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Exploring Algebraic Identities

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Chapter Notes

Overview
This chapter builds on the concept of linear polynomials and equations to introduce algebraic identities—special equations that are true for all values of the variables. The big idea is that identities are powerful tools for simplifying calculations, factorising algebraic expressions, and solving problems efficiently. Students must understand that an identity differs from an ordinary equation because it holds universally, and they must master the standard identities to manipulate expressions with speed and accuracy in board exams.
Key Concepts (10)
Algebraic Identity
An algebraic identity is an equation that is true for all values of the variables occurring in it.
e.g. (x + y)² = x² + 2xy + y² is true for all values of x and y, e.g., x = 2, y = 3 gives 25 = 25.
Equation
An equation is a mathematical statement that is true only for specific values of the variables.
e.g. x² – 1 = 24 is true only for x = 5 or x = –5.
Factorisation
Factorisation is the process of expressing an algebraic expression as a product of its factors.
e.g. x² + 4x + 4 = (x + 2)², so (x + 2) is a factor of x² + 4x + 4.
Distributive Property
The distributive property states that a(b + c) = ab + ac, used to expand or factor algebraic expressions.
e.g. (a + b)² = (a + b)(a + b) = a(a + b) + b(a + b) = a² + ab + ba + b² = a² + 2ab + b².
Algebra Tiles
Algebra tiles are geometric models (squares and rectangles) used to visualise algebraic expressions and their factorisation.
e.g. x² + 7x + 12 is represented by one x²-tile, seven x-tiles, and twelve unit tiles arranged as a rectangle of sides (x + 3) and (x + 4).
Rational Algebraic Expression
A rational algebraic expression is an expression of the form numerator/denominator, where both numerator and denominator are polynomials and denominator ≠ 0.
e.g. (x² – 7x + 12)/(5x² + 5x – 100) is a rational expression that can be simplified by cancelling common factors.
Consecutive Square Numbers
Consecutive square numbers are squares of three consecutive integers, e.g., (n – 1)², n², (n + 1)².
e.g. For n = 3, the consecutive squares are 4, 9, 16. Their sum of smallest and largest minus twice the middle gives 2.
Splitting the Middle Term
Splitting the middle term is a method of factorising a quadratic expression x² + (a + b)x + ab by writing the x-term as ax + bx, where a and b satisfy a + b = coefficient of x and ab = constant term.
e.g. x² + 7x + 12 is split as x² + 3x + 4x + 12, then grouped as (x + 3)(x + 4).
Common Factor
A common factor is a number or algebraic term that divides each term of an expression exactly.
e.g. In 50p² + 60pq + 18q², 2 is a common factor, giving 2(25p² + 30pq + 9q²).
Volume of a Cube
The volume of a cube of side (a + b) is (a + b)³, which can be split into smaller cubes and cuboids to derive the identity (a + b)³ = a³ + 3a²b + 3ab² + b³.
e.g. A cube of edge (p + 2q) has volume p³ + 6p²q + 12pq² + 8q³.
Formulas & Equations (12)
(a + b)² = a² + 2ab + b²
Reads as: a plus b whole square equals a square plus 2ab plus b square
Variables: a and b are any numbers or algebraic terms
Use when: Expanding (a + b)², factorising a² + 2ab + b², or calculating squares of numbers like 43² = (40 + 3)²
⚠ Students often write (a + b)² = a² + b², forgetting the 2ab term.
(a – b)² = a² – 2ab + b²
Reads as: a minus b whole square equals a square minus 2ab plus b square
Variables: a and b are any numbers or algebraic terms
Use when: Expanding (a – b)², factorising a² – 2ab + b², or calculating squares like 29² = (30 – 1)²
⚠ Students forget the minus sign on 2ab or write a² – b² instead.
