📑 Table of Contents
📖 About Class 9 Maths Chapter 4 Exercise 4.4 Solutions
Class 9 Maths Chapter 4 Exercise 4.4 Solutions help you understand how algebraic identities can make mathematical calculations simpler, faster, and more accurate. In this exercise, you learn to recognise patterns, choose suitable identities, and apply them correctly instead of using lengthy multiplication or expansion. These methods are useful for solving Class 9 Maths Exercise 4.4 questions and developing strong problem-solving skills in algebra.
This Exercise 4.4 Class 9 Maths Solutions page covers important applications of identities and factorisation of algebraic expressions. You will also learn how suitable identities can be used for quick calculations and how algebraic expressions can be simplified using the correct method. The explanations are written in simple, student-friendly language so that you can follow each step with confidence.
All the Ganita Manjari Class 9 Exercise 4.4 Solutions on this page are prepared according to the latest NCERT Ganita Manjari (2026–27) and the CBSE Class 9 Mathematics syllabus. Each answer follows a clear, step-by-step approach designed for homework, classroom learning, revision, and examination preparation. Whether you are looking for Ex 4.4 Class 9 Maths Solutions or revising Chapter 4 algebraic identities, you can use this page to understand the method before moving to the next question.
🎯 Your Learning Mission in Exercise 4.4
The goal of Class 9 Maths Chapter 4 Exercise 4.4 is not just to find the correct answer. You should learn how to identify a suitable algebraic identity, decide which method fits the given expression, and apply it correctly. Once you recognise the pattern, many questions involving identities and factorisation become much easier to solve.
- Recognise the correct algebraic identity.
- Choose and apply suitable identities.
- Use identities for faster calculations.
- Reduces lengthy algebraic calculations.
- Strengthens algebraic problem-solving skills.
- Builds confidence for CBSE Class 9 examinations.
- Choose the correct identity independently.
- Factorise algebraic expressions accurately.
- Avoid common mistakes while solving.
🎯 Today You Will Learn
🔄 From Exercise 4.3 to Exercise 4.4
| Exercise 4.3 | Exercise 4.4 |
|---|---|
| Expand Expressions | Factorise Expressions |
| Apply Identities | Choose the Correct Identity |
| Simplify | Reverse the Process |
📒 Formula Revision Sheet
👀 Step 1 : Observe the Expression
📒 Which Identity Should You Use?
| If You See… | Use This Identity |
|---|---|
| (a + b)² | Square of Sum |
| (a − b)² | Square of Difference |
| a² − b² | Difference of Squares |
| a² + 2ab + b² | Perfect Square Trinomial |
| a² − 2ab + b² | Perfect Square Trinomial |
🧠 Can You Identify the Identity?
💡 One Concept to Remember
📊 What’s the Difference?
| Expansion | Factorisation |
|---|---|
| Start with Factors | Start with Expression |
| Multiply | Split into Factors |
| Final Answer is an Expression | Final Answer is Product of Factors |
⭐ Remember This
📊 Compare Both Processes
| Expansion | Factorisation |
|---|---|
| Multiply the brackets | Split into brackets |
| (x+2)(x+3) | x²+5x+6 |
| ↓ | ↑ |
| x²+5x+6 | (x+2)(x+3) |
🔁 Think in Reverse
⚡ Quick Examples
|
Given Expression
x2 + 5x + 6
|
⇔ |
Standard Form
x2 + (a+b)x + ab
|
| From the Given Expression | We Get |
|---|---|
| 5x | a+b = 5 |
| 6 | ab = 6 |
We are looking for two numbers whose:
- Sum is 5.
- Product is 6.
- Sum is 5.
- Product is 6.
| Numbers | Sum | Product | Result |
|---|---|---|---|
| 1 and 6 | 7 | 6 | ✕ |
| 2 and 3 | 5 | 6 | ✓ |
↓
x2 + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
Take it common.
