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📐 Chapter 13 · NCERT 2025-26

🔢 Algebra Play

Factorisation · Division of Expressions · Linear Equations · Word Problems

x
y²
=
a+b
3ab
÷
📐 Introduction to Algebra Play

Algebra is the branch of mathematics where we use letters (variables) like x, y, a, b to represent unknown numbers and form expressions and equations. In this chapter, we will learn to factorise algebraic expressions (break them into simpler parts), divide one expression by another, and solve linear equations in one variable.

Think of factorisation as the reverse of multiplication. Just as we can multiply 3 and 5 to get 15, we can factorise 15 back into 3 × 5. In algebra, we do the same thing with expressions like 6x² + 3x = 3x(2x + 1).

🇮🇳 Indian Mathematical Heritage

Indian mathematicians made foundational contributions to algebra centuries before the rest of the world. The very word "algebra" comes from the Arabic al-jabr, but many key ideas originated in India.

📜 Brahmagupta (598–668 CE)

Brahmagupta's Brahmasphutasiddhanta contains the first systematic rules for operating with zero and negative numbers, and solving linear and quadratic equations. He used the concept of factorisation in solving equations centuries before European mathematicians.

🌟 Bhaskaracharya (1114–1185 CE)

Also known as Bhaskara II, his masterpiece Lilavati and Bijaganita contain elegant methods for solving equations, working with identities, and factorising expressions. Many techniques taught today trace back to his work.

💫 Mahavira (c. 850 CE)

The Jain mathematician Mahavira wrote Ganita Sara Sangraha, which contained advanced algebraic methods including systematic approaches to factorisation and division of expressions.
📚 What You Will Learn
Topic Key Concepts
Algebraic Expressions Terms, factors, coefficients, types of expressions
Factorisation Common factor, regrouping, using identities
Division Monomial ÷ monomial, polynomial ÷ monomial
Linear Equations Solving by transposing, cross-multiplication, word problems
💡 Why Learn This? Factorisation and equations are used in physics (calculating forces, speed), economics (profit/loss formulas), engineering (structural design), and computer science (algorithms). Mastering algebra in Class 8 is crucial for Class 9 and 10 mathematics!
🔢 Algebraic Expressions Review
📚 Key Terminology

Before we factorise or solve equations, let us revise the basic building blocks of algebra.

🔢 Variable

A letter representing an unknown number. Examples: x, y, a, b, p. The value of a variable can change.

🔢 Constant

A fixed number that does not change. Examples: 5, −3, ½, π. Constants are the "settled" parts of expressions.

🔢 Term

A single part of an expression, separated by + or − signs. In 3x² + 5x − 7, the terms are 3x², 5x, and −7.

🔢 Coefficient

The numerical part of a term. In 7xy, the coefficient is 7. In −3a²b, the coefficient is −3.
📝 Types of Algebraic Expressions
Type No. of Terms Examples
Monomial 1 5x, −3a²b, 7, xy²z
Binomial 2 x + 3, 2a − 5b, x² + y²
Trinomial 3 x² + 5x + 6, a + b + c
Polynomial 1 or more Any expression with terms using whole-number exponents
🔧 Factors of a Term

Every term in an algebraic expression is a product of its factors. For example:

  • 5xy = 5 × x × y (three factors)
  • 3a²b = 3 × a × a × b (four factors)
  • −2x²y³ = −2 × x × x × y × y × y (six factors)
💡 Memory Tip: Think of a term like a necklace made of beads. Each bead is a "factor." Breaking the necklace (factorising) means separating it into individual beads. Putting beads together (multiplying) gives back the full necklace.
📋 Like Terms and Unlike Terms

✅ Like Terms

Terms that have the same variable parts (same letters raised to the same powers). Like terms can be added or subtracted.
Example: 3x² and −7x² are like terms.

❌ Unlike Terms

Terms with different variable parts. Unlike terms cannot be combined by addition or subtraction.
Example: 3x² and 5x are unlike terms.
🔑 Factorisation of Algebraic Expressions

Factorisation means expressing an algebraic expression as a product of two or more simpler expressions (factors). It is the reverse of expansion (multiplication).