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Reads as: a plus b plus c whole square equals a square plus b square plus c square plus 2ab plus 2bc plus 2ca
Variables: a, b, c are any numbers or algebraic terms
Use when: Expanding (a + b + c)² or factorising expressions like a² + b² + c² + 2ab + 2bc + 2ca
⚠ Students miss the 2ca term or write 2ac incorrectly as 2ac instead of 2ca (order does not matter, but they often forget one term).
a² – b² = (a + b)(a – b)
Reads as: a square minus b square equals a plus b times a minus b
Variables: a and b are any numbers or algebraic terms
Use when: Factorising difference of squares or calculating products like 23 × 17 = (20 + 3)(20 – 3)
⚠ Students write a² – b² = (a – b)², which is incorrect.
(x + a)(x + b) = x² + (a + b)x + ab
Reads as: x plus a times x plus b equals x square plus a plus b times x plus ab
Variables: x is a variable; a and b are constants
Use when: Multiplying two linear binomials or factorising quadratics like x² + 7x + 12 = (x + 3)(x + 4)
⚠ Students forget the middle term coefficient is (a + b), not just a or b.
(ax + b)(cx + d) = acx² + (ad + bc)x + bd
Reads as: ax plus b times cx plus d equals ac x square plus ad plus bc times x plus bd
Variables: a, b, c, d are constants; x is a variable
Use when: Multiplying binomials with coefficients other than 1, e.g., (2x + 3)(3x + 1)
⚠ Students misapply the distributive property and forget the cross terms ad and bc.
(a + b)³ = a³ + 3a²b + 3ab² + b³
Reads as: a plus b whole cube equals a cube plus 3 a square b plus 3 a b square plus b cube
Variables: a and b are any numbers or algebraic terms
Use when: Expanding (a + b)³ or factorising expressions like p³ + 6p²q + 12pq² + 8q³
⚠ Students write (a + b)³ = a³ + b³, omitting the 3a²b and 3ab² terms.
(a – b)³ = a³ – 3a²b + 3ab² – b³
Reads as: a minus b whole cube equals a cube minus 3 a square b plus 3 a b square minus b cube
Variables: a and b are any numbers or algebraic terms
Use when: Expanding (a – b)³ or factorising expressions like 8n³ – 60n²m + 150nm² – 125m³
⚠ Students get the signs wrong; they often write all positive or miss the alternating signs.
x³ – y³ = (x – y)(x² + xy + y²)
Reads as: x cube minus y cube equals x minus y times x square plus xy plus y square
Variables: x and y are any numbers or algebraic terms
Use when: Factorising difference of cubes or simplifying rational expressions
⚠ Students write x³ – y³ = (x – y)(x² – xy + y²), confusing it with sum of cubes.
x³ + y³ = (x + y)(x² – xy + y²)
Reads as: x cube plus y cube equals x plus y times x square minus xy plus y square
Variables: x and y are any numbers or algebraic terms
Use when: Factorising sum of cubes or simplifying rational expressions
⚠ Students write x³ + y³ = (x + y)(x² + xy + y²), which is incorrect.
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
Reads as: x cube plus y cube plus z cube minus 3xyz equals x plus y plus z times x square plus y square plus z square minus xy minus yz minus zx
Variables: x, y, z are any numbers or algebraic terms
Use when: Finding sum of cubes when sum, product, and sum of squares are given, or factorising expressions like x³ + y³ + z³ – 3xyz
⚠ Students forget the minus signs in the second factor or write +xy instead of –xy.
a² = (a + b)(a – b) + b²
Reads as: a square equals a plus b times a minus b plus b square
Variables: a and b are any numbers
Use when: Quickly computing squares of numbers, e.g., 55² = (55 + 5)(55 – 5) + 5² = 60 × 50 + 25 = 3025
⚠ Students forget to add b² at the end, giving only (a + b)(a – b).
How It Works
Visualising (a + b)² using a square
  1. Step 1: Draw a square of side (a + b) units.
  2. Step 2: Partition it into a square of side a (area a²), a square of side b (area b²), and two rectangles each of area ab.
  3. Step 3: The total area is a² + ab + ab + b² = a² + 2ab + b².
