THE RECTANGLE

Arrange the Algebra Tiles

Don’t guess the factors. Arrange the tiles into one complete rectangle and let the rectangle reveal the answer.

🟧

1 Square Tile

Forms the largest part of the rectangle.

🟩

5 Rectangle Tiles

Fill the sides of the square.

🟨

6 Unit Tiles

Complete the remaining empty spaces.

💡 The Secret

When every tile fits perfectly into a single rectangle, the rectangle is no longer just a picture. Its two side lengths become the two factors of the expression.

✨ Coming Next

Now read the two side lengths of the rectangle and discover (x + 2)(x + 3) without guessing.

📏 STEP 4 • READ THE FACTORS

The Rectangle Reveals the Factors

Once every tile fits perfectly, you don’t have to guess the answer. Simply read the two side lengths of the rectangle.

📏

Length

x + 3

📐

Width

x + 2

🎉 The Rectangle Has Solved It!

x² + 5x + 6

↓

(x + 2)(x + 3)

You didn’t memorise the factors. You discovered them by building a rectangle.

🧩 Tiles → 📐 Rectangle → 📏 Side Lengths → ✂️ Factors → ✅ Solution

🧠 CHOOSE THE RIGHT METHOD

How to Factorise Any Expression

Don’t Guess the Method. Ask the Right Question.

Every algebraic expression can be factorised, but the first step is choosing the correct method. This simple decision guide for Class 9 Maths Chapter 4 – Exploring Algebraic Identities helps you decide whether to use the Common Factor, Grouping, Split the Middle Term, or an Algebraic Identity.

🟧 Common Factor

→

🟩 Grouping

→

🟦 Split Middle Term

→

🟪 Identity

🚦 STEP 1 • FIND THE RIGHT METHOD

Ask This First Question

Before solving any factorisation question, don’t start multiplying or guessing. Simply follow this decision guide to choose the correct method.

START

↓

❓ Question 1

Is there a Common Factor?

✅

YES

Take the Common Factor first.

Go to Method 1

❌

NO

Move to the next question.

Continue ↓

➡️ Next Question

If there is no common factor, check whether the expression has 4 or more terms.

🟩 METHOD 1 • COMMON FACTOR

Always Check for a Common Factor First

This is the quickest method of factorisation. If every term has a common number or variable, take it outside first.

🔍 Ask Yourself

Can every term be divided by the same number or same variable?

Example

6x + 12

Factorisation

6x + 12

↓

6(x + 2)

🧠 Golden Rule

Never skip this step.

If a common factor exists, take it out before trying any other method.

🟩 METHOD 2 • GROUPING

Factorise by Grouping

If there is no common factor and the expression has four or more terms, group the terms in pairs and look for a common factor again.

🔍 Ask Yourself

Does the expression have 4 or more terms? Can I make two groups?

Example

ax + ay + bx + by

Factorise

a(x+y)

b(x+y)

↓

(a+b)(x+y)

🧠 Memory Trick

Group the terms. Take the common factor from each group. If the same bracket appears twice, take that bracket outside.

➡️ Next Method

If the expression has 3 terms and starts with x², learn Split the Middle Term.

🟦 METHOD 3 • SPLIT THE MIDDLE TERM

Split the Middle Term

Use this method when the expression has three terms, no common factor, and begins with a squared term.

🔍 Ask Yourself

Is it of the form ax² + bx + c with no common factor?

Example

x² + 5x + 6

Find two numbers whose Product = 6 and Sum = 5

Solution

x² + 2x + 3x + 6

↓

x(x+2)+3(x+2)

↓

(x+2)(x+3)

🧠 Memory Trick

Product → Sum → Split → Group → Factorise

➡️ Next Method

If the expression matches a known pattern like a² − b² or a² ± 2ab + b², use an Algebraic Identity.

🟪 METHOD 4 • IDENTITY BASED FACTORISATION

Factorise Using Algebraic Identities

Some expressions already match a well-known identity. Instead of splitting the middle term, apply the identity directly.

🔍 Ask Yourself

Does the expression look exactly like a known algebraic identity?

Example 1

a² − b²

↓

(a+b)(a−b)

Example 2

a²+2ab+b²

↓

(a+b)²

Example 3

a²−2ab+b²

↓

(a−b)²

Example 4

9x²−25

↓

(3x−5)(3x+5)

Example 5

4x²+12x+9

↓

(2x+3)²

🧠 Golden Rule

👀 Pattern Matches? → ✅ Apply Identity → ❌ Don’t Split Middle Term

Whenever an expression exactly matches a known identity, apply the identity directly. It is faster, simpler and avoids unnecessary calculations.

