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    HomeClass 9 Maths SolutionsClass 9 Maths Chapter 4 – Exploring Algebraic Identities
    Class 9 Maths Chapter 4 End of Chapter Exercise Solutions
    R K Singh 08/08/2026 No Comments

    Class 9 Maths Chapter 4 End of Chapter Exercise Solutions | Ganita Manjari 2026 – Exploring Algebraic Identities

    📘 NCERT Ganita Manjari (2026) 📚 CBSE 2026–27 🏆 Step-by-Step Solutions

    Class 9 Maths Chapter 4 End of Chapter Exercise Solutions

    Complete Chapter Revision with Step-by-Step NCERT Solutions

    Prepare for your Class 9 Maths examination with complete Class 9 Maths Chapter 4 End of Chapter Exercise Solutions based on the latest NCERT Ganita Manjari (2026). This chapter-end exercise brings together important concepts from Chapter 4 – Exploring Algebraic Identities, including algebraic identities, expansions, factorisation, cube identities, rational expressions and application-based problems. Each question is explained in a clear, step-by-step CBSE answer-writing style to help students choose the correct method, understand the algebraic process and revise the chapter confidently.

    📖
    Exercise
    End of Chapter Exercise
    ❓
    Questions
    13 Questions
    📚
    Coverage
    Entire
    Chapter
    🧠
    Skills
    Mixed Problem Solving
    ⏱️
    Study Time
    90–120 Min
    ⭐
    Difficulty
    Moderate to Challenging
    🎯
    Exam Importance
    ★★★★★
    🎯 By the End of This Exercise, You Will Be Able To…
    ✅ Choose the Correct Algebraic Identity
    ✅ Expand Algebraic Expressions Accurately
    ✅ Factorise Expressions Completely
    ✅ Simplify Rational Expressions Correctly
    ✅ Solve Application-Based Problems
    ✅ Write CBSE Exam-Ready Solutions

    📑 Table of Contents

    📖 About Class 9 Maths Chapter 4 End of Chapter Exercise

    Class 9 Maths Chapter 4 End of Chapter Exercise Solutions provide complete, step-by-step answers to the questions given at the end of Chapter 4 – Exploring Algebraic Identities. This chapter-end exercise brings together important concepts studied throughout the chapter and gives students an opportunity to apply algebraic identities, expansions, factorisation and related algebraic methods in different types of questions.

    These NCERT Class 9 Maths solutions are prepared according to the latest NCERT Ganita Manjari (2026) textbook and are aligned with the CBSE Class 9 Mathematics curriculum. Each answer is presented in a clear, student-friendly, step-by-step format to support classroom practice, homework, self-study, chapter revision and examination preparation.

    🎯 Exercise Snapshot

    📘 What You’ll Find
    • Complete Chapter 4 end-of-chapter revision.
    • NCERT questions with question-wise solutions.
    • Clear, step-by-step CBSE answer-writing approach.
    📚 Page Includes
    • Complete End of Chapter Exercise solutions.
    • Important Chapter 4 concepts for revision.
    • Simple and exam-ready explanations.
    🏆 Best For
    • Students following Ganita Manjari (2026).
    • Homework, classroom practice and self-study.
    • CBSE Class 9 exam preparation and revision.

    🔎 Jump to Any Question

    Quickly jump to any question from the 13 End-of-Chapter Exercises of Class 9 Maths Chapter 4.

    Regular Question ★ Higher-Order Question

    📚 Learn Before You Solve

    🎯 Are You Ready for the End-of-Chapter Exercise?

    Before solving the Class 9 Maths Chapter 4 End-of-Chapter Exercise, quickly revise the important methods and concepts from Chapter 4 – Exploring Algebraic Identities. This revision guide connects you directly to the important Class 9 Maths Chapter 4 learning sections so that you can review a method before attempting the End of Chapter Exercise.

    🧠 What should you revise before solving?
    🔹 How to Solve a Quadratic Equation? — revise how to form, factorise and solve a quadratic equation step by step
    🔹 Which Method Should I Use? — use the decision tree to identify the suitable method for algebraic expressions and factorisation
    🔹 Strategy to Solve Any Question — follow a simple step-by-step approach to read, identify, choose, solve and verify
    🔹 Quick Revision Dashboard — revise the important Chapter 4 methods, key ideas and common mistakes before solving
    🧠 Need a Quick Revision? — open the Quick Revision Dashboard for a final review of Class 9 Maths Chapter 4 before attempting the End-of-Chapter Exercise
    👆 Click any topic above to jump directly to the relevant learning section.

    Most questions in the Class 9 Maths Chapter 4 End of Chapter Exercise are based on the concepts you have already learned throughout this chapter. However, a few questions introduce new concepts for the first time. Don’t worry! Before solving the exercise, let’s understand these concepts in a simple classroom style with easy explanations and worked examples.

    🔢 How to Solve a Quadratic Equation?

    A quadratic equation is an equation in which the highest power of the variable is 2. In this chapter, we will solve quadratic equations using the factorisation method.

    Example
    x² − 5x + 6 = 0
    👨‍🏫 Teacher Explains

    Don’t try to factorise the expression immediately. First, carefully observe the middle term. We need to split the middle term into two parts so that we can take common factors from each group.

    Step 1 — Split the Middle Term

    Find two numbers whose product is +6 and whose sum is −5. These numbers are −2 and −3.

    x² − 5x + 6 = 0
    x² − 2x − 3x + 6 = 0
    👉 What Next?

    Now that we have split the middle term, we can group the terms and take common factors. This is the key idea behind the factorisation method.

    Step 2 — Group the Terms

    Now group the first two terms and the last two terms. This helps us take a common factor from each group separately.

    (x² − 2x) − (3x − 6) = 0
    Step 3 — Take the Common Factor

    From the first group, the common factor is x. From the second group, the common factor is −3. Taking these common factors gives both groups the same bracket.

    x(x − 2) − 3(x − 2) = 0
    💡 Observe Carefully

    Look at both terms carefully.

    x(x − 2)
    −3(x − 2)

    Both terms contain the common factor (x − 2). Since the same bracket appears in both terms, we can take it common.

    Step 4 — Take the Common Bracket

    Take (x − 2) common from both terms.

    (x − 2)(x − 3) = 0

    Excellent! We have successfully factorised the quadratic equation.

    Step 5 — Find the Value of x

    The product of two factors is zero.

    (x − 2)(x − 3)=0

    This is possible only when at least one factor is equal to zero.

    x − 2 = 0
    or
    x − 3 = 0

    x = 2
    or
    x = 3
    ✍️ Student Notebook Solution
    x² − 5x + 6 = 0
    x² − 2x − 3x + 6 = 0
    x(x − 2) − 3(x − 2) = 0
    (x − 2)(x − 3) = 0
    x − 2 = 0   or   x − 3 = 0
    x = 2   or   x = 3
    ✅ Remember
    ✔ Split the middle term correctly.
    ✔ Take common factors carefully.
    ✔ Factorise completely before solving.
    ✔ Write both values of the variable.

    📝 How to Solve Word Problems?

    Word problems may look lengthy, but they become easy if we solve them step by step. We simply convert the given information into Mathematics and then solve the equation carefully.

    👨‍🏫 Teacher Explains

    Let’s understand the complete method through a simple example. Read every step carefully and notice why we perform each operation.

    Example

    The product of two consecutive positive integers is 56. Find the integers.

    Step 1 — Assume the Variable

    Suppose the first integer is x. Since the integers are consecutive, the next integer will be x + 1.