Expansion: a(b + c) = ab + ac
Factorisation: ab + ac = a(b + c) Factorisation is the reverse process of expansion
① Method 1: Taking Out the Common Factor

When all terms of an expression share a common factor, we "take it out" (extract it) and write the remaining parts inside brackets.

Example 1: Factorise 6x + 12

Step 1: Find the HCF of all terms.
6x = 6 × x
12 = 6 × 2
HCF = 6
Step 2: Take out the common factor.
6x + 12 = 6(x) + 6(2) = 6(x + 2)

Example 2: Factorise 12a²b + 18ab²

Step 1: Find the HCF of the terms.
12a²b = 2 × 2 × 3 × a × a × b
18ab² = 2 × 3 × 3 × a × b × b
HCF = 2 × 3 × a × b = 6ab
Step 2: Divide each term by 6ab.
12a²b ÷ 6ab = 2a
18ab² ÷ 6ab = 3b
Step 3: Write the factorised form.
12a²b + 18ab² = 6ab(2a + 3b)
② Method 2: Factorisation by Regrouping

When there is no single common factor for all terms, we rearrange and group the terms so that each group has a common factor. Then we factorise each group and look for a common binomial factor.

Example 3: Factorise ax + bx + ay + by

Step 1: Group the terms in pairs.
(ax + bx) + (ay + by)
Step 2: Take out the common factor from each group.
x(a + b) + y(a + b)
Step 3: Take out the common binomial factor (a + b).
ax + bx + ay + by = (a + b)(x + y)

Example 4: Factorise 6xy − 4y + 6 − 9x

Step 1: Rearrange and group.
(6xy − 9x) + (−4y + 6)
= (6xy − 9x) − (4y − 6)
Step 2: Factor each group.
3x(2y − 3) − 2(2y − 3)
Step 3: Take out the common factor (2y − 3).
6xy − 4y + 6 − 9x = (2y − 3)(3x − 2)
💡 Regrouping Tip: There may be more than one correct way to group terms. Try different groupings until you find one where a common binomial factor emerges. If one grouping doesn't work, rearrange and try again!
💡 Verification: You can always check your factorisation by expanding (multiplying) the factors. If you get back the original expression, your factorisation is correct!
💡 Factorisation Using Identities

Algebraic identities are equations that are true for all values of the variables. They are powerful shortcuts for factorisation. Let us recall the key identities.

📚 Standard Algebraic Identities
(a + b)² = a² + 2ab + b² Identity I — Square of a sum
(a − b)² = a² − 2ab + b² Identity II — Square of a difference
a² − b² = (a + b)(a − b) Identity III — Difference of two squares
(x + a)(x + b) = x² + (a + b)x + ab Identity IV — Product of two binomials
✏️ Factorisation Using Identity I

Example 5: Factorise x² + 6x + 9

Step 1: Compare with a² + 2ab + b².
x² = (x)² → a = x
9 = (3)² → b = 3
Check middle term: 2ab = 2(x)(3) = 6x ✅
Step 2: Apply the identity.
x² + 6x + 9 = (x)² + 2(x)(3) + (3)² = (x + 3)²
✏️ Factorisation Using Identity III

Example 6: Factorise 49p² − 36q²

Step 1: Express as a² − b².
49p² = (7p)² → a = 7p
36q² = (6q)² → b = 6q
Step 2: Apply Identity III: a² − b² = (a + b)(a − b).
49p² − 36q² = (7p + 6q)(7p − 6q)
✏️ Factorisation Using Identity IV

Example 7: Factorise x² + 7x + 12

Step 1: Compare with x² + (a + b)x + ab.
We need two numbers whose sum = 7 and product = 12.
Step 2: Find the numbers.
3 + 4 = 7 ✅
3 × 4 = 12 ✅
So a = 3, b = 4.
Step 3: Write the factorised form.
x² + 7x + 12 = (x + 3)(x + 4)

