Special Algebraic Identities

What Are Special Algebraic Identities?

A special algebraic identity is an equation that is always true, regardless of the values substituted for the variables. Instead of expanding brackets the long way every time, these identities give you a shortcut — provided you can recognise the pattern.

There are three standard identities you must know for Elementary Mathematics:

Identity	Name
(a + b)² = a² + 2ab + b²	Perfect Square (Sum)
(a - b)² = a² - 2ab + b²	Perfect Square (Difference)
(a + b)(a - b) = a² - b²	Difference of Two Squares

Section 1 — Proof and Derivation

Identity 1: (a + b)² = a² + 2ab + b²

Expand from first principles:

(a + b)² means (a + b) multiplied by itself.

(a + b)²
= (a + b)(a + b)
= a(a + b) + b(a + b)        [distribute the first bracket]
= a² + ab + ba + b²           [expand each term]
= a² + ab + ab + b²           [since ba = ab]
= a² + 2ab + b²              ✓

Geometric proof (optional but memorable):

Imagine a square with side length (a + b). Its area is (a + b)². Split the square into 4 regions:

Top-left: a × a = a²
Top-right: a × b = ab
Bottom-left: b × a = ab
Bottom-right: b × b = b²

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

Identity 2: (a - b)² = a² - 2ab + b²

Expand from first principles:

(a - b)²
= (a - b)(a - b)
= a(a - b) - b(a - b)        [distribute the first bracket]
= a² - ab - ba + b²           [expand each term; note: -b × -b = +b²]
= a² - ab - ab + b²
= a² - 2ab + b²              ✓

Key point: Notice that the last term is +b², not -b². A negative number squared is always positive.

Identity 3: (a + b)(a - b) = a² - b²

Expand from first principles:

(a + b)(a - b)
= a(a - b) + b(a - b)        [distribute the first bracket]
= a² - ab + ba - b²
= a² - ab + ab - b²           [since ba = ab]
= a²       - b²              ✓

Why does the middle disappear? The terms -ab and +ab cancel each other out. This only happens because one bracket has (+b) and the other has (-b). This is the special structure to recognise.

Section 2 — Recognising the Identity: Pattern Tips

Before you can apply an identity, you need to spot the pattern. Here are the visual clues:

For (a + b)² = a² + 2ab + b²

When expanding: You have a binomial squared.

Both terms are positive.
Result: first term squared, plus twice the product, plus second term squared.

When factorising: You have a trinomial. Check:

Is the first term a perfect square?
Is the last term a perfect square?
Is the middle term exactly twice the product of the square roots of the first and last terms?

Checklist mnemonic: “Square, Double, Square”

For (a - b)² = a² - 2ab + b²

When expanding: You have a binomial squared with a minus sign.

The middle term of the result is always negative.
The last term of the result is always positive (b² is always positive).

When factorising: Same checklist as above, but the middle term is negative.

For (a + b)(a - b) = a² - b²

When expanding: You have two brackets that are identical except for the sign between the two terms.

Clue: “conjugate pair” — same two terms, opposite signs.

When factorising: You have exactly two terms, both perfect squares, separated by a minus sign.

Clue: No middle term. No xy-type term.

Section 3 — Expansion Using Identities: 5 Worked Examples

Example 1: Expand (3x + 4)²

Identify: This matches (a + b)² with a = 3x and b = 4.

Apply identity: (a + b)² = a² + 2ab + b²

(3x + 4)²
= (3x)² + 2(3x)(4) + (4)²
= 9x² + 24x + 16

Answer: 9x² + 24x + 16

Example 2: Expand (2p - 5q)²

Identify: This matches (a - b)² with a = 2p and b = 5q.

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

(2p - 5q)²
= (2p)² - 2(2p)(5q) + (5q)²
= 4p² - 20pq + 25q²

Answer: 4p² - 20pq + 25q²

Example 3: Expand (7 + 3x)(7 - 3x)

Identify: This matches (a + b)(a - b) with a = 7 and b = 3x.

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

(7 + 3x)(7 - 3x)
= 7² - (3x)²
= 49 - 9x²

Answer: 49 - 9x²

Example 4: Expand (4x² - 3y)²

Identify: This matches (a - b)² with a = 4x² and b = 3y.

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

(4x² - 3y)²
= (4x²)² - 2(4x²)(3y) + (3y)²
= 16x⁴ - 24x²y + 9y²

Answer: 16x⁴ - 24x²y + 9y²

Example 5: Expand (½x + 6)(½x - 6)

Identify: This matches (a + b)(a - b) with a = ½x and b = 6.