    First Integer = x
    Second Integer = x + 1
    Step 2 — Form the Equation

    The product of the two integers is 56.

    x(x + 1)=56
    Step 3 — Convert into a Quadratic Equation

    First expand the brackets and then bring every term to one side.

    x² + x = 56
    x² + x − 56 = 0
    Step 4 — Solve by Factorisation

    Now solve this quadratic equation exactly as we learned earlier.

    x² + 8x − 7x − 56 = 0
    x(x + 8) − 7(x + 8) = 0
    (x + 8)(x − 7)=0
    x + 8 =0   or   x − 7 =0
    x = −8   or   x = 7
    👉 Final Observation

    The question asks for positive integers. Therefore, we reject −8 and take only 7.
    ∴ x = 7 and other integer be x + 1 = 7 + 1 = 8

    Required Integers = 7 and 8
    ✍️ Student Notebook Solution
    Let the first integer = x
    Second integer = x + 1

    x(x + 1)=56
    x² + x − 56 =0
    x² + 8x − 7x − 56 =0
    x(x + 8) − 7(x + 8)=0
    (x + 8)(x − 7)=0
    x = −8   or   x =7
    The question asks for positive integers.
    we reject −8 and take only 7
    ∴ x = 7 and other integer be x + 1 = 7 + 1 = 8
    Required integers are 7 and 8
    ✅ Remember
    ✔ Read the question carefully.
    ✔ Assume the unknown quantity.
    ✔ Form the equation correctly.
    ✔ Reject impossible values.
    ✔ Always answer in words.

    🧠 Strategy to Solve Any Question

    Feeling confused about where to begin? Follow this simple five-step strategy whenever you solve an algebra question. It will help you identify the correct concept, choose the appropriate method, and avoid unnecessary mistakes while solving the Class 9 Maths Chapter 4 questions.

    📖
    Read the Question
    ➡️
    🔍
    Identify the Concept
    ➡️
    🧩
    Choose the Correct Method
    ➡️
    ✍️
    Solve Step by Step
    ➡️
    ✅
    Verify the Answer
    💡 Teacher’s Advice

    Never rush to solve the question immediately. First identify which concept the question is testing. Once you choose the correct method, solving the question becomes much easier. This simple strategy works for almost every algebra question in Class 9 Maths Chapter 4.

    🌳 Which Method Should I Use?

    Not sure how to start a question? Don’t worry! Follow this simple decision tree. Answer each question one by one, and you’ll know exactly which method to use while solving Class 9 Maths Chapter 4 questions.

    📖 Read the Question Carefully
    ⬇
    🔍 What Type of Question Is It?
    ⬇

    🧩 Is it an Expression?

    ⬇
    Does it match an Algebraic Identity?
    ⬇
    YES ✅

    Use the suitable algebraic identity.

    NO ❌

    Can it be factorised?

    ⬇
    Choose the Correct Method
    • Take Common Factor
    • Use Identity
    • Middle-term Splitting

    ➗ Is it a Rational Expression?

    ⬇
    Follow these Steps
    1️⃣ Factorise Numerator
    2️⃣ Factorise Denominator
    3️⃣ Cancel Common Factors
    4️⃣ Simplify

    📝 Is it a Word Problem?

    ⬇
    Follow these Steps
    1️⃣ Assume the Variable
    2️⃣ Form the Equation
    3️⃣ Is it Quadratic?
    4️⃣ Factorise
    5️⃣ Find the Roots
    6️⃣ Verify the Answer
    ⬇
    🎯 Final Answer

    Choose the correct method first, then solve the question step by step. This simple strategy works for almost every question in Chapter 4 – Exploring Algebraic Identities.

    📒 Complete Formula Revision Sheet
    Revise every important algebraic identity from Chapter 4 before solving the End Exercise.

    ⭐ Basic Algebraic Identities

    ① Square of a Sum
    (a + b)² = a² + 2ab + b²
    ② Square of a Difference
    (a − b)² = a² − 2ab + b²
    ③ Difference of Squares
    a² − b² = (a + b)(a − b)
    ④ Square of Three Terms
    (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

    ⭐ Product Formulae

    ⑤ Product of Two Binomials
    (x + a)(x + b) = x² + (a + b)x + ab
    ⑥ General Product Formula
    (ax + b)(cx + d) = acx² + (ad + bc)x + bd

    ⭐ Cube Identities

    ⑦ Cube of a Sum
    (a + b)³ = a³ + 3a²b + 3ab² + b³
    ⑧ Cube of a Difference
    (a − b)³ = a³ − 3a²b + 3ab² − b³
    ⑨ Sum of Cubes
    a³ + b³ = (a + b)(a² − ab + b²)
    ⑩ Difference of Cubes
    a³ − b³ = (a − b)(a² + ab + b²)

    ⭐ Special Identities

    ⑪ Three Variable Cube Identity
    x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − xz)
    ⑫ Important Result
    If  x + y + z = 0,

    then  x³ + y³ + z³ = 3xyz

    or

    x³ + y³ + z³ − 3xyz = 0

    ⭐ Useful Factorisation Formula

    Product of Two Binomials
    (x + a)(x + b) = x² + (a + b)x + ab

    ⭐ Middle-Term Splitting Pattern

    x² + bx + c

    Find two numbers whose

    ➕ Sum = b
    ✖ Product = c
    Then split the middle term.

    ⭐ Rational Expression Reminder

    1️⃣ Factorise the Numerator
    ⬇
    2️⃣ Factorise the Denominator
    ⬇
    3️⃣ Cancel only Common Factors
    ⬇
    4️⃣ Write the Simplified Expression

    ⭐ Quadratic Equation Reminder

    ax² + bx + c = 0
    Split the Middle Term
    ⬇
    Factorise Completely
    ⬇
    Apply the Zero Product Property
    ⬇
    Find Both Roots

    ⚡ Quick Revision Dashboard

    Before solving the Class 9 Maths Chapter 4 End Exercise, quickly revise the key identities, factorisation methods, rational expressions, and applications you have studied throughout Exploring Algebraic Identities.

    📘 1. Identity Toolkit — Know the Patterns

    Square of a Sum
    (a + b)2 = a2 + 2ab + b2
    Use when a binomial is squared.
    Square of a Difference
    (a − b)2 = a2 − 2ab + b2
    Remember: the middle term is negative.
    Square of Three Terms
    (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
    Difference of Squares
    a2 − b2 = (a + b)(a − b)
    Look for two perfect squares with a subtraction sign.
    Cubes
    (a + b)3 = a3 + 3a2b + 3ab2 + b3
    (a − b)3 = a3 − 3a2b + 3ab2 − b3
    Sum & Difference of Cubes
    a3 − b3 = (a − b)(a2 + ab + b2)
    a3 + b3 = (a + b)(a2 − ab + b2)
    Three-Variable Cube Identity
    a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)

    🔍 2. Recognise the Pattern Before You Calculate

    Binomial Squared?

    Think of (a + b)2 or (a − b)2.

    Two Squares with −?

    Check for a2 − b2.

    Cubic Expression?

    Check for sum or difference of cubes.

    Three Terms with Squares?

    Check whether (a + b + c)2 applies.

    🧩 3. Factorisation Toolkit — Work from Structure

    Step 1 — Common Factor

    First check whether every term has a common factor.

    Step 2 — Identity

    Check whether the expression matches a known algebraic identity.