💡 How to Identify Which Identity to Use

• Three terms with a² + 2ab + b² pattern → Identity I
• Three terms with a² − 2ab + b² pattern → Identity II
• Two terms as difference of squares → Identity III
• x² + (sum)x + (product) → Identity IV

🔧 Quick Check Method

For Identity I or II, check: Is the middle term = 2 × (square root of first) × (square root of last)?
If yes, it fits Identity I (if +) or Identity II (if −).
💡 Memory Aid for Identity IV: To factorise x² + bx + c, think: "What two numbers add to give b and multiply to give c?" Write them as (x + first number)(x + second number).
✖ Division of Algebraic Expressions

Division of algebraic expressions is the reverse of multiplication. We can divide a monomial by a monomial, or a polynomial by a monomial.

① Monomial ÷ Monomial

To divide one monomial by another, divide the coefficients and subtract the exponents of like variables.

am ÷ an = am−n Law of exponents for division (m ≥ n)

Example 8: Divide 18x³y² by 6xy

Step 1: Divide coefficients: 18 ÷ 6 = 3
Step 2: Divide variable parts using exponent laws.
x³ ÷ x = x3−1 = x²
y² ÷ y = y2−1 = y
Answer: 18x³y² ÷ 6xy = 3x²y
② Polynomial ÷ Monomial

To divide a polynomial by a monomial, divide each term of the polynomial separately by the monomial.

(a + b + c) ÷ d = a/d + b/d + c/d Divide each term individually

Example 9: Divide (12x³ + 8x² − 4x) by 4x

Step 1: Divide each term by 4x.
12x³ ÷ 4x = 3x²
8x² ÷ 4x = 2x
−4x ÷ 4x = −1
Answer: (12x³ + 8x² − 4x) ÷ 4x = 3x² + 2x − 1

Example 10: Divide (15a²b³ − 10a³b² + 25ab) by 5ab

Step 1: Divide each term by 5ab.
15a²b³ ÷ 5ab = 3ab²
−10a³b² ÷ 5ab = −2a²b
25ab ÷ 5ab = 5
Answer: (15a²b³ − 10a³b² + 25ab) ÷ 5ab = 3ab² − 2a²b + 5
③ Division of Polynomial by Polynomial (Using Factorisation)

To divide one polynomial by another, we first factorise both the dividend and divisor, then cancel the common factors.

Example 11: Divide (x² + 5x + 6) by (x + 2)

Step 1: Factorise the dividend.
x² + 5x + 6 = (x + 2)(x + 3) [since 2 + 3 = 5 and 2 × 3 = 6]
Step 2: Divide by cancelling common factors.
(x + 2)(x + 3) ÷ (x + 2) = (x + 3)
💡 Important Rule: Division and factorisation work hand in hand. To divide polynomials, always try to factorise first, then cancel common factors. This is much simpler than long division for the expressions in this chapter.
💡 Quick Check: After dividing, multiply the quotient by the divisor. You should get back the dividend. This is the best way to verify your answer!
Quotient × Divisor = Dividend
✍️ Linear Equations in One Variable

A linear equation in one variable is an equation where the highest power of the variable is 1. The general form is ax + b = c, where a, b, c are constants and x is the variable.

📚 Key Concepts

🔢 What is an Equation?

An equation is a mathematical statement that two expressions are equal. It always has an = sign. For example: 2x + 3 = 11.

🔢 What is a Solution?

The solution (or root) of an equation is the value of the variable that makes both sides equal. For 2x + 3 = 11, the solution is x = 4 because 2(4) + 3 = 11.

🔢 LHS and RHS

LHS (Left Hand Side) is the expression on the left of the = sign. RHS (Right Hand Side) is the expression on the right. A solution makes LHS = RHS.

