Introduction & Visualising Identities — Part 1

Chapter 4: Exploring Algebraic Identities

Page 1 of 8: Introduction & Visualising Identities — Part 1

Welcome to the world of algebraic identities! You've worked with algebraic expressions and equations before. Equations are like puzzles that are true for only certain values. For instance, x + 5 = 8 is only true when x is 3. But what if we find a special type of equality that is always true, no matter what value we substitute for the variable?

These universal truths of algebra are called identities. They are powerful tools that help us simplify complex expressions, solve problems faster, and even understand the geometry behind the numbers.

Imagine a city planner designing a square park. The initial plan is for a park with side length a. Later, they decide to expand the park by adding b meters to both its length and width, keeping it a square. The new side length is a+b. To find the new area, we calculate (a+b)². An identity gives us a direct way to understand the components of this new area without complex multiplication, showing us it's the original area (a²), plus the area of the two new rectangular strips (2ab), plus the new corner square (b²).

{{VISUAL: diagram: a square plot of land with side 'a' being extended by 'b' on two adjacent sides, forming a larger square of side 'a+b'. The extension strips are highlighted.}}

This chapter is your guide to mastering these fundamental tools. We will not just learn them by heart; we will see why they work, both algebraically and visually.

{{FORMULA: expr=(a+b)² = a² + 2ab + b² | symbols=a:first term, b:second term}}

Definitions & Formulas

Before we derive our first identity, let's be clear on the terminology we'll be using. Understanding these distinctions is crucial.

The key identity we will explore on this page is the square of a sum.

Derivation & Logic of (a+b)²

Why is (a+b)² equal to a² + 2ab + b²? Let's explore this truth in two ways: through pure algebra and through a visual, geometric proof.

Method 1: Algebraic Derivation

This method uses the distributive property of multiplication that you're already familiar with.

1. Start with the expression (a+b)². The square of any number means multiplying it by itself.

   (a+b)² = (a+b) × (a+b)
   

2. Now, we apply the distributive property. We will multiply the first term a from the first bracket with the entire second bracket (a+b).

   = a(a+b) + b(a+b)
   

3. Next, distribute a into its bracket and b into its bracket.

   = (a × a) + (a × b) + (b × a) + (b × b)
   

4. Simplify the terms. Remember that b × a is the same as a × b.

   = a² + ab + ab + b²
   

5. Combine the like terms (ab + ab) to get the final identity.

   = a² + 2ab + b²
   

Method 2: Geometric Visualisation

This is where the identity truly comes to life! Let's build a square with a side length of (a+b). The area of this square will be (a+b)².

{{VISUAL: diagram: a large square of side (a+b) divided into four quadrilaterals. A top-left square is labeled with side 'a' and area 'a²'. A bottom-right square is labeled with side 'b' and area 'b²'. Two rectangles are labeled with sides 'a' and 'b' and area 'ab'.}}

1. Imagine a line segment of length (a+b). We can think of it as a segment of length a joined to a segment of length b.

2. Now, construct a square using this line segment (a+b) as its side. The total area of this large square is (a+b) × (a+b), or (a+b)².

3. Let's divide this large square based on the inner lengths a and b. This dissection creates four smaller shapes inside the large square:

   • One square with side length a. Its area is a × a = a².
   • One smaller square with side length b. Its area is b × b = b².
   • Two rectangles, each with length a and width b. The area of each is a × b = ab.

4. The total area of the large square must be the sum of the areas of these four smaller parts.

   Total Area = (Area of big square) + (Area of small square) + (Area of two rectangles)
   

5. Substituting the values we found:

   (a+b)² = a² + b² + ab + ab
   

6. Combining the like terms, we arrive at the same identity.

   (a+b)² = a² + 2ab + b²
   

This visual proof shows that the formula isn't just an abstract rule; it's a description of a geometric reality.

Solved Examples

Let's apply this identity to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Direct Expansion

Given: The expression (x + 5)².

To Find: The expanded form of the expression.

Solution:

1. Identify the terms a and b in the identity (a+b)² = a² + 2ab + b². Here, a = x and b = 5.

2. Substitute these values into the identity.

   (x+5)² = (x)² + 2(x)(5) + (5)²
   

3. Simplify each term in the expression.

   = x² + 10x + 25
   

Final Answer: x² + 10x + 25

Example 2: Numerical Calculation

Given: The number 103.

To Find: The value of 103² using a suitable identity.

Solution:

1. We can write 103 as a sum of two numbers that are easy to square. The best choice is 100 + 3.

   103² = (100 + 3)²
   

2. This expression is in the form (a+b)², where a = 100 and b = 3. Apply the identity (a+b)² = a² + 2ab + b².

   (100 + 3)² = (100)² + 2(100)(3) + (3)²
   

3. Calculate the value of each term.

   = 10000 + 600 + 9
   

4. Add the numbers together to get the final result.

   = 10609
   

Final Answer: 10609

Example 3: Working Backwards

Given: x + 1/x = 6.

To Find: The value of x² + 1/x².

Solution:

1. The given equation is x + 1/x = 6. We need to find an expression with x² terms. This suggests we should square both sides of the given equation.

   (x + 1/x)² = 6²
   

2. Apply the identity (a+b)² = a² + 2ab + b² to the left-hand side (LHS). Here, a = x and b = 1/x.

   (x)² + 2(x)(1/x) + (1/x)² = 36
   

3. Simplify the LHS. Notice that in the middle term 2(x)(1/x), the x in the numerator and denominator cancel out.

   x² + 2(1) + 1/x² = 36
   

   x² + 2 + 1/x² = 36
   

4. To isolate x² + 1/x², subtract 2 from both sides of the equation.

   x² + 1/x² = 36 - 2
   

   x² + 1/x² = 34
   

Final Answer: 34

Example 4: Tricky Application (Factorisation)

Given: The expression 9x² + 30xy + 25y².

To Find: The factorised form of the expression.

Solution:

1. We need to check if this expression fits the pattern a² + 2ab + b². First, look at the first and last terms.

   {{VISUAL: diagram: a square with area labeled '9x² + 30xy + 25y²', indicating that the expression represents the area of a square whose side length needs to be found.}}

2. The first term is 9x². Can this be written as a square? Yes, 9x² = (3x)². So, our potential a is 3x.

3. The last term is 25y². Can this be written as a square? Yes, 25y² = (5y)². So, our potential b is 5y.

4. Now, we must verify the middle term. The middle term in the identity is 2ab. Let's calculate 2ab using our potential a and b.

   2ab = 2(3x)(5y)
   

   = 2(15xy)
   

   = 30xy
   

5. This matches the middle term of the given expression exactly. Therefore, the expression is a perfect square trinomial.

6. We can now write it in the factorised form (a+b)².

   9x² + 30xy + 25y² = (3x + 5y)²
   

Final Answer: (3x + 5y)²

{{KEY: type=concept | title=Recognising the Pattern | text=To check if a trinomial Ax² + Bx + C is a perfect square, confirm two things: 1. The first term A and the last term C must be perfect squares (let their roots be 'a' and 'b'). 2. The middle term B must be equal to 2 × a × b.}}

Tips & Tricks

Here are some shortcuts and mental models to help you master this identity.

Common Mistakes

Many students stumble on the same points when first learning this identity. Here’s how to avoid those pitfalls.

Brain-Teaser Questions

Ready to test your understanding at a higher level? Try these HOTS (Higher Order Thinking Skills) questions.