    Step 3 — Middle-Term Splitting

    For expressions such as x2 + (a+b)x + ab, find two numbers whose sum gives the coefficient of x and whose product gives the constant term.

    Step 4 — Verify

    Multiply the factors when needed to check that the original expression is obtained.

    ➗ 4. Simplifying Rational Expressions

    1️⃣ Factorise the numerator
    2️⃣ Factorise the denominator
    3️⃣ Identify common factors
    4️⃣ Cancel common factors
    5️⃣ Write the simplest form
    Important: Cancel only common factors, not individual terms. The denominator must not be equal to zero.

    📝 5. Apply Factorisation to Problems

    Rectangle / Area

    Express the area, factorise it, and interpret the factors as possible dimensions.

    Cuboid / Volume

    Factorise the volume expression and identify possible length, breadth and height.

    Word Problem

    Assume the variable, form the equation, factorise, find the possible values, and reject values that do not make physical sense.

    🚫 6. Check These Before Moving On

    ✘ Choosing the wrong identity
    ✘ Losing the sign of the middle term
    ✘ Incorrect middle-term splitting
    ✘ Cancelling terms instead of factors
    ✘ Forgetting the denominator condition
    ✘ Accepting an impossible value in an application

    🎯 7. Can You Recall the Chapter?

    Identity:
    Can I recognise the pattern before expanding?
    Factorisation:
    Can I choose the correct method?
    Rational Expression:
    Can I factorise before cancelling?
    Application:
    Can I interpret the factors in the problem?
    Study Tip: Do not just read the formulas. Look at an expression, identify its pattern, recall the suitable identity or factorisation method, and then solve it step by step. You are ready for the End Exercise when you can choose the method before starting the calculation.

    📝 Class 9 Maths Chapter 4 End Exercise Solutions

    Solve every question of the Class 9 Maths Chapter 4 End Exercise with easy, step-by-step NCERT Ganita Manjari (2026) solutions. This End Exercise combines all the important concepts of the chapter, including algebraic identities, expanding expressions, factorisation, middle-term splitting, rational expressions, cube identities, quadratic equations, and word problems. Every solution follows the latest CBSE answer-writing style with complete working, making it ideal for homework, revision, and exam preparation.

    📝 Step-by-Step Solutions 🎯 NCERT & CBSE Aligned ⭐ Complete End Exercise
    Question 1 9 Parts (i–ix)

    📝 Question 1

    Use suitable identities to find the following products.

