1. Difference of Two Squares

Difference of Two Squares

Introduction

In algebra, some expressions can be factored much more quickly when we recognise special patterns.

One of the most useful patterns is called the difference of two squares.

Consider the expression:

x² − 25

At first, this may look like an ordinary subtraction problem. However, both terms are perfect squares:

• x² = x × x

• 25 = 5 × 5

Because the two squared quantities are being subtracted, we can use a special algebraic identity:

a² − b² = (a − b)(a + b)

For example:

x² − 25 = (x − 5)(x + 5)

This identity is important because it allows us to factor expressions, simplify algebraic fractions, solve equations, and perform calculations efficiently.

Understanding Perfect Squares

Before factoring a difference of two squares, we must recognise perfect squares.

A perfect square is a number or algebraic expression obtained by multiplying a quantity by itself.

Numerical Perfect Squares

Algebraic Perfect Squares

• x² = x × x

• 4x² = (2x)²

• 9y² = (3y)²

• 16a⁴ = (4a²)²

• 25m⁶ = (5m³)²

Worked Example

Determine whether 49x² is a perfect square.

49 = 7²

x² = (x)²

Therefore:

49x² = (7x)²

49x² is a perfect square.

Important Observation

A term can be a perfect square even when it contains variables and coefficients.

Recognising the Difference of Two Squares

A difference of two squares has three important features:

• There are two terms.

• The terms are separated by subtraction.

• Both terms are perfect squares.

Examples

x² − 16

Both x² and 16 are perfect squares.

9x² − 25

Both 9x² and 25 are perfect squares.

4a² − 49b²

Both 4a² and 49b² are perfect squares.

Non-Examples

x² + 16

This is a sum of squares, not a difference.

x² − 8

The number 8 is not a perfect square of an integer, so this is not a difference of two integer perfect squares. It can still be factored over the real numbers using √8.

x² − 4x + 4

This contains three terms. It is a perfect-square trinomial rather than a two-term difference of squares.

Important Observation

The subtraction sign is essential to recognising the difference-of-squares pattern.

The Difference of Two Squares Identity

The identity is:

a² − b² = (a − b)(a + b)

This tells us that subtracting two squares is equivalent to multiplying the difference and sum of their bases.

Understanding the Geometry

Imagine a large square with side length a.

Its area is:

a²

Now imagine removing a smaller square with side length b.

The removed area is:

b²

The remaining area is:

a² − b²

For a ≥ b ≥ 0, the remaining pieces can be rearranged to form a rectangle with dimensions:

• Length = a + b

• Width = a − b

The area of this rectangle is:

(a + b)(a − b)

Therefore, the two expressions represent the same area.

Important Observation

The identity is not simply a rule to memorise. It describes a relationship that can be understood geometrically.

Deriving the Identity Algebraically

We can verify the identity using the distributive property.

Start with:

(a − b)(a + b)

Expand:

= a(a + b) − b(a + b)

= a² + ab − ab − b²

The middle terms cancel:

+ab − ab = 0

Therefore:

(a − b)(a + b) = a² − b²

This proves the identity.

Important Observation

The middle terms cancel because the two factors contain opposite signs.

Factoring Simple Expressions

Worked Example: x² − 9

Identify the squares:

x² = (x)²

9 = 3²

Apply the identity:

x² − 9 = (x − 3)(x + 3)

Final Answer: (x − 3)(x + 3)

Worked Example: x² − 64

64 = 8²

Therefore:

x² − 64 = (x − 8)(x + 8)

Final Answer: (x − 8)(x + 8)

Worked Example: y² − 100

100 = 10²

Therefore:

y² − 100 = (y − 10)(y + 10)

Final Answer: (y − 10)(y + 10)

Factoring Expressions with Coefficients

Sometimes both squared terms contain numerical coefficients.

Worked Example: 4x² − 25

Identify the square roots of the terms:

4x² = (2x)²

25 = 5²

Apply the identity:

4x² − 25 = (2x − 5)(2x + 5)

Final Answer: (2x − 5)(2x + 5)

Worked Example: 9a² − 16

9a² = (3a)²

16 = 4²

Therefore:

9a² − 16 = (3a − 4)(3a + 4)

Final Answer: (3a − 4)(3a + 4)

Worked Example: 49m² − 81

49m² = (7m)²

81 = 9²

Therefore:

49m² − 81 = (7m − 9)(7m + 9)

Final Answer: (7m − 9)(7m + 9)

Important Observation

When factoring expressions with coefficients, take the square root of the entire squared term.

Factoring Expressions with Two Variables

The identity also works when both terms contain variables.

Worked Example: x² − y²

x² = (x)²

y² = ²

Therefore:

x² − y² = (x − y)(x + y)

Worked Example: 16a² − 9b²

16a² = (4a)²

9b² = (3b)²

Therefore:

16a² − 9b² = (4a − 3b)(4a + 3b)

Worked Example: 25p⁴ − 36q²

25p⁴ = (5p²)²

36q² = (6q)²

Therefore:

25p⁴ − 36q² = (5p² − 6q)(5p² + 6q)

Important Observation

The identity applies to any expressions that can be written as squares, not only individual variables.

Verifying Factorisation

Factoring should always be checked by multiplying the factors.

Worked Example

Verify:

x² − 36 = (x − 6)(x + 6)

Expand the right-hand side:

(x − 6)(x + 6)

= x² + 6x − 6x − 36

= x² − 36

The expanded expression matches the original.

Therefore, the factorisation is correct.

Another Example

Verify:

9x² − 4 = (3x − 2)(3x + 2)

Expand:

= 9x² + 6x − 6x − 4

= 9x² − 4

The factorisation is correct.

Important Observation

Expansion and factorisation are inverse processes.

Checking Factorisation by Substitution

Another useful checking method is to substitute a numerical value.

Example

Check:

x² − 16 = (x − 4)(x + 4)

Choose x = 6.

Original expression:

6² − 16 = 36 − 16 = 20

Factored expression:

(6 − 4)(6 + 4)

= 2 × 10

= 20

Both sides give the same result.

This supports the factorisation.

However, testing one value does not prove that two expressions are identical for every value. Expanding the factors provides an algebraic proof.

Factoring Completely

Sometimes an expression has a common factor that must be removed first.

Worked Example: 2x² − 50

Both terms have a common factor of 2.

Factor out 2:

2(x² − 25)

Now apply the difference-of-squares identity:

x² − 25 = (x − 5)(x + 5)

Therefore:

2x² − 50 = 2(x − 5)(x + 5)

Worked Example: 3a² − 75b²

Factor out 3:

3(a² − 25b²)

Apply the identity:

3(a − 5b)(a + 5b)

Final Answer: 3(a − 5b)(a + 5b)

Important Observation

Always check for a greatest common factor before applying special factorisation identities.

Repeated Difference of Squares

Some expressions can be factored more than once.

Worked Example: x⁴ − 16

Recognise:

x⁴ = (x²)²

16 = 4²

Therefore:

x⁴ − 16 = (x² − 4)(x² + 4)

The first factor is another difference of squares:

x² − 4 = (x − 2)(x + 2)

Therefore:

x⁴ − 16 = (x − 2)(x + 2)(x² + 4)

This is completely factored over the real numbers using polynomial factors.

Worked Example: x⁴ − 81

x⁴ − 81 = (x² − 9)(x² + 9)

Factor x² − 9:

x⁴ − 81 = (x − 3)(x + 3)(x² + 9)

Important Observation

After factoring once, examine the resulting factors to see whether any can be factored further.