1. If a+b = 12 and ab = 32, what is the value of a² + b²?

💡 Answer: We know (a+b)² = a² + b² + 2ab. Substitute the given values: (12)² = a² + b² + 2(32). 144 = a² + b² + 64. a² + b² = 144 - 64 = 80.

2. A large square has an area of (49x² + 28x + 4) square units. A smaller square with an area of 4x² is cut out from it. What is the remaining area in terms of x?

💡 Answer: First, recognize that 49x² + 28x + 4 = (7x)² + 2(7x)(2) + (2)² = (7x+2)². The area of the large square is (7x+2)². The area of the smaller square is 4x². The remaining area is (Large Area) - (Small Area) = (49x² + 28x + 4) - 4x². Combine like terms: (49x² - 4x²) + 28x + 4 = 45x² + 28x + 4.

3. The sum of two numbers is 10. The sum of their squares is 58. Find the product of the two numbers.

💡 Answer: Let the two numbers be a and b. Given: a+b = 10 and a² + b² = 58. We use the identity (a+b)² = a² + b² + 2ab. Substitute the known values into the identity: (10)² = 58 + 2ab. 100 = 58 + 2ab. 2ab = 100 - 58 = 42. ab = 42 ÷ 2 = 21. The product of the numbers is 21.

Mini Cheatsheet

Here is a summary of the key concepts from this page. Screenshot this for quick revision!

Visualising Identities — Part 2

Page 2: Visualising Identities — Part 2

Concept Introduction

Imagine you are helping your family design a new square garden. Your mother wants to add a border of 2 meters on each side of an existing 5m × 5m plot. Instead of calculating the new area by measuring the whole garden again, you can use algebra!

The new side is (5 + 2) meters, so the area is (5 + 2)² square meters. Using the identity we'll learn today, you can instantly expand this as 5² + 2(5)(2) + 2² = 25 + 20 + 4 = 49 m² — no need for physical measurement!

This is the power of algebraic identities — they give us shortcuts to expand expressions, perform calculations quickly, and factor complex polynomials. In this page, we'll distinguish identities from equations, master the (a + b)² identity through visual models, and apply it to both algebraic expansions and numerical computations.

{{FORMULA: expr=(a + b)² = a² + 2ab + b² | symbols=a:first term, b:second term, 2ab:middle term (twice the product)}}

Definitions & Key Terms

Visualising the Identity (a + b)² = a² + 2ab + b²

The NCERT text shows us how to see this identity using a square. Let's walk through the derivation step-by-step.

Step 1: Take two line segments of lengths a and b units. Join them end-to-end to create a line segment of total length (a + b) units.

Step 2: Construct a square with side length (a + b) units. The area of this outer square is (a + b)² square units.

{{VISUAL: diagram: a large square with side labeled (a + b), subdivided into four regions — top-left square labeled a×a, bottom-right square labeled b×b, and two identical rectangles labeled a×b each, with all sides clearly marked}}

Step 3: Partition the square into four regions:

• A square of side a (area = a²)
• A square of side b (area = b²)
• Two identical rectangles, each with dimensions a × b (total area = 2ab)

Step 4: Since the four regions completely fill the outer square, their areas must add up:

(a + b)² = a² + ab + ab + b²

Step 5: Combine the two ab terms:

(a + b)² = a² + 2ab + b²

This visual proof works whenever a and b represent positive lengths. But what about negative numbers or fractions?

Algebraic Verification (For All Numbers)

The geometric model only works for positive values. To prove the identity for all real numbers (negative, fractions, irrationals), we use the distributive property:

Step 1: Write (a + b)² as a product:

(a + b)² = (a + b)(a + b)

Step 2: Distribute the first term a:

= a(a + b) + b(a + b)

Step 3: Expand each bracket:

= a² + ab + ba + b²

Step 4: Use commutativity (ab = ba):

= a² + ab + ab + b²

Step 5: Combine like terms:

= a² + 2ab + b²

This proves the identity algebraically — it holds for negative numbers, fractions, and all real numbers!

{{KEY: type=concept | title=Identity vs Equation | text=An identity is true for ALL values (e.g., (x + y)² = x² + 2xy + y²), while an equation is true only for specific values (e.g., x² = 25 is true only for x = ±5).}}

Solved Examples

Example 1: Expanding a Binomial (Easy)

Given: (7x + 4y)²

To Find: Expanded form using the identity (a + b)² = a² + 2ab + b²

Solution:

1. Identify a = 7x and b = 4y.

2. Apply the identity:

(7x + 4y)² = (7x)² + 2(7x)(4y) + (4y)²

3. Evaluate each term:

= 49x² + 56xy + 16y²

Final Answer: 49x² + 56xy + 16y²

Example 2: Fractional Binomial (Medium)

Given: (7x/5 + 3y/2)²

To Find: Expanded form

Solution:

1. Identify a = 7x/5 and b = 3y/2.

2. Apply the identity:

(7x/5 + 3y/2)² = (7x/5)² + 2(7x/5)(3y/2) + (3y/2)²

3. Calculate (7x/5)² = 49x²/25.

4. Calculate 2(7x/5)(3y/2) = 2 × (21xy/10) = 42xy/10 = 21xy/5.

5. Calculate (3y/2)² = 9y²/4.

6. Combine all terms:

= 49x²/25 + 21xy/5 + 9y²/4

Final Answer: 49x²/25 + 21xy/5 + 9y²/4

Example 3: Numerical Calculation (Medium)

Given: Calculate 43² using the identity

To Find: The value of 43² without direct multiplication

Solution:

1. Write 43 = 40 + 3, so we need (40 + 3)².

2. Identify a = 40, b = 3.

3. Apply the identity:

(40 + 3)² = 40² + 2(40)(3) + 3²

4. Calculate each term:

= 1600 + 240 + 9

5. Add:

= 1849

Final Answer: 1849

{{VISUAL: diagram: a square subdivided to show (40 + 3)² = 40² + 2(40×3) + 3², with each region labeled with its area — 1600, 240, 240, and 9}}

Example 4: Complex Mixed Term (Hard)

Given: (2.5p + 1.5q)²

To Find: Expanded form (keeping decimals or converting to fractions)

Solution:

1. We can work directly with decimals or convert: 2.5 = 5/2, 1.5 = 3/2.

2. Using decimals, identify a = 2.5p and b = 1.5q.

3. Apply the identity:

(2.5p + 1.5q)² = (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²

4. Calculate (2.5p)² = 6.25p².

5. Calculate 2(2.5p)(1.5q) = 2 × 3.75pq = 7.5pq.

6. Calculate (1.5q)² = 2.25q².

7. Combine:

= 6.25p² + 7.5pq + 2.25q²

Final Answer: 6.25p² + 7.5pq + 2.25q² (or 25p²/4 + 15pq/2 + 9q²/4 in fraction form)

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Q1: If (x + 5)² = x² + kx + 25, what is the value of k?

💡 Answer: Using the identity (a + b)² = a² + 2ab + b², here a = x and b = 5. The middle term is 2ab = 2(x)(5) = 10x. Therefore, k = 10.

Q2: A square garden has side (a + b) meters. If a = 12 m and b = 3 m, and the cost of tiling is ₹50 per m², what is the total cost?

💡 Answer: Area = (a + b)² = (12 + 3)² = 15² = 225 m². Total cost = 225 × 50 = ₹11,250. Alternatively, using the identity: 12² + 2(12)(3) + 3² = 144 + 72 + 9 = 225 m².