🔢 Linear vs Non-linear

Linear: highest power = 1 (e.g., 3x + 5 = 20).
Non-linear: power > 1 (e.g., x² + 3 = 12). We only solve linear equations in this chapter.
📋 Examples of Linear Equations
Equation Linear? Reason
3x + 7 = 22 ✅ Yes Highest power of x is 1
5y − 3 = 2y + 9 ✅ Yes Variables on both sides but all power 1
x² + 2x = 8 ❌ No Highest power is 2 (quadratic)
(x + 3)/2 = 5 ✅ Yes When simplified, x + 3 = 10 → linear
1/x + 2 = 5 ❌ No x is in the denominator (power −1)
💡 Golden Rule of Equations: Whatever you do to one side of an equation, you must do the same thing to the other side. Add, subtract, multiply, or divide — but always on both sides!
🔧 Solving Linear Equations
① Method 1: Transposing

Transposing means moving a term from one side of the equation to the other. When we transpose, the sign changes:

  • + becomes −
  • − becomes +
  • × becomes ÷
  • ÷ becomes ×

Example 12: Solve 3x + 5 = 20

Step 1: Transpose +5 to RHS (it becomes −5).
3x = 20 − 5
3x = 15
Step 2: Divide both sides by 3.
x = 15 ÷ 3 = 5
Verification: LHS = 3(5) + 5 = 15 + 5 = 20 = RHS ✅

Example 13: Solve 7x − 4 = 2x + 11

Step 1: Transpose 2x to LHS and −4 to RHS.
7x − 2x = 11 + 4
5x = 15
Step 2: Divide both sides by 5.
x = 15 ÷ 5 = 3
Verification: LHS = 7(3) − 4 = 21 − 4 = 17. RHS = 2(3) + 11 = 6 + 11 = 17 ✅
② Method 2: Cross-Multiplication

When an equation has fractions on both sides, we use cross-multiplication.

If a/b = c/d, then a × d = b × c Cross-multiplication — multiply diagonally

Example 14: Solve (2x + 1)/3 = (x + 4)/5

Step 1: Cross-multiply.
5(2x + 1) = 3(x + 4)
Step 2: Expand both sides.
10x + 5 = 3x + 12
Step 3: Transpose and solve.
10x − 3x = 12 − 5
7x = 7
x = 1
Verification: LHS = (2(1) + 1)/3 = 3/3 = 1. RHS = (1 + 4)/5 = 5/5 = 1 ✅
③ Equations with Variables on Both Sides

Example 15: Solve 5(x − 2) = 3(x + 4) − 2

Step 1: Expand both sides.
5x − 10 = 3x + 12 − 2
5x − 10 = 3x + 10
Step 2: Transpose variable terms to LHS, constants to RHS.
5x − 3x = 10 + 10
2x = 20
Step 3: Solve.
x = 20 ÷ 2 = 10
Verification: LHS = 5(10 − 2) = 5(8) = 40. RHS = 3(10 + 4) − 2 = 3(14) − 2 = 42 − 2 = 40 ✅

🔧 Solving Strategy

1. Remove brackets (expand).
2. Collect variable terms on LHS.
3. Collect constant terms on RHS.
4. Simplify both sides.
5. Divide to isolate the variable.
6. Always verify your answer!

⚠️ Common Mistakes

• Forgetting to change sign when transposing.
• Not distributing the negative sign correctly: −(2x − 3) = −2x + 3, not −2x − 3.
• Dividing only one side of the equation.
💡 Transposing Mantra: "When a term crosses the equals sign, its sign flips!" Positive becomes negative, multiplication becomes division. Think of the = sign as a mirror that reverses operations.
📚 Word Problems Leading to Linear Equations

Many real-life problems can be solved by translating them into linear equations. The key is to identify the unknown, represent it as a variable, form an equation, and solve it.

📋 Steps for Solving Word Problems
  1. Read the problem carefully (twice if needed!).
  2. Identify what is unknown — let it be x (or any variable).
  3. Translate the English sentences into a mathematical equation.
  4. Solve the equation.
  5. Check if the answer makes sense in the context of the problem.
✏️ Word Problem Examples

Example 16 (Age Problem): The present age of Rohan's mother is three times the present age of Rohan. After 5 years, their ages will add up to 66 years. Find their present ages.