🎯 FACTORISATION MASTER CHECKLIST

Now You Know Which Method to Choose

Every factorisation question becomes easier when you ask the right question first instead of guessing the method.

🟧

Common Factor

Check this first for every expression.

🟩

Grouping

Use when the expression has 4 or more terms.

🟦

Split Middle Term

Use for quadratic trinomials.

🟪

Identity

Apply directly when the pattern matches.

⭐ Golden Order to Remember

🟧 Common Factor → 🟩 Grouping → 🟦 Split Middle Term → 🟪 Identity

🚀 You’re Ready!

Now that you know how to choose the correct factorisation method, you’re ready to solve the NCERT exercises with greater speed, accuracy and confidence.

📘 UNDERSTAND THE IDEA

What is a Rational Expression?

A Rational Expression is an algebraic fraction in which both the numerator and the denominator are algebraic expressions.

(x + 3)
──────────
(x − 2)

A simple example of a Rational Expression.

⬆️

Numerator

x + 3

The algebraic expression written above the fraction bar.

⬇️

Denominator

x − 2

The algebraic expression written below the fraction bar.

💡 Important Rule

A Rational Expression is defined only when the denominator is not equal to 0.

Always check the denominator before solving or simplifying.

✂️ FACTORISE • CANCEL • SIMPLIFY

How to Simplify a Rational Expression

Always factorise first, then cancel the common factors. Never cancel individual terms.

①

Factorise

Factorise the numerator and denominator.

②

Cancel

Cancel only the common factor.

③

Simplify

Write the remaining expression.

Worked Example

(x² + 5x + 6)
──────────────
(x + 2)

↓

((x + 2)(x + 3))
────────────────
(x + 2)

↓

x + 3

⚠️ Golden Rule

Never cancel terms.

Always factorise first, then cancel only the common factors.

⚡ 1-MINUTE REVISION

Chapter 4 Revision Hub

Before solving NCERT questions, spend one minute revising the five most important ideas from this chapter.

🟧

Algebraic Identities

Expand and factorise expressions using standard identities.

🟩

Factorisation

Use the correct method instead of guessing.

🟦

Algebra Tiles

Visualise expressions as shapes and read factors from rectangles.

🟪

Rational Expressions

Factorise first, then cancel only common factors.

🟨

Visual Proofs

Understand the picture before remembering the formula.

👀 Observe → 📐 Visualise → 🧮 Apply → ✂️ Factorise → ✅ Solve

🎯 FINAL SELF CHECK

Can You Answer These Questions?

If your answer is “Yes” to every question below, you’re ready to solve the NCERT exercises confidently.

✅ Can I identify the correct algebraic identity?

✅ Can I choose the correct factorisation method?

✅ Can I factorise an algebraic expression correctly?

✅ Can I simplify a rational expression?

✅ Can I explain an identity using a visual proof?

✅ Can I solve NCERT questions without memorising steps?

🚀 You’re Ready!

You have revised the complete chapter. Now it’s time to strengthen your understanding by solving the NCERT Exercise Questions step by step.

💡 NCERT 2026 • THINK & REFLECT SOLUTIONS

Class 9 Maths Chapter 4 Think & Reflect Solutions

Understand the Concept Before You Solve the Exercises.

Master the Class 9 Maths Chapter 4 Solutions through these Think & Reflect Solutions based on the latest NCERT Ganita Manjari (2026). Each solution is explained step by step to help you understand Exploring Algebraic Identities, Algebra Tiles, Factorisation and Rational Expressions before solving the NCERT exercises with confidence.

💡 Think

→

🔍 Understand

→

✍️ Solve

→

🎯 Master

📖 Chapter 4 • Page 65

Think and Reflect – 1 Solution

🔍 First Recall the Previous Pattern

On the previous page, we observed a surprising pattern for three consecutive square numbers.

1 + 9 − 2 × 4 = 2
25 + 49 − 2 × 36 = 2

In both examples, we added the smallest and largest square numbers and then subtracted twice the middle square(s). The result was always the same. Now NCERT asks us to investigate whether a similar pattern exists for four consecutive square numbers.

❓ Question

Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.

✍️ Solution

Instead of guessing the answer directly, let us try one example and carefully observe what happens.