Difference of Squares with Binomials

The quantities being squared can themselves be algebraic expressions.

Worked Example

Factor:

(x + 3)² − 16

Identify:

a = x + 3

b = 4

Apply the identity:

[(x + 3) − 4][(x + 3) + 4]

Simplify:

(x − 1)(x + 7)

Final Answer: (x − 1)(x + 7)

Worked Example

Factor:

(2x − 1)² − 9

Identify:

a = 2x − 1

b = 3

Apply the identity:

[(2x − 1) − 3][(2x − 1) + 3]

Simplify:

(2x − 4)(2x + 2)

Factor out common factors:

4(x − 2)(x + 1)

Important Observation

The identity works even when a and b represent entire algebraic expressions.

Simplifying Algebraic Fractions

The difference-of-squares identity is particularly useful when simplifying rational expressions.

Worked Example

Simplify:

(x² − 25)/(x − 5)

Factor the numerator:

(x − 5)(x + 5)/(x − 5)

Cancel the common factor:

x + 5

However, the original denominator cannot equal zero.

Therefore:

Final Answer: x + 5, where x ≠ 5

Important Observation

Cancelling a common factor does not remove restrictions from the original denominator.

Simplifying More Complex Fractions

Worked Example

Simplify:

(4x² − 9)/(2x − 3)

Factor the numerator:

(2x − 3)(2x + 3)/(2x − 3)

Cancel the common factor:

2x + 3

Restriction:

2x − 3 ≠ 0

x ≠ 3/2

Final Answer

2x + 3, where x ≠ 3/2

Worked Example

Simplify:

(x² − 49)/(x² + 14x + 49)

Factor the numerator:

(x − 7)(x + 7)

Factor the denominator:

(x + 7)²

Therefore:

[(x − 7)(x + 7)]/[(x + 7)(x + 7)]

Cancel one common factor:

(x − 7)/(x + 7), where x ≠ −7

Important Observation

Only common factors may be cancelled. Individual terms connected by addition or subtraction cannot simply be cancelled.

Solving Equations Using Difference of Squares

The identity can help solve quadratic equations.

Worked Example: x² − 25 = 0

Factor:

(x − 5)(x + 5) = 0

Apply the zero-product property.

If two factors multiply to zero, at least one factor must equal zero.

Therefore:

x − 5 = 0

or

x + 5 = 0

Solutions:

x = 5 or x = −5

Verification

For x = 5:

5² − 25 = 0

For x = −5:

(−5)² − 25 = 0

Both solutions are correct.

Solving Equations with Coefficients

Worked Example

Solve:

9x² − 16 = 0

Factor:

(3x − 4)(3x + 4) = 0

Set each factor equal to zero:

3x − 4 = 0

3x = 4

x = 4/3

Or:

3x + 4 = 0

3x = −4

x = −4/3

Final Answer

x = ±4/3

Important Observation

A difference-of-squares equation set equal to zero often produces two solutions that are opposites.

Equations Requiring Rearrangement

Sometimes the equation must be rearranged before applying the identity.

Worked Example

Solve:

x² = 64

Subtract 64 from both sides:

x² − 64 = 0

Factor:

(x − 8)(x + 8) = 0

Therefore:

x = 8 or x = −8

Worked Example

Solve:

4x² = 49

Rearrange:

4x² − 49 = 0

Factor:

(2x − 7)(2x + 7) = 0

Therefore:

x = ±7/2

Applications in Geometry

The difference of two squares has a direct geometric interpretation involving areas.

Worked Example: Square Garden

A square garden has side length 12 m.

A smaller square flower bed with side length 5 m is removed from the centre.

Calculate the remaining area.

Method

Area of large square:

12² = 144 m²

Area of smaller square:

5² = 25 m²

Remaining area:

144 − 25 = 119 m²

Using the identity:

12² − 5² = (12 − 5)(12 + 5)

= 7 × 17

Remaining area = 119 m²

Important Observation

The identity can make calculations involving differences in square areas more efficient.

Applications in Mental Arithmetic

The difference-of-squares identity can simplify numerical calculations.

Worked Example: 51 × 49

Notice:

51 = 50 + 1

49 = 50 − 1

Therefore:

51 × 49 = (50 + 1)(50 − 1)

= 50² − 1²

= 2500 − 1

51 × 49 = 2499

Worked Example: 103 × 97

103 = 100 + 3

97 = 100 − 3

Therefore:

103 × 97 = 100² − 3²

= 10000 − 9

103 × 97 = 9991

Important Observation

The identity is especially useful when multiplying two numbers equally spaced above and below a convenient number.

Applications in Physics

Difference-of-squares expressions also appear in scientific equations.

Example: Kinematics

One equation of motion is:

v² = u² + 2as

where:

• v = final velocity

• u = initial velocity

• a = acceleration

• s = displacement

Rearranging:

v² − u² = 2as

Factor the left side:

(v − u)(v + u) = 2as

This shows how the difference of squared velocities can be represented as a product.

Important Observation

Algebraic identities are useful beyond mathematics because they help rearrange and simplify scientific equations.

Recognising the Pattern in More Advanced Expressions

Example: 64x⁶ − 121y⁴

Identify:

64x⁶ = (8x³)²

121y⁴ = (11y²)²

Therefore:

64x⁶ − 121y⁴ = (8x³ − 11y²)(8x³ + 11y²)

Example: 100a⁸ − b¹⁰

100a⁸ = (10a⁴)²

b¹⁰ = (b⁵)²

Therefore:

100a⁸ − b¹⁰ = (10a⁴ − b⁵)(10a⁴ + b⁵)

Important Observation

Higher powers can still form perfect squares when their exponents are even.

Common Mistakes

Confusing difference and sum of squares

x² − 9 = (x − 3)(x + 3)

But x² + 9 cannot be factored using the same identity over the real numbers.

Forgetting opposite signs

Incorrect:

(x − 5)(x − 5)

Correct:

(x − 5)(x + 5)

Forgetting to square coefficients

For 9x² − 16, the correct factors are:

(3x − 4)(3x + 4)

Failing to factor out a common factor

For 2x² − 18:

2(x² − 9) = 2(x − 3)(x + 3)

Stopping too early

For x⁴ − 16, the expression x² − 4 can be factored further.

Cancelling terms instead of factors

In algebraic fractions, cancellation is valid only for common multiplicative factors.

Forgetting denominator restrictions

When simplifying fractions, retain values excluded by the original denominator.

Forgetting negative solutions

For x² = 25:

x = 5 or x = −5

Not checking the answer

Expand the factors to verify the original expression is recovered.

Key Terms

• Perfect Square: A number or expression obtained by multiplying a quantity by itself.

• Difference of Two Squares: An expression involving the subtraction of two squared quantities.

• Identity: An equation that is true for all permitted values of its variables.

• Factorisation: Rewriting an expression as a product of factors.

• Factor: A quantity multiplied by another quantity to form a product.

• Binomial: An algebraic expression containing two terms.

• Conjugate Binomials: Binomials with identical terms but opposite signs between them.

• Expansion: Multiplying factors to produce an equivalent expression.

• Distributive Property: Rule used to multiply expressions across addition or subtraction.

• Greatest Common Factor: Largest factor common to all terms of an expression.

• Zero-Product Property: If a product equals zero, at least one factor must equal zero.

• Quadratic Equation: Equation involving a variable raised to the second power as its highest power.

• Rational Expression: Fraction containing algebraic expressions.

• Excluded Value: A value that makes an original denominator equal to zero.

• Simplification: Rewriting an expression in an equivalent, often more convenient form.

• Verification: Checking that a mathematical result is correct.

