📘 NCERT Ganita Manjari (2026)
📚 CBSE 2026–27
✅ Step-by-Step Solutions
Class 9 Maths Chapter 4 Exercise 4.4 Solutions
Learn Class 9 Maths Chapter 4 Exercise 4.4 Solutions with easy,
step-by-step NCERT solutions based on the latest
Ganita Manjari (2026). In this exercise, you will learn
how to identify the correct algebraic identity, perform
mental calculations using identities, understand the
difference of squares, and factorise algebraic expressions
using simple CBSE methods. Every solution is explained in easy language to help
Class 9 students build strong concepts, improve problem-solving
skills, and prepare confidently for school and board examinations.
🧠
Concept
Identities & Factorisation
⭐
Difficulty
Easy to Moderate
📌 Main Algebraic Identities Used in Exercise 4.4
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(a + b)(a − b) = a2 − b2
a2 − b2 = (a + b)(a − b)
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
💡 Student Tip: Before solving any question, carefully observe the given expression. First identify whether it involves a square, a difference of two squares, or factorisation. Choosing the correct algebraic identity before starting the calculation makes every solution faster, easier, and more accurate.
🎯 What You’ll Master in Exercise 4.4
✅ Choosing the correct algebraic identity
✅ Difference of Squares concept
✅ Mental calculations using identities
✅ Factorising algebraic expressions
✅ Applying identities correctly
✅ Avoiding common algebraic mistakes
📖 About Class 9 Maths Chapter 4 Exercise 4.4
Class 9 Maths Chapter 4 Exercise 4.4 Solutions help you understand how algebraic identities can make mathematical calculations simpler, faster, and more accurate. Instead of solving every question through lengthy multiplication or expansion, this exercise teaches you to recognise patterns and apply the most suitable method at the right time. These skills not only help in solving NCERT questions but also strengthen your logical thinking and problem-solving ability.
All the solutions on this page are prepared according to the latest NCERT Ganita Manjari (2026) textbook and the CBSE Class 9 syllabus. Every solution follows a simple, step-by-step approach that is easy to understand, making this page useful for homework, classroom learning, revision, and exam preparation.
🎯 Your Learning Mission
By the time you complete this exercise, your goal is not just to get the correct answers but to understand how to identify the correct algebraic identity and apply it confidently in different situations. Once you learn this skill, solving similar questions becomes much easier.
📘 You Will Learn
- Recognise the correct algebraic identity.
- Apply identities to simplify calculations.
- Know when to expand and when to factorise.
🚀 Why It Matters
- Reduces lengthy calculations.
- Improves mathematical thinking.
- Builds confidence for CBSE examinations.
🏆 Your Goal
- Choose the correct method independently.
- Solve every NCERT question with confidence.
- Avoid common mistakes while solving.
💡 Maths Gurukulam Tip: Don’t try to memorise identities blindly. Learn to recognise the pattern first. Once you identify the correct pattern, choosing the right identity becomes easy, and solving the question takes much less time.
🎯 Jump to Any Question
Click any question below to jump directly to its complete step-by-step solution.
📚 Learn Before You Solve
Revise the important concepts before solving Exercise 4.4.
🎯 Today You Will Learn
🧩
Choose the Correct Identity
🔄 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
① (a + b)² = a² + 2ab + b²
② (a − b)² = a² − 2ab + b²
③ a² − b² = (a + b)(a − b)
④ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
⭐ Ready for the Next Step?
✔ Revise the identities
✔ Observe the pattern
✔ Choose the correct identity
✔ Start solving confidently
🔍 Identity Selection Guide
Learn how to recognise the correct algebraic identity by observing the pattern.
👀 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?
m²−n²
👉 Difference of Squares
p²−2pq+q²
👉 Square of Difference
💡 Maths Gurukulam Smart Tip
✔ Don’t memorise identities blindly.
✔ Observe the pattern first.
✔ Pattern → Identity → Solution.