Take four consecutive square numbers:

1, 4, 9, 16

Now add the first and last square numbers.

1 + 16 = 17

Also add the two middle square numbers.

4 + 9 = 13

Now compare these two sums.

17 − 13 = 4

Try another set of four consecutive squares:

4, 9, 16, 25

(4 + 25) − (9 + 16)
29 − 25
= 4

Again, we get the same answer.

🎯 Observation

For these examples, we observe the pattern:

(First Square + Last Square)
−
(Two Middle Squares)

= 4

So, for four consecutive square numbers, the difference between the sum of the first and last squares and the sum of the two middle squares is always 4.

Chapter 4 • Page 71 | Think and Reflect – 2

📖 Chapter 4 • Page 71

Think and Reflect – 2

❓ Question

What can you say about a and b if
(a + b)² < a² + b²
What can you say about a and b if
(a + b)² > a² + b²
When will
(a + b)² = a² + b²
Did you observe that a + b and a² + b² are both positive? What term will decide which is larger? Use the expansion of
(a + b)²
to decide.

✍️ Solution

To answer all four questions, first recall the algebraic identity

(a + b)² = a² + 2ab + b²

Now compare this with

a² + b²

Notice that the only extra term is 2ab. Therefore, the comparison depends only on whether 2ab is positive, negative or zero.

① When will (a + b)² be less than a² + b²?

(a+b)² < a²+b²
a²+2ab+b² < a²+b²
2ab < 0

This is possible only when a and b have opposite signs.

② When will (a + b)² be greater than a² + b²?

(a+b)² > a²+b²
a²+2ab+b² > a²+b²
2ab > 0

This happens when a and b have the same sign (both positive or both negative).

③ When will they be equal?

(a+b)² = a²+b²
2ab = 0

This is possible only when a = 0 or b = 0.

④ Which term decides the comparison?

The deciding term is

2ab

If 2ab is positive, then (a+b)² is larger. If 2ab is negative, then a²+b² is larger. If 2ab = 0, both expressions are equal.

✅ Final Answer

Condition	Conclusion
ab < 0	a and b have opposite signs, so (a+b)² < a² + b²
ab > 0	a and b have the same sign, so (a+b)² > a² + b²
ab = 0	Either a = 0 or b = 0, so (a+b)² = a² + b²
Deciding Term	The comparison depends only on the term 2ab.

📖 Chapter 4 • Page 73

Think and Reflect – 3

❓ Question

What if we replace b by −b in the identity

(a + b)² = a² + 2ab + b²

Solution

We already know the identity

(a + b)² = a² + 2ab + b²

Now replace b by −b everywhere.

Step 1

(a + (−b))²

a² + 2a(−b) + (−b)²

Step 2

Now simplify each term.

2a(−b) = −2ab
(−b)² = b²

Step 3

(a − b)² = a² − 2ab + b²

✅ Final Answer

(a − b)² = a² − 2ab + b²

Only the middle term changes its sign because replacing b by −b makes 2ab → −2ab, while (−b)² = b² remains positive.

📖 Chapter 4 • Page 76

Think and Reflect – 4 Solution

❓ Question

Label the squares and rectangles in Fig. 4.4 so that it represents the identity

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

Solution

The large square has side (a + b + c), so its total area is (a + b + c)². Divide this square into smaller squares and rectangles as shown below.

a²

b²

c²

Observe the Figure

There are three squares: a², b² and c².
There are two rectangles of area ab, so together they give 2ab.
There are two rectangles of area bc, so together they give 2bc.
There are two rectangles of area ac, so together they give 2ac.

✅ Final Answer

The figure contains a², b², c², two rectangles of area ab, two rectangles of area bc, and two rectangles of area ac. Therefore, the total area of the large square is

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

📖 Chapter 4 • Page 78

Think and Reflect – 5 Solution

❓ Question

Try to evaluate the following using a suitable identity:

(i) 35² (ii) 65² (iii) 85² (iv) 105²

Do you observe any interesting pattern?

Solution

Notice that every number ends with 5. Therefore, we can use the identity

(a+b)² = a² + 2ab + b²

Here, take b = 5.

(i) Find 35²

35² = (30 + 5)²
= 30² + 2(30)(5) + 5²
= 900 + 300 + 25
= 1225

(ii) Find 65²

65² = (60 + 5)²
= 60² + 2(60)(5) + 5²
= 3600 + 600 + 25
= 4225

(iii) Find 85²

85² = (80 + 5)²
= 80² + 2(80)(5) + 5²
= 6400 + 800 + 25
= 7225

(iv) Find 105²

105² = (100 + 5)²
= 100² + 2(100)(5) + 5²
= 10000 + 1000 + 25
= 11025

What Pattern Do You Observe?