Q3: Without expanding, determine which is greater: (100 + 1)² or 100² + 1²?

{{VISUAL: diagram: two side-by-side squares — one labeled (100 + 1)² subdivided into four regions, the other showing only two squares 100² and 1² with the missing 2×100×1 region highlighted}}

💡 Answer: (100 + 1)² = 100² + 2(100)(1) + 1² = 10000 + 200 + 1 = 10201, while 100² + 1² = 10000 + 1 = 10001. The identity version is greater by 200 (the 2ab term). In general, (a + b)² > a² + b² whenever both a and b are positive (because 2ab > 0).

Mini Cheatsheet

Next Steps: In the coming pages, we'll explore the companion identity (a − b)² = a² − 2ab + b², learn how to use identities for factorisation, and discover the difference-of-squares identity a² − b² = (a + b)(a − b). These identities are your power tools for solving equations, simplifying expressions, and acing your exams!

Factorisation of Algebraic Expressions Using Identities

Page 3: Factorisation of Algebraic Expressions Using Identities

Concept Introduction

Factorisation is the reverse process of expanding algebraic expressions. Instead of multiplying factors to get an expanded form, we break down a complex expression into simpler multiplicative components. This skill is fundamental in solving equations, simplifying fractions, and understanding polynomial behaviour.

Real-Life Application: Imagine you're designing a square garden with a walking path around it. The total area including the path is (x + 3)² square meters, where x is the side of the inner garden. To calculate the area of just the path, you need to expand this expression and subtract the garden area x². Understanding factorisation helps reverse this process — if you know the total area is x² + 6x + 9, you can immediately identify it as (x + 3)², revealing the structure of your design. This principle applies in architecture, engineering designs, and financial planning where compound relationships exist.

{{FORMULA: expr=(a + b)² = a² + 2ab + b² | symbols=a:first term, b:second term, 2ab:middle term (twice the product)}}

Core Identities for Factorisation

The following identities are our primary tools for factorising quadratic expressions:

Recognition Strategy: To use these identities for factorisation, identify whether an expression matches the pattern a² + 2ab + b² or a² - 2ab + b².

Logic Behind Factorisation Using Identities

The key insight is that factorisation is expansion in reverse. Here's the systematic approach:

Step 1: Identify if the expression is a trinomial (three-term polynomial) of degree 2.

Step 2: Check if the first and last terms are perfect squares. Find their square roots — these become your 'a' and 'b'.

Step 3: Verify the middle term. It must equal 2 × (first square root) × (second square root).

Step 4: Determine the sign. If the middle term is positive, use (a + b)². If negative, use (a - b)².

Step 5: Write the factored form as the square of a binomial.

Step 6: Verify by expanding your answer to ensure it matches the original expression.

{{KEY: type=strategy | title=The Square Root Test | text=An expression x² + px + q can be factorised as a perfect square ONLY if q is a perfect square AND p = ±2√q times the coefficient of x}}

Geometric Interpretation

Understanding these identities visually strengthens algebraic manipulation skills.

Visualising (a + b)²

Consider a square with side length (a + b). Its total area represents (a + b)².

{{VISUAL: diagram: a square with side (a + b) divided into four regions — top-left square labeled a², top-right rectangle labeled ab, bottom-left rectangle labeled ab, bottom-right square labeled b², with all sides clearly marked showing how a² + ab + ab + b² = a² + 2ab + b²}}

When we divide this square using vertical and horizontal lines at distances 'a' and 'b' from the edges, we create four regions:

• One square of area a² (top-left)
• One square of area b² (bottom-right)
• Two rectangles each of area ab (top-right and bottom-left)

Total area: a² + ab + ab + b² = a² + 2ab + b²

Visualising (a - b)²

For this identity, imagine a square of side 'a' with a smaller square of side 'b' removed from one corner.

{{VISUAL: diagram: a large square with side 'a' containing a smaller square of side 'b' cut from the bottom-right corner, with the remaining L-shaped region divided to show two rectangles of area b(a-b) and ab, and a square of area (a-b)², demonstrating that a² minus the removed portions equals (a-b)²}}

The remaining area after removing b² can be rearranged to show: (a - b)² = a² - 2ab + b²

Solved Examples

Example 1: Basic Perfect Square Recognition (Easy)

Given: x² + 6x + 9

To Find: Factorise the expression using algebraic identities.

Solution:

1. Identify the first term's square root: √(x²) = x, so a = x.

2. Identify the last term's square root: √9 = 3, so b = 3.

3. Verify the middle term pattern:

2ab = 2 × x × 3 = 6x ✓

4. Since the middle term is positive, use the identity (a + b)²:

x² + 6x + 9 = (x + 3)²

Final Answer: (x + 3)²

Example 2: Factorising with Larger Coefficients (Medium)

Given: 49y² - 84y + 36

To Find: Express in factored form.

Solution:

1. Check if the first term is a perfect square: √(49y²) = 7y, so a = 7y.

2. Check if the last term is a perfect square: √36 = 6, so b = 6.

3. Calculate what the middle term should be:

2ab = 2 × 7y × 6 = 84y

4. The actual middle term is -84y (negative), so we use (a - b)²:

49y² - 84y + 36 = (7y)² - 2(7y)(6) + 6²

5. Apply the identity:

= (7y - 6)²

Final Answer: (7y - 6)²

Example 3: Factorising with Common Factors First (Hard)

Given: 18p² + 60pq + 50q²

To Find: Completely factorise the expression.

Solution:

1. Notice that all coefficients are even. Extract the common factor 2:

18p² + 60pq + 50q² = 2(9p² + 30pq + 25q²)

2. Now work with the trinomial 9p² + 30pq + 25q². Find the square roots: √(9p²) = 3p and √(25q²) = 5q.

3. Verify the middle term:

2 × 3p × 5q = 30pq ✓

4. Since the middle term is positive, apply (a + b)²:

9p² + 30pq + 25q² = (3p + 5q)²

5. Combine with the common factor:

18p² + 60pq + 50q² = 2(3p + 5q)²

Final Answer: 2(3p + 5q)²

Example 4: Algebraic Fractions (Tricky)

Given: (4s²/9) + (8st/3) + 4t²

To Find: Factorise completely.

Solution:

1. First identify the structure. Notice that 4s²/9 = (2s/3)² and 4t² = (2t)².

2. Express all terms with a common denominator concept to check the middle term:

(2s/3)² + 2(2s/3)(2t) + (2t)²

3. Calculate the middle term pattern:

2 × (2s/3) × (2t) = 8st/3 ✓

4. This matches our given middle term exactly. Apply the identity:

(4s²/9) + (8st/3) + 4t² = (2s/3 + 2t)²

5. We can factor out 2 from the binomial for a cleaner form:

= [2(s/3 + t)]² = 4(s/3 + t)²

Final Answer: (2s/3 + 2t)² or equivalently 4(s/3 + t)²

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Q1: If x² + 2kx + 49 is a perfect square trinomial, find the value of k (k > 0).

💡 Answer: For x² + 2kx + 49 to be a perfect square, we need 2kx = 2 × x × 7 (since √49 = 7). Therefore, 2k = 14, giving k = 7. The expression becomes (x + 7)².

Q2: Factorise: (9a²/16) - (3ab/2) + b²

💡 Answer: Recognise that 9a²/16 = (3a/4)² and b² = b². Check middle term: 2(3a/4)(b) = 3ab/2. Since the given middle term is negative, this factors as [(3a/4) - b]² or equivalently (3a - 4b)²/16.