    (i)   (−3x + 4)²
    (ii)   (2s + 7)(2s − 7)
    (iii)   ( p² + 1 2 ) ( p² − 1 2 )
    (iv)   (2n + 7)(2n − 7)
    (v)   (s − 2t)(s² + 2st + 4t²)
    (vi)   ( 1 2r − 4r )²
    (vii)   (−3m + 4k − l)²
    (viii)   (x − ⅓y)³
    (ix)   ( 7r − 2m 3 )³
    Question 1 (i)
    Use Suitable Identity
    Find:
    (-3x + 4)²
    ✍️ Solution
    Given,
    (-3x + 4)²
    Using the identity,
    (a + b)² = a² + 2ab + b²
    Here,
    a = -3x,    b = 4
    ∴ (-3x + 4)²
    = (-3x)² + 2 × (-3x) × 4 + 4²
    = 9x² + 2 × (-3x) × 4 + 16
    = 9x² + (-24x) + 16
    = 9x² − 24x + 16
    Therefore,
    (-3x + 4)² = 9x² − 24x + 16
    ✅ Final Answer
    (-3x + 4)² = 9x² − 24x + 16
    Question 1 (ii)
    Use Suitable Identity
    Find:
    (2s + 7)(2s − 7)
    ✍️ Solution
    Given,
    (2s + 7)(2s − 7)
    Using the identity,
    (a + b)(a − b) = a² − b²
    Here,
    a = 2s,    b = 7
    ∴ (2s + 7)(2s − 7)
    = (2s)² − 7²
    = 4s² − 49
    Therefore,
    (2s + 7)(2s − 7) = 4s² − 49
    ✅ Final Answer
    (2s + 7)(2s − 7) = 4s² − 49
    Question 1 (iii)
    Use Suitable Identity
    Find:
    ( p² + 1 2 ) ( p² − 1 2 )
    ✍️ Solution
    Given,
    ( p² + 1 2 ) ( p² − 1 2 )
    Using the identity,
    (a + b)(a − b) = a² − b²
    Here,
    a = p²,    b = 1 2
    ∴ ( p² + 1 2 ) ( p² − 1 2 )
    = (p²)² − ( 1 2 )²
    = p⁴ − 1² 2²
    = p⁴ − 1 4
    Therefore,
    ( p² + 1 2 ) ( p² − 1 2 ) = p⁴ − 1 4
    ✅ Final Answer
    ( p² + 1 2 ) ( p² − 1 2 ) = p⁴ − 1 4
    Question 1 (iii)
    Use Suitable Identity
    Find:
    ( p² + 1 2 ) ( p² − 1 2 )
    ✍️ Solution
    Given,
    ( p² + 1 2 ) ( p² − 1 2 )
    Using the identity,
    (a + b)(a − b) = a² − b²
    Here,
    a = p²,   b = 1 2
    ∴ ( p² + 1 2 ) ( p² − 1 2 )
    = (p²)² − ( 1 2 )²
    = p⁴ − 1² 2²
    = p⁴ − 1 4
    ✅ Final Answer
    ( p² + 1 2 ) ( p² − 1 2 ) = p⁴ − 1 4
    Question 1 (iv)
    Use Suitable Identity
    Find:
    (2n + 7)(2n − 7)
    ✍️ Solution
    Given,
    (2n + 7)(2n − 7)
    Using the identity,
    (a + b)(a − b) = a² − b²
    Here,
    a = 2n,    b = 7
    ∴ (2n + 7)(2n − 7)
    = (2n)² − 7²
    = 4n² − 49
    ✅ Final Answer
    (2n + 7)(2n − 7)
    = 4n² − 49
    Question 1 (v)
    Use Suitable Identity
    Find:
    (s − 2t)(s² + 2st + 4t²)
    ✍️ Solution
    Given,
    (s − 2t)(s² + 2st + 4t²)
    Using the identity,
    (a − b)(a² + ab + b²) = a³ − b³
    Here,
    a = s,    b = 2t
    ∴ (s − 2t)(s² + 2st + 4t²)
    = s³ − (2t)³
    = s³ − 2³t³
    = s³ − 8t³
    ✅ Final Answer
    (s − 2t)(s² + 2st + 4t²) = s³ − 8t³
    Question 1 (vi)
    Use Suitable Identity
    Find:
    ( 1 2r − 4r )²
    ✍️ Solution
    Given,
    ( 1 2r − 4r )²
    Using the identity,
    (a − b)² = a² − 2ab + b²
    Here,
    a = 1 2r ,    b = 4r
    ∴ ( 1 2r − 4r )²
    = 1 2r ² − 2 × 1 2r × 4r + (4r)²
    = 1² (2r)² − 8r 2r + 16r²
    = 1 4r² − 4 + 16r²
    ✅ Final Answer
    ( 1 2r − 4r )² = 1 4r² − 4 + 16r²
    Question 1 (vii)
    Use Suitable Identity
    Find:
    (-3m + 4k − l)²
    ✍️ Solution
    Given,
    (-3m + 4k − l)²
    Using the identity,
    (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
    Here,
    a = -3m,   b = 4k,   c = -l
    ∴ (-3m + 4k − l)²
    = (-3m)² + (4k)² + (-l)²
      + 2 × (-3m) × 4k
      + 2 × 4k × (-l)
      + 2 × (-3m) × (-l)
    = 9m² + 16k² + l²
      − 24mk − 8kl + 6ml
    = 9m² + 16k² + l² − 24mk + 6ml − 8kl
    ✅ Final Answer
    (-3m + 4k − l)² = 9m² + 16k² + l² − 24mk + 6ml − 8kl
    Question 1 (viii)
    Use Suitable Identity
    Find:
    ( x − 1 3 y )³
    ✍️ Solution
    Given,
    ( x − 1 3 y )³
    Using the identity,
    (a − b)³ = a³ − 3a²b + 3ab² − b³
    Here,
    a = x,   b = 1 3 y
    ∴ ( x − 1 3 y )³
    = x³ − 3 × x² × 1 3 y + 3 × x × ( 1 3 y )² − ( 1 3 y )³
    = x³ − x²y + 3x × y² 9 − y³ 27
    = x³ − x²y + xy² 3 − y³ 27
    ✅ Final Answer
    ( x − 1 3 y )³ = x³ − x²y + xy² 3 − y³ 27
    Question 1 (ix)
    Use Suitable Identity
    Find:
    ( 7 2 k − 2 3 m )³
    ✍️ Solution
    Given,
    ( 7 2 k − 2 3 m )³
    Using the identity,
    (a − b)³ = a³ − 3a²b + 3ab² − b³
    Here,
    a = 7 2 k,    b = 2 3 m
    ∴ ( 7 2 k − 2 3 m )³
    = ( 7 2 k )³ − 3 × ( 7 2 k )² × 2 3 m + 3 × ( 7 2 k ) × ( 2 3 m )² − ( 2 3 m )³
    = 343 8 k³ − 147 4 k²m + 28 3 km² − 8 27 m³
    ✅ Final Answer
    ( 7 2 k − 2 3 m )³ = 343 8 k³ − 147 4 k²m + 28 3 km² − 8 27 m³
    Question 2
    Find the Values Using Suitable Identities
    Find the values using suitable identities:
    (i)  17 × 21
    (ii)  104 × 96
    (iii)  24 × 16
    (iv)  147³
    (v)  199³
    (vi)  127³
    (vii)  (−107)³
    (viii)  (−299)³
    Question 2 (i)
    Use Suitable Identity
    Find:
    17 × 21
    ✍️ Solution
    Given,
    17 × 21
    Write the numbers as,
    17 = 19 − 2
    21 = 19 + 2
    ∴ 17 × 21
    = (19 − 2)(19 + 2)
    Using the identity,
    (a − b)(a + b) = a² − b²
    = 19² − 2²
    = 361 − 4
    = 357
    ✅ Final Answer
    17 × 21 = 357
    Question 2 (ii)
    Use Suitable Identity
    Find:
    104 × 96
    ✍️ Solution
    Given,
    104 × 96
    Write the numbers as,
    104 = 100 + 4
    96 = 100 − 4
    ∴ 104 × 96
    = (100 + 4)(100 − 4)
    Using the identity,
    (a + b)(a − b) = a² − b²
    = 100² − 4²
    = 10000 − 16
    = 9984
    ✅ Final Answer
    104 × 96 = 9984
    Question 2 (iii)
    Use Suitable Identity
    Find:
    24 × 16
    ✍️ Solution
    Given,
    24 × 16
    Write the numbers as,
    24 = 20 + 4
    16 = 20 − 4
    ∴ 24 × 16
    = (20 + 4)(20 − 4)
    Using the identity,
    (a + b)(a − b) = a² − b²
    = 20² − 4²
    = 400 − 16
    = 384
    ✅ Final Answer
    24 × 16 = 384
    Question 2 (iv)
    Use Suitable Identity
    Find:
    147³
    ✍️ Solution
    Given,
    147³
    Write the number as,
    147 = 150 − 3
    ∴ 147³
    = (150 − 3)³
    Using the identity,
    (a − b)³ = a³ − 3a²b + 3ab² − b³
    = 150³ − 3 × 150² × 3 + 3 × 150 × 3² − 3³
    = 3375000 − 202500 + 4050 − 27
    = 3176523
    ✅ Final Answer
    147³ = 3176523
    Question 2 (v)
    Use Suitable Identity
    Find:
    199³
    ✍️ Solution
    Given,
    199³
    Write the number as,
    199 = 200 − 1
    ∴ 199³
    = (200 − 1)³
    Using the identity,
    (a − b)³ = a³ − 3a²b + 3ab² − b³
    = 200³ − 3 × 200² × 1
      + 3 × 200 × 1² − 1³
    = 8000000 − 120000 + 600 − 1
    = 7880599
    ✅ Final Answer
    199³ = 7880599
    Question 2 (vi)
    Use Suitable Identity
    Find:
    127³
    ✍️ Solution
    Given,
    127³
    Write the number as,
    127 = 120 + 7
    ∴ 127³
    = (120 + 7)³
    Using the identity,
    (a + b)³ = a³ + 3a²b + 3ab² + b³
    = 120³ + 3 × 120² × 7