• Square Root: A value that produces a given quantity when squared.

Key Takeaways

• The difference-of-squares identity is a² − b² = (a − b)(a + b).

• Both terms must be squares, and they must be separated by subtraction.

• The two factors contain the same terms with opposite signs.

• Perfect squares can include coefficients, variables, and higher even powers.

• Factorisation can be verified by expansion.

• Always check for common factors before applying the identity.

• Some expressions require repeated factorisation.

• The identity can simplify algebraic fractions.

• Denominator restrictions must be preserved when simplifying rational expressions.

• Difference-of-squares factorisation helps solve quadratic equations.

• The identity has geometric applications involving areas.

• It can simplify mental arithmetic.

• Recognising algebraic patterns makes mathematical problem-solving more efficient.

Check Your Understanding

Recognising the Pattern

1. What is a perfect square?

2. Write the difference-of-squares identity.

3. Is x² − 49 a difference of squares?

4. Is x² + 25 a difference of squares?

5. Explain why 16a² − 81b² is a difference of squares.

Basic Factorisation

6. Factor x² − 16.

7. Factor y² − 64.

8. Factor 4x² − 9.

9. Factor 25a² − 36.

10. Factor 49m² − 100n².

Advanced Factorisation

11. Factor completely: 2x² − 32.

12. Factor completely: x⁴ − 81.

13. Factor: 16a⁴ − 25b².

14. Factor: (x + 2)² − 9.

15. Verify by expansion that (x − 7)(x + 7) = x² − 49.

Solving and Simplifying

16. Solve x² − 36 = 0.

17. Solve 4x² − 25 = 0.

18. Simplify (x² − 9)/(x − 3), stating any restrictions.

19. Calculate 52 × 48 using the difference-of-squares identity.

20. Explain why x² + 16 cannot be factored using the difference-of-squares identity over the real numbers.

Answer Key

Extended Challenge

1. Factor completely: 3x² − 147.

2. Factor completely: x⁴ − 625.

3. Factor: 81a⁶ − 64b⁴.

4. Factor completely: (2x + 1)² − 25.

5. Solve: 16x² − 81 = 0.

6. Solve: (x − 2)² − 49 = 0.

7. Simplify (9x² − 16)/(3x − 4), stating restrictions.

8. Simplify (x² − 64)/(x² + 16x + 64), stating restrictions.

9. Calculate 1004 × 996 using the difference-of-squares identity.

10. Explain why x⁴ − 16 can be factored into three real polynomial factors, while x⁴ + 16 cannot be factored directly using the same identity.

Extended Challenge Answers

Final Investigation: Exploring Square Areas

A student is investigating the relationship between two square gardens.

The larger square has side length x metres.

The smaller square has side length 4 metres.

The smaller square is removed from the larger square.

Assume x is greater than 4.

Part A: Building the Model

1. Write an expression for the area of the larger square.

2. Write an expression for the area of the smaller square.

3. Write an expression for the remaining area.

4. Factor the remaining-area expression.

5. Explain why the expression is a difference of squares.

Part B: Creating a Table

6. Calculate the remaining area when x = 5, 6, 7, 8, 9, and 10.

7. Organise the results in a table.

8. Calculate the difference in remaining area between consecutive values.

9. Determine whether the relationship between x and remaining area is linear.

10. Explain your conclusion.

Part C: Comparing Two Expressions

11. Evaluate x² − 16 when x = 7.

12. Evaluate (x − 4)(x + 4) when x = 7.

13. Compare the results.

14. Repeat for x = 10.

15. Explain why the two expressions always produce the same result.

Part D: Solving a Practical Problem

16. The remaining area is 65 m². Write an equation.

17. Solve the equation using factorisation.

18. Determine the side length of the larger square.

19. Explain why a negative solution would not be meaningful.

20. Calculate the total area of the larger square.

Investigation Answer Check

Area Model

Area of larger square:

A₁ = x²

Area of smaller square:

A₂ = 16

Remaining area:

A = x² − 16

Factor:

A = (x − 4)(x + 4)

Completed Table

The area increases by 11, 13, 15, 17, and 19 m² between consecutive side lengths.

The differences are not constant.

Therefore, the relationship is nonlinear.

Solving for 65 m²

x² − 16 = 65

x² − 81 = 0

(x − 9)(x + 9) = 0

x = 9 or x = −9

Since a physical side length cannot be negative:

x = 9 m

Area of larger square:

9² = 81 m²

Final Reflection

The difference-of-squares identity is a powerful algebraic tool that connects multiplication, factorisation, geometry, and equation solving.

By recognising when two perfect squares are separated by subtraction, we can rewrite expressions as products, simplify calculations, verify algebraic results, and solve practical problems.

The essential identity is a² − b² = (a − b)(a + b). Recognising when and how to apply it is an important foundation for more advanced algebra.

2. Perfect Square Trinomials

Perfect Square Trinomials

Introduction

Some algebraic expressions follow special patterns that make them easier to expand, factor, and simplify.

One of the most important patterns is the perfect square trinomial.

Consider the expression:

x² + 6x + 9

This expression contains three terms, but it can be rewritten as the square of a binomial:

x² + 6x + 9 = (x + 3)²

Similarly:

x² − 10x + 25 = (x − 5)²

Recognising these patterns allows us to work more efficiently with algebraic expressions, solve quadratic equations, simplify calculations, and understand the structure of quadratic functions.

Perfect square trinomials are especially important when studying:

• Polynomial factorisation

• Quadratic equations

• Completing the square

• Graphing parabolas

• Algebraic identities

• Geometry and area

• Mathematical problem-solving

What Is a Perfect Square Trinomial?

A trinomial is an algebraic expression containing three terms.

For example:

x² + 5x + 6

A perfect square trinomial is a trinomial that can be written as the square of a binomial.

For example:

x² + 8x + 16 = (x + 4)²

The expression on the right is a binomial multiplied by itself:

(x + 4)(x + 4)

When expanded, it produces the original trinomial.

Important Observation

Every perfect square trinomial can be expressed as the square of a binomial, but not every trinomial is a perfect square trinomial.

Understanding the Square of a Binomial

A binomial contains two terms.

Examples include:

• x + 3

• x − 5

• 2a + 1

• 3m − 4n

Squaring a binomial means multiplying it by itself.

For example:

(x + 3)² = (x + 3)(x + 3)

It does not mean squaring each term separately.

Expanding Step by Step

(x + 3)(x + 3)

= x² + 3x + 3x + 9

Combine like terms:

(x + 3)² = x² + 6x + 9

Notice that the expanded expression contains three terms.

Important Observation

The middle term comes from adding the two cross-products.

The Two Perfect Square Identities

There are two important perfect square identities.

Square of a Sum

(a + b)² = a² + 2ab + b²

Square of a Difference

(a − b)² = a² − 2ab + b²

These identities are true for all real values of a and b.

Comparing the Patterns

The first and last terms are squares in both identities.

The sign of the middle term depends on whether the original binomial contains addition or subtraction.

Important Observation

The last term is positive in both identities because b² is nonnegative for real b.

Geometric Meaning of a Perfect Square Trinomial

Perfect square trinomials can be understood using areas.

Imagine a square with side length a + b.

Its total area is:

A = (a + b)²

The square can be divided into four regions:

• A square with area a²

• Two rectangles, each with area ab

• A smaller square with area b²

  

  Adding the areas:

  A = a² + ab + ab + b²

  Combine the rectangles:

  A = a² + 2ab + b²

  Therefore:

  (a + b)² = a² + 2ab + b²

  Important Observation

  The middle term 2ab represents the combined area of two rectangles.