Q3: The area of a square garden is (x² + 14x + 49) m². What is the length of one side?

💡 Answer: Factorise x² + 14x + 49: √(x²) = x, √49 = 7, and 2(x)(7) = 14x ✓. So x² + 14x + 49 = (x + 7)². Since area = (side)², the side length is (x + 7) meters.

Mini Cheatsheet: Factorisation Formulas

More Identities

More Identities

Concept Introduction

Imagine you're designing a modular storage cube system where you can stack three different sizes of boxes together. If each dimension is a, b, and c, how do you quickly calculate the total volume when packed into a single larger cube? Or consider a carpenter who needs to quickly calculate areas when working with composite measurements like 103 cm (100 + 3) without a calculator.

Algebraic identities serve as mathematical shortcuts that transform complex calculations into manageable steps. In this section, we'll explore two powerful identities: the expansion of (a + b + c)² for handling three-term expressions, and the difference of squares (a² - b²) for rapid mental calculations. These identities not only simplify algebraic manipulations but also reveal elegant patterns in arithmetic that mathematicians have used for centuries.

{{FORMULA: expr=(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca | symbols=a:first term, b:second term, c:third term}}

Definitions & Formulas

Derivation of (a + b + c)²

Understanding the Three-Term Square Identity

Let's build this identity step-by-step using a clever substitution method:

Step 1: Start with a familiar two-term identity.

We know from previous study:

(x + y)² = x² + 2xy + y²

Step 2: Replace one variable with a composite expression.

Let's set x = a and y = (b + c). Now substitute:

(a + (b + c))² = a² + 2a(b + c) + (b + c)²

Step 3: Expand the middle term using distributive property.

2a(b + c) = 2ab + 2ac

Step 4: Expand the last term (b + c)² using the two-term identity.

(b + c)² = b² + 2bc + c²

Step 5: Combine all expanded terms.

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

Step 6: Rearrange terms for the standard form.

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

{{KEY: type=concept | title=Pattern Recognition | text=Notice that the expansion contains: all three squares (a², b², c²) plus twice the product of each pair (2ab, 2bc, 2ca). This pattern makes it easy to remember!}}

Derivation of a² - b²

The Difference of Squares Identity

This elegant identity connects multiplication and subtraction:

Step 1: Consider the product of a sum and difference.

(a + b)(a - b)

Step 2: Apply distributive property (FOIL method).

= a·a + a·(-b) + b·a + b·(-b)

Step 3: Simplify each term.

= a² - ab + ab - b²

Step 4: Cancel the middle terms (they're opposites).

= a² - b²

Therefore:

a² - b² = (a + b)(a - b)

Solved Examples

Example 1: Basic Expansion of Three Terms (Easy)

Given: Expand (x + 2y + 3z)²

To Find: The complete expanded form

Solution:

1. Identify the three terms: a = x, b = 2y, c = 3z

2. Apply the identity formula.

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

3. Calculate the square terms.

a² = x², b² = (2y)² = 4y², c² = (3z)² = 9z²

4. Calculate the cross-product terms.

2ab = 2(x)(2y) = 4xy
2bc = 2(2y)(3z) = 12yz
2ca = 2(3z)(x) = 6xz

5. Combine all terms.

(x + 2y + 3z)² = x² + 4y² + 9z² + 4xy + 12yz + 6xz

Final Answer: x² + 4y² + 9z² + 4xy + 12yz + 6xz

Example 2: Numerical Calculation Using Identity (Medium)

Given: Calculate (119)² without direct multiplication

To Find: The value using (a + b + c)² identity

Solution:

1. Break 119 into convenient parts.

119 = 100 + 10 + 9

2. Apply the three-term square identity with a = 100, b = 10, c = 9.

(100 + 10 + 9)² = 100² + 10² + 9² + 2(100)(10) + 2(10)(9) + 2(9)(100)

3. Calculate the square terms.

100² = 10000, 10² = 100, 9² = 81

4. Calculate the cross-product terms.

2(100)(10) = 2000
2(10)(9) = 180
2(9)(100) = 1800

5. Sum all components.

10000 + 100 + 81 + 2000 + 180 + 1800 = 14161

Final Answer: 14161

Example 3: Factorization Using Difference of Squares (Medium-Hard)

Given: Factorize 81x² - 49y²

To Find: The factored form

Solution:

1. Recognize this as a difference of squares pattern.

81x² - 49y² = (9x)² - (7y)²

2. Identify a = 9x and b = 7y.

3. Apply the identity a² - b² = (a + b)(a - b).

(9x)² - (7y)² = (9x + 7y)(9x - 7y)

Final Answer: (9x + 7y)(9x - 7y)

Example 4: Mixed Application with Substitution (Hard)

Given: If x + y + z = 12 and xy + yz + zx = 30, find x² + y² + z²

To Find: The value of sum of squares

Solution:

1. Start with the three-term square identity.

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

2. Substitute the given value (x + y + z) = 12.

(12)² = x² + y² + z² + 2(xy + yz + zx)

3. Calculate the left side.

144 = x² + y² + z² + 2(xy + yz + zx)

4. Substitute the second given value xy + yz + zx = 30.

144 = x² + y² + z² + 2(30)

5. Simplify and solve.

144 = x² + y² + z² + 60

x² + y² + z² = 144 - 60 = 84

Final Answer: 84

{{KEY: type=technique | title=Reverse Engineering Identities | text=When given sums and products of variables, work backwards from expanded identities to find unknown quantities. This technique is crucial for competitive exams!}}

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Question 1

Without direct calculation, determine which is larger: (51)² or (50 + 51)²/4?

💡 Answer: Using the identity, (50 + 51)²/4 = (101)²/4 = 10201/4 = 2550.25, while 51² = 2601. So 51² is larger. Alternatively, recognize that (a + b)²/4 is the square of the average, which is always less than or equal to the average of squares when a ≠ b.

Question 2

If p + q + r = 0, find the value of p² + q² + r² in terms of pq + qr + rp.

💡 Answer: Using (p + q + r)² = p² + q² + r² + 2(pq + qr + rp), substitute p + q + r = 0: 0 = p² + q² + r² + 2(pq + qr + rp) Therefore: p² + q² + r² = -2(pq + qr + rp)

Question 3

Factor completely: (x + y)² - (x - y)²

💡 Answer: Let a = x + y and b = x - y. Then we have a² - b²: (x + y)² - (x - y)² = [(x + y) + (x - y)][(x + y) - (x - y)] = (2x)(2y) = 4xy Shortcut: Recognize this pattern directly expands the cross-terms!

Mini Cheatsheet

Practice Note: Master these identities through pattern recognition rather than rote memorization. Try calculating squares of numbers like 35, 65, 105 using different identity forms to see which method feels most natural to you. The best mathematicians choose the path of least calculation!

Factorisation Using Algebra Tiles

Page 5: Factorisation Using Algebra Tiles

Concept Introduction

Imagine you're designing a rectangular patio using three types of tiles: large square tiles, long rectangular tiles, and small unit square tiles. You are given a specific number of each: one large square, a handful of rectangles, and a pile of small squares. Your challenge is to arrange all of them, with no gaps, to form one perfect, large rectangle.