🔄 Reverse Maths
Factorisation is the Reverse Journey of Expansion
💡 One Concept to Remember
Expansion
(a+3)(a+5)
⬇
a² + 8a + 15
⇄
Factorisation
a² + 8a + 15
⬇
(a+3)(a+5)
📊 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
✖
Expansion
Multiply the factors.
➗
Factorisation
Split the expression into factors.
🔄
Easy Trick
Factorisation is simply the reverse of expansion.
🧠 Mini Challenge
Expand it first. Then try to factorise your answer again.
🔄 Expansion vs Factorisation
Factorisation is simply the reverse process of expansion.
📊 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
①
Observe
Look carefully at the expression.
②
Recognise
Find the matching identity.
③
Factorise
Write the expression as factors.
💡 Maths Gurukulam Tip
✔ Expansion → Multiply
✔ Factorisation → Reverse the process
✔ Same identity, opposite direction
Factorisation
Method 1
Middle Term Splitting Method
🎯 When should we use this method?
Use this method when the algebraic expression is in the standard form
The coefficient of x² is 1.
💡 Learn Through an Example
Instead of remembering the rule, let’s understand the method by solving one example.
👀 Step 1 : Observe the Expression
First compare the given expression with the standard form.
|
Given Expression
x2 + 5x + 6
|
⇔
|
Standard Form
x2 + (a+b)x + ab
|
🔍 Step 2 : Identify the Values
Now compare the corresponding terms.
|
From the Given Expression
|
We Get
|
|
5x
|
a+b = 5
|
|
6
|
ab = 6
|
Think like a mathematician 💡
We are looking for
two numbers whose:
These numbers will help us split the middle term.
🎯 Step 3 : Find the Two Numbers
We need two numbers whose
| Numbers |
Sum |
Product |
Result |
| 1 and 6 |
7 |
6 |
✕ |
| 2 and 3 |
5 |
6 |
✓ |
Therefore, a = 2 and b = 3
✏️ Step 4 : Split the Middle Term
Replace 5x by 2x + 3x because
2 + 3 = 5
x2 + 5x + 6
↓
x2 + 2x + 3x + 6
🧩 Step 5 : Make Two Groups
Group the first two terms together and the last two terms together.
(x2 + 2x)
+
(3x + 6)
🟢 Step 6 : Take the Common Factor
Take the common factor from each group.
(x2 + 2x)
+
(3x + 6)
=
x(x + 2)
+
3(x + 2)
📦 Step 7 : Take the Common Bracket
Both terms contain the common bracket (x + 2).
Take it common.
x(x + 2)
+
3(x + 2)
=
(x + 2)(x + 3)
📝 Complete Solution
x2 + 5x + 6
= x2 + 2x + 3x + 6
= (x2 + 2x) + (3x + 6)
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
✅ Final Answer
x2 + 5x + 6
=
(x + 2)(x + 3)
💡 Remember This Method
Whenever an expression is in the form
| 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. |
📌 Shortcut to Remember
Identify → Find Numbers → Split → Group → Common Factor → Common Bracket
Factorisation
Method 2
Using Identity (a² − b²)
🎯 When Should We Use This Method?
Use this method when
- the first term is a perfect square,
- the second term is also a perfect square, and
- there is a minus (−) sign between them.
💡 Observation
Whenever you see
Perfect Square − Perfect Square
there is a high chance that the identity
(a² − b²) can be used.
📘 Identity Used
This identity is called the Difference of Two Squares Identity.
📝 Learn Through an Example
Factorise
🔍 Step 1 : Check Whether Both Terms Are Perfect Squares
Write each term as a square.
| Given Term |
Can be Written As |
Perfect Square? |
|
x²
|
(x)²
|
✅ Yes
|
|
49
|
(7)²
|
✅ Yes
|
🧠 Think Like a Mathematician
We found that
- x² = (x)²
- 49 = (7)²
- There is a minus (−) sign between them.
Therefore, this expression matches the identity
a² − b².