Step 1: Let Rohan's present age = x years.
Then mother's present age = 3x years.
Step 2: After 5 years:
Rohan's age = (x + 5) years
Mother's age = (3x + 5) years
Step 3: Form the equation.
(x + 5) + (3x + 5) = 66
4x + 10 = 66
Step 4: Solve.
4x = 66 − 10 = 56
x = 56 ÷ 4 = 14
Answer: Rohan's age = 14 years, Mother's age = 3 × 14 = 42 years.
Check: After 5 years: 19 + 47 = 66 ✅

Example 17 (Number Problem): The sum of three consecutive even numbers is 78. Find the numbers.

Step 1: Let the three consecutive even numbers be x, x + 2, and x + 4.
Step 2: Form the equation.
x + (x + 2) + (x + 4) = 78
3x + 6 = 78
Step 3: Solve.
3x = 78 − 6 = 72
x = 72 ÷ 3 = 24
Answer: The numbers are 24, 26, and 28.
Check: 24 + 26 + 28 = 78 ✅

Example 18 (Perimeter Problem): The length of a rectangle is 5 cm more than its breadth. If the perimeter is 50 cm, find the length and breadth.

Step 1: Let breadth = x cm.
Then length = (x + 5) cm.
Step 2: Perimeter of rectangle = 2(length + breadth).
2(x + 5 + x) = 50
2(2x + 5) = 50
Step 3: Solve.
4x + 10 = 50
4x = 40
x = 10
Answer: Breadth = 10 cm, Length = 10 + 5 = 15 cm.
Check: Perimeter = 2(15 + 10) = 2(25) = 50 cm ✅

Example 19 (Distribution Problem): A sum of ₹480 is divided among A, B, and C such that A gets ₹20 more than B, and C gets ₹40 more than B. Find each person's share.

Step 1: Let B's share = ₹x.
A's share = ₹(x + 20)
C's share = ₹(x + 40)
Step 2: Form the equation.
(x + 20) + x + (x + 40) = 480
3x + 60 = 480
Step 3: Solve.
3x = 480 − 60 = 420
x = 420 ÷ 3 = 140
Answer: B = ₹140, A = ₹160, C = ₹180.
Check: 160 + 140 + 180 = 480 ✅

💡 Translation Guide

• "sum of" → addition (+)
• "difference" → subtraction (−)
• "product" → multiplication (×)
• "is" / "was" / "will be" → equals (=)
• "more than" → add
• "less than" → subtract
• "times" / "twice" / "thrice" → multiply

🔢 Consecutive Numbers

• Consecutive integers: x, x+1, x+2
• Consecutive even: x, x+2, x+4
• Consecutive odd: x, x+2, x+4
(Note: consecutive even AND odd both differ by 2!)
💡 Word Problem Tip: If you are stuck, try working backwards. Assume an answer, check if it satisfies all conditions, and adjust. This "guess and check" approach can also help you set up the correct equation.
📊 Chapter Summary
📋 All Key Formulas
(a + b)² = a² + 2ab + b² Identity I
(a − b)² = a² − 2ab + b² Identity II
a² − b² = (a + b)(a − b) Identity III
(x + a)(x + b) = x² + (a + b)x + ab Identity IV
📚 Methods Summary
Method When to Use Example
Common Factor All terms share a factor 6x + 12 = 6(x + 2)
Regrouping 4 terms, no single common factor ax + bx + ay + by = (a+b)(x+y)
Identity I a² + 2ab + b² pattern x² + 6x + 9 = (x+3)²
Identity III Difference of two perfect squares 49 − x² = (7+x)(7−x)
Identity IV x² + bx + c pattern x² + 7x + 12 = (x+3)(x+4)
Transposing Solving linear equations 3x + 5 = 20 → x = 5
Cross-multiplication Equations with fractions a/b = c/d → ad = bc
⚠️ Common Mistakes to Avoid

❌ Incomplete Factorisation

Always check if factors can be further factorised. For example, 2x(4x² − 9) should be written as 2x(2x + 3)(2x − 3).

❌ Sign Errors in Transposing

When moving a term across =, the sign MUST change. +5 on LHS becomes −5 on RHS. This is the most common error!