Number	Square
35²	1225
65²	4225
85²	7225
105²	11025

✅ Final Answer

35² = 1225
65² = 4225
85² = 7225
105² = 11025

All these numbers end with 5, and their squares always end with 25. This pattern helps us calculate such squares quickly using the identity (a + b)².

📖 Chapter 4 • Page 78

Think and Reflect – 6 Solution

❓ Question

Observe the two rows of figures below. They represent an algebraic identity. Try to identify the identity.

💡 How Will We Find the Identity?

Instead of guessing the algebraic identity, we shall prove it using the areas of different coloured parts of the figure.

✅ First, find the dimensions of each coloured region.
✅ Next, calculate the area of every rectangle and square.
✅ Then, identify the common overlapping region.
✅ Finally, compare the total area with the area of the complete square to obtain the required algebraic identity.

In the next step, we shall begin by finding the dimensions and area of Rectangle R₁.

🟥 Step 1 : Find the Dimensions of Rectangle R₁

Before finding the area of Rectangle R₁, we must first determine its length and breadth from the given figure.

① Find the Width of Rectangle R₁

The total length of the upper side of the large square is

a + b + c

The white square occupies a length of

Therefore, the remaining horizontal length gives the width of Rectangle R₁.

Width of R₁
= (a + b + c) − 2c
= a + b − c

✔ Width of Rectangle R₁

a + b − c

② Find the Height of Rectangle R₁

Now observe the left side of the large square. The total height of the square is

a + b + c

The green square occupies a height of

Therefore, the remaining vertical length gives the height of Rectangle R₁.

Height of R₁
= (a + b + c) − 2b
= a − b + c

✔ Height of Rectangle R₁

a − b + c

We have now found both the dimensions of Rectangle R₁. In the next step, we shall use these dimensions to calculate its area.

③ Find the Area of Rectangle R₁

Now we know both the dimensions of Rectangle R₁.

Length (Height) of Rectangle R₁ = a − b + c

Breadth (Width) of Rectangle R₁ = a + b − c

We know that,

Area of Rectangle
=
Length × Breadth

Substituting the dimensions of Rectangle R₁,

Area of R₁
=
(a − b + c)(a + b − c)

✔ Area of Rectangle R₁

(a − b + c)(a + b − c)

We have successfully found the dimensions and area of Rectangle R₁. Next, we shall calculate the dimensions and area of Rectangle R₂ in the same way.

🟥 Step 2 : Find the Dimensions of Rectangle R₂

Now let us find the dimensions of Rectangle R₂. We shall begin by finding its horizontal side.

① Find the Horizontal Side of Rectangle R₂

The total length of the upper side of the large square is

a + b + c

The green square occupies a length of

Therefore, the remaining horizontal length gives the horizontal side of Rectangle R₂.

Horizontal Side of R₂
= (a + b + c) − 2b
= a − b + c

✔ Horizontal Side of Rectangle R₂

a − b + c

② Find the Vertical Side of Rectangle R₂

Now observe the right side of the large square. The total height of the square is

a + b + c

The white square occupies a height of

Therefore, the remaining vertical length gives the vertical side of Rectangle R₂.

Vertical Side of R₂
= (a + b + c) − 2c
= a + b − c

✔ Vertical Side of Rectangle R₂

a + b − c

We have now found both the dimensions of Rectangle R₂:
Horizontal Side = a − b + c
Vertical Side = a + b − c

In the next step, we shall use these dimensions to calculate the area of Rectangle R₂.

③ Find the Area of Rectangle R₂

Now we know both the dimensions of Rectangle R₂.

Horizontal Side of Rectangle R₂ = a − b + c

Vertical Side of Rectangle R₂ = a + b − c

We know that,

Area of Rectangle
=
Length × Breadth

Substituting the dimensions of Rectangle R₂,

Area of R₂
=
(a − b + c)(a + b − c)

✔ Area of Rectangle R₂

(a − b + c)(a + b − c)

We have now found the areas of both Rectangle R₁ and Rectangle R₂. In the next step, we shall find the common overlapping region, which has been counted in both rectangles.