  4. Step 4: Conclude that (a + b)² = a² + 2ab + b².
💡 The two rectangles each have area ab, not a × b separately; students often count only one rectangle.
Visualising (a – b)² using a square
  1. Step 1: Draw a square of side a units.
  2. Step 2: Mark a segment of length b from one corner, leaving a smaller square of side (a – b).
  3. Step 3: The remaining area consists of two rectangles: one of area ab and one of area b(a – b).
  4. Step 4: Subtract these areas from a²: (a – b)² = a² – ab – b(a – b) = a² – 2ab + b².
💡 The subtraction is of two rectangles, not just one; students often subtract only ab once.
Factorisation by splitting the middle term
  1. Step 1: Write the quadratic in the form x² + (a + b)x + ab.
  2. Step 2: Find two numbers a and b such that their sum equals the coefficient of x and their product equals the constant term.
  3. Step 3: Split the middle term into ax + bx.
  4. Step 4: Group terms: (x² + ax) + (bx + ab).
  5. Step 5: Factor each group: x(x + a) + b(x + a).
  6. Step 6: Take common factor (x + a): (x + a)(x + b).
💡 The signs of a and b must satisfy both sum and product conditions; students often check only product.
Factorisation using algebra tiles
  1. Step 1: Represent the quadratic expression using tiles: one x²-tile, x-tiles for the x-term, and unit tiles for the constant.
  2. Step 2: Arrange the x²-tile at the top-left corner.
  3. Step 3: Place x-tiles along the right and bottom edges to form a rectangle, splitting the x-term as needed.
  4. Step 4: Fill the remaining area with unit tiles in a rectangular array.
  5. Step 5: Read the side lengths of the rectangle as the factors.
💡 The splitting of the x-term must allow a complete rectangle; not all splits work.
Deriving (a + b)³ using distributive property
  1. Step 1: Write (a + b)³ = (a + b)(a + b)².
  2. Step 2: Expand (a + b)² = a² + 2ab + b².
  3. Step 3: Multiply (a + b)(a² + 2ab + b²) using distributive property.
  4. Step 4: Compute: a(a² + 2ab + b²) + b(a² + 2ab + b²) = a³ + 2a²b + ab² + a²b + 2ab² + b³.
  5. Step 5: Combine like terms: a³ + 3a²b + 3ab² + b³.
💡 There are 6 terms before combining; students often miss the cross terms and combine incorrectly.
Simplifying rational algebraic expressions
  1. Step 1: Factorise the numerator completely using identities.
  2. Step 2: Factorise the denominator completely using identities.
  3. Step 3: Identify and cancel common factors between numerator and denominator, provided they are not zero.
  4. Step 4: Write the simplified expression.
💡 Cancellation is only valid if the common factor is not equal to zero; students often cancel without checking the condition.
Using the identity x³ + y³ + z³ – 3xyz
  1. Step 1: Identify given values: sum (x + y + z), product (xyz), and sum of squares (x² + y² + z²).
  2. Step 2: Use (x + y + z)² = x² + y² + z² + 2(xy + yz + zx) to find (xy + yz + zx).
  3. Step 3: Substitute into the identity: (x + y + z)(x² + y² + z² – xy – yz – zx) = x³ + y³ + z³ – 3xyz.
  4. Step 4: Solve for x³ + y³ + z³.
💡 Students often forget to compute (xy + yz + zx) first and directly substitute.
Exam Traps — Don't Make These Mistakes
⚠ Confusing identity with equation
✗ x² – 1 = 24 is an identity because it has an equals sign.
→ An identity must be true for all values of the variable; x² – 1 = 24 is only true for x = 5 or –5, so it is an equation, not an identity.
✓ x² – 1 = 24 is an equation because it holds only for specific values of x, whereas (x + y)² = x² + 2xy + y² is an identity because it holds for all x and y.
⚠ Forgetting the 2ab term in (a + b)²
✗ (a + b)² = a² + b²
→ The expansion of (a + b)² includes the cross term 2ab from the distributive property; omitting it gives an incorrect result.
✓ (a + b)² = a² + 2ab + b²
⚠ Incorrect sign pattern in (a – b)³