   

  Expanding the Square of a Sum

  Worked Example: (x + 5)²

  Use:

  (a + b)² = a² + 2ab + b²

  Identify:

  a = x

  b = 5

  Substitute:

  (x + 5)² = x² + 2(x)(5) + 5²

  Simplify:

  = x² + 10x + 25

  Final Answer: x² + 10x + 25

  Worked Example: (y + 7)²

  (y + 7)² = y² + 2(7) + 7²

  = y² + 14y + 49

  Final Answer: y² + 14y + 49

  Important Observation

  The middle term is twice the product of the two terms in the original binomial.

   

  Expanding the Square of a Difference

  Worked Example: (x − 4)²

  Use:

  (a − b)² = a² − 2ab + b²

  Identify:

  a = x

  b = 4

  Substitute:

  (x − 4)² = x² − 2(x)(4) + 4²

  Simplify:

  (x − 4)² = x² − 8x + 16

  Worked Example: (m − 9)²

  (m − 9)² = m² − 2(m)(9) + 9²

  = m² − 18m + 81

  Final Answer: m² − 18m + 81

  Important Observation

  The final term is positive even when the binomial contains subtraction.

   

  Recognising Perfect Square Trinomials

  To determine whether a trinomial is a perfect square, look for three features.

  • The first term is a perfect square.

  • The last term is a perfect square.

  • The middle term equals twice the product of the square roots of the first and last terms, with the appropriate sign.

  Worked Example

  Determine whether:

  x² + 12x + 36

  is a perfect square trinomial.

  First term:

  x² = (x)²

  Last term:

  36 = 6²

  Expected middle term:

  2(x)(6) = 12x

  The middle term matches.

  Therefore:

  x² + 12x + 36 = (x + 6)²

  Important Observation

  Checking only the first and last terms is not enough. The middle term must also match.

   

  Recognising a Negative Middle Term

  Worked Example

  Factor:

  x² − 14x + 49

  First term:

  x² = (x)²

  Last term:

  49 = 7²

  Expected middle term:

  −2(x)(7) = −14x

  The middle term matches.

  Therefore:

  x² − 14x + 49 = (x − 7)²

  Important Observation

  A negative middle term indicates subtraction inside the squared binomial when the square roots of the outer terms are taken as positive quantities.

   

  Identifying Expressions That Are Not Perfect Squares

  Not every trinomial with square first and last terms is a perfect square.

  Example

  Consider:

  x² + 8x + 25

  The first and last terms are squares:

  x² = (x)²

  25 = 5²

  However, the expected middle term would be:

  2(x)(5) = 10x

  The actual middle term is 8x.

  Therefore, this is not a perfect square trinomial.

  Comparison

  Expression	Perfect Square?	Reason
  x² + 6x + 9	Yes	Middle term = 2(x)(3)
  x² − 10x + 25	Yes	Middle term = −2(x)(5)
  x² + 8x + 16	Yes	Middle term = 2(x)(4)
  x² + 8x + 25	No	Middle term does not match
  x² − 12x + 36	Yes	Middle term = −2(x)(6)

   

  Factoring Perfect Square Trinomials

  Factoring is the reverse of expanding.

  General Patterns

  a² + 2ab + b² = (a + b)²

  a² − 2ab + b² = (a − b)²

  Worked Example: x² + 18x + 81

  Find the square roots of the first and last terms:

  √x² = x

  √81 = 9

  Check the middle term:

  2(x)(9) = 18x

  Therefore:

  x² + 18x + 81 = (x + 9)²

  Worked Example: x² − 20x + 100

  √x² = x

  √100 = 10

  Check:

  −2(x)(10) = −20x

  Therefore:

  x² − 20x + 100 = (x − 10)²

   

  Factoring with Coefficients

  Some perfect square trinomials have coefficients greater than 1.

  Worked Example: 4x² + 12x + 9

  First term:

  4x² = (2x)²

  Last term:

  9 = 3²

  Check the middle term:

  2(2x)(3) = 12x

  Therefore:

  4x² + 12x + 9 = (2x + 3)²

  Worked Example: 9x² − 30x + 25

  First term:

  9x² = (3x)²

  Last term:

  25 = 5²

  Check:

  −2(3x)(5) = −30x

  Therefore:

  9x² − 30x + 25 = (3x − 5)²

  Important Observation

  When coefficients are present, take the square roots of the complete first and last terms.

   

  Factoring with Two Variables

  The identities also apply to expressions containing more than one variable.

  Worked Example

  Factor:

  a² + 10ab + 25b²

  First term:

  a² = (a)²

  Last term:

  25b² = (5b)²

  Check:

  2(a)(5b) = 10ab

  Therefore:

  a² + 10ab + 25b² = (a + 5b)²

  Worked Example

  Factor:

  16x² − 24xy + 9y²

  First term:

  16x² = (4x)²

  Last term:

  9y² = (3y)²

  Check:

  −2(4x)(3y) = −24xy

  Therefore:

  16x² − 24xy + 9y² = (4x − 3y)²

   

  Expanding More Complex Binomials

  Worked Example: (3x + 4)²

  Use:

  (a + b)² = a² + 2ab + b²

  Identify:

  a = 3x

  b = 4

  Expand:

  (3x + 4)² = (3x)² + 2(3x)(4) + 4²

  = 9x² + 24x + 16

  Final Answer: 9x² + 24x + 16

  Worked Example: (5a − 2b)²

  (5a − 2b)² = (5a)² − 2(5a)(2b) + (2b)²

  = 25a² − 20ab + 4b²

  Final Answer: 25a² − 20ab + 4b²

   

  Factoring Out Common Factors First

  Sometimes a trinomial contains a common factor.

  Worked Example

  Factor completely:

  2x² + 12x + 18

  All terms share a factor of 2.

  Factor out 2:

  2(x² + 6x + 9)

  Recognise the perfect square:

  x² + 6x + 9 = (x + 3)²

  Therefore:

  2x² + 12x + 18 = 2(x + 3)²

  Worked Example

  Factor:

  3a² − 24a + 48

  Factor out 3:

  3(a² − 8a + 16)

  Recognise:

  a² − 8a + 16 = (a − 4)²

  Therefore:

  3a² − 24a + 48 = 3(a − 4)²

  Important Observation

  Always check for a greatest common factor before applying a special factorisation identity.

   

  Verifying Factorisation

  Factorisation can be checked by expanding the result.

  Worked Example

  Verify:

  x² + 16x + 64 = (x + 8)²

  Expand:

  (x + 8)(x + 8)

  = x² + 8x + 8x + 64

  = x² + 16x + 64

  The result matches the original expression.

  Therefore, the factorisation is correct.

  Another Example

  Verify:

  4x² − 20x + 25 = (2x − 5)²

  Expand:

  (2x − 5)(2x − 5)

  = 4x² − 10x − 10x + 25

  = 4x² − 20x + 25

  The factorisation is correct.

   

  The Relationship Between Perfect Squares and Quadratic Equations

  Perfect square trinomials are especially useful when solving quadratic equations.

  Worked Example

  Solve:

  x² + 10x + 25 = 0

  Factor:

  (x + 5)² = 0

  Take the square root or apply the zero-product property:

  x + 5 = 0

  Therefore:

  x = −5

  This equation has one distinct real solution.

  Important Observation

  A perfect square quadratic equation set equal to zero has a repeated root.