🟪 Step 3 : Find the Side of the Common Overlapping Region

The two rectangles R₁ and R₂ overlap each other. To find the area of this common region, we must first determine its side length.

Figure 4.6 of Think and Reflect 6 from Class 9 Maths Chapter 4 Page 78 showing a square of side (a+b+c) and labelled regions | NCERT Ganita Manjari 2026

(All dimensions marked)

① Find the Horizontal Side of the Overlapping Region

The horizontal side of Rectangle R₂ is

a − b + c

Out of this length, the white square occupies

Therefore,

Horizontal Side
= (a − b + c) − 2c
= a − b − c

✔ Horizontal Side of the Common Region

a − b − c

② Find the Vertical Side of the Common Overlapping Region

Now let us find the vertical side of the common overlapping region. The vertical side of Rectangle R₁ is

a + b − c

Out of this length, the green square occupies

Therefore,

Vertical Side
= (a + b − c) − 2b
= a − b − c

✔ Vertical Side of the Common Region

a − b − c

💡 Important Observation

The horizontal side and the vertical side of the common region are both equal to

a − b − c

Since both sides are equal, the common overlapping region is a square. Therefore, its area is

✔ Area of the Common Overlapping Square

(a − b − c)²

🟩 Step 4 : Find the Areas of the Green and White Squares

Now let us calculate the areas of the remaining two squares shown in the figure.

① Area of the Green Square

The side of the green square is

Area of a square = Side × Side

Area
= (2b)²
= 4b²

✔ Area of the Green Square

4b²

② Area of the White Square

The side of the white square is

Area of a square = Side × Side

Area
= (2c)²
= 4c²

✔ Area of the White Square

4c²

🧮 Step 5 : Form the Area Equation

The side of the complete outer square is

a + b + c

Therefore,

Area of Complete Square
= (a+b+c)²

The same figure can also be obtained by combining:

Rectangle R₁ + Rectangle R₂ + Green Square + White Square − Common Overlapping Square

Required Area Equation

(a+b+c)² = R₁ + R₂ + 4b² + 4c² − (a−b−c)²

✅ Step 6 : Final Simplification & Algebraic Identity

Expanding and simplifying the area equation step by step:

✅ Final Answer

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

🎉 We have successfully proved this identity by comparing the area of the complete square with the areas of its different parts.

📖 Chapter 4 • Page 79

Think and Reflect – 7

❓ Question

Suppose 7x is split as 2x + 5x. Can a similar rectangular arrangement be formed? Consider other possibilities and check.

📚 Recall the Previous Figure

In the previous example, the algebra tiles were arranged by splitting 7x = 3x + 4x, which formed a rectangle of dimensions (x + 3) and (x + 4). Now, we shall investigate whether a similar rectangle can also be formed by splitting 7x = 2x + 5x.

In the next step, we will rearrange the algebra tiles by splitting 7x = 2x + 5x and observe whether a new rectangle can be formed.

🧩 Step 1 : Split 7x as 2x + 5x

In the previous arrangement, the 7x tiles were divided into 3x and 4x. Now, split them differently as

7x = 2x + 5x

Arrange the tiles again so that they form a complete rectangle without leaving any gaps.

Rectangular arrangement after splitting 7x into 2x and 5x

Observe the new arrangement carefully. In the next step, we will identify the length and breadth of the rectangle formed.

📏 Step 2 : Find the Dimensions of the Rectangle

From the new arrangement of algebra tiles, observe the sides of the rectangle carefully.

Horizontal Side

= x + 2

Vertical Side

= x + 5

Since the figure is a rectangle, its area is equal to

Area
=
Length × Breadth
=
(x + 2)(x + 5)

In the next step, we shall expand (x + 2)(x + 5) and compare the result with the original polynomial.

🧮 Step 3 : Find the Polynomial Represented by the Rectangle

The dimensions of the rectangle are (x + 2) and (x + 5). Therefore,

Area
= (x + 2)(x + 5)

= x(x + 5) + 2(x + 5)

= x² + 5x + 2x + 10

= x² + 7x + 10

Now compare this result with the previous arrangement.

Split of 7x	Rectangle	Polynomial
3x + 4x	(x + 3)(x + 4)	x² + 7x + 12
2x + 5x	(x + 2)(x + 5)	x² + 7x + 10

✅ Conclusion

Yes, a similar rectangular arrangement can be formed by splitting 7x = 2x + 5x. However, the rectangle now represents the polynomial x² + 7x + 10, which is different from the previous polynomial x² + 7x + 12. Hence, different ways of splitting the middle term can produce different rectangular arrangements and different polynomials.