❌ Forgetting to Distribute

When expanding −2(x − 3), both terms get multiplied: −2x + 6, NOT −2x − 6.

❌ Not Verifying the Answer

Always substitute your answer back into the original equation. It takes 30 seconds and can save you from losing marks!
💡 Exam Day Checklist:
✅ For factorisation: Check by expanding your answer.
✅ For division: Verify that Quotient × Divisor = Dividend.
✅ For equations: Substitute the answer back into the original equation.
✅ For word problems: Check if the answer makes sense (e.g., age cannot be negative).
✅ Show all steps — marks are given for working!
🧠 Multiple Choice Questions (10 MCQs)

Click on an option to see if your answer is correct. The correct option will turn green.

  • Q1. The factorised form of 6x + 18 is:
    • a) 2(3x + 18)
    • b) 6(x + 3)
    • c) 3(2x + 18)
    • d) 6(x + 18)
    ✅ Answer: (b) 6(x + 3) — HCF of 6x and 18 is 6. So 6x + 18 = 6(x + 3).
  • Q2. Which identity is used to factorise x² − 25?
    • a) (a + b)²
    • b) (a − b)²
    • c) a² − b² = (a + b)(a − b)
    • d) (x + a)(x + b)
    ✅ Answer: (c) a² − b² = (a + b)(a − b) — x² − 25 = x² − 5² = (x + 5)(x − 5).
  • Q3. The solution of 5x − 3 = 17 is:
    • a) x = 3
    • b) x = 4
    • c) x = 5
    • d) x = 2
    ✅ Answer: (b) x = 4 — 5x = 17 + 3 = 20, so x = 20/5 = 4.
  • Q4. The factorised form of x² + 10x + 25 is:
    • a) (x + 10)(x + 15)
    • b) (x + 25)²
    • c) (x + 5)²
    • d) (x − 5)²
    ✅ Answer: (c) (x + 5)² — Using Identity I: x² + 2(x)(5) + 5² = (x + 5)².
  • Q5. 12x³ ÷ 3x equals:
    • a) 4x³
    • b) 4x²
    • c) 9x²
    • d) 4x
    ✅ Answer: (b) 4x² — (12/3)(x³/x) = 4 × x² = 4x².
  • Q6. The factorised form of x² − 8x + 16 is:
    • a) (x + 4)²
    • b) (x − 4)²
    • c) (x + 4)(x − 4)
    • d) (x − 8)(x − 2)
    ✅ Answer: (b) (x − 4)² — Using Identity II: x² − 2(x)(4) + 4² = (x − 4)².
  • Q7. If 3(x − 2) = 12, then x equals:
    • a) 2
    • b) 4
    • c) 6
    • d) 8
    ✅ Answer: (c) 6 — 3x − 6 = 12, so 3x = 18, x = 6.
  • Q8. The HCF of 8a²b and 12ab² is:
    • a) 24ab
    • b) 4ab
    • c) 8ab
    • d) 2ab
    ✅ Answer: (b) 4ab — HCF of 8 and 12 is 4. Common variable factors: a (minimum power 1) and b (minimum power 1). So HCF = 4ab.
  • Q9. The factorised form of x² + 5x + 6 is:
    • a) (x + 2)(x + 3)
    • b) (x + 1)(x + 6)
    • c) (x + 5)(x + 1)
    • d) (x − 2)(x − 3)
    ✅ Answer: (a) (x + 2)(x + 3) — We need two numbers with sum 5 and product 6: 2 + 3 = 5, 2 × 3 = 6.
  • Q10. The sum of three consecutive odd numbers is 51. The middle number is:
    • a) 15
    • b) 17
    • c) 19
    • d) 13
    ✅ Answer: (b) 17 — Let the numbers be x, x+2, x+4. Then 3x + 6 = 51, 3x = 45, x = 15. The middle number is 15 + 2 = 17.
✍️ Short Answer Questions (NCERT Style)
  • Q1. Factorise: 15xy + 25x
    HCF = 5x. So 15xy + 25x = 5x(3y + 5).
  • Q2. Factorise using an identity: 4x² − 9
    This is a² − b² where a = 2x and b = 3.
    4x² − 9 = (2x)² − (3)² = (2x + 3)(2x − 3).
  • Q3. Divide: (8x² − 4x) ÷ 2x
    Divide each term by 2x:
    8x² ÷ 2x = 4x, and −4x ÷ 2x = −2.
    Answer: 4x − 2.
  • Q4. Solve: 2x + 7 = 19
    Transpose: 2x = 19 − 7 = 12.
    x = 12 ÷ 2 = 6.
    Check: 2(6) + 7 = 12 + 7 = 19 ✅
  • Q5. Factorise by regrouping: am + bm + an + bn
    Group: (am + bm) + (an + bn) = m(a + b) + n(a + b) = (a + b)(m + n).
  • Q6. Solve: (x + 3)/4 = (x − 1)/3
    Cross-multiply: 3(x + 3) = 4(x − 1).
    3x + 9 = 4x − 4.
    9 + 4 = 4x − 3x → x = 13.
    Check: LHS = 16/4 = 4, RHS = 12/3 = 4 ✅
  • Q7. Factorise: x² − 14x + 49
    Using Identity II: x² − 2(x)(7) + 7² = (x − 7)².
  • Q8. What is a monomial? Give two examples.
    A monomial is an algebraic expression containing only one term. Examples: 5x² and −3ab.
  • Q9. Divide: (x² − 9) ÷ (x + 3)
    Factorise the numerator: x² − 9 = (x + 3)(x − 3).
    (x + 3)(x − 3) ÷ (x + 3) = (x − 3).
  • Q10. A number is 12 more than twice another number. If their sum is 45, find the numbers.
    Let the smaller number = x. Then larger = 2x + 12.
    x + 2x + 12 = 45 → 3x = 33 → x = 11.
    The numbers are 11 and 2(11) + 12 = 34.
    Check: 11 + 34 = 45 ✅
📖 Long Answer Questions (NCERT Style)
Q1. Factorise the following expressions completely:
(i) 3x² + 9x + 6    (ii) 2x² − 8    (iii) x² − 3x − 10
(i) 3x² + 9x + 6
First, take out common factor 3: 3(x² + 3x + 2).
Now factorise x² + 3x + 2: We need two numbers with sum = 3 and product = 2 → 1 and 2.
x² + 3x + 2 = (x + 1)(x + 2).
Answer: 3(x + 1)(x + 2).