   

  Solving Equations with Repeated Roots

  Worked Example

  Solve:

  4x² − 12x + 9 = 0

  Recognise the perfect square:

  (2x − 3)² = 0

  Therefore:

  2x − 3 = 0

  2x = 3

  x = 3/2

  Why Is This Called a Repeated Root?

  The equation can be written:

  (2x − 3)(2x − 3) = 0

  Both factors produce the same solution.

  The root x = 3/2 has multiplicity 2.

   

  Perfect Square Trinomials and Graphs

  A quadratic function produces a parabola.

  Consider:

  y = (x − 3)²

  Expanding:

  y = x² − 6x + 9

  The graph is a parabola with its minimum at:

  (3, 0)

  

  Important Observation

  A quadratic function written as y = (x − h)² has its vertex at (h, 0).

   

  Connecting Perfect Squares to Completing the Square

  Completing the square is a method used to rewrite quadratic expressions.

  Worked Example

  Rewrite:

  x² + 6x + 5

  as a squared binomial plus or minus a constant.

  Step 1: Focus on the First Two Terms

  x² + 6x

  Step 2: Take Half the Coefficient of x

  6/2 = 3

  Step 3: Square the Result

  3² = 9

  Step 4: Add and Subtract 9

  x² + 6x + 9 − 9 + 5

  Group the first three terms:

  (x + 3)² − 9 + 5

  Simplify:

  x² + 6x + 5 = (x + 3)² − 4

  Important Observation

  Completing the square uses the perfect square trinomial identity to rewrite quadratic expressions in a useful form.

   

  Applications in Geometry

  Perfect square trinomials can represent areas of squares.

  

   

   

   

  torrisonstone.com

  Worked Example

  A square garden has side length:

  x + 4 metres

  Find an expression for its area.

  Area of square = Side × Side

  A = (x + 4)²

  Expand:

  A = x² + 8x + 16

  Area = x² + 8x + 16 m²

  Interpretation

  The expression can be understood as:

  • A square with area x².

  • Two rectangular regions with total area 8x.

  • A smaller square with area 16.

   

  Applications in Mental Arithmetic

  Perfect square identities can make numerical calculations easier.

  Worked Example: 103²

  Rewrite:

  103 = 100 + 3

  Therefore:

  103² = (100 + 3)²

  Apply the identity:

  = 100² + 2(100)(3) + 3²

  = 10000 + 600 + 9

  103² = 10609

  Worked Example: 98²

  Rewrite:

  98 = 100 − 2

  Therefore:

  98² = (100 − 2)²

  = 100² − 2(100)(2) + 2²

  = 10000 − 400 + 4

  98² = 9604

  Important Observation

  Squaring numbers close to 10, 100, or 1000 can often be simplified using perfect square identities.

   

  Applications in Physics

  Perfect square expressions appear in many physical relationships.

  Example: Kinetic Energy

  The kinetic energy of an object is:

  Eₖ = ½mv²

  Suppose the velocity is represented by:

  v = u + at

  Substituting:

  Eₖ = ½m(u + at)²

  Expand the square:

  (u + at)² = u² + 2uat + a²t²

  Therefore:

  Eₖ = ½m(u² + 2uat + a²t²)

  This can be expanded further:

  Eₖ = ½mu² + muat + ½ma²t²

  Important Observation

  Perfect square identities are useful when substituting expressions into scientific equations.

   

  Comparing Perfect Square Trinomials with Difference of Squares

  These two special algebraic identities are related but different.

  Perfect Square Trinomial

  (a + b)² = a² + 2ab + b²

  (a − b)² = a² − 2ab + b²

  Difference of Squares

  a² − b² = (a − b)(a + b)

  Comparison

  Perfect Square Trinomial	Difference of Squares
  Three terms when expanded	Two terms when expanded
  Square of one binomial	Product of conjugate binomials
  Middle term is ±2ab	Middle terms cancel
  Example: x² + 6x + 9	Example: x² − 9
  Factors to (x + 3)²	Factors to (x − 3)(x + 3)

  Important Observation

  Recognising which pattern is present helps us choose the correct factorisation method.

   

  Common Mistakes

  Forgetting the middle term

  Incorrect:

  (x + 5)² = x² + 25

  Correct:

  (x + 5)² = x² + 10x + 25

  Using the wrong sign

  Incorrect:

  (x − 4)² = x² + 8x + 16

  Correct:

  (x − 4)² = x² − 8x + 16

  Forgetting to square coefficients

  Incorrect:

  (3x + 2)² = 3x² + 12x + 4

  Correct:

  (3x + 2)² = 9x² + 12x + 4

  Assuming all trinomials are perfect squares

  The middle term must match twice the product of the square roots of the first and last terms.

  Confusing perfect squares with difference of squares

  x² − 16 is not a perfect square trinomial.

  Failing to factor out a common factor

  Always check for a greatest common factor first.

  Stopping before simplifying completely

  A factored expression may still contain common factors or other opportunities for simplification.

  Forgetting to verify

  Expand the factored expression to confirm that it matches the original trinomial.

   

  Key Terms

  • Trinomial: Algebraic expression containing three terms.

  • Binomial: Algebraic expression containing two terms.

  • Perfect Square: Expression obtained by multiplying a quantity by itself.

  • Perfect Square Trinomial: Trinomial that can be written as the square of a binomial.

  • Algebraic Identity: Equation true for all permitted values of its variables.

  • Square of a Sum: Identity (a + b)² = a² + 2ab + b².

  • Square of a Difference: Identity (a − b)² = a² − 2ab + b².

  • Factorisation: Rewriting an expression as a product of factors.

  • Expansion: Multiplying factors to obtain an equivalent expression.

  • Coefficient: Numerical factor multiplying a variable.

  • Middle Term: Term between the first and last terms of a trinomial.

  • Cross-Product: Product formed by multiplying terms from different factors.

  • Like Terms: Terms with identical variable parts and exponents.

  • Greatest Common Factor: Largest factor shared by all terms.

  • Quadratic Expression: Polynomial expression with highest variable power 2.

  • Quadratic Equation: Equation involving a quadratic expression.

  • Repeated Root: Solution that occurs more than once as a factor of a polynomial.

  • Parabola: U-shaped or inverted U-shaped graph of a quadratic function.

  • Vertex: Turning point of a parabola.

  • Completing the Square: Method for rewriting quadratic expressions using perfect square identities.

  • Verification: Checking a mathematical result for correctness.

   

  Key Takeaways

  • Perfect square trinomials can be written as the square of a binomial.

  • The two identities are (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².

  • The first and last terms are perfect squares.

  • The middle term is twice the product of the square roots of the first and last terms.

  • A positive middle term corresponds to the square of a sum when positive square roots are used.

  • A negative middle term corresponds to the square of a difference.

  • Factoring is the reverse of expanding.

  • Expansion can be used to verify factorisation.

  • Always check for common factors first.

  • Perfect square trinomials can help solve quadratic equations.

  • They are useful for completing the square.

  • They have geometric interpretations involving areas.

  • They can simplify mental arithmetic and scientific calculations.

  • Recognising special algebraic patterns improves efficiency and accuracy.