This is exactly what we do when we factorise with algebra tiles! The total area of all your tiles represents a quadratic expression like x² + 7x + 12. Finding the length and width of the final, assembled rectangle is like finding the factors of that expression. This hands-on method turns abstract algebra into a visual puzzle, making it easier to see how (x + 3) and (x + 4) are the dimensions that build the area x² + 7x + 12.

{{FORMULA: expr=(x + a)(x + b) = x² + (a + b)x + ab | symbols=x:variable, a:constant term in first factor, b:constant term in second factor}}

Definitions & Formulas

Understanding the tools and terms is the first step. Algebra tiles are physical or visual blocks that give shape to algebraic terms.

The Logic: Building Rectangles from Expressions

Factorising x² + 7x + 12 using algebra tiles is a step-by-step construction project. The goal is to arrange the given tiles into a solid rectangle. The dimensions of that rectangle will be our factors.

1. Represent the Expression: Gather the tiles for x² + 7x + 12. This means you need:

   • One x²-tile
   • Seven x-tiles
   • Twelve 1-tiles

2. Start with the Corner: Place the x²-tile in the top-left corner of your workspace. This tile will anchor your rectangle.

{{VISUAL: diagram: An x²-tile placed in the top-left corner. To its right and below are empty spaces where other tiles will be arranged.}}

3. Arrange the Tiles: Now, the crucial step. You need to arrange the seven x-tiles and twelve 1-tiles around the x²-tile to form a larger rectangle. How you split the x-tiles is the key. You must split the seven x-tiles into two groups.

   • Let's try splitting them as 3x and 4x. Place three x-tiles along one side of the x²-tile and four x-tiles along the other side.

4. Fill the Gap: You'll notice a rectangular gap has formed. This gap needs to be filled perfectly by the twelve 1-tiles.

   • The space created by the three x-tiles on one side and four x-tiles on the other has dimensions of 3 units by 4 units.
   • The area of this gap is 3 × 4 = 12. This perfectly matches our twelve 1-tiles! We can arrange them in a 3-by-4 grid to complete the rectangle.

5. Find the Dimensions: Look at the overall shape. It's a perfect rectangle.

   • The length along the top is the side of the x²-tile (x) plus the width of the four 1-tiles (4). So, the length is x + 4.
   • The length along the side is the side of the x²-tile (x) plus the width of the three 1-tiles (3). So, the width is x + 3.

6. State the Factors: The dimensions of the rectangle are the factors of the expression.

Area = Length × Width
x² + 7x + 12 = (x + 4)(x + 3)

The magic was in splitting 7x into 3x + 4x. Notice that the numbers 3 and 4 have a special relationship: they add up to 7 (the coefficient of x) and multiply to 12 (the constant term). This is the core principle of factoring these trinomials.

Solved Examples

Let's apply this visual method to a few problems, moving from simple to more complex.

Example 1: Factoring a Simple Positive Trinomial

Given: The expression x² + 5x + 6.

To Find: The factors of the expression using the algebra tile method.

Solution:

1. Identify the required tiles: We need one x²-tile, five x-tiles, and six 1-tiles.

2. We need to split the 5x term. Let's look for two numbers that add up to 5 and multiply to 6.

   • Factors of 6: (1, 6) and (2, 3).
   • Sum check: 1 + 6 = 7 (Incorrect). 2 + 3 = 5 (Correct!).
   • So, we split 5x into 2x and 3x.

3. Arrange the tiles. Place the x²-tile. Arrange two x-tiles along one side and three x-tiles along the other.

{{VISUAL: diagram: An algebra tile arrangement for x² + 5x + 6. An x² tile is in the top-left. Three x-tiles are placed horizontally to its right. Two x-tiles are placed vertically below it. The bottom-right corner is filled with a 2x3 grid of six unit tiles.}}

4. Fill the remaining space with the six 1-tiles. This will form a 2-by-3 grid.

5. Determine the dimensions of the final rectangle.

   • The top edge has length x + 3.
   • The left edge has length x + 2.

6. The factors are the dimensions of the rectangle.

x² + 5x + 6 = (x + 2)(x + 3)

Final Answer: (x + 2)(x + 3)

Example 2: Factoring a Trinomial with Larger Numbers

Given: The expression x² + 11x + 30.

To Find: The factors of the expression.

Solution:

1. This corresponds to x² + (a + b)x + ab. We need to find a and b such that a + b = 11 and ab = 30.

2. List the pairs of factors for the constant term, 30:

   • 1 and 30 (Sum = 31)
   • 2 and 15 (Sum = 17)
   • 3 and 10 (Sum = 13)
   • 5 and 6 (Sum = 11) - This is the correct pair!

3. The numbers are 5 and 6. This means we split the middle term 11x into 5x + 6x.

4. Visually, this means we would arrange five x-tiles along one side of the x²-tile and six x-tiles along the other. The remaining space would be a 5 × 6 rectangle, which requires exactly thirty 1-tiles.

5. The dimensions of this conceptual rectangle would be (x + 5) and (x + 6).

x² + 11x + 30 = (x + 5)(x + 6)

Final Answer: (x + 5)(x + 6)

Example 3: Factoring a Trinomial with a Negative Term

Given: The expression x² – 5x + 6.

To Find: The factors of the expression.

Solution:

1. Compare with x² + (a + b)x + ab. We need to find a and b such that a + b = -5 and ab = 6.

2. The product ab is positive (6), which means a and b must have the same sign.

3. The sum a + b is negative (-5), which means both a and b must be negative.

4. List the pairs of negative factors for 6:

   • -1 and -6 (Sum = -7)
   • -2 and -3 (Sum = -5) - This is our pair!

5. So, we can split the middle term -5x as -2x - 3x. The factors are based on the numbers a = -2 and b = -3.

x² – 5x + 6 = (x - 2)(x - 3)

Final Answer: (x - 2)(x - 3)

Example 4: Factoring with a Leading Coefficient

Given: The expression 2x² + 7x + 3.

To Find: The factors of the expression.

Solution:

1. This expression needs two x²-tiles, seven x-tiles, and three 1-tiles. The general form is (px + a)(qx + b).

2. Here, pq = 2 (coefficient of x²), ab = 3 (constant), and the middle term (pb + aq) = 7.

3. Let's break it down:

   • Factors of pq = 2 are p=2, q=1.
   • Factors of ab = 3 are a=3, b=1 OR a=1, b=3.

4. We need to test which combination gives us the middle term 7x. Let's try p=2, q=1.

   • Case 1: a=3, b=1. Middle term is pb + aq = 2(1) + 1(3) = 2 + 3 = 5. Incorrect.
   • Case 2: a=1, b=3. Middle term is pb + aq = 2(3) + 1(1) = 6 + 1 = 7. Correct!

5. So the components of our factors are p=2, q=1, a=1, b=3. This gives the factors (px + a) and (qx + b).

(2x + 1)(1x + 3) or (2x + 1)(x + 3)

{{VISUAL: diagram: An algebra tile arrangement for 2x² + 7x + 3. Two x² tiles are arranged vertically. To their right, one column of two x-tiles is placed. Below the x² tiles, three rows of two x-tiles are placed. The bottom right corner is a 1x3 grid of unit tiles. The total dimensions are (x+x+1) by (x+3), which is incorrect. Let's rethink the visual. The two x² tiles are arranged in a 2x1 block. One x-tile is placed to the right of the top x²-tile. Six x-tiles are placed below the two x² tiles in a 2x3 grid. The bottom right is a 1x3 rectangle of unit tiles. Dimensions are (2x+1) and (x+3).}}

Final Answer: (2x + 1)(x + 3)

{{KEY: type=concept | title=The Golden Rule of Factoring | text=To factor a trinomial x² + sx + p, find two numbers that multiply to the constant term p and add to the middle coefficient s. These two numbers will be the constants in your binomial factors.}}

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. If you use algebra tiles to form a rectangle for the expression x² + (m + n)x + mn, what will the dimensions of the rectangle be?