📖 Chapter 4 • Page 79

Think and Reflect Question 8 Solution

🧩 What Will We Learn?

In this activity, we will use Algebra Tiles to understand how multiplication and factorisation work visually. Instead of memorising formulas, we will arrange tiles to form complete rectangles and discover the answers ourselves.

❓ NCERT Question

Figure out the product of (x + 2) and (x + 3) using Algebra Tiles.
Lay out Algebra Tiles for x² + 11x + 30 in such a way that you will see its factors.

🚀 Our Learning Journey

🟧 Represent Tiles

→

🟩 Build Rectangle

→

🟦 Read the Answer

→

🟪 Understand Why

👉 Next Step

We will first solve Part (i) by arranging the Algebra Tiles for (x + 2) and (x + 3). As the rectangle is formed, the product will appear automatically.

✍️ Solution (i)

Find the Product of (x + 2) and (x + 3) using Algebra Tiles

📌 Given

(x + 2)(x + 3)

We shall represent both expressions using Algebra Tiles and then arrange them into one complete rectangle.

Step 1 • Represent the Tiles

The first factor (x + 2) contains

🟦 One x tile
🟥 Two unit tiles

The second factor (x + 3) contains

🟦 One x tile
🟥 Three unit tiles

Step 2 • Arrange the Rectangle

Arrange all the Algebra Tiles to form one complete rectangle.

The completed rectangle clearly shows how multiplication takes place visually.

Step 3 • Count the Tiles

After arranging the rectangle, we observe:

🟨 1 square tile = x²
🟦 5 rectangle tiles = 5x
🟥 6 unit tiles = 6

✅ Final Answer

(x + 2)(x + 3)
=
x² + 5x + 6

💡 Observation

Instead of multiplying term by term, Algebra Tiles allow us to see the product visually. Once the rectangle is complete, the total area automatically gives the expanded algebraic expression.

✍️ Solution (ii)

Factorise x² + 11x + 30 Using Algebra Tiles

📌 Given

x² + 11x + 30

Represent this expression using Algebra Tiles and arrange them into one complete rectangle.

🧩 Step 1 • Collect the Tiles

🟧

1 x² Tile

🟩

11 x Tiles

🟨

30 Unit Tiles

📏 Step 3 • Read the Side Lengths

Length = x + 5

Width = x + 6

Therefore, the required factors are (x + 5) and (x + 6).

✅ Final Answer

x² + 11x + 30

↓

(x + 5)(x + 6)

✔ Verification

(x + 5)(x + 6)
↓
x² + 6x + 5x + 30
↓
x² + 11x + 30

💡 Key Learning

When Algebra Tiles form one complete rectangle, the length and width of the rectangle directly give the factors of the algebraic expression. This visual method makes factorisation easy to understand without memorising rules.

📖 Chapter 4 • Page 80

Think and Reflect Question 9 Solution

❓ Question

We have seen that

(x + 3)(x + 4)=x²+7x+12
(x + 6)(x + 7)=x²+13x+42

Generalise the above pattern to obtain an expression for (x + a)(x + b).

✍ Solution

Observe the two given examples carefully.

Expression	Expanded Form
(x+3)(x+4)	x²+7x+12
(x+6)(x+7)	x²+13x+42

From these examples we observe that:

The coefficient of x is obtained by adding the two numbers.
The constant term is obtained by multiplying the two numbers.

Now replace the numbers by two general numbers a and b.

(x+a)(x+b)
= x(x+b)+a(x+b)
= x²+bx+ax+ab
= x²+(a+b)x+ab

Hence, for every expression of the form (x+a)(x+b), the coefficient of x is always (a+b), and the constant term is always ab.

✅ General Rule

(x+a)(x+b) = x²+(a+b)x+ab

Coefficient of x = a+b

Constant Term = ab

📖 Chapter 4 • Page 80

Think and Reflect Question 10 Solution

❓ Question

James and Reshma are simplifying the algebraic expression

(a − b)²(a + b)

James first expands (a − b)², whereas Reshma first uses the identity (a-b)(a+b)=a²−b². Determine whether both students are correct.

✍ Solution

Let us verify both methods one by one.