   

  Check Your Understanding

  Recognising Perfect Squares

  1. What is a perfect square trinomial?

  2. Write the identity for the square of a sum.

  3. Write the identity for the square of a difference.

  4. Is x² + 10x + 25 a perfect square trinomial?

  5. Is x² + 6x + 16 a perfect square trinomial?

  Expanding Binomials

  6. Expand (x + 4)².

  7. Expand (x − 6)².

  8. Expand (2x + 3)².

  9. Expand (3a − 5)².

  10. Expand (x + 2y)².

  Factoring Perfect Square Trinomials

  11. Factor x² + 8x + 16.

  12. Factor x² − 12x + 36.

  13. Factor 4x² + 20x + 25.

  14. Factor 9a² − 24a + 16.

  15. Factor 16x² + 40xy + 25y².

  Applications

  16. Solve x² + 6x + 9 = 0.

  17. Solve 4x² − 12x + 9 = 0.

  18. Calculate 102² using a perfect square identity.

  19. Explain why (x + 5)² is not equal to x² + 25.

  20. Explain how perfect square trinomials are related to completing the square.

  Answer Key

  Question	Answer
  1	A trinomial that is the square of a binomial
  2	(a + b)² = a² + 2ab + b²
  3	(a − b)² = a² − 2ab + b²
  4	Yes, (x + 5)²
  5	No
  6	x² + 8x + 16
  7	x² − 12x + 36
  8	4x² + 12x + 9
  9	9a² − 30a + 25
  10	x² + 4xy + 4y²
  11	(x + 4)²
  12	(x − 6)²
  13	(2x + 5)²
  14	(3a − 4)²
  15	(4x + 5y)²
  16	x = −3
  17	x = 3/2
  18	10404
  19	The expansion must include the middle term 10x
  20	Completing the square rewrites quadratic expressions using perfect square identities

   

  Extended Challenge

  1. Expand (4x + 7)².

  2. Expand (5a − 3b)².

  3. Factor 25x² + 70x + 49.

  4. Factor 36m² − 60mn + 25n².

  5. Factor completely: 3x² + 18x + 27.

  6. Solve: 9x² + 30x + 25 = 0.

  7. Rewrite x² + 8x + 7 by completing the square.

  8. Rewrite x² − 10x + 16 by completing the square.

  9. Calculate 997² using a perfect square identity.

  10. Explain why the graph of y = (x − 4)² touches the x-axis at exactly one point.

  Extended Challenge Answers

  Question	Answer
  1	16x² + 56x + 49
  2	25a² − 30ab + 9b²
  3	(5x + 7)²
  4	(6m − 5n)²
  5	3(x + 3)²
  6	x = −5/3
  7	(x + 4)² − 9
  8	(x − 5)² − 9
  9	994009
  10	The squared expression is zero only when x = 4, giving a repeated root

   

  Final Investigation: Designing a Square Garden

  A school is designing a square garden.

  The original side length is x metres.

  The school plans to increase each side length by 3 metres.

  Part A: Creating an Expression

  1. Write an expression for the original area.

  2. Write an expression for the new side length.

  3. Write an expression for the new area.

  4. Expand the expression using a perfect square identity.

  5. Explain the meaning of each term in the expanded expression.

  Part B: Creating a Table

  6. Calculate the new garden area when x = 2, 3, 4, 5, and 6.

  7. Organise the results in a table.

  8. Calculate the increase in area for each additional metre of original side length.

  9. Determine whether the relationship is linear or nonlinear.

  10. Explain your conclusion.

  Part C: Using the Identity

  11. Write the new area as (x + 3)².

  12. Expand the expression.

  13. Identify the first square term.

  14. Identify the middle term.

  15. Identify the final square term.

  16. Explain why the middle term is 6x rather than 3x.

  Part D: Solving a Practical Problem

  17. The new garden has an area of 100 m². Write an equation.

  18. Solve the equation.

  19. Determine the original side length.

  20. Calculate the original area.

  Investigation Answer Check

  Original Area

  A₁ = x²

  New Side Length

  x + 3

  New Area

  A₂ = (x + 3)²

  Expand:

  A₂ = x² + 6x + 9

  Completed Table

  Original Side Length (m)	New Side Length (m)	New Area (m²)
  2	5	25
  3	6	36
  4	7	49
  5	8	64
  6	9	81

  The increases in area are:

  11, 13, 15, 17 m²

  The differences are not constant.

  Therefore, the relationship is nonlinear.

  Solving for an Area of 100 m²

  (x + 3)² = 100

  Take square roots:

  x + 3 = ±10

  Therefore:

  x = 7 or x = −13

  Since x represents a physical side length, the negative solution is rejected.

  Original side length = 7 m

  Original area:

  7² = 49 m²

  New side length:

  7 + 3 = 10 m

  New area:

  10² = 100 m²

  Final Reflection

  Perfect square trinomials are important algebraic patterns that connect expansion, factorisation, quadratic equations, and geometry.

  By recognising the relationship between the first term, middle term, and last term, we can identify and factor these expressions efficiently.

  The two essential identities are:

  (a + b)² = a² + 2ab + b²

  (a − b)² = a² − 2ab + b²

  Understanding these identities provides a strong foundation for more advanced algebra, including completing the square, solving quadratic equations, and analysing parabolic graphs.

3. Expanding Binomials

Expanding Binomials

Introduction

Algebra often involves expressions containing two or more terms. Learning how to multiply these expressions is an essential skill for working with polynomials, solving equations, and understanding quadratic functions.

A binomial is an algebraic expression containing two terms.

Examples include:

• x + 3

• x − 5

• 2x + 7

• 3a − 4b

When two binomials are multiplied, we use a process called expansion.

For example:

(x + 2)(x + 3)

Expanding this expression gives:

x² + 5x + 6

The expanded expression is equivalent to the original product, but it is written as a sum of terms rather than a multiplication of factors.

Expanding binomials is important because it allows us to:

• Simplify algebraic expressions.

• Develop quadratic equations.

• Understand polynomial multiplication.

• Calculate areas of geometric shapes.

• Verify factorisation.

• Solve practical problems involving changing quantities.

What Is a Binomial?

A binomial is a polynomial containing exactly two terms.

Examples:

x + 4

Terms: x and 4

3x − 7

Terms: 3x and −7

2a + 5b

Terms: 2a and 5b

x² − 9

Terms: x² and −9

A binomial may contain variables, coefficients, constants, and powers.

Important Observation

A binomial contains two terms, but those terms do not have to be simple numbers or variables.

Understanding Multiplication of Binomials

Consider:

(x + 2)(x + 3)

This means:

(x + 2) multiplied by (x + 3).

Each term in the first binomial must multiply each term in the second binomial.

There are four individual products:

• x × x = x²

• x × 3 = 3x

• 2 × x = 2x

• 2 × 3 = 6

Add the results:

x² + 3x + 2x + 6

Combine like terms:

x² + 5x + 6

Important Observation

Multiplying two binomials normally produces four intermediate products, although some terms can be combined.

The Distributive Property

The distributive property is the foundation of expanding algebraic expressions.

For numbers:

a(b + c) = ab + ac

For example:

3(4 + 5) = 3 × 4 + 3 × 5

= 12 + 15

= 27

The same rule applies to algebra.

Example

Expand:

2x(x + 4)

Distribute 2x:

2x × x + 2x × 4

= 2x² + 8x

Important Observation

Every term inside the brackets must be multiplied by the term outside.

The Double Distributive Property

When multiplying two binomials, we apply the distributive property twice.

Worked Example

Expand:

(x + 2)(x + 5)

Step 1: Distribute the First Term

x(x + 5) = x² + 5x

Step 2: Distribute the Second Term

2(x + 5) = 2x + 10

Step 3: Combine

x² + 5x + 2x + 10

Step 4: Collect Like Terms

x² + 7x + 10

This method works for all products of binomials.

The Area Model for Expanding Binomials

Multiplication of binomials can also be understood geometrically.

Imagine a rectangle with side lengths:

x + 3 and x + 2

Its area is:

A = (x + 3)(x + 2)

We can divide the rectangle into four smaller regions.