🧩 Step 2 : Compare with the Identity
Compare the given expression with the standard identity.
| Given Expression |
Identity |
|
x² − 49
|
a² − b²
|
📌 Observation
Both expressions have the same pattern.
So, we can directly use the identity
a² − b² = (a + b)(a − b)
🎯 Step 3 : Identify a and b
Compare each square with the identity.
| Identity |
Given Expression |
Value |
|
a²
|
x²
|
a = x
|
|
b²
|
49 = 7²
|
b = 7
|
✏️ Step 4 : Apply the Identity
Substitute a = x and b = 7 into the identity.
a² − b²
=
(a + b)(a − b)
x² − 7²
=
(x + 7)(x − 7)
x² − 49
=
(x + 7)(x − 7)
💡 Think Like a Mathematician
There is no need to split terms or make groups.
As soon as you recognize the pattern Perfect Square − Perfect Square, you can directly apply the identity.
Factorisation
Method 3
Using Three-Term Identity
🎯 When Should We Use This Method?
Use this method when an algebraic expression contains
- three perfect square terms,
- three cross-product terms, and
- all six terms match the pattern of a perfect square identity.
💡 Observation
Whenever you find
3 Square Terms + 3 Cross Terms
the expression may be factorised using the identity
(a + b + c)².
📘 Identity Used
(a + b + c)2
=
a2 + b2 + c2
+ 2ab + 2bc + 2ca
📝 Learn Through an Example
Factorise
x2 + 4xy + 4y2
+ 6xz + 12yz + 9z2
👀 Step 1 : Recognize the Pattern
Observe each term carefully and identify whether it is a square term or a cross-product term.
🟩 Square Terms
| Given Term |
Recognise It As |
Perfect Square? |
| x2 |
(x)2 |
✅ Yes |
| 4y2 |
(2y)2 |
✅ Yes |
| 9z2 |
(3z)2 |
✅ Yes |
🟧 Cross Terms
| Given Term |
Recognise It As |
| 4xy |
2 × x × 2y |
| 6xz |
2 × x × 3z |
| 12yz |
2 × 2y × 3z |
🧠 Think Like a Mathematician
We found three perfect square terms and the remaining terms are of the form 2ab, 2bc and 2ca.
Therefore, this expression follows the identity (a + b + c)².
🔍 Step 2 : Identify a, b and c
From the three square terms,
| Square Term |
Recognise It As |
Value |
| x2 |
(x)2 |
a = x |
| 4y2 |
(2y)2 |
b = 2y |
| 9z2 |
(3z)2 |
c = 3z |
✏️ Step 3 : Apply the Identity
Substitute a = x, b = 2y and c = 3z into the identity.
(a + b + c)2
=
(x + 2y + 3z)2
∴
x2 + 4xy + 4y2
+ 6xz + 12yz + 9z2
=
(x + 2y + 3z)2
🧠 Think Like a Mathematician
After identifying the values of a, b and c, there is no need for further calculation.
Simply replace them in the identity to obtain the factorised form.
How to Choose the Correct Factorisation Method?
Before solving any question, first observe the expression carefully. Then choose the correct 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
|
Quick Decision Guide
Look at the Expression
│
▼
Is it of the form
a² − b² ?
│Yes │No
▼ ▼
Difference Is it a
of Squares Perfect Square?
│Yes │No
▼ ▼
Perfect Square Middle-Term
Identity Splitting
💡 Remember
Don’t start solving immediately.
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
Many students know the method but lose marks because of small mistakes. Check these points before writing your final answer.
| # |
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.
|
✔ Final Exam Tip
Don’t start solving immediately.
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?
If all four answers are
Yes, your solution is most likely correct.
Final Revision Before Solving NCERT Questions
Spend one minute revising these important points. They will help you solve the questions correctly and confidently.
| ✓ |
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. |
🎉 You Are Ready!
You have learned:
- ✔ 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.
Now you are ready to solve the NCERT Exercise with confidence.
📝 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.