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

(½x + 6)(½x - 6)
= (½x)² - 6²
= ¼x² - 36

Answer: ¼x² - 36

Section 4 — Factorisation Using Identities: 5 Worked Examples

Factorisation is the reverse process — you start with an expanded expression and rewrite it as a product.

Example 1: Factorise x² + 10x + 25

Check for perfect square trinomial:

First term: x² = (x)² ✓
Last term: 25 = (5)² ✓
Middle term: 10x = 2 × x × 5 ✓

This matches (a + b)² with a = x, b = 5.

x² + 10x + 25 = (x + 5)²

Answer: (x + 5)²

Example 2: Factorise 9y² - 12y + 4

Check for perfect square trinomial:

First term: 9y² = (3y)² ✓
Last term: 4 = (2)² ✓
Middle term: 12y = 2 × 3y × 2 ✓ (and it is negative)

This matches (a - b)² with a = 3y, b = 2.

9y² - 12y + 4 = (3y - 2)²

Answer: (3y - 2)²

Example 3: Factorise 25x² - 49

Check for difference of two squares:

First term: 25x² = (5x)² ✓
Second term: 49 = (7)² ✓
Separated by a minus sign, no middle term ✓

This matches a² - b² with a = 5x, b = 7.

25x² - 49 = (5x + 7)(5x - 7)

Answer: (5x + 7)(5x - 7)

Example 4: Factorise 4p² - 20pq + 25q²

Check for perfect square trinomial:

First term: 4p² = (2p)² ✓
Last term: 25q² = (5q)² ✓
Middle term: 20pq = 2 × 2p × 5q ✓ (and it is negative)

This matches (a - b)² with a = 2p, b = 5q.

4p² - 20pq + 25q² = (2p - 5q)²

Answer: (2p - 5q)²

Example 5: Factorise 3x² - 48

This does not immediately look like a standard identity. First, take out the common factor.

3x² - 48
= 3(x² - 16)        [common factor of 3]

Now check x² - 16:

x² = (x)² ✓
16 = (4)² ✓
Separated by minus ✓ → Difference of two squares

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

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

Tip: Always look for a common factor first before applying an identity.

Section 5 — Applications

Application 1: Simplifying Numerical Expressions

Special identities can make mental arithmetic much faster.

Example: Calculate 103² without a calculator.

Write 103 = 100 + 3, so this is (100 + 3)².

(100 + 3)²
= 100² + 2(100)(3) + 3²
= 10000 + 600 + 9
= 10609

Example: Calculate 99² without a calculator.

Write 99 = 100 - 1, so this is (100 - 1)².

(100 - 1)²
= 100² - 2(100)(1) + 1²
= 10000 - 200 + 1
= 9801

Example: Calculate 97 × 103.

Notice 97 = 100 - 3 and 103 = 100 + 3. This is (100 - 3)(100 + 3).

(100 - 3)(100 + 3)
= 100² - 3²
= 10000 - 9
= 9991

Application 2: Solving Equations Using Difference of Two Squares

Example: Solve x² - 81 = 0

x² - 81 = 0
(x + 9)(x - 9) = 0        [factorise using difference of two squares]

x + 9 = 0  or  x - 9 = 0
x = -9      or  x = 9

Answer: x = -9 or x = 9

Example: Solve 4x² = 25

4x² - 25 = 0              [rearrange]
(2x + 5)(2x - 5) = 0      [difference of two squares: a = 2x, b = 5]

2x + 5 = 0  or  2x - 5 = 0
x = -5/2    or  x = 5/2

Answer: x = -5/2 or x = 5/2

Application 3: Simplifying Algebraic Fractions

Example: Simplify (x² - 9) / (x + 3)

(x² - 9) / (x + 3)
= (x + 3)(x - 3) / (x + 3)    [factorise numerator]
= (x - 3)                      [cancel the common factor (x + 3)]

Answer: x - 3 (valid for x ≠ -3)

Section 6 — Mixed Problems: 5 Harder Examples

Mixed Example 1

Expand and simplify: (2x + 3)² - (2x - 3)²

Step 1: Expand (2x + 3)²
= (2x)² + 2(2x)(3) + 3²
= 4x² + 12x + 9

Step 2: Expand (2x - 3)²
= (2x)² - 2(2x)(3) + 3²
= 4x² - 12x + 9

Step 3: Subtract
(4x² + 12x + 9) - (4x² - 12x + 9)
= 4x² + 12x + 9 - 4x² + 12x - 9
= 24x

Answer: 24x

Shortcut insight: You could also notice (A)² - (B)² = (A+B)(A-B) where A = (2x+3) and B = (2x-3):

= [(2x+3) + (2x-3)][(2x+3) - (2x-3)]
= [4x][6]
= 24x  ✓

Mixed Example 2

Expand and simplify: (x + y + z)(x + y - z)

Treat (x + y) as a single unit. Let u = x + y.