✗ (a – b)³ = a³ – 3a²b – 3ab² – b³
→ The signs in (a – b)³ alternate: positive, negative, positive, negative. The 3ab² term must be positive, not negative.
✓ (a – b)³ = a³ – 3a²b + 3ab² – b³
⚠ Mixing up sum and difference of cubes formulas
✗ x³ – y³ = (x – y)(x² – xy + y²)
→ The difference of cubes formula has a plus sign in the middle term of the second factor: x² + xy + y². The sum of cubes formula has minus: x² – xy + y².
✓ x³ – y³ = (x – y)(x² + xy + y²) and x³ + y³ = (x + y)(x² – xy + y²)
⚠ Cancelling without checking denominator ≠ 0
✗ Simplify (x² – 7x + 12)/(5x² + 5x – 100) = (x – 3)/(5(x + 5)) without stating the condition.
→ Cancellation of (x – 4) is valid only if x – 4 ≠ 0, i.e., x ≠ 4. The problem statement assumes denominator ≠ 0, but students must note this in their working.
✓ Simplify assuming denominator ≠ 0: (x² – 7x + 12)/(5x² + 5x – 100) = (x – 3)(x – 4)/[5(x – 4)(x + 5)] = (x – 3)/[5(x + 5)], provided x ≠ 4.
⚠ Incorrect splitting of middle term for negative constant
✗ Factor x² – 5x + 6: find a and b such that a + b = –5 and ab = 6. Students often take a = 2, b = 3.
→ 2 + 3 = 5, not –5. To get sum –5 and product 6, both a and b must be negative: a = –2, b = –3.
✓ a = –2, b = –3, so x² – 5x + 6 = (x – 2)(x – 3).
Compare & Contrast
Truth for variable values
Algebraic Identity
Equation
An identity is true for all values of the variables, while an equation is true only for specific values.
Value comparison
(a + b)²
a² + b²
(a + b)² = a² + 2ab + b² is always greater than or equal to a² + b²; they are equal only when ab = 0 (i.e., a = 0 or b = 0).
Factorisation formula
x³ – y³
x³ + y³
x³ – y³ = (x – y)(x² + xy + y²) has a plus sign in the middle term of the second factor, while x³ + y³ = (x + y)(x² – xy + y²) has a minus sign.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 Very Short Answer (1 mark), 1 Short Answer (3 marks) on factorisation, 1 Long Answer (5 marks) on applying multiple identities or simplifying rational expressions.
Most asked: Expanding and factorising using (a + b)² and (a – b)² · Factorisation by splitting the middle term · Using identities for numerical calculations (e.g., 105², 79²) · Simplifying rational algebraic expressions · Applying (a + b)³ and (a – b)³
Don't skip: Factorisation of quadratic expressions using identities and splitting the middle term (3–5 marks); Simplifying rational expressions by factorising numerator and denominator (3–5 marks); Word problems involving area/volume leading to factorisation (3–5 marks)
One-Line Revision (read aloud before the exam)
  1. An algebraic identity is an equation true for all values of the variables.
  2. (a + b)² = a² + 2ab + b² is the fundamental square of sum identity.
  3. (a – b)² = a² – 2ab + b² is the square of difference identity.
  4. (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
  5. a² – b² = (a + b)(a – b) is the difference of squares identity.
  6. (x + a)(x + b) = x² + (a + b)x + ab is used for factorising quadratics.
  7. (a + b)³ = a³ + 3a²b + 3ab² + b³ is the cube of sum identity.
  8. (a – b)³ = a³ – 3a²b + 3ab² – b³ is the cube of difference identity.
  9. x³ – y³ = (x – y)(x² + xy + y²) factorises difference of cubes.
  10. x³ + y³ = (x + y)(x² – xy + y²) factorises sum of cubes.
  11. x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx).
  12. To factor a quadratic, find two numbers whose sum is the coefficient of x and product is the constant term.
  13. When simplifying rational expressions, cancel common factors only if the denominator is not zero.
  14. The identity a² = (a + b)(a – b) + b² helps compute squares quickly.
  15. Three consecutive square numbers (n – 1)², n², (n + 1)² always give 2 when smallest and largest are added and twice the middle is subtracted.