📖 NCERT Solutions
📝 Step-by-Step
🎯 CBSE Ready
Question 1
Fill in the blanks to complete the following identities:
(i)
s2 − 11s + 24
=
()
()
(ii)
()
(x + 1)
=
(3x2 − 4x − 7)
(iii)
10x2 − 11x − 6
=
(2x − )
( + 2)
(iv)
6x2 + 7x + 2
=
()
()
✅ Solution (i)
Sub Question
s2 − 11s + 24 = ( ______ ) ( ______ )
Step 1: Find two numbers
We need two numbers whose:
Step 2: Choose the numbers
The numbers −3 and −8 satisfy both conditions because
(−3) × (−8) = 24
(−3) + (−8) = −11
Step 3: Write the factors
s2 − 11s + 24
=
(s − 3)(s − 8)
Final Answer
s2 − 11s + 24
=
(s − 3)(s − 8)
✅ Solution (ii)
Sub Question
( ______ ) (x + 1)
=
3x2 − 4x − 7
Step 1: Find two numbers
We need two numbers whose
- Product = 3 × (−7) = −21
- Sum = −4
The required numbers are −7 and 3.
Step 2: Split the middle term
3x2 − 4x − 7
=
3x2 − 7x + 3x − 7
Step 3: Factorise by grouping
=
x(3x − 7)
+
1(3x − 7)
=
(3x − 7)(x + 1)
Step 4: Fill in the blank
Comparing
( ______ )(x + 1)
=
(3x − 7)(x + 1)
The missing factor is 3x − 7.
Final Answer
(3x − 7)(x + 1)
=
3x2 − 4x − 7
✅ Solution (iii)
Sub Question
10x2 − 11x − 6
=
(2x − ____)
(____ + 2)
Step 1: Find two numbers
We need two numbers whose
- Product = 10 × (−6) = −60
- Sum = −11
The required numbers are −15 and 4.
Step 2: Split the middle term
10x2 − 11x − 6
=
10x2 − 15x + 4x − 6
Step 3: Factorise by grouping
=
5x(2x − 3)
+
2(2x − 3)
=
(2x − 3)(5x + 2)
Step 4: Fill in the blanks
Comparing
(2x − ____)
(____ + 2)
=
(2x − 3)
(5x + 2)
The blanks are:
- First blank = 3
- Second blank = 5x
Final Answer
10x2 − 11x − 6
=
(2x − 3)(5x + 2)
✅ Solution (iv)
Sub Question
6x2 + 7x + 2
=
( ______ )
( ______ )
Step 1: Find two numbers
We need two numbers whose
- Product = 6 × 2 = 12
- Sum = 7
The required numbers are 3 and 4.
Step 2: Split the middle term
6x2 + 7x + 2
= 6x2 + 3x + 4x + 2
Step 3: Factorise by grouping
= 3x(2x + 1) + 2(2x + 1)
= (2x + 1)(3x + 2)
Step 4: Fill in the blanks
Comparing both sides,
( ______ )( ______ )
= (2x + 1)(3x + 2)
Therefore, the blanks are:
Final Answer
6x2 + 7x + 2
=
(2x + 1)(3x + 2)
Question 2
Select and use the identity that will help you to find the following products without multiplying directly.
(i)
(41)2
(ii)
(27)2
(iii)
(23 × 17)
(iv)
(135)2
(v)
(97)2
(vi)
(18 × 29)
(vii)
(34 × 43)
(viii)
(205)2
✅ Solution (i)
Step 1: Rewrite the number
41 is 1 more than 40.
41
=
40 + 1
Step 2: Choose the identity
Since the number is written as the sum of two terms, we use
(a + b)2.
(a + b)2
=
a2
+
2ab
+
b2
Step 3: Substitute the values
Here,
a = 40
and
b = 1.
(41)2
=
(40 + 1)2
=
402
+
2 × 40 × 1
+
12
=
1600
+
80
+
1
=
1681
Final Answer
(41)2
=
1681
✅ Solution (ii)
Step 1: Rewrite the number
27 is 3 less than 30.