   💡 Answer: The dimensions will be (x + m) and (x + n). The two numbers that add to the middle coefficient (m + n) and multiply to the constant (mn) are m and n themselves.

2. A rectangular mural has an area represented by the expression x² – 3x – 10. What are the possible expressions for its length and width?

   💡 Answer: We need two numbers that multiply to -10 and add to -3. The pairs of factors for -10 are (1,-10), (-1,10), (2,-5), (-2,5). The pair that sums to -3 is 2 and -5. So the dimensions are (x + 2) and (x – 5).

3. Can you form a single rectangle using four x²-tiles, eight x-tiles, and three 1-tiles? If so, what are its dimensions?

   💡 Answer: Yes. The expression is 4x² + 8x + 3. We look for factors (px+a)(qx+b). Here pq=4, ab=3, pb+aq=8. Let's try p=2, q=2. Then 2b+2a=8, which simplifies to a+b=4. The factors of ab=3 are 1 and 3. Since 1+3=4, this works! The dimensions are (2x + 1) and (2x + 3).

Mini Cheatsheet

Factorisation Without Using Algebra Tiles

Page 6 of 8: Factorisation Without Using Algebra Tiles

Concept Introduction

Imagine you're a mobile phone app developer who needs to optimize screen space. You have a rectangular display area of (x² + 7x + 12) square pixels. To fit two side-by-side app widgets perfectly, you need to factor this area into two linear dimensions — like (x + 3) pixels wide and (x + 4) pixels tall.

While algebra tiles helped us visualize factorisation physically, they're not always practical. What if your expression involves large coefficients like x² + 57x + 782? You'd need hundreds of tiles! This is where algebraic factorisation becomes essential.

By identifying patterns in coefficients and using strategic splitting of the middle term, we can factor quadratic expressions purely through reasoning. This method is faster, works for any numbers, and is the foundation for solving quadratic equations later in your mathematical journey.

{{FORMULA: expr=x² + (a+b)x + ab = (x+a)(x+b) | symbols=a:first factor constant, b:second factor constant, ab:product equals constant term, a+b:sum equals middle coefficient}}

Definitions & Formulas

Derivation / Logic: The Middle Term Splitting Method

Core Principle: For any quadratic x² + px + q, we need two numbers whose sum is p and product is q.

Step 1: Start with the general form of factored quadratic:

(x + a)(x + b) = x² + (a+b)x + ab

Step 2: Compare this with your target expression x² + px + q. We observe:

Coefficient of x: a + b = p
Constant term: a × b = q

Step 3: Find two numbers a and b that satisfy BOTH conditions simultaneously. List factor pairs of q, then check which pair adds to p.

Step 4: Once identified, split the middle term px as ax + bx:

x² + px + q = x² + ax + bx + ab

Step 5: Group the terms and factor by common extraction:

= x(x + a) + b(x + a) = (x + a)(x + b)

Step 6: Verify by expanding the factors using distributive property.

{{KEY: type=strategy | title=The Two-Number Hunt | text=Factorising x² + px + q means finding two numbers that ADD to p and MULTIPLY to q. This is the golden rule of simple quadratic factorisation.}}

Solved Examples

Example 1: Basic Factorisation (Easy)

Given: Expression x² + 7x + 12

To Find: Factors in the form (x + a)(x + b)

Solution:

1. Identify what we need: two numbers that add to 7 and multiply to 12.

Sum required: a + b = 7
Product required: a × b = 12

2. List factor pairs of 12 and check their sums:

(1, 12) → sum = 13 ✗
(2, 6) → sum = 8 ✗
(3, 4) → sum = 7 ✓

3. We found a = 3 and b = 4. Split the middle term 7x as 3x + 4x:

x² + 7x + 12 = x² + 3x + 4x + 12

4. Group and factor:

= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)

5. Verify by expansion:

(x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓

Final Answer: (x + 3)(x + 4)

Example 2: Factorisation with Larger Coefficients (Medium)

Given: Expression x² + 11x + 30

To Find: Complete factorisation

Solution:

1. Set up the requirement:

Two numbers with: sum = 11, product = 30

2. Systematically check factor pairs of 30:

(1, 30) → sum = 31 ✗
(2, 15) → sum = 17 ✗
(3, 10) → sum = 13 ✗
(5, 6) → sum = 11 ✓

3. Split 11x using a = 5 and b = 6:

x² + 11x + 30 = x² + 5x + 6x + 30

4. Factor by grouping:

= x(x + 5) + 6(x + 5)
= (x + 5)(x + 6)

Final Answer: (x + 5)(x + 6)

Example 3: Negative Middle Term (Hard)

Given: Expression x² − 5x + 6

To Find: Factors

Solution:

1. Note the negative coefficient. We need:

Two numbers with: sum = −5, product = +6

2. Since product is positive and sum is negative, BOTH numbers must be negative:

(−1, −6) → sum = −7 ✗
(−2, −3) → sum = −5 ✓

3. Split the middle term using a = −2, b = −3:

x² − 5x + 6 = x² − 2x − 3x + 6

4. Group and factor:

= x(x − 2) − 3(x − 2)
= (x − 2)(x − 3)

5. Verification:

(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓

Final Answer: (x − 2)(x − 3)

{{KEY: type=warning | title=Sign Patterns Matter | text=When constant term is POSITIVE: both factors have the SAME sign (both + or both −). When constant is NEGATIVE: factors have OPPOSITE signs.}}

Example 4: Expression in Different Variables (Tricky)

Given: Expression s² − s − 42

To Find: Complete factorisation using variable s

Solution:

1. Apply the same logic with variable s instead of x:

Two numbers with: sum = −1, product = −42

2. Product is negative, so factors have opposite signs. List pairs of 42:

(1, −42) → sum = −41 ✗
(2, −21) → sum = −19 ✗
(3, −14) → sum = −11 ✗
(6, −7) → sum = −1 ✓

3. Use a = 6 and b = −7 to split −s as 6s − 7s:

s² − s − 42 = s² + 6s − 7s − 42

4. Factor by grouping:

= s(s + 6) − 7(s + 6)
= (s + 6)(s − 7)

5. Double-check:

(s + 6)(s − 7) = s² − 7s + 6s − 42 = s² − s − 42 ✓

Final Answer: (s + 6)(s − 7)

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Q1: Factor the expression p² + 13p + 36 and then find the value when p = 7.

💡 Answer: Factor pairs of 36 that add to 13: (4, 9). So p² + 13p + 36 = (p + 4)(p + 9). When p = 7: (7 + 4)(7 + 9) = 11 × 16 = 176

Q2: If x² + kx + 15 = (x + 3)(x + 5), find the value of k.

💡 Answer: Expand (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15. Comparing with x² + kx + 15, we get k = 8

Q3: The area of a rectangular garden is (m² + 12m + 35) square meters. If one side is (m + 5) meters, what is the other side's length?