Method 1 (James)

James first uses the identity

(a − b)² = a² − 2ab + b²

Substituting this into the given expression,

(a − b)²(a + b)
= (a² − 2ab + b²)(a + b)
= a³ + a²b − 2a²b − 2ab² + ab² + b³
= a³ − a²b − ab² + b³

Method 2 (Reshma)

Reshma groups one factor of (a-b) and first applies the identity

(a − b)(a + b)=a²−b²

Therefore,

(a − b)²(a + b)
= (a − b)(a² − b²)
= a³ − ab² − a²b + b³
= a³ − a²b − ab² + b³

Both methods give exactly the same simplified expression. This happens because both students used valid algebraic identities correctly. Although the order of applying the identities is different, the final result remains the same.

✅ Final Answer

Both James and Reshma are correct. They used different algebraic identities, but both methods simplify the expression correctly.

(a − b)²(a + b) = a³ − a²b − ab² + b³

📖 Chapter 4 • Page 85

Think and Reflect Question 11 Solution

❓ Question

We already know that

x² − y² = (x − y)(x + y)

We have also verified that

x³ − y³ = (x − y)(x² + xy + y²)

Observe that (x−y) is a common factor in both identities.

Do you think (x−y) is also a factor of x⁴−y⁴?

Can you predict whether it is also a factor of x⁵−y⁵?

✍️ Solution

Instead of solving immediately, let us first compare the two identities.

Identity	Factorised Form
x² − y²	(x−y)(x+y)
x³ − y³	(x−y)(x²+xy+y²)

👀 Compare Carefully

Now compare the two factorised forms.

Which factor appears in the first identity?
Which factor appears in the second identity?

Both identities contain the same factor

(x − y)

Did the first factor change? No. Only the second factor changed. This is the important pattern that NCERT wants us to observe.

🤔 Make Your Prediction

Since x²−y² contains (x−y) and x³−y³ also contains (x−y), what do you think about x⁴−y⁴?

Before reading further, pause for a moment and make your own prediction. Do you think (x−y) will again be a common factor?

✔ Let’s Verify Our Prediction

To check whether our prediction is correct, let us factorise x⁴ − y⁴.

Notice that x⁴ and y⁴ are perfect squares. So, we can write

x⁴ − y⁴
= (x²)² − (y²)²

Now apply the identity

a² − b² = (a − b)(a + b)

We get

x⁴ − y⁴
= (x² − y²)(x² + y²)

But we already know that

x² − y² = (x − y)(x + y)

Substituting this result,

x⁴ − y⁴
= (x − y)(x + y)(x² + y²)

✅ Our prediction was correct.

The expression x⁴ − y⁴ also contains the common factor (x − y).

🤔 Can You Predict Again?

Now compare all three identities.

Expression	Common Factor
x² − y²	(x − y)
x³ − y³	(x − y)
x⁴ − y⁴	(x − y)

What do you think about x⁵ − y⁵? Will it also contain the factor (x − y)? Take a moment to think before reading the answer.

In the next step, we will verify our prediction and then discover a beautiful new algebraic identity involving three variables.

✔ Let’s Check Our Prediction

If the same pattern continues, then x⁵ − y⁵ should also contain the common factor (x−y). Let us verify it.

x⁵−y⁵
=(x−y) (x⁴+x³y+x²y²+xy³+y⁴)

Our prediction is correct. The factor (x−y) appears once again.

🌟 What Have We Learnt?

From all four identities, we notice one beautiful mathematical pattern.

Expression	Common Factor
x²−y²	(x−y)
x³−y³	(x−y)
x⁴−y⁴	(x−y)
x⁵−y⁵	(x−y)

Whenever you see an expression of the form xⁿ−yⁿ, always check whether (x−y) is a common factor. This simple observation helps us factorise many algebraic expressions.

🚀 Let’s Explore One More Identity

So far, we have explored identities involving two variables. Now NCERT introduces a new identity involving three variables. Instead of guessing, let us verify it by multiplying the two expressions step by step.

✍️ Let’s Verify the New Identity

NCERT now introduces a beautiful identity involving three variables. Our goal is to check whether multiplying the following two expressions really gives the required result.

We need to verify

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

To verify this identity, multiply each term of (x+y+z) with every term of the second bracket.

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

After multiplying, several positive and negative terms cancel each other. Only four terms remain.

x³+y³+z³−3xyz

Hence, the identity is verified.

Identity Verified

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

📖 Chapter 4 • Page 87

Think and Reflect Question 12 Solution

❓ Question

Simplify the following rational expression.