27
=
30 − 3
Step 2: Choose the identity
Since the number is written as the difference of two terms, we use
(a − b)2.
(a − b)2
=
a2
−
2ab
+
b2
Step 3: Substitute the values
Here,
a = 30
and
b = 3.
(27)2
=
(30 − 3)2
=
302
−
2 × 30 × 3
+
32
=
900
−
180
+
9
=
729
✅ Solution (iii)
Step 1: Rewrite the numbers
23 and 17 are equally distant from 20.
23 = 20 + 3
17 = 20 − 3
Step 2: Choose the identity
Since the numbers are of the form (a + b) and (a − b), we use the identity
(a + b)(a − b)
=
a2
−
b2
Step 3: Substitute the values
Here,
a = 20
and
b = 3.
23 × 17
=
(20 + 3)(20 − 3)
=
202
−
32
=
400
−
9
=
391
Final Answer
23 × 17
=
391
✅ Solution (iv)
Step 1: Rewrite the number
Write 135 as the sum of three numbers.
135 = 100 + 30 + 5
Step 2: Choose the identity
Since the number has three terms, we use
(a + b + c)2.
(a + b + c)2
=
a2
+
b2
+
c2
+
2ab
+
2bc
+
2ca
Step 3: Substitute the values
Here,
a = 100,
b = 30
and
c = 5.
(135)2
=
(100 + 30 + 5)2
=
1002
+
302
+
52
+
2×100×30
+
2×30×5
+
2×100×5
=
10000
+
900
+
25
+
6000
+
300
+
1000
=
18225
Final Answer
(135)2
=
18225
✅ Solution (v)
Step 1: Rewrite the number
97 is 3 less than 100.
97 = 100 − 3
Step 2: Use the identity
(a − b)2
=
a2
−
2ab
+
b2
Step 3: Substitute the values
Here,
a = 100
and
b = 3.
(97)2
=
(100 − 3)2
=
1002
−
2 × 100 × 3
+
32
=
10000
−
600
+
9
=
9409
Final Answer
(97)2
=
9409
✅ Solution (vi)
Step 1: Rewrite the numbers
18 = 20 − 2
29 = 20 + 9
Step 2: Use the identity
(x + a)(x + b)
=
x²
+
(a + b)x
+
ab
Step 3: Substitute the values
Here,
x = 20,
a = −2
and
b = 9.
18 × 29
=
(20 − 2)(20 + 9)
=
20²
+
(−2 + 9) × 20
+
(−2 × 9)
=
400
+
7 × 20
−
18
=
400
+
140
−
18
=
522
Final Answer
18 × 29 = 522
✅ Solution (vii)
Step 1: Rewrite the numbers
34 = 30 + 4
43 = 30 + 13
Step 2: Use the identity
(x + a)(x + b)
=
x² + (a + b)x + ab
Step 3: Substitute the values
Here,
x = 30,
a = 4
and
b = 13.
34 × 43
=
(30 + 4)(30 + 13)
=
30²
+
(4 + 13) × 30
+
4 × 13
=
900
+
17 × 30
+
52
=
900
+
510
+
52
=
1462
Final Answer
34 × 43 = 1462
✅ Solution (viii)
Step 1: Rewrite the number
205 = 200 + 5
Step 2: Use the identity
(a + b)2
=
a2
+
2ab
+
b2
Step 3: Substitute the values
Here,
a = 200
and
b = 5.