= (x + 2)(x + 3)
= x2 + 2x + 3x + 6
= (x2 + 2x) + (3x + 6)
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
| Step | What to Do |
|---|---|
| 1 | Identify a+b and ab. |
| 2 | Find two numbers whose sum = a+b and product = ab. |
| 3 | Split the middle term using those two numbers. |
| 4 | Make two groups. |
| 5 | Take the common factor from each group. |
| 6 | Take the common bracket to get the final factorised form. |
- the first term is a perfect square,
- the second term is also a perfect square, and
- there is a minus (−) sign between them.
| Given Term | Can be Written As | Perfect Square? |
|---|---|---|
| x² | (x)² | ✅ Yes |
| 49 | (7)² | ✅ Yes |
- x² = (x)²
- 49 = (7)²
- There is a minus (−) sign between them.
| Given Expression | Identity |
|---|---|
| x² − 49 | a² − b² |
So, we can directly use the identity
a² − b² = (a + b)(a − b)
| Identity | Given Expression | Value |
|---|---|---|
| a² | x² | a = x |
| b² | 49 = 7² | b = 7 |
b = 7
x² − 7² = (x + 7)(x − 7)
x² − 49 = (x + 7)(x − 7)
As soon as you recognize the pattern Perfect Square − Perfect Square, you can directly apply the identity.
- three perfect square terms,
- three cross-product terms, and
- all six terms match the pattern of a perfect square identity.
=
a2 + b2 + c2
+ 2ab + 2bc + 2ca
+ 6xz + 12yz + 9z2
| Given Term | Recognise It As | Perfect Square? |
|---|---|---|
| x2 | (x)2 | ✅ Yes |
| 4y2 | (2y)2 | ✅ Yes |
| 9z2 | (3z)2 | ✅ Yes |
| Given Term | Recognise It As |
|---|---|
| 4xy | 2 × x × 2y |
| 6xz | 2 × x × 3z |
| 12yz | 2 × 2y × 3z |
Therefore, this expression follows the identity (a + b + c)².
| Square Term | Recognise It As | Value |
|---|---|---|
| x2 | (x)2 | a = x |
| 4y2 | (2y)2 | b = 2y |
| 9z2 | (3z)2 | c = 3z |
b = 2y
c = 3z
= (x + 2y + 3z)2
∴
x2 + 4xy + 4y2
+ 6xz + 12yz + 9z2
= (x + 2y + 3z)2
Simply replace them in the identity to obtain the factorised form.
How to Choose the Correct Factorisation Method?
| Observe the Expression | Use This Method |
|---|---|
|
x² − 25 49a² − 16 |
✅ Difference of Squares |
|
x² + 6x + 9 4x² + 12x + 9 |
✅ Perfect Square Identity |
|
x² + 7x + 12 2x² + 5x + 3 |
✅ Middle-Term Splitting |
Always spend a few seconds observing the expression first. Choosing the correct factorisation method makes the solution easier and saves time in exams.
Common Mistakes to Avoid in Factorisation
| # | Common Mistake | Correct Practice |
|---|---|---|
| 1 | Choosing wrong numbers while splitting the middle term. | Choose numbers whose sum is the middle coefficient and whose product is the product of the first and last coefficients. |
| 2 | Ignoring negative signs. | Always check the signs before selecting the numbers. |
| 3 | Stopping after splitting the middle term. | Group the terms and take the common factor from each group. |
| 4 | Not checking the answer. | Multiply the factors again to verify the original expression. |
Before writing your final answer, ask yourself:
- Did I choose the correct factorisation method?
- Did I split the middle term correctly?
- Did I take the common factor properly?
- Does multiplying the factors give the original expression?
Final Revision Before Solving NCERT Questions
| ✓ | Remember This |
|---|---|
| ✔ | Factorisation is the reverse process of multiplication. |
| ✔ | Always observe the expression first, then choose the correct factorisation method. |
| ✔ | For middle-term splitting, choose two numbers whose sum is the middle coefficient and whose product is the product of the first and last coefficients. |
| ✔ | Take the common factor carefully after splitting the middle term. |
| ✔ | Multiply the factors once to check your final answer. |
- ✔ What factorisation means.
- ✔ Important algebraic identities.
- ✔ How to choose the correct factorisation method.
- ✔ Why the middle term is split.
- ✔ How algebra tiles explain factorisation.
- ✔ Common mistakes to avoid.
📝 Class 9 Maths Chapter 4 Exercise 4.1 Solutions
Solve every question of Class 9 Maths Chapter 4 Exercise 4.1 with easy, step-by-step NCERT Ganita Manjari (2026) solutions. Every answer is prepared according to the latest CBSE answer-writing format to help students understand the concepts of Factorisation clearly and score better in exams.
Fill in the blanks to complete the following identities:
We need two numbers whose:
- Product = 24
- Sum = −11
The numbers −3 and −8 satisfy both conditions because
(−3) + (−8) = −11
We need two numbers whose
- Product = 3 × (−7) = −21
- Sum = −4
The required numbers are −7 and 3.