(ii) 2x² − 8
Take out common factor 2: 2(x² − 4).
x² − 4 = x² − 2² = (x + 2)(x − 2) [using Identity III].
Answer: 2(x + 2)(x − 2).

(iii) x² − 3x − 10
We need two numbers with sum = −3 and product = −10.
−5 + 2 = −3 ✅ and (−5)(2) = −10 ✅
Answer: (x − 5)(x + 2).
Q2. The denominator of a fraction is 4 more than the numerator. If 1 is added to both the numerator and denominator, the fraction becomes 1/2. Find the fraction.
Let numerator = x. Then denominator = x + 4.
Original fraction = x/(x + 4).

Given: (x + 1)/(x + 4 + 1) = 1/2
(x + 1)/(x + 5) = 1/2

Cross-multiply: 2(x + 1) = 1(x + 5)
2x + 2 = x + 5
2x − x = 5 − 2
x = 3

Answer: The fraction is 3/7.
Check: Adding 1 to both: 4/8 = 1/2 ✅
Q3. Divide (x³ + 2x² − x − 2) by (x + 1) using factorisation.
Step 1: Factorise the dividend by regrouping.
x³ + 2x² − x − 2
= (x³ + 2x²) + (−x − 2)
= x²(x + 2) − 1(x + 2)
= (x + 2)(x² − 1)

Step 2: Further factorise x² − 1 using Identity III.
x² − 1 = (x + 1)(x − 1)

So the dividend = (x + 2)(x + 1)(x − 1).