The total area is:

A = x² + 3x + 2x + 6

Combine like terms:

A = x² + 5x + 6

Therefore:

(x + 3)(x + 2) = x² + 5x + 6

Important Observation

The area model helps explain why each term in one binomial must multiply every term in the other binomial.

The FOIL Method

FOIL is a useful memory aid for multiplying two binomials.

FOIL stands for:

• F — First: Multiply the first terms.

• O — Outer: Multiply the outer terms.

• I — Inner: Multiply the inner terms.

• L — Last: Multiply the last terms.

Worked Example

Expand:

(x + 4)(x + 7)

First:

x × x = x²

Outer:

x × 7 = 7x

Inner:

4 × x = 4x

Last:

4 × 7 = 28

Combine:

x² + 7x + 4x + 28

Final Answer: x² + 11x + 28

Important Observation

FOIL is a convenient method for multiplying two binomials. The distributive property is more general and also works for expressions containing more than two terms.

Organising Terms Systematically

A systematic approach reduces errors.

Worked Example

Expand:

(2x + 3)(x + 5)

Add:

2x² + 10x + 3x + 15

Combine like terms:

2x² + 13x + 15

Important Observation

Writing all four products before simplifying helps prevent missing terms.

Expanding Binomials with Positive Terms

Worked Example

Expand:

(x + 6)(x + 2)

First:

x × x = x²

Outer:

x × 2 = 2x

Inner:

6 × x = 6x

Last:

6 × 2 = 12

Combine:

x² + 2x + 6x + 12

Final Answer: x² + 8x + 12

Another Example

Expand:

(3x + 4)(2x + 5)

First:

3x × 2x = 6x²

Outer:

3x × 5 = 15x

Inner:

4 × 2x = 8x

Last:

4 × 5 = 20

Combine:

6x² + 23x + 20

Expanding Binomials with Negative Terms

Negative signs require careful attention.

Worked Example

Expand:

(x − 3)(x + 5)

First:

x × x = x²

Outer:

x × 5 = 5x

Inner:

−3 × x = −3x

Last:

−3 × 5 = −15

Combine:

x² + 5x − 3x − 15

Final Answer: x² + 2x − 15

Important Observation

Treat subtraction as addition of a negative term.

This helps avoid sign errors.

Expanding Two Binomials with Negative Terms

Worked Example

Expand:

(x − 4)(x − 6)

First:

x × x = x²

Outer:

x × (−6) = −6x

Inner:

−4 × x = −4x

Last:

−4 × (−6) = +24

Combine:

x² − 6x − 4x + 24

Final Answer: x² − 10x + 24

Important Observation

The product of two negative numbers is positive.

Sign Rules in Binomial Expansion

Remember the multiplication rules:

Example

Expand:

(2x − 5)(3x − 4)

Products:

2x × 3x = 6x²

2x × (−4) = −8x

−5 × 3x = −15x

−5 × (−4) = +20

Combine:

6x² − 8x − 15x + 20

Final Answer: 6x² − 23x + 20

Combining Like Terms

After expansion, some terms can be combined.

Like terms have the same variables raised to the same powers.

Examples:

• 3x and 7x are like terms.

• 5x² and −2x² are like terms.

• 4ab and 9ab are like terms.

However:

• x and x² are not like terms.

• x and y are not like terms.

• xy and x² are not like terms.

Worked Example

Simplify:

3x² + 5x + 2x² − 7x + 4

Group like terms:

(3x² + 2x²) + (5x − 7x) + 4

Simplify:

5x² − 2x + 4

Important Observation

Only terms with identical variable parts can be combined by addition or subtraction.

Expanding Binomials with Different Variables

Binomials may contain more than one variable.

Worked Example

Expand:

(a + b)(a + 3b)

First:

a × a = a²

Outer:

a × 3b = 3ab

Inner:

b × a = ab

Last:

b × 3b = 3b²

Combine:

a² + 3ab + ab + 3b²

Final Answer: a² + 4ab + 3b²

Another Example

Expand:

(2x − y)(3x + 4y)

Products:

2x × 3x = 6x²

2x × 4y = 8xy

−y × 3x = −3xy

−y × 4y = −4y²

Combine:

6x² + 5xy − 4y²

Expanding Binomials Containing Powers

The terms inside a binomial may already contain powers.

Worked Example

Expand:

(x² + 3)(x + 4)

First:

x² × x = x³

Outer:

x² × 4 = 4x²

Inner:

3 × x = 3x

Last:

3 × 4 = 12

Therefore:

x³ + 4x² + 3x + 12

Worked Example

Expand:

(2x² − 3x)(x + 5)

Products:

2x² × x = 2x³

2x² × 5 = 10x²

−3x × x = −3x²

−3x × 5 = −15x

Combine:

2x³ + 7x² − 15x

Important Observation

When multiplying powers with the same base, add their exponents.

For example:

x² × x³ = x⁵

The General Binomial Product

The distributive property can be expressed using a general formula.

(a + b)(c + d) = ac + ad + bc + bd

This formula works for numbers, variables, and algebraic expressions.

Example

Expand:

(2x + 5)(3x + 7)

Using the general formula:

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

= 6x² + 14x + 15x + 35

= 6x² + 29x + 35

Important Observation

FOIL is simply a specific way of applying this general distributive pattern.

Using an Algebraic Grid

An algebraic grid provides another systematic method for multiplication.

Worked Example

Expand:

(2x + 3)(x + 4)

Add the four products:

2x² + 8x + 3x + 12

Combine:

2x² + 11x + 12

Advantages of the Grid Method

• Organises every multiplication.

• Reduces the chance of missing terms.

• Makes signs easier to track.

• Extends naturally to larger polynomials.

Important Observation

The grid method and FOIL produce identical results because both use the distributive property.

Special Case: Squaring a Binomial

Sometimes the two binomials are identical.

Example

(x + 4)(x + 4)

This can be written:

(x + 4)²

Expand:

x² + 4x + 4x + 16

= x² + 8x + 16

This is a perfect square trinomial.

General Identity

(a + b)² = a² + 2ab + b²

Another Example

(x − 5)² = x² − 10x + 25

Important Observation

The square of a binomial includes a middle term. It is not simply the sum of the squares of its terms.

Special Case: Conjugate Binomials

Conjugate binomials contain the same terms but opposite signs.

For example:

(x + 4)(x − 4)

Expand:

x² − 4x + 4x − 16

The middle terms cancel.

= x² − 16

This is called the difference of two squares.

General Identity

(a + b)(a − b) = a² − b²

Another Example

(3x + 5)(3x − 5)

= 9x² − 25

Important Observation

Recognising conjugate binomials allows us to expand without writing all four intermediate products.

Comparing Common Expansion Patterns

These patterns are useful shortcuts, but they all come from the distributive property.

Verifying Expansions by Factorisation

Expansion and factorisation are inverse operations.

Worked Example

Expand:

(x + 3)(x + 5)

Result:

x² + 8x + 15

To verify, factor the expanded expression.

We need two numbers that:

• Multiply to 15.

• Add to 8.

The numbers are 3 and 5.

Therefore:

x² + 8x + 15 = (x + 3)(x + 5)

The expansion is correct.

Important Observation

Factoring the result back into the original binomials is one way to verify an expansion.

Verifying Expansions by Substitution

A numerical check can help detect mistakes.

Worked Example

Verify:

(x + 2)(x + 6) = x² + 8x + 12

Choose x = 3.

Original Expression

(3 + 2)(3 + 6)

= 5 × 9

= 45

Expanded Expression

3² + 8(3) + 12

= 9 + 24 + 12

= 45

Both expressions produce the same result.