= 3x2 − 7x + 3x − 7
= (3x − 7)(x + 1)
Comparing
The missing factor is 3x − 7.
We need two numbers whose
- Product = 10 × (−6) = −60
- Sum = −11
The required numbers are −15 and 4.
= 10x2 − 15x + 4x − 6
= (2x − 3)(5x + 2)
Comparing
= (2x − 3) (5x + 2)
The blanks are:
- First blank = 3
- Second blank = 5x
We need two numbers whose
- Product = 6 × 2 = 12
- Sum = 7
The required numbers are 3 and 4.
Comparing both sides,
Therefore, the blanks are:
- 2x + 1
- 3x + 2
Select and use the identity that will help you to find the following products without multiplying directly.
41 is 1 more than 40.
Since the number is written as the sum of two terms, we use (a + b)2.
Here, a = 40 and b = 1.
27 is 3 less than 30.
Since the number is written as the difference of two terms, we use (a − b)2.
Here, a = 30 and b = 3.
23 and 17 are equally distant from 20.
Since the numbers are of the form (a + b) and (a − b), we use the identity
Here, a = 20 and b = 3.
Write 135 as the sum of three numbers.
Since the number has three terms, we use (a + b + c)2.
Here, a = 100, b = 30 and c = 5.
97 is 3 less than 100.
Here, a = 100 and b = 3.
Here, x = 20, a = −2 and b = 9.
Here, x = 30, a = 4 and b = 13.
Here, a = 200 and b = 5.
2ab = 40st
b2 = 25t2
b = 5t
=
(4s − 5t)2
2ab = 40st
b2 = 25t2
b = 5t
Product = −42
Sum = −1
The required numbers are
−7 and 6
= r2 − 7r + 6r − 42
= (r − 7)(r + 6)
2ab = 14gh
b2 = h2
b = h
📚 Continue Learning
Continue exploring Class 9 Maths Chapter 4 – Factorisation by revising the previous exercise or moving to the chapter-end exercise.
📖 Explore More Class 9 Maths Chapters
🚀 Quick Revision Dashboard
Quickly revise the important concepts of Class 9 Maths Chapter 4 Exercise 4.4. This one-minute dashboard will help you remember the correct factorisation method, important identities, common mistakes, and useful CBSE exam tips.
📌 Method Selection Guide
| Common Factor | Take HCF first |
| x² + bx + c | Split the middle term |
| a² − b² | Difference of Squares |
| Perfect Square | Use Identity |
📖 Formula & Identity Sheet
(a − b)2 = a2 − 2ab + b2
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
a2 − b2 = (a + b)(a − b)
❌ Common Mistakes
- Ignore the common factor.
- Choose wrong numbers.
- Make sign mistakes.
- Skip answer verification.
🎯 CBSE Exam Tips
- Observe the expression first.
- Choose the correct method.
- Write every step neatly.
- Verify by multiplying the factors.
❓ Frequently Asked Questions
Find answers to the most common questions related to Class 9 Maths Chapter 4 Exercise 4.1 Solutions. These FAQs will help you understand factorisation methods, algebraic identities, and important CBSE exam concepts.
What is the main concept of Class 9 Maths Chapter 4 Exercise 4.1?
Exercise 4.1 introduces the basic methods of factorisation. Students learn how to factorise algebraic expressions using common factors, algebraic identities, middle-term splitting, and the difference of squares.
How do I choose the correct factorisation method?
First observe the given expression carefully. If there is a common factor, take it out first. Then check whether the expression matches an algebraic identity, the difference of squares, or requires middle-term splitting.
Why is middle-term splitting important in factorisation?
Middle-term splitting helps factorise quadratic expressions of the form ax² + bx + c. Choose two numbers whose sum equals the middle coefficient and whose product equals the product of the first and last coefficients.
Are these Exercise 4.1 solutions based on the latest NCERT book?
Yes. These solutions are prepared according to the latest NCERT Ganita Manjari (2026) textbook and follow the current CBSE Class 9 Mathematics syllabus and answer-writing format.
How can I check whether my factorisation is correct?
Multiply the obtained factors. If the product is exactly the same as the original algebraic expression, then your factorisation is correct.
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These Class 9 Maths Chapter 4 Exercise 4.4 Solutions are carefully prepared according to the latest NCERT Ganita Manjari (2026) and the CBSE curriculum. Every solution follows a clear, step-by-step approach designed to strengthen conceptual understanding, improve problem-solving skills, and help students perform confidently in school and board examinations.