      + 3 × 120 × 7² + 7³
    = 1728000 + 302400 + 17640 + 343
    = 2048383
    ✅ Final Answer
    127³ = 2048383
    Question 2 (vii)
    Use Suitable Identity
    Find:
    (−107)³
    ✍️ Solution
    Given,
    (−107)³
    Write the number as,
    −107 = (−100 − 7)
    ∴ (−107)³
    = (−100 − 7)³
    Using the identity,
    (a − b)³ = a³ − 3a²b + 3ab² − b³
    = (−100)³ − 3 × (−100)² × 7
      + 3 × (−100) × 7² − 7³
    = −1000000 − 210000 − 14700 − 343
    = −1225043
    ✅ Final Answer
    (−107)³ = −1225043
    Question 2 (viii)
    Use Suitable Identity
    Find:
    (−299)³
    ✍️ Solution
    Given,
    (−299)³
    Write the number as,
    −299 = (−300 + 1)
    ∴ (−299)³
    = (−300 + 1)³
    Using the identity,
    (a + b)³ = a³ + 3a²b + 3ab² + b³
    = (−300)³ + 3 × (−300)² × 1
      + 3 × (−300) × 1² + 1³
    = −27000000 + 270000 − 900 + 1
    = −26730999
    ✅ Final Answer
    (−299)³ = −26730999
    Question 3
    Factor the Following Algebraic Expressions
    Factor the following algebraic expressions:
    (i)   4y² + 1 + 1 16y²
    (ii)   9m² − 1 25n²
    (iii)   27b³ − 1 64b³
    (iv)   x² + 5x 6 + 1 6
    (v)   27u³ − 1 125 − 27u² 5 + 9u 25
    (vi)   64y³ + 1 125 z³
    (vii)   p³ + 27q³ + r³ − 9pqr
    (viii)   9m² − 12m + 4
    (ix)   9x³ − 8 3 y³ + z³ 3 + 6xyz
    (x)   4x² + 9y² + 36z²
    + 12xz + 36yz + 24xy
    (xi)   27u³ − 1 216 − 9u² 2 + u 4
    Question 3
    Factor the Following Algebraic Expressions
    Factor the following algebraic expressions:
    (i)   4y² + 1 + 1 16y²
    (ii)   9m² − 1 25n²
    (iii)   27b³ − 1 64b³
    (iv)   x² + 5x 6 + 1 6
    (v)   27u³ − 1 125 − 27u² 5 + 9u 25
    (vi)   64y³ + 1 125 z³
    (vii)   p³ + 27q³ + r³ − 9pqr
    (viii)   9m² − 12m + 4
    (ix)   9x³ − 8 3 y³ + z³ 3 + 6xyz
    (x)   4x² + 9y² + 36z² + 12xz + 36yz + 24xy
    (xi)   27u³ − 1 216 − 9u² 2 + u 4
    Question 3 (i)
    Factorise
    Factorise:
    4y² + 1 + 1 16y²
    ✍️ Solution
    Given,
    4y² + 1 + 1 16y²
    Rewrite each term as,
    4y² = (2y)²
    1 = 2 × (2y) × 1 4y
    1 16y² = ( 1 4y )²
    ∴ 4y² + 1 + 1 16y²
    = (2y)² + 2 × (2y) × 1 4y + ( 1 4y )²
    Using the identity,
    a² + 2ab + b² = (a + b)²
    = ( 2y + 1 4y )²
    ✅ Final Answer
    4y² + 1 + 1 16y² = ( 2y + 1 4y )²
    Question 3 (ii)
    Factorise
    Factorise:
    9m² − 1 25n²
    ✍️ Solution
    Given,
    9m² − 1 25n²
    Observe that,
    9m² = (3m)²
    1 25n² = ( 1 5n )²
    ∴ 9m² − 1 25n²
    = (3m)² − ( 1 5n )²
    Using the identity,
    a² − b² = (a + b)(a − b)
    Here,
    a = 3m,   b = 1 5n
    ∴ = ( 3m + 1 5n ) ( 3m − 1 5n )
    ✅ Final Answer
    9m² − 1 25n² = ( 3m + 1 5n ) ( 3m − 1 5n )
    Question 3 (iii)
    Factorise
    Factorise:
    27b³ − 1 64b³
    ✍️ Solution
    Given,
    27b³ − 1 64b³
    Observe that,
    27b³ = (3b)³
    1 64b³ = ( 1 4b )³
    ∴ 27b³ − 1 64b³
    = (3b)³ − ( 1 4b )³
    Using the identity,
    a³ − b³ = (a − b)(a² + ab + b²)
    Here,
    a = 3b,   b = 1 4b
    ∴ = ( 3b − 1 4b ) [ (3b)² + (3b) ( 1 4b ) + ( 1 4b )² ]
    = ( 3b − 1 4b ) ( 9b² + 3b 4 + 1 16b² )
    ✅ Final Answer
    27b³ − 1 64b³ = ( 3b − 1 4b ) ( 9b² + 3b 4 + 1 16b² )
    Question 3 (iv)
    Factorise
    Factorise:
    x² + 5x 6 + 1 6
    ✍️ Solution
    Given,
    x² + 5x 6 + 1 6
    Taking LCM of 6,
    = 6x² + 5x + 1 6
    Now factorise the numerator:
    6x² + 5x + 1
    = 6x² + 3x + 2x + 1
    = 3x(2x + 1) + 1(2x + 1)
    = (3x + 1)(2x + 1)
    Therefore,
    6x² + 5x + 1 6
    = (3x + 1)(2x + 1) 6
    ✅ Final Answer
    x² + 5x 6 + 1 6 = (3x + 1)(2x + 1) 6
    Q.3 (iv) 2nd Method
    Factorise:
    x² + 5x 6 + 1 6
    ✍️ Solution
    Given,
    x² + 5x 6 + 1 6
    To factorise, we need two numbers whose
    Sum = 5 6
    Product = 1 6
    These numbers are
    1 and 1 6
    because,
    1 + 1 6 = 7 6
    So these numbers are not correct.
    Instead, take
    1 2 and 1 3
    because, 1 2 + 1 3 = 5 6
    and 1 2 × 1 3 = 1 6
    ∴ x² + 5x 6 + 1 6
    = x² + x 2 + x 3 + 1 6
    = x² + x ( 1 2 + 1 3 ) + 1 6
    = x² + x 2 + x 3 + 1 6
    = ( x + 1 2 ) ( x + 1 3 )
    ✅ Final Answer
    x² + 5x 6 + 1 6 = ( x + 1 2 ) ( x + 1 3 )
    Question 3 (v)
    Factorise
    Factorise:
    27u³ − 1 125 − 27u² 5 + 9u 25
    ✍️ Solution
    Given,
    27u³ − 1 125 − 27u² 5 + 9u 25
    Observe that,
    27u³ = (3u)³
    1 125 = ( 1 5 )³
    27u² 5 = 3(3u)² × 1 5
    9u 25 = 3(3u) × 1 25
    Therefore, the given expression can be written as
    (3u)³ − 3(3u)² 1 5 + 3(3u) 1 5 2 − ( 1 5 ) 3
    Using the identity,
    a³ − 3a²b + 3ab² − b³ = (a − b)³
    Here,
    a = 3u,    b = 1 5
    Therefore,
    = (a − b)³
    = ( 3u − 1 5 )³
    ✅ Final Answer
    27u³ − 1 125 − 27u² 5 + 9u 25 = ( 3u − 1 5 )³
    Q 3 (v) 2nd method
    Factorise:
    27u³ − 1 125 − 27u² 5 + 9u 25
    ✍️ Solution
    Given,
    27u³ − 1 125 − 27u² 5 + 9u 25
    Taking LCM of 125,
    = 3375u³ − 675u² + 45u − 1 125
    Now factorise the numerator:
    3375u³ − 675u² + 45u − 1
    = (15u)³ − 3(15u)²(1) + 3(15u)(1)² − 1³
    Using the identity,
    a³ − 3a²b + 3ab² − b³ = (a − b)³
    Here,
    a = 15u,    b = 1
    ∴ 3375u³ − 675u² + 45u − 1
    = (15u − 1)³
    Therefore,
    = (15u − 1)³ 125
    = ( 15u − 1 5 )³
    ✅ Final Answer
    27u³ − 1 125 − 27u² 5 + 9u 25 = ( 15u − 1 5 )³
    Question 3 (vi)
    Factorise
    Factorise:
    64y³ + 1 125 z³
    ✍️ Solution
    Given,
    64y³ + 1 125 z³
    Observe that,
    64y³ = (4y)³
    1 125 z³ = ( z 5 )³
    Therefore,
    64y³ + 1 125 z³ = (4y)³ + ( z 5 )³
    Using the identity,
    a³ + b³ = (a + b)(a² − ab + b²)
    Here,
    a = 4y
    b = z 5
    ∴
    = ( 4y + z 5 ) [ (4y)² − (4y) z 5 + ( z 5 )² ]
    = ( 4y + z 5 ) ( 16y² − 4yz 5 + z² 25 )
    ✅ Final Answer
    64y³ + 1 125 z³ = ( 4y + z 5 ) ( 16y² − 4yz 5 + z² 25 )
    Question 3 (vii)
    Factorise
    Factorise:
    p³ + 27q³ + r³ − 9pqr
    ✍️ Solution
    Given,
    p³ + 27q³ + r³ − 9pqr
    Observe that,
    27q³ = (3q)³
    9pqr = 3(p)(3q)(r)
    Therefore,
    p³ + (3q)³ + r³ − 3(p)(3q)(r)
    Using the identity,
    a³ + b³ + c³ − 3abc
    = (a + b + c)(a² + b² + c² − ab − bc − ca)
    Here,
    a = p,   b = 3q,   c = r
    ∴
    = (p + 3q + r)
    × [p² + (3q)² + r² − p(3q) − (3q)r − rp]
    = (p + 3q + r)
    × (p² + 9q² + r² − 3pq − 3qr − pr)
    ✅ Final Answer
    p³ + 27q³ + r³ − 9pqr
    = (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr)
    Question 3 (viii)
    Factorise
    Factorise:
    9m² − 12m + 4
    ✍️ Solution
    Given,
    9m² − 12m + 4
    Using the identity,
    (a − b)² = a² − 2ab + b²
    Here,
    a = 3m,    b = 2