36s² − 12st + t²

t² + 2ts − 48s²

Hint: Start by factorising the denominator t² + 2ts − 48s².

✍️ Solution

Whenever we simplify a rational expression, we should first factorise the denominator and the numerator. Since NCERT has already given us a hint, let us begin with the denominator.

👀 Step 1 • Factorise the Denominator

We have

t² + 2ts − 48s²

Find two numbers whose

Product is −48.
Sum is 2.

The required numbers are 8 and −6.

t² + 2ts − 48s²
=
(t + 8s)(t − 6s)

✅ The denominator has now been factorised.

🤔 Before Moving Ahead…

Now look carefully at the numerator.

36s² − 12st + t²

Can you recognise it as a familiar algebraic identity? Try rewriting it in descending powers of t before moving to the next step.

✔ Step 2 • Factorise the Numerator

To recognise the algebraic identity more easily, rewrite the numerator in descending powers of t.

36s² − 12st + t²
=
t² − 12st + 36s²

Now compare it with the identity

a² − 2ab + b² = (a − b)²

Here, a=t and b=6s. Therefore,

t² − 12st + 36s²
=
(t−6s)²

✅ Now both the numerator and denominator have been factorised.

✔ Step 3 • Substitute the Factors

Replace the numerator and denominator with their factorised forms.

(t−6s)²

(t+8s)(t−6s)

The factor (t−6s) appears in both the numerator and the denominator. So, we can cancel one common factor.

(t−6s)

t+8s

✅ Final Answer

(t−6s)

t+8s

💡 What Did We Learn?

To simplify any rational expression, always remember these four steps:

Factorise the denominator.
Factorise the numerator.
Cancel only the common factors (never individual terms).
Write the simplified expression.

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Class 9 MathsChapter 4Exploring Algebraic Identities

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Learn Chapter 4 Like a Classroom

Table of Contents

Exercise Navigation Class 9 Maths Chapter 4 Solutions

Algebraic Identities

Applications of Identities

Factorisation

Split the Middle Term

Rational Expressions

Complete Chapter Revision

Before You Learn Algebraic Identities

One Idea. Many Uses.

What Every Student Should Know

Patterns Create Identities

Identity ≠ Equation

Why NCERT Uses Diagrams

Expansion & Factorisation

Why Learn Identities?

Where Will You Use Them?

Understand the Pattern,Not the Formula

👨‍🏫 Study This Chapter Like a Topper

Chapter 4 Formula Wall

Your One-Minute Formula Revision Sheet

Square of a Sum

Square of a Difference

Difference of Squares

Product of Two Binomials

Square of Three Terms

Cube of a Sum

Cube of a Difference

Sum of Two Cubes

Difference of Two Cubes

Three Cubes Identity

Product of Two Binomials

General Product of Two Binomials

Visual Proof Gallery

Square of a Sum

Square of a Difference

Difference of Two Squares

Square of Three Terms

👀 What Can You See?

3 Squares

2ab

2bc

2ca

💡 The Secret

Algebra Tiles Learning Lab

Every Algebraic Expression Starts With Three Simple Tiles

🟧 Square Tile

🟩 Rectangle Tile

🟨 Unit Tile

Build an Expression

Expression

Collect the Tiles

🤔 Think Before Moving On

Arrange the Algebra Tiles

1 Square Tile

5 Rectangle Tiles

6 Unit Tiles

💡 The Secret

The Rectangle Reveals the Factors

Length

Width

🎉 The Rectangle Has Solved It!

How to Factorise Any Expression

Ask This First Question

YES

NO

Always Check for a Common Factor First

🧠 Golden Rule

Factorise by Grouping

Example

Factorise

🧠 Memory Trick

Split the Middle Term

Example

Solution

🧠 Memory Trick

Factorise Using Algebraic Identities

Class 9 Maths
Chapter 4Exploring Algebraic Identities

Understand the Pattern,
Not the Formula

Algebra Tiles
Learning Lab

① When will (a + b)² be less than a² + b²?

② When will (a + b)² be greater than a² + b²?

🟥 Step 1 : Find the Dimensions of Rectangle R₁

① Find the Width of Rectangle R₁

② Find the Height of Rectangle R₁

③ Find the Area of Rectangle R₁

🟥 Step 2 : Find the Dimensions of Rectangle R₂

① Find the Horizontal Side of Rectangle R₂

② Find the Vertical Side of Rectangle R₂

③ Find the Area of Rectangle R₂