Step 3: Divide by (x + 1).
(x + 2)(x + 1)(x − 1) ÷ (x + 1) = (x + 2)(x − 1) = x² + x − 2.

Verification: (x² + x − 2)(x + 1) = x³ + x² − 2x + x² + x − 2 = x³ + 2x² − x − 2 ✅
Q4. Explain the four standard algebraic identities with examples of both expansion and factorisation for each.
Identity I: (a + b)² = a² + 2ab + b²
Expansion: (x + 4)² = x² + 2(x)(4) + 16 = x² + 8x + 16
Factorisation: 9x² + 12x + 4 = (3x)² + 2(3x)(2) + 2² = (3x + 2)²

Identity II: (a − b)² = a² − 2ab + b²
Expansion: (y − 5)² = y² − 2(y)(5) + 25 = y² − 10y + 25
Factorisation: 4a² − 20ab + 25b² = (2a)² − 2(2a)(5b) + (5b)² = (2a − 5b)²

Identity III: a² − b² = (a + b)(a − b)
Expansion: (m + 7)(m − 7) = m² − 49
Factorisation: 16x² − 81 = (4x)² − 9² = (4x + 9)(4x − 9)

Identity IV: (x + a)(x + b) = x² + (a + b)x + ab
Expansion: (x + 3)(x + 5) = x² + 8x + 15
Factorisation: x² + 11x + 30 = (x + 5)(x + 6) [since 5 + 6 = 11, 5 × 6 = 30]

These identities are powerful tools that allow us to quickly expand and factorise expressions without having to multiply or find factors through trial and error every time.
Q5. A father is 30 years older than his son. In 12 years, the father's age will be twice the son's age. Find their present ages. Also, find the son's age when the father is exactly three times his age.
Part 1:
Let son's present age = x years. Father's present age = (x + 30) years.

After 12 years:
Son's age = (x + 12), Father's age = (x + 30 + 12) = (x + 42).

Given: Father's age = 2 × Son's age
x + 42 = 2(x + 12)
x + 42 = 2x + 24
42 − 24 = 2x − x
x = 18

Answer: Son's present age = 18 years, Father's present age = 18 + 30 = 48 years.
Check: After 12 years: Son = 30, Father = 60 = 2 × 30 ✅

Part 2:
Let the number of years ago when father was 3 times son's age = y years ago.
Father's age y years ago: (48 − y). Son's age: (18 − y).
48 − y = 3(18 − y)
48 − y = 54 − 3y
3y − y = 54 − 48
2y = 6 → y = 3.

So 3 years ago, son was 18 − 3 = 15 years old when father was 45 = 3 × 15.
🃏 Quick Revision Flashcards

Use these for a quick last-minute revision before your exam!

Q: What is factorisation?

A: Expressing an expression as a product of its factors. It is the reverse of expansion.

Q: Factorise a² − b²

A: (a + b)(a − b) — difference of two squares.

Q: HCF of 12x²y and 8xy²?

A: HCF of 12, 8 = 4. Common vars: xy. Answer: 4xy.

Q: Solve 4x − 7 = 13

A: 4x = 20, x = 5.

Q: What changes when transposing?

A: The sign changes. + becomes −, × becomes ÷, and vice versa.

Q: Expand (a + b)²

A: a² + 2ab + b²

Q: Cross-multiplication rule?

A: If a/b = c/d, then ad = bc.

Q: 10x² ÷ 2x = ?

A: 5x (divide coefficients, subtract exponents).

Q: Factorise x² + 9x + 20

A: 4 + 5 = 9, 4 × 5 = 20 → (x + 4)(x + 5).

Q: Consecutive even numbers?

A: x, x + 2, x + 4 (each differs by 2).
💡 Final Revision Mantra:
• Factorisation → Reverse of expansion. Check: HCF method, regrouping, identities.
• Division → Factorise first, then cancel common factors.
• Equations → Transpose, simplify, solve, verify.
• Word Problems → Read, Let x, Form equation, Solve, Check.
• Identities → I: sum squared, II: difference squared, III: diff of squares, IV: split middle term.

You've got this! Go ace that exam! 💪
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