    ∴
    9m² − 12m + 4
    = (3m)² − 2(3m)(2) + 2²
    = (3m − 2)²
    ✅ Final Answer
    9m² − 12m + 4 = (3m − 2)²
    Question 3 (ix)
    Factorise
    Factorise:
    9x³ − 8 3 y³ + z³ 3 + 6xyz
    ✍️ Solution
    Given,
    9x³ − 8 3 y³ + z³ 3 + 6xyz
    Taking 1 3 common,
    = 1 3 [27x³ − 8y³ + z³ + 18xyz]
    Using the identity,
    a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
    Here,
    a = 3x,   b = −2y,   c = z
    ∴
    = 1 3 (3x − 2y + z)
    × [9x² + 4y² + z² + 6xy + 2yz − 3xz]
    ✅ Final Answer
    9x³ − 8 3 y³ + z³ 3 + 6xyz
    = 1 3 (3x − 2y + z)
    × (9x² + 4y² + z² + 6xy + 2yz − 3xz)
    Question 3 (x)
    Factorise
    Factorise:
    4x² + 9y² + 36z² + 12xz + 36yz + 24xy
    ✍️ Solution
    Given,
    4x² + 9y² + 36z² + 12xz + 36yz + 24xy
    Using the identity,
    (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
    Here,
    a = 2x,   b = 3y,   c = 6z
    Therefore,
    4x² + 9y² + 36z² + 12xz + 36yz + 24xy
    = (2x)² + (3y)² + (6z)²
      + 2(2x)(3y) + 2(3y)(6z) + 2(6z)(2x)
    = (2x + 3y + 6z)²
    ✅ Final Answer
    4x² + 9y² + 36z² + 12xz + 36yz + 24xy
    = (2x + 3y + 6z)²
    Question 3 (xi)
    Factorise
    Factorise:
    27u³ − 1 216 − 9u² 2 + u 4
    ✍️ Solution
    Given,
    27u³ − 1 216 − 9u² 2 + u 4
    Observe that,
    27u³ = (3u)³
    1 216 = ( 1 6 )³
    9u² 2 = 3(3u)² × 1 6
    u 4 = 3(3u) × 1 36
    Therefore, the expression can be written as
    (3u)³ − 3(3u)² 1 6 + 3(3u) 1 6 ² − ( 1 6 )³
    Using the identity,
    a³ − 3a²b + 3ab² − b³ = (a − b)³
    Here,
    a = 3u
    b = 1 6
    ∴
    = (a − b)³
    = ( 3u − 1 6 )³
    ✅ Final Answer
    27u³ − 1 216 − 9u² 2 + u 4
    = ( 3u − 1 6 )³
    Question 4
    Simplify the following:
    (i) 4x² + 4x + 1 4x² − 1
    (ii) 9(3a³ − 24b³) 9a² − 36b²
    (iii) s³ + 125t³ s² − 2st − 35t²
    Question 4 (i)
    ✍️ Solution
    Given,
    = 4x² + 4x + 1 4x² − 1
    Factorising the numerator,
    4x² + 4x + 1
    = (2x)² + 2(2x)(1) + 1²
    = (2x + 1)²
    Factorising the denominator,
    4x² − 1
    = (2x)² − 1²
    = (2x − 1)(2x + 1)
    Therefore,
    = (2x + 1)² (2x − 1)(2x + 1)
    Cancelling the common factor (2x + 1),
    = (2x + 1) (2x − 1) (2x + 1)
    = 2x + 1 2x − 1
    Therefore,
    = 2x + 1 2x − 1
    ✅ Final Answer
    4x² + 4x + 1 4x² − 1 = 2x + 1 2x − 1
    Question 4 (ii)
    ✍️ Solution
    Given,
    = 9(3a³ − 24b³) 9a² − 36b²
    Factorising the numerator,
    9(3a³ − 24b³)
    = 9 × 3(a³ − 8b³)
    = 27(a³ − (2b)³)
    Using the identity,
    a³ − b³ = (a − b)(a² + ab + b²)
    a³ − (2b)³ = (a − 2b)(a² + 2ab + 4b²)
    Therefore,
    9(3a³ − 24b³) = 27(a − 2b)(a² + 2ab + 4b²)
    Factorising the denominator,
    9a² − 36b²
    = 9(a² − 4b²)
    = 9(a² − (2b)²)
    Using the identity,
    a² − b² = (a − b)(a + b)
    a² − (2b)² = (a − 2b)(a + 2b)
    Therefore,
    9a² − 36b² = 9(a − 2b)(a + 2b)
    Therefore,
    = 27(a − 2b)(a² + 2ab + 4b²) 9(a − 2b)(a + 2b)
    Cancelling the common factors 9 and (a − 2b),
    = 27 (a − 2b) (a² + 2ab + 4b²) 9 (a − 2b) (a + 2b)
    = 3 a² + 2ab + 4b² a + 2b
    Therefore,
    = 3 a² + 2ab + 4b² a + 2b
    ✅ Final Answer
    3 a² + 2ab + 4b² a + 2b
    Question 4 (iii)
    ✍️ Solution
    Given,
    = s³ + 125t³ s² − 2st − 35t²
    Factorising the numerator,
    s³ + 125t³
    = s³ + (5t)³
    Using the identity,
    a³ + b³ = (a + b)(a² − ab + b²)
    s³ + (5t)³ = (s + 5t)(s² − 5st + 25t²)
    Factorising the denominator,
    s² − 2st − 35t²
    = s² − 7st + 5st − 35t²
    = s(s − 7t) + 5t(s − 7t)
    = (s + 5t)(s − 7t)
    Therefore,
    = (s + 5t)(s² − 5st + 25t²) (s + 5t)(s − 7t)
    Cancelling the common factor (s + 5t),
    = (s + 5t) (s² − 5st + 25t²) (s + 5t) (s − 7t)
    = s² − 5st + 25t² s − 7t
    Therefore,
    = s² − 5st + 25t² s − 7t
    ✅ Final Answer
    s² − 5st + 25t² s − 7t
    Question 5
    Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
    (i) 25a² − 30ab + 9b²
    Area = 25a² − 30ab + 9b² square units
    (ii) 36s² − 49t²
    Area = 36s² − 49t² square units
    Question 5 (i)
    ✍️ Solution
    Given,
    Area = 25a² − 30ab + 9b² square units
    We need to find the possible expressions for the length and breadth.
    Factorising the given expression,
    25a² − 30ab + 9b²
    = (5a)² − 2(5a)(3b) + (3b)²
    Using the identity,
    a² − 2ab + b² = (a − b)²
    Here,
    a = 5a,   b = 3b
    Therefore,
    25a² − 30ab + 9b² = (5a − 3b)²
    Since,
    Area = Length × Breadth
    = (5a − 3b)(5a − 3b)
    Therefore, possible expressions for the length and breadth are
    Length = 5a − 3b
    Breadth = 5a − 3b
    ✅ Final Answer
    Length = 5a − 3b
    Breadth = 5a − 3b
    Question 5 (ii)
    ✍️ Solution
    Given,
    Area = 36s² − 49t² square units
    We need to find the possible expressions for the length and breadth.
    Factorising the given expression,
    36s² − 49t²
    = (6s)² − (7t)²
    Using the identity,
    a² − b² = (a − b)(a + b)
    Here,
    a = 6s,   b = 7t
    Therefore,
    36s² − 49t²
    = (6s − 7t)(6s + 7t)
    Since,
    Area = Length × Breadth
    = (6s − 7t)(6s + 7t)
    Therefore, possible expressions for the length and breadth are
    Length = 6s − 7t
    Breadth = 6s + 7t
    ✅ Final Answer
    Length = 6s − 7t
    Breadth = 6s + 7t
    Question 6
    Find possible expressions for the length, breadth, and height of each of the following cuboids whose volumes are given by the following expressions in cubic units.
    (i) 6a² − 24b²
    Volume = 6a² − 24b² cubic units
    (ii) 3ps² − 15ps + 12p
    Volume = 3ps² − 15ps + 12p cubic units
    Question 6 (i)
    ✍️ Solution
    Given,
    Volume = 6a² − 24b² cubic units
    We know that,
    Volume = Length × Breadth × Height
    Factorising the given expression,
    6a² − 24b²
    = 6(a² − 4b²)
    = 6[a² − (2b)²]
    Using the identity,
    a² − b² = (a − b)(a + b)
    Here,
    a = a,   b = 2b
    Therefore,
    6a² − 24b²
    = 6(a − 2b)(a + 2b)
    Hence, the volume can be written as
    Volume = 6 × (a − 2b) × (a + 2b)
    Therefore, possible expressions for the three dimensions are:
    Length = 6
    Breadth = a − 2b
    Height = a + 2b
    ✅ Final Answer
    Length = 6
    Breadth = a − 2b
    Height = a + 2b
    Question 6 (ii)
    ✍️ Solution
    Given,
    Volume = 3ps² − 15ps + 12p cubic units
    We know that,
    Volume = Length × Breadth × Height
    Factorising the given expression,
    3ps² − 15ps + 12p
    = 3p(s² − 5s + 4)
    = 3p(s² − s − 4s + 4)
    = 3p[s(s − 1) − 4(s − 1)]
    = 3p(s − 1)(s − 4)
    Hence, the volume can be written as
    Volume = 3p × (s − 1) × (s − 4)