= (u + z)(u - z)
= u² - z²
= (x + y)² - z²
= x² + 2xy + y² - z²

Answer: x² + 2xy + y² - z²

Mixed Example 3

Factorise completely: 2x³ - 8x

Step 1: Take out common factor
= 2x(x² - 4)

Step 2: Factorise (x² - 4) using difference of two squares
= 2x(x + 2)(x - 2)

Answer: 2x(x + 2)(x - 2)

Mixed Example 4

Given that (x + y) = 7 and (x - y) = 3, find x² - y².

Recognise the pattern: x² - y² = (x + y)(x - y)

x² - y² = (x + y)(x - y)
         = 7 × 3
         = 21

Answer: 21

This avoids solving for x and y separately, saving time in an exam.

Mixed Example 5

Simplify: [(3a + 2b)² - (3a - 2b)²] / (48ab)

Step 1: Let A = 3a + 2b, B = 3a - 2b
Numerator = A² - B² = (A + B)(A - B)

A + B = (3a + 2b) + (3a - 2b) = 6a
A - B = (3a + 2b) - (3a - 2b) = 4b

Step 2:
Numerator = 6a × 4b = 24ab

Step 3:
24ab / 48ab = 1/2

Answer: 1/2

Section 7 — Exam Strategies

When to Use an Identity vs Direct Expansion

Situation	Recommendation
Binomial is squared: (expression)²	Use identity — faster and fewer errors
Two brackets differ only by a sign: (A+B)(A-B)	Use identity — the middle terms cancel
Two different binomials with no pattern	Direct expansion using FOIL
Factorising a trinomial	Check identity first; if middle term ≠ 2 × √(first) × √(last), use other methods
Factorising two terms with minus	Always check difference of two squares first

How to Spot the Pattern Quickly (Step-by-Step)

For expansion questions:

Count the brackets. Two identical brackets being multiplied? → Squared identity.
Are the brackets the same except for the sign in the middle? → Difference of two squares.

For factorisation questions:

Two terms only, both perfect squares, minus sign? → Difference of two squares: a² - b² = (a+b)(a-b)
Three terms? → Check if it is a perfect square trinomial:
- √(first term) × √(last term) × 2 = middle term?
- Middle term positive? → (a + b)²
- Middle term negative? → (a - b)²
If none of the above, extract a common factor first, then re-check.

Exam Time-Saving Tips

Write out a = ___ and b = ___ clearly before substituting. This catches sign errors.
For numerical calculations (e.g. 198 × 202), immediately look for conjugate pairs.
In show/prove questions, expand the identity step-by-step — never skip lines.
If asked to “factorise completely”, always check if there is a common factor before applying an identity.

Section 8 — Common Mistakes and How to Avoid Them

Mistake 1: Forgetting the 2ab Middle Term

Wrong: (x + 5)² = x² + 25

Right: (x + 5)² = x² + 10x + 25

Why it happens: Students treat the square as distributing to each term separately — this is incorrect. The square applies to the entire binomial.

How to avoid: Always write out all three terms. Never skip straight to the squared terms.

Mistake 2: Confusing (a + b)² with a² + b²

(a + b)² ≠ a² + b²

This is one of the most common errors in algebra. The expression (a + b)² always has a middle term 2ab.

Memory hook: Think of the geometric proof — the two off-diagonal rectangles (each with area ab) are being ignored when students write just a² + b². Those rectangles exist and must be counted.

Mistake 3: Sign Errors in (a - b)²

Wrong: (x - 3)² = x² - 6x - 9

Right: (x - 3)² = x² - 6x + 9

Why it happens: Students write b² as -b² because they see a minus sign.

How to avoid: Remember that b² is always positive — squaring removes any negative sign. The only negative term is the middle term (-2ab).