MCQs — 10 Questions with Answers

Q1. Which of the following is an algebraic identity?

  • A. (a + b)² = a² + 2ab + b² ✓
  • B. x² - 1 = 24
  • C. 3x + 5 = 20
  • D. 2y - 3 = 7

Answer: A — Option A is true for all values of a and b, making it an identity; options B, C, and D are equations true for only specific values.

Q2. Expand (7x + 4y)² using the identity (a + b)² = a² + 2ab + b².

  • A. 49x² + 16y²
  • B. 49x² + 56xy + 16y² ✓
  • C. 7x² + 4y²
  • D. 49x + 56xy + 16y

Answer: B — (7x)² + 2(7x)(4y) + (4y)² = 49x² + 56xy + 16y²; option A forgets the 2ab term.

Q3. Using the identity, calculate 105².

  • A. 11025 ✓
  • B. 11000
  • C. 10900
  • D. 11050

Answer: A — 105² = (100 + 5)² = 10000 + 1000 + 25 = 11025.

Q4. Factor the expression x² + 8x + 16 using identities.

  • A. (x + 4)² ✓
  • B. (x + 8)²
  • C. (x + 2)²
  • D. (x + 16)²

Answer: A — x² + 8x + 16 = (x)² + 2(x)(4) + (4)² matches the pattern a² + 2ab + b², so it equals (x + 4)².

Q5. Which of the following is NOT correct?

  • A. (a - b)² = a² - 2ab + b²
  • B. (a + b)² = a² + b² + 2ab
  • C. An identity is true for all values of variables ✓
  • D. (a + b)² = (a + b)(a + b) = a² + 2ab + b²

Answer: C — Option C is incorrect because an identity is true for all values, but an equation is only true for some values—the statement confuses the two concepts; all other options are correct statements about identities.

Q6. Priya observes that (10 + 3)² = 169 and 10² + 3² = 109. She concludes that (a + b)² is always greater than a² + b². Which statement best explains why this conclusion is not universally true?

  • A. The identity works only for whole numbers
  • B. The difference (a + b)² - (a² + b²) = 2ab, which can be negative if one variable is negative ✓
  • C. The equation (a + b)² = a² + 2ab + b² is not an identity
  • D. Squaring changes the sign of the numbers

Answer: B — (a + b)² - (a² + b²) = 2ab; if a and b have opposite signs, 2ab < 0, making (a + b)² < a² + b², so the conclusion is not universally true.

Q7. What is the value of 2ab in the expansion of (4p + 5q)²?

  • A. 20pq
  • B. 40pq ✓
  • C. 8p + 10q
  • D. 9pq

Answer: B — In (a + b)² = a² + 2ab + b², with a = 4p and b = 5q, the term 2ab = 2(4p)(5q) = 40pq.

Q8. Factorize 36x² + 12x + 1 and identify its factors.

  • A. (6x - 1)²
  • B. (6x + 1)² ✓
  • C. (3x + 1)²
  • D. (12x + 1)²

Answer: B — 36x² + 12x + 1 = (6x)² + 2(6x)(1) + 1² matches a² + 2ab + b², so it factors as (6x + 1)².

Q9. Consider three consecutive square numbers: 16, 25, 36. If you add the smallest and largest, then subtract twice the middle square, what do you get?

  • A. 2 ✓
  • B. 4
  • C. 6
  • D. 0

Answer: A — Using the pattern: (16 + 36) - 2(25) = 52 - 50 = 2; this follows from (n-1)² + (n+1)² - 2n² = 2 for any n.

Q10. Which identity would you use to factor the expression 25p² + 30pq + 9q²?

  • A. (a + b)² = a² + 2ab + b² ✓
  • B. (a - b)² = a² - 2ab + b²
  • C. Both identities apply equally
  • D. Neither identity applies

Answer: A — The expression 25p² + 30pq + 9q² = (5p)² + 2(5p)(3q) + (3q)² matches (a + b)² with positive middle term, so (a + b)² is the correct identity.

Flashcards

What is an algebraic identity?

An equation that is true for all values of the variables occurring in it.

State the identity for (a + b)².

(a + b)² = a² + 2ab + b²

What is the difference between an identity and an equation?

An identity is true for all values of variables, while an equation is true for only some specific values.

Expand (5x + 2y)² using the identity.

(5x + 2y)² = 25x² + 20xy + 4y²

State the identity for (a - b)².

(a - b)² = a² - 2ab + b²

How can you factor x² + 4x + 4?

Recognize it as (x)² + 2(x)(2) + (2)², so x² + 4x + 4 = (x + 2)².

When is (a + b)² equal to a² + b²?

Only when 2ab = 0, which means either a = 0 or b = 0 (or both).

Calculate 43² using the identity (a + b)² = a² + 2ab + b².

43² = (40 + 3)² = 1600 + 240 + 9 = 1849

What must you first check before factoring 50p² + 60pq + 18q² using identities?