    Therefore, possible expressions for the three dimensions are:
    Length = 3p
    Breadth = s − 1
    Height = s − 4
    ✅ Final Answer
    Length = 3p
    Breadth = s − 1
    Height = s − 4
    Question 7
    The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
    ✍️ Solution
    Given,
    Side of the square playground = 40 m
    Width of the path = s m
    Class 9 Maths Chapter 4 End Exercise Question 7 square playground of side 40 metres with path of width s metres
    Square playground of side 40 m surrounded by a path of width s m.
    We have to find the area of the path.
    The path is created around the playground.
    Therefore, the side of the outer square
    = 40 + s + s
    = 40 + 2s m
    Area of the outer square
    = (40 + 2s)²
    Area of the playground
    = 40²
    Therefore,
    Area of the path
    = Area of outer square − Area of playground
    = (40 + 2s)² − 40²
    Using the identity,
    a² − b² = (a − b)(a + b)
    We get,
    = [(40 + 2s) − 40][(40 + 2s) + 40]
    = (2s)(80 + 2s)
    = 2s × 2(40 + s)
    = 4s(s + 40)
    Therefore, the required area of the path is
    ✅ Final Answer
    Area of the path = 4s(s + 40) m²
    Question 8
    If a number plus its reciprocal equals 10 3 , find the number.
    ✍️ Solution
    Let the number be x.
    Then its reciprocal = 1 x
    According to the question,
    x + 1 x = 10 3
    Multiplying both sides by x,
    x² + 1 = 10x 3
    3x² + 3 = 10x
    3x² − 10x + 3 = 0
    Factorising,
    3x² − 9x − x + 3 = 0
    3x(x − 3) − 1(x − 3) = 0
    (3x − 1)(x − 3) = 0
    Therefore,
    3x − 1 = 0    or    x − 3 = 0
    x = 1 3    or    x = 3
    ✅ Final Answer
    The numbers are 1 3 and 3.
    Question 9
    A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length.
    ✍️ Solution
    Given,
    Area of the rectangular pool = 2x² + 7x + 3 square hastas
    Width = 2x + 1 hastas
    We know that,
    Area = Length × Width
    Therefore,
    Length = 2x² + 7x + 3 2x + 1 hastas
    Factorising the numerator,
    2x² + 7x + 3
    = 2x² + 6x + x + 3
    = 2x(x + 3) + 1(x + 3)
    = (2x + 1)(x + 3)
    Hence,
    Length = (2x + 1) (x + 3) (2x + 1)
    = x + 3 hastas
    ✅ Final Answer
    Length = (x + 3) hastas
    Question 10 — Solution
    ✍️ Solution
    Given,
    x − 2 and x − 1 2 are factors of px² + 5x + r.
    Therefore,
    px² + 5x + r = k(x − 2) ( x − 1 2 )
    Now,
    (x − 2) ( x − 1 2 ) = x² − 1 2 x − 2x + 1
    = x² − 5 2 x + 1
    Multiplying k,
    px² + 5x + r = k ( x² − 5 2 x + 1 )
    px² + 5x + r = kx² − 5k 2 x + k
    Since both expressions are equal, comparing the corresponding terms on both sides,
    Comparing the coefficients of x²:
    p = k
    Comparing the coefficients of x:
    5 = − 5k 2
    k = −2
    Therefore, p = −2
    Comparing the constant terms:
    r = k
    r = −2
    Hence,
    p = r
    ✅ Final Answer
    p = r = −2
    Question 11
    If a + b + c = 5 and ab + bc + ca = 10, then prove that
    a³ + b³ + c³ − 3abc = −25
    ✍️ Solution
    Given,
    a + b + c = 5
    ab + bc + ca = 10
    First, find a² + b² + c².
    Using the identity,
    (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
    Substituting the given values,
    5² = a² + b² + c² + 2(10)
    25 = a² + b² + c² + 20
    a² + b² + c² = 25 − 20
    a² + b² + c² = 5
    Now, using the identity,
    a³ + b³ + c³ − 3abc
    = (a + b + c) (a² + b² + c² − ab − bc − ca)
    Substituting the values,
    = 5(5 − 10)
    = 5(−5)
    = −25
    Hence,
    a³ + b³ + c³ − 3abc = −25
    Hence proved.
    ✅ Final Answer
    a³ + b³ + c³ − 3abc = −25
    Question 12 — Solution
    By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.
    ✍️ Solution
    Factorising,
    n³ − n = n(n² − 1)
    = n(n² − 1²)
    = n(n − 1)(n + 1)
    Thus, n − 1, n, and n + 1 are three consecutive integers.
    Divisibility by 2:
    Among any three consecutive integers, at least one is even.
    ∴ n(n − 1)(n + 1) is divisible by 2.
    Divisibility by 3:
    When n is divided by 3, the remainder can only be 0, 1 or 2.
    If remainder = 0, then n is divisible by 3.
    If remainder = 1, then n − 1 is divisible by 3.
    If remainder = 2, then n + 1 is divisible by 3.
    Hence, in all cases, one of n − 1, n, n + 1 is divisible by 3.
    ∴ n(n − 1)(n + 1) is divisible by 3.
    Therefore, the product is divisible by both 2 and 3.
    2 × 3 = 6
    Hence,
    n³ − n is divisible by 6.
    Hence proved for all natural numbers n.
    ✅ Final Answer
    n³ − n is always divisible by 6.
    💡 Alternative Method
    We have,
    n³ − n = n(n − 1)(n + 1)
    By Euclid Division Lemma,
    n = 3q + r,    r = 0, 1 or 2
    Thus, when n is divided by 3, there are only three possible remainders: 0, 1 and 2.
    If r = 0:
    Then n is divisible by 3.
    If r = 1:
    Then n − 1 = 3q so n − 1 is divisible by 3.
    If r = 2:
    Then n + 1 = 3q + 3 = 3(q + 1) so n + 1 is divisible by 3.
    Therefore, in all three cases, one of n − 1, n, n + 1 is divisible by 3.
    Also, n − 1, n, and n + 1 are three consecutive integers. Therefore, one of them is even.
    Hence, the product
    n(n − 1)(n + 1)
    is divisible by both 2 and 3.
    2 × 3 = 6
    Therefore,
    n³ − n is divisible by 6.
    Hence proved.
    Question 13
    Find the value of
    (i)
    x³ + y³ − 12xy + 64 , when  x + y = −4
    (ii)
    x³ − 8y³ − 36xy − 216 , when  x = 2y + 6
    Question 13 (i)
    Find the value of
    x³ + y³ − 12xy + 64
    when  x + y = −4
    ✍️ Solution
    We have,
    x³ + y³ − 12xy + 64
    Since,
    64 = 4³
    12xy = 3(x)(y)(4)
    Therefore,
    x³ + y³ + 4³ − 3(x)(y)(4)
    Using the identity,
    a³ + b³ + c³ − 3abc
    = (a + b + c) (a² + b² + c² − ab − bc − ca)
    Here,
    a = x,   b = y,   c = 4
    Hence,
    = (x + y + 4) (x² + y² + 16 − xy − 4y − 4x)
    Given,
    x + y = −4
    x + y + 4 = 0
    = 0 × (x² + y² + 16 − xy − 4y − 4x)
    = 0
    Hence, the required value is 0.
    ✅ Final Answer
    0
    Question 13 (ii)
    Find the value of
    x³ − 8y³ − 36xy − 216
    when  x = 2y + 6
    ✍️ Solution
    We have,
    x³ − 8y³ − 36xy − 216
    Since,
    8y³ = (2y)³
    216 = 6³
    36xy = 3(x)(−2y)(−6)
    Therefore,
    x³ + (−2y)³ + (−6)³ − 3(x)(−2y)(−6)
    Using the identity,
    a³ + b³ + c³ − 3abc
    = (a + b + c) (a² + b² + c² − ab − bc − ca)
    Here,
    a = x,   b = −2y,   c = −6
    Hence,
    = (x − 2y − 6) [x² + (−2y)² + (−6)² − x(−2y) − (−2y)(−6) − (−6)x]
    Given,
    x = 2y + 6
    x − 2y − 6 = 0
    = 0 × [expression]
    = 0
    Hence, the required value is 0.
    ✅ Final Answer
    0