Mistake 4: Incorrect Identification of a and b

Wrong approach to (5 - 2x)²: treating a = 5 and b = 2x but then writing 2ab = 2(5)(2x) = 10x instead of 20x.

How to avoid: Write a = 5, b = 2x, then compute 2ab = 2 × 5 × 2x = 20x. Do not skip the multiplication step.

Mistake 5: Incomplete Factorisation

Wrong: 3x² - 12 = (x + 2)(x - 2) [forgot to take out the common factor 3]

Check:

(x + 2)(x - 2) = x² - 4

This is not equal to 3x² - 12.

Right:

3x² - 12
= 3(x² - 4)         [take out common factor 3 first]
= 3(x + 2)(x - 2)

How to avoid: Always check for a common factor before applying an identity.

Section 9 — Practice Questions

Questions

Basic Level

Q1. Expand (x + 7)².

Q2. Expand (3a - 2b)².

Q3. Expand (4x + 5y)(4x - 5y).

Q4. Factorise x² - 14x + 49.

Q5. Factorise 16m² - 9n².

Intermediate Level

Q6. Expand and simplify (3x + 1)² - (3x - 1)².

Q7. Factorise completely 5x² - 45.

Q8. Given that p + q = 8 and p - q = 2, find p² - q².

Harder Level

Q9. Expand and simplify (x + y)² + (x - y)².

Q10. Factorise (2a + b)² - (a - 2b)² completely.

Full Answers

A1. Expand (x + 7)²

(x + 7)²
= x² + 2(x)(7) + 7²
= x² + 14x + 49

A2. Expand (3a - 2b)²

(3a - 2b)²
= (3a)² - 2(3a)(2b) + (2b)²
= 9a² - 12ab + 4b²

A3. Expand (4x + 5y)(4x - 5y)

(4x + 5y)(4x - 5y)
= (4x)² - (5y)²
= 16x² - 25y²

A4. Factorise x² - 14x + 49

Check: √(x²) = x, √49 = 7, middle term check: 2 × x × 7 = 14x ✓ (negative middle)

x² - 14x + 49 = (x - 7)²

A5. Factorise 16m² - 9n²

Check: 16m² = (4m)², 9n² = (3n)², separated by minus ✓

16m² - 9n² = (4m + 3n)(4m - 3n)

A6. Expand and simplify (3x + 1)² - (3x - 1)²

(3x + 1)² = 9x² + 6x + 1
(3x - 1)² = 9x² - 6x + 1

(3x + 1)² - (3x - 1)²
= (9x² + 6x + 1) - (9x² - 6x + 1)
= 9x² + 6x + 1 - 9x² + 6x - 1
= 12x

Answer: 12x

A7. Factorise completely 5x² - 45

5x² - 45
= 5(x² - 9)         [common factor of 5]
= 5(x + 3)(x - 3)   [difference of two squares]

A8. Given p + q = 8 and p - q = 2, find p² - q²

p² - q² = (p + q)(p - q)
         = 8 × 2
         = 16

A9. Expand and simplify (x + y)² + (x - y)²

(x + y)² = x² + 2xy + y²
(x - y)² = x² - 2xy + y²

Sum = (x² + 2xy + y²) + (x² - 2xy + y²)
    = 2x² + 2y²
    = 2(x² + y²)

Answer: 2(x² + y²)

Note: The 2xy and -2xy terms cancel. This is a useful identity in its own right: (x+y)² + (x-y)² = 2(x² + y²)

A10. Factorise (2a + b)² - (a - 2b)² completely

Recognise this as a difference of two squares where A = (2a + b) and B = (a - 2b):

= [(2a + b) + (a - 2b)][(2a + b) - (a - 2b)]
= [2a + b + a - 2b][2a + b - a + 2b]
= [3a - b][a + 3b]

Answer: (3a - b)(a + 3b)

Quick Reference Summary

Identity	Use when expanding	Use when factorising
(a + b)² = a² + 2ab + b²	Binomial with + is squared	Trinomial, all positive, middle = 2√(first)√(last)
(a - b)² = a² - 2ab + b²	Binomial with - is squared	Trinomial, last term positive, middle negative
(a + b)(a - b) = a² - b²	Two brackets, same terms, opposite signs	Two terms only, both squares, minus between them

Golden rule: Always extract any common factor before applying an identity.

Exam golden rule: When you see two terms being multiplied and they look almost the same, ask yourself — “Are these conjugates?” If yes, use the third identity.