Check if there is a common factor first; here 2 is common, leaving 25p² + 30pq + 9q² = (5p + 3q)².

In the pattern of three consecutive squares (n-1)², n², (n+1)², why does (n-1)² + (n+1)² - 2n² always equal 2?

Because (n-1)² + (n+1)² = n² - 2n + 1 + n² + 2n + 1 = 2n² + 2, and subtracting 2n² leaves 2.

Important Board Questions

Define an algebraic identity and distinguish it from an algebraic equation with one example each. [2 marks]

An identity is true for all variable values; an equation is true for specific values only. Example of identity: (a + b)² = a² + 2ab + b²; example of equation: x² - 1 = 24 (true only for x = ±5).

Using the identity (a + b)² = a² + 2ab + b², expand (3x + 4y)² and verify your answer by direct multiplication. [3 marks]

Apply the formula with a = 3x and b = 4y to get 9x² + 24xy + 16y². Then verify by multiplying (3x + 4y)(3x + 4y) term by term to confirm the result is identical.

Prove that for any three consecutive numbers n – 1, n, and n + 1, the sum of their squares (n – 1)² + (n + 1)² minus twice the middle square 2n² always equals 2. Use this to explain the pattern observed in Example 1 of the chapter. [5 marks]

Expand (n – 1)² and (n + 1)² using the identity (a – b)² = a² – 2ab + b². Add them and subtract 2n² to simplify and prove the result is always 2. This algebraic proof explains why the numerical pattern holds for any set of three consecutive square numbers.

Assertion-Reason Questions for Exploring Algebraic Identities

Q1. Assertion (A):

The expression x² + 8x + 16 can be factored as (x + 4)² using the algebraic identity.

Reason (R):

In x² + 8x + 16, we have x² = (x)², 16 = 4², and 8x = 2(x)(4), which matches the form a² + 2ab + b².

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The expression perfectly matches the pattern (a + b)² = a² + 2ab + b² with a = x and b = 4, and the reason correctly explains why the assertion is valid.

Q2. Assertion (A):

(a + b)² is always greater than a² + b² for all positive real numbers a and b.

Reason (R):

(a + b)² = a² + 2ab + b², and since 2ab > 0 for positive a and b, (a + b)² exceeds a² + b² by the term 2ab.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both the assertion and reason are true and correct. However, the reason is not the primary explanation for why the assertion holds—it's more of a restatement. The real explanation is that the identity shows the extra term 2ab makes the difference, which is what the reason states, making them both true but reason not being the causal explanation.

Q3. Assertion (A):

The identity (a - b)² = a² - 2ab + b² is obtained by replacing b with -b in (a + b)² = a² + 2ab + b².

Reason (R):

When b is replaced by -b, the term 2ab becomes 2a(-b) = -2ab, and (-b)² = b².

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true and correct. The reason properly explains the mathematical substitution process that leads to the assertion, showing how one identity can be derived from another through variable substitution.

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Case-Based Questions (CBQ) for Exploring Algebraic Identities

Case Study:

A carpenter is designing a square platform with side length (3x + 2) meters. She needs to calculate the total area to determine the amount of material needed. Using algebraic identities, she can expand (3x + 2)² to find the area. The carpenter also notices that if she reduces each side by 2 meters, the new side length becomes (3x) meters, and the new area is (3x)². She wonders how much material is saved by reducing the side length.

Q1. What is the expanded form of the original platform area (3x + 2)²?

  • A. 9x² + 12x + 4 ✓
  • B. 9x² + 6x + 4
  • C. 9x² + 12x + 2
  • D. 3x² + 12x + 4

Answer: A

Q2. What is the difference in area between the original platform and the reduced platform?

  • A. 12x + 4 ✓
  • B. 6x + 2
  • C. 12x + 2
  • D. 6x + 4

Answer: A

Q3. If x = 2, what is the area of the original platform in square meters?

  • A. 48 square meters
  • B. 64 square meters ✓
  • C. 40 square meters
  • D. 56 square meters

Answer: B

Q4. Using the identity (a + b)² = a² + 2ab + b², identify which term represents the middle term in the expansion of (3x + 2)².

  • A. 2ab = 6x
  • B. 2ab = 12x ✓
  • C. 2ab = 4
  • D. 2ab = 9x²

Answer: B

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