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    Congratulations! You have completed the Class 9 Maths Chapter 4 End of Chapter Exercise Solutions. This page provides step-by-step solutions for the complete end-of-chapter exercise based on NCERT Ganita Manjari (2026). Use these solutions to revise algebraic identities, factorisation, rational expressions, cube identities, and quadratic equations, or revisit any Chapter 4 exercise for additional practice before your Class 9 Maths examination.

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    Explore the complete Class 9 Maths Chapter 4 learning resources, including algebraic identities, visual proofs, algebra tiles, factorisation methods, chapter exercises, and revision material based on NCERT Ganita Manjari (2026).

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    🟪 Exercise 4.4 — Factorisation

    Revise factorisation methods and practise applying suitable algebraic identities and factorisation techniques to algebraic expressions.

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    🟧 Exercise 4.1

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    🟦 Exercise 4.2

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    🟩 Exercise 4.3

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    🟨 Exercise 4.5

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    ❓ Frequently Asked Questions

    Find quick answers to common questions about Class 9 Maths Chapter 4 End of Chapter Exercise Solutions. These FAQs help students understand the important topics covered in the chapter, choose suitable methods, and use the solutions effectively for revision, practice, and CBSE examination preparation.

    What is covered in the Class 9 Maths Chapter 4 End of Chapter Exercise?

    The End of Chapter Exercise brings together important concepts from Chapter 4 – Exploring Algebraic Identities. The questions require students to apply ideas related to algebraic identities, expansion, factorisation, rational expressions, cube identities, quadratic equations, and application-based problems.

    Which topics should I revise before solving Chapter 4 End Exercise?

    Before attempting the Class 9 Maths Chapter 4 End of Chapter Exercise, revise algebraic identities, expansion of expressions, factorisation methods, middle-term splitting, rational expressions, cube identities, quadratic equations, and word problems. Revising these concepts makes it easier to identify the correct method for each question.

    How do I choose the correct factorisation method in Chapter 4?

    First observe the algebraic expression carefully. Check whether a common factor can be taken out, whether an algebraic identity can be applied, or whether middle-term splitting is required. Selecting the correct factorisation method first makes the remaining steps simpler and helps reduce calculation errors.

    How are quadratic equations approached in Class 9 Chapter 4?

    In the relevant Chapter 4 questions, quadratic expressions can be handled by factorisation. After factorising the expression, use the appropriate equality condition to obtain the required values and check the result. Writing each step clearly is important for a complete CBSE-style solution.

    What common mistakes should students avoid in the Chapter 4 End Exercise?

    Students should avoid choosing the wrong identity, making errors while splitting the middle term, cancelling terms instead of common factors in rational expressions, and skipping important algebraic steps. Always identify the method first and write the solution systematically.

    Are these Class 9 Maths Chapter 4 End of Chapter Exercise Solutions based on NCERT Ganita Manjari 2026?

    Yes. These Class 9 Maths Chapter 4 End of Chapter Exercise Solutions are prepared according to the latest NCERT Ganita Manjari (2026) textbook and are presented in a clear, step-by-step format for Class 9 Mathematics. They are designed to support classroom practice, self-study, chapter revision, homework, and examination preparation.

    How can these Chapter 4 End of Chapter solutions help with exam preparation?

    The solutions show the questions in a step-by-step CBSE answer-writing style. Students can use them to revise important Chapter 4 concepts, understand how different algebraic methods are applied, identify common errors, and practise writing complete mathematical solutions before their examination.

    Are these solutions useful for complete Chapter 4 revision?

    Yes. The Class 9 Maths Chapter 4 End of Chapter Exercise Solutions are useful for revising the chapter as a whole because the End Exercise requires students to apply concepts from different parts of Exploring Algebraic Identities. Students can first revise the required concepts and then use the step-by-step solutions for practice and examination preparation.

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    These Class 9 Maths Chapter 4 End of Chapter Exercise Solutions are carefully prepared using the latest NCERT Ganita Manjari (2026) and aligned with the CBSE Class 9 Mathematics curriculum. The solutions are presented in a clear, step-by-step format to help students understand algebraic identities, factorisation, rational expressions, quadratic equations and other concepts covered in Chapter 4, while developing accurate mathematical answer-writing skills.

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