💡 Answer: Factor m² + 12m + 35: pairs of 35 are (5, 7), sum = 12 ✓ So area = (m + 5)(m + 7). If one side is (m + 5), other side is (m + 7) meters

Mini Cheatsheet: Factorisation Essentials

Next Steps: Once you master this technique, you'll apply it to solve quadratic equations (finding values of x where the expression equals zero) and handle more complex factorisations involving coefficients greater than 1 for x². Practice identifying factor pairs quickly — it's a skill that accelerates with repetition!

Finding New Identities

Page 7: Finding New Identities

Introduction: Building Blocks of Advanced Mathematics

Cubic identities are powerful algebraic tools that extend our understanding beyond quadratic expressions. Just as (a + b)² helped us expand binomial squares, cubic identities like (a + b)³ and (a - b)³ unlock the ability to handle three-dimensional problems efficiently.

Imagine a water tank company that designs cubic tanks. If a tank has edge length (x + 5) meters, calculating its volume requires expanding (x + 5)³. Without cubic identities, you'd need to multiply (x + 5)(x + 5)(x + 5) step by step—tedious and error-prone. With the identity, you can instantly write the expansion as x³ + 15x² + 75x + 125 cubic meters. These identities appear everywhere: in volume calculations, physics formulas for work and energy, and even in computer graphics algorithms for 3D modeling.

{{FORMULA: expr=(a + b)³ = a³ + 3a²b + 3ab² + b³ | symbols=a:first term, b:second term}}

Definitions & Formulas

Derivation of Cubic Identities

Deriving (a + b)³

Step 1: Start with the definition of a cube.

(a + b)³ = (a + b)(a + b)²

Step 2: Expand (a + b)² using the known quadratic identity.

(a + b)² = a² + 2ab + b²

Step 3: Multiply (a + b) with the expanded form.

(a + b)³ = (a + b)(a² + 2ab + b²)

Step 4: Apply the distributive property—distribute a first, then b.

= a(a² + 2ab + b²) + b(a² + 2ab + b²)

Step 5: Multiply each term and collect like terms.

= a³ + 2a²b + ab² + a²b + 2ab² + b³

Step 6: Combine similar terms (2a²b + a²b = 3a²b and ab² + 2ab² = 3ab²).

(a + b)³ = a³ + 3a²b + 3ab² + b³

Deriving (a - b)³

Replace b with -b in the (a + b)³ identity:

(a - b)³ = a³ + 3a²(-b) + 3a(-b)² + (-b)³

= a³ - 3a²b + 3ab² - b³

Notice the alternating signs: positive, negative, positive, negative.

{{KEY: type=pattern | title=Sign Pattern in Cubic Identities | text=In (a + b)³, all terms are positive. In (a - b)³, signs alternate: + - + -. This pattern helps verify your expansions quickly.}}

Factorization Identities

Deriving x³ - y³

Step 1: Use the distributive property on (x - y)(x² + xy + y²).

(x - y)(x² + xy + y²) = x(x² + xy + y²) - y(x² + xy + y²)

Step 2: Distribute x into each term.

= x³ + x²y + xy² - yx² - xy² - y³

Step 3: Cancel terms that appear with opposite signs (x²y - yx² = 0, xy² - xy² = 0).

x³ - y³ = (x - y)(x² + xy + y²)

Deriving x³ + y³

Step 1: Use the distributive property on (x + y)(x² - xy + y²).

(x + y)(x² - xy + y²) = x(x² - xy + y²) + y(x² - xy + y²)

Step 2: Distribute both terms.

= x³ - x²y + xy² + yx² - xy² + y³

Step 3: Cancel opposite terms (-x²y + yx² = 0, xy² - xy² = 0).

x³ + y³ = (x + y)(x² - xy + y²)

The Three-Variable Identity

Deriving x³ + y³ + z³ - 3xyz

Multiply (x + y + z) with (x² + y² + z² - xy - xz - yz):

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

This beautiful identity connects sums of cubes with products of three variables.

Solved Examples

Example 1: Basic Expansion (Easy)

Given: Expand (2x + 3)³

To Find: The complete expanded form

Solution:

1. Identify a = 2x and b = 3 in the identity (a + b)³ = a³ + 3a²b + 3ab² + b³.

2. Calculate each term separately:

   • a³ = (2x)³ = 8x³
   • 3a²b = 3(2x)²(3) = 3(4x²)(3) = 36x²
   • 3ab² = 3(2x)(3²) = 3(2x)(9) = 54x
   • b³ = 3³ = 27

3. Combine all terms.

(2x + 3)³ = 8x³ + 36x² + 54x + 27

Final Answer: 8x³ + 36x² + 54x + 27

Example 2: Expansion with Subtraction (Medium)

Given: Expand (3m - 2n)³

To Find: The complete expanded form

Solution:

1. Identify a = 3m and b = 2n in the identity (a - b)³ = a³ - 3a²b + 3ab² - b³.

2. Calculate each term with careful attention to signs:

   • a³ = (3m)³ = 27m³
   • 3a²b = 3(3m)²(2n) = 3(9m²)(2n) = 54m²n
   • 3ab² = 3(3m)(2n)² = 3(3m)(4n²) = 36mn²
   • b³ = (2n)³ = 8n³

3. Apply the alternating sign pattern.

(3m - 2n)³ = 27m³ - 54m²n + 36mn² - 8n³

Final Answer: 27m³ - 54m²n + 36mn² - 8n³

Example 3: Factorization Application (Hard)

Given: The volume of a cube is 8p³ - 60p²q + 150pq² - 125q³ cubic units.

To Find: The length of one edge of the cube

Solution:

1. Compare the given expression with the identity (a - b)³ = a³ - 3a²b + 3ab² - b³.

2. Rewrite 8p³ as (2p)³ to identify the first term.

8p³ = (2p)³

3. Check if -60p²q matches -3a²b pattern where a = 2p.

-3(2p)²b = -3(4p²)b = -12p²b

For this to equal -60p²q, we need -12p²b = -60p²q, so b = 5q.

4. Verify all terms with a = 2p and b = 5q:

   • 3ab² = 3(2p)(5q)² = 3(2p)(25q²) = 150pq² ✓
   • b³ = (5q)³ = 125q³ ✓

5. Factor the complete expression.

8p³ - 60p²q + 150pq² - 125q³ = (2p - 5q)³

Final Answer: Edge length = (2p - 5q) units

Example 4: Three-Variable Identity (Tricky)

Given: Three numbers x, y, z satisfy: x + y + z = 12, xyz = 48, and x² + y² + z² = 74

To Find: The value of x³ + y³ + z³

Solution:

1. Use the three-variable identity.

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

2. Rearrange to isolate x³ + y³ + z³.

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

3. First find (xy + xz + yz) using the quadratic identity expansion.

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

12² = 74 + 2(xy + xz + yz)

144 = 74 + 2(xy + xz + yz)

xy + xz + yz = 35

4. Calculate (x² + y² + z² - xy - xz - yz).

x² + y² + z² - xy - xz - yz = 74 - 35 = 39

5. Substitute into the main identity.

x³ + y³ + z³ = (12)(39) + 3(48)

= 468 + 144 = 612

Final Answer: x³ + y³ + z³ = 612

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Q1: If a + b = 5 and ab = 6, find the value of a³ + b³ without finding a and b individually.

💡 Answer: Use the identity a³ + b³ = (a + b)(a² - ab + b²). First find a² + b²: (a + b)² = a² + 2ab + b² → 25 = a² + 12 + b² → a² + b² = 13. Then a² - ab + b² = 13 - 6 = 7. Finally, a³ + b³ = 5 × 7 = 35.