(205)2
=
(200 + 5)2
=
2002
+
2 × 200 × 5
+
52
=
40000
+
2000
+
25
=
42025
Final Answer
(205)2 = 42025
Question 3
Factor the following:
(i)
9a² + b² + 4c² − 6ab + 12ac − 4bc
(v)
64u² + 121v² + 4w² − 176uv − 32uw + 44vw
✅ Solution (i)
Sub Question
(i)
9a2 + b2 + 4c2 − 6ab + 12ac − 4bc
Given
9a2 + b2 + 4c2 − 6ab + 12ac − 4bc
Identity Used
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Step 1: Compare with the identity
x2 = 9a2
y2 = b2
z2 = 4c2
2xy = −6ab
2yz = −4bc
2zx = 12ac
Step 2: Identify the terms
x = 3a
y = −b
← Terms containing b are negative, so y = −b.
z = 2c
Step 3: Verify the middle terms
2 × 3a × (−b) = −6ab ✓
2 × (−b) × 2c = −4bc ✓
2 × 2c × 3a = 12ac ✓
Step 4: Write the factorised form
9a2 + b2 + 4c2 − 6ab + 12ac − 4bc
= (x + y + z)2
= (3a − b + 2c)2
Final Answer
9a
2 + b
2 + 4c
2 − 6ab + 12ac − 4bc
= (3a − b + 2c)2
✅ Solution (ii)
Sub Question
(ii) 16s2 + 25t2 − 40st
Given
The algebraic expression is
16s2 + 25t2 − 40st
Identity Used
(a − b)2 = a2 − 2ab + b2
Step 1: Compare with the identity
a2 = 16s2
2ab = 40st
b2 = 25t2
Step 2: Identify the terms
a = 4s
b = 5t
Step 3: Write the factorised form
16s2 + 25t2 − 40st
=
(4s − 5t)2
Final Answer
16s2 + 25t2 − 40st
=
(4s − 5t)2
✅ Solution (ii)
Sub Question
(ii) 16s2 + 25t2 − 40st
Given
The algebraic expression is
16s2 + 25t2 − 40st
Identity Used
(a − b)2 = a2 − 2ab + b2
Step 1: Compare with the identity
a2 = 16s2
2ab = 40st
b2 = 25t2
Step 2: Identify the terms
a = 4s
b = 5t
Step 3: Write the factorised form
16s2 + 25t2 − 40st
= (4s − 5t)2
Final Answer
16s2 + 25t2 − 40st
= (4s − 5t)2
✅ Solution (iii)
Sub Question
(iii) r2 − r − 42
Given
The algebraic expression is
r2 − r − 42
Step 1: Find two numbers
We need two numbers whose
Product = −42
Sum = −1
The required numbers are
−7 and 6
Step 2: Split the middle term
r2 − r − 42
= r2 − 7r + 6r − 42
Step 3: Factor by grouping
= r(r − 7) + 6(r − 7)
= (r − 7)(r + 6)
Final Answer
r2 − r − 42
= (r − 7)(r + 6)
✅ Solution (iv)
Sub Question
(iv) 49g2 + 14gh + h2
Given
The algebraic expression is
49g2 + 14gh + h2
Identity Used
(a + b)2 = a2 + 2ab + b2
Step 1: Compare with the identity
a2 = 49g2
2ab = 14gh
b2 = h2
Step 2: Identify the terms
a = 7g
b = h
Step 3: Write the factorised form
49g2 + 14gh + h2
= (7g + h)2
Final Answer
49g2 + 14gh + h2
= (7g + h)2
✅ Solution (v)
Sub Question
(v)
64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw
Given
The algebraic expression is
64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw
Identity Used
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Step 1: Compare with the identity
x2 = 64u2
y2 = 121v2
z2 = 4w2
2xy = −176uv
2yz = 44vw
2zx = −32uw
Step 2: Identify the terms
x = −8u
y = 11v
z = 2w
← Terms containing u are negative, so x = −8u.
Step 3: Verify the middle terms
2 × (−8u) × 11v = −176uv ✓
2 × 11v × 2w = 44vw ✓
2 × 2w × (−8u) = −32uw ✓
Step 4: Write the factorised form
64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw
= (x + y + z)2
= (−8u + 11v + 2w)2
Final Answer
64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw
= (−8u + 11v + 2w)2
📚 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)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.
📘 NCERT Ganita Manjari 2026
🎯 CBSE Aligned
📝 Step-by-Step Solutions
💡 Concept-Based Learning
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