Q2: Simplify: (x + y + z)³ - x³ - y³ - z³

💡 Answer: Use (x + y + z)³ = x³ + y³ + z³ + 3(x + y)(y + z)(z + x) is incorrect. The correct approach: expand (x + y + z)³ using the identity, then subtract x³ + y³ + z³. The result simplifies to 3(x + y)(y + z)(z + x) or equivalently 3xyz + 3x²y + 3xy² + 3y²z + 3yz² + 3z²x + 3zx².

Q3: If x³ + y³ = 35 and x + y = 5, find xy.

💡 Answer: Use x³ + y³ = (x + y)(x² - xy + y²). We have 35 = 5(x² - xy + y²), so x² - xy + y² = 7. Also, (x + y)² = x² + 2xy + y² → 25 = x² + 2xy + y². Subtracting: 25 - 7 = (x² + 2xy + y²) - (x² - xy + y²) → 18 = 3xy → xy = 6.

Mini Cheatsheet

{{KEY: type=exam-tip | title=Memory Aid for Cubic Identities | text=Remember the coefficient pattern 1-3-3-1 for expansions. For factorizations, the middle term in the trinomial has OPPOSITE sign to the binomial: (x - y) pairs with (x² + xy + y²), while (x + y) pairs with (x² - xy + y²).}}

Summary & Quick Revision

Summary & Quick Revision

Introduction: The Power of Algebraic Identities in Real Life

Imagine you are a construction engineer designing a square plaza with side length x metres. The client asks you to expand it by adding a border of width 2 metres on all sides. What will be the new area? You could multiply (x + 2)(x + 2), or you could instantly use the identity (a + b)² = a² + 2ab + b² to get x² + 4x + 4 square metres — much faster!

Algebraic identities are pre-proven formulas that help us expand, factorise, and simplify expressions without repetitive calculations. This chapter explored how these identities — from simple squares to cubes and beyond — unlock powerful problem-solving shortcuts in mathematics, physics, economics, and engineering. We learned not just the formulas, but also why they work and how to apply them to factorise polynomials, simplify rational expressions, and solve real-world problems efficiently.

This summary page recaps every key identity, factorisation technique, and application covered in Chapter 4, giving you a complete revision toolkit.

Key Algebraic Identities at a Glance

{{FORMULA: expr=(a + b)² = a² + 2ab + b² | symbols=a:first term, b:second term, ab:cross-product term}}

Below is the complete list of identities from this chapter:

Understanding the Core Logic: Why Do These Identities Work?

Let's revisit the geometric proof of (a + b)²:

1. Start with a square: Imagine a square with side length (a + b).

2. Calculate total area: The area of the entire square is (a + b) × (a + b) = (a + b)².

3. Divide the square: Split the square into four regions — a square of side a (area a²), a square of side b (area b²), and two identical rectangles each with dimensions a × b (total area 2ab).

4. Add all regions:

Total area = a² + ab + ab + b² = a² + 2ab + b²

5. Equate both expressions:

(a + b)² = a² + 2ab + b²

This visual proof shows that algebraic identities are not arbitrary — they represent geometric truths and logical consistency in mathematics.

Solved Examples: Mastery Through Practice

Example 1: Basic Identity Application (Easy)

Given: Expand (3x + 5)² using algebraic identity.

To Find: The expanded form of the expression.

Solution:

1. Identify the identity: We use (a + b)² = a² + 2ab + b² where a = 3x and b = 5.

2. Calculate a²:

a² = (3x)² = 9x²

3. Calculate 2ab:

2ab = 2 × 3x × 5 = 30x

4. Calculate b²:

b² = 5² = 25

5. Combine all terms:

(3x + 5)² = 9x² + 30x + 25

Final Answer: 9x² + 30x + 25

Example 2: Factorisation Using Difference of Squares (Medium)

Given: Factorise 49m² - 64n².

To Find: Factors of the expression.

Solution:

1. Recognize the pattern: This is a difference of squares a² - b² where a = 7m and b = 8n.

2. Apply the identity a² - b² = (a + b)(a - b):

49m² - 64n² = (7m)² - (8n)²

3. Substitute into the identity:

= (7m + 8n)(7m - 8n)

Final Answer: (7m + 8n)(7m - 8n)

Example 3: Numerical Calculation Using Cubic Identity (Hard)

Given: Calculate 103³ without a calculator.

To Find: The exact value.

Solution:

1. Express 103 as sum: Write 103 = 100 + 3, so 103³ = (100 + 3)³.

2. Use identity (a + b)³ = a³ + 3a²b + 3ab² + b³ where a = 100, b = 3.

3. Calculate each term:

a³ = 100³ = 1,000,000

3a²b = 3 × 100² × 3 = 3 × 10,000 × 3 = 90,000

3ab² = 3 × 100 × 3² = 3 × 100 × 9 = 2,700

b³ = 3³ = 27

4. Add all terms:

103³ = 1,000,000 + 90,000 + 2,700 + 27 = 1,092,727

Final Answer: 1,092,727

Example 4: Simplifying Rational Expressions (Tricky)

Given: Simplify the expression:

(x² - 9)/(x² + 6x + 9)

Assume the denominator is not zero.

To Find: The simplified form.

Solution:

1. Factorise the numerator using difference of squares:

x² - 9 = (x + 3)(x - 3)

2. Factorise the denominator — it's a perfect square trinomial:

x² + 6x + 9 = (x + 3)²

3. Rewrite the expression:

(x² - 9)/(x² + 6x + 9) = (x + 3)(x - 3)/(x + 3)(x + 3)

4. Cancel common factor (x + 3):

= (x - 3)/(x + 3)

Final Answer: (x - 3)/(x + 3)

Tips & Tricks for Exam Success

{{KEY: type=concept | title=Golden Rule for Identities | text=Always verify which identity fits BEFORE expanding. Check the signs (+ or -), the powers (² or ³), and whether it's a product or sum. This saves calculation time and prevents errors.}}

Common Mistakes: Wrong vs Right

Brain-Teaser Questions

Question 1: If x + (1/x) = 5, find the value of x² + (1/x²) without finding x.

💡 Answer: Square both sides: (x + 1/x)² = 5² → x² + 2×x×(1/x) + 1/x² = 25 → x² + 2 + 1/x² = 25 → x² + 1/x² = 23.

Question 2: Factorise completely: a⁴ - 16b⁴.

💡 Answer: First use difference of squares: a⁴ - 16b⁴ = (a²)² - (4b²)² = (a² + 4b²)(a² - 4b²). Then factor a² - 4b² again: (a² + 4b²)(a + 2b)(a - 2b).

Question 3: Without direct calculation, prove that 37³ + 63³ + 9261 is divisible by 100.

💡 Answer: Use identity a³ + b³ + c³ - 3abc = (a+b+c)(...). Notice 37 + 63 = 100, so let c = -100. Then 37³ + 63³ + (-100)³ - 3(37)(63)(-100) = 0. Rearranging: 37³ + 63³ = 100³ - 3(37)(63)(100). Since 3(37)(63)(100) = 9×37×7×100 = 9261×100, we get 37³ + 63³ = 100(100² - 9261), proving divisibility.

Mini Cheatsheet: Formulas to Remember

You've now mastered Chapter 4! These identities are foundational tools you'll use throughout algebra, calculus, and applied mathematics. Practice recognizing patterns instantly, and you'll solve problems faster than ever. Keep this page bookmarked for quick revision before exams! 🎯