Important Observation

Substitution is useful for checking an expansion, but a single numerical test does not prove an identity.

Algebraic expansion provides a complete verification.

Verifying with a Second Method

A particularly reliable strategy is to expand using one method and check using another.

Example

Expand:

(3x − 2)(2x + 5)

Using FOIL:

6x² + 15x − 4x − 10

= 6x² + 11x − 10

Check with an algebraic grid:

The same four terms appear.

Therefore, the expansion is correct.

Applications in Geometry

Binomial multiplication is useful for calculating the areas of rectangles.

Worked Example

A rectangular garden has:

Length = x + 7 metres

Width = x + 3 metres

Find an expression for its area.

Area = Length × Width

A = (x + 7)(x + 3)

Expand:

A = x² + 3x + 7x + 21

A = x² + 10x + 21

Interpretation

The area depends on the value of x.

If x = 5:

A = 5² + 10(5) + 21

= 25 + 50 + 21

A = 96 m²

Applications in Number Calculations

Binomial multiplication can simplify mental arithmetic.

Worked Example: 23 × 14

Rewrite:

23 = 20 + 3

14 = 10 + 4

Therefore:

23 × 14 = (20 + 3)(10 + 4)

Expand:

= 200 + 80 + 30 + 12

= 322

Worked Example: 48 × 52

Rewrite:

48 = 50 − 2

52 = 50 + 2

Apply the difference-of-squares identity:

(50 − 2)(50 + 2)

= 50² − 2²

= 2500 − 4

= 2496

Important Observation

Recognising useful binomial patterns can make calculations faster and more efficient.

Applications in Science

Binomial expansion appears when scientific equations contain sums or differences.

Example: Kinetic Energy

The kinetic energy of an object is:

Eₖ = ½mv²

Suppose velocity is:

v = u + at

Substitute:

Eₖ = ½m(u + at)²

Expand:

(u + at)² = u² + 2uat + a²t²

Therefore:

Eₖ = ½mu² + muat + ½ma²t²

This expression shows how kinetic energy depends on initial velocity, acceleration, and time when acceleration is constant.

Important Observation

Algebraic expansion is essential for rearranging and analysing many scientific equations.

Applications in Quadratic Functions

Multiplying two linear factors often produces a quadratic expression.

Example

Consider:

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

Expand:

y = x² + 6x + 8

This is a quadratic function.

Its graph is a parabola.

The original factored form shows that the function equals zero when:

x = −2 or x = −4

The expanded form shows the coefficients of the quadratic.

Important Observation

Factored and expanded forms provide different useful information about the same quadratic function.

Common Mistakes

Forgetting One of the Four Products

Incorrect:

(x + 2)(x + 3) = x² + 6

Correct:

(x + 2)(x + 3) = x² + 5x + 6

Multiplying Only the First and Last Terms

Every term in the first binomial must multiply every term in the second binomial.

Incorrect Sign Rules

Incorrect:

(x − 4)(x − 2) = x² − 6x − 8

Correct:

(x − 4)(x − 2) = x² − 6x + 8

Combining Unlike Terms

Incorrect:

x² + 3x = 4x²

Correct:

x² + 3x cannot be combined into one term.

Forgetting Exponent Rules

Incorrect:

x × x = 2x

Correct:

x × x = x²

Forgetting the Middle Term When Squaring

Incorrect:

(x + 5)² = x² + 25

Correct:

(x + 5)² = x² + 10x + 25

Not Simplifying Completely

An expansion should normally be simplified by combining all like terms.

Failing to Verify

Use substitution, factorisation, or a second multiplication method to check your result.

Key Terms

• Binomial: Algebraic expression containing two terms.

• Polynomial: Algebraic expression formed from terms containing variables with nonnegative integer powers.

• Expansion: Rewriting a product as a sum of terms.

• Distributive Property: Rule that distributes multiplication across addition or subtraction.

• Double Distribution: Applying the distributive property to multiply two binomials.

• FOIL: Method involving First, Outer, Inner, and Last products.

• Term: Part of an algebraic expression separated by addition or subtraction.

• Coefficient: Numerical factor multiplying a variable.

• Constant: Term without a variable.

• Like Terms: Terms with identical variable parts and exponents.

• Simplification: Rewriting an expression in an equivalent, more convenient form.

• Algebraic Grid: Table used to organise polynomial multiplication.

• Area Model: Geometric representation of multiplication using rectangular regions.

• Perfect Square Trinomial: Trinomial formed by squaring a binomial.

• Conjugate Binomials: Binomials with the same terms and opposite signs.

• Difference of Squares: Identity (a + b)(a − b) = a² − b².

• Factorisation: Rewriting an expression as a product.

• Verification: Checking that a mathematical result is correct.

• Quadratic Expression: Polynomial expression with highest variable power 2.

• Quadratic Function: Function represented by a second-degree polynomial.

Key Takeaways

• A binomial contains two terms.

• Expanding means multiplying algebraic expressions and rewriting the result as a sum.

• Every term in one binomial must multiply every term in the other.

• FOIL is a useful method for multiplying two binomials.

• The distributive property is the foundation of binomial expansion.

• Algebraic grids help organise multiplication systematically.

• Negative signs must be handled carefully.

• Like terms should be combined after expansion.

• When multiplying powers with the same base, add the exponents.

• Squaring a binomial produces a perfect square trinomial.

• Multiplying conjugate binomials produces a difference of squares.

• Expansion and factorisation are inverse processes.

• Substitution can help check an expansion.

• Binomial expansion is useful in geometry, arithmetic, science, and quadratic functions.

Check Your Understanding

Basic Expansion

1. What is a binomial?

2. What does FOIL stand for?

3. Expand (x + 2)(x + 4).

4. Expand (x + 3)(x + 7).

5. Expand (x + 5)(x + 6).

Negative Terms

6. Expand (x − 2)(x + 5).

7. Expand (x − 4)(x − 3).

8. Expand (x + 6)(x − 2).

9. Expand (2x − 3)(x + 4).

10. Expand (3x − 2)(2x − 5).

Coefficients and Variables

11. Expand (2x + 3)(3x + 4).

12. Expand (4x + 1)(2x + 5).

13. Expand (a + b)(a + 2b).

14. Expand (2x − y)(x + 3y).

15. Expand (x² + 2)(x + 3).

Special Patterns and Verification

16. Expand (x + 5)².

17. Expand (x − 7)².

18. Expand (x + 4)(x − 4).

19. Explain why (x + 3)² is not equal to x² + 9.

20. Describe two methods for verifying an expansion.

Answer Key

Extended Challenge

1. Expand (4x + 7)(3x + 2).

2. Expand (5a − 3b)(2a + 4b).

3. Expand (x² + 3x)(2x − 5).

4. Expand (2x² − 3)(x² + 4).

5. Expand (3x − 4)².

6. Expand (5a + 2b)².

7. Expand (4x + 7)(4x − 7).

8. Verify by expansion that (x + 2)(x + 6) = x² + 8x + 12.

9. A rectangle has dimensions 2x + 3 and x + 5. Find an expression for its area.

10. Explain why multiplying two first-degree binomials usually produces a quadratic expression.

Extended Challenge Answers

Final Investigation: Designing a Rectangular Garden

A school plans to create a rectangular garden.

The length is:

x + 6 metres

The width is:

x + 4 metres

Part A: Building the Expression

1. Write an expression for the garden's area.

2. Expand the expression using FOIL.

3. Simplify the result.

4. Expand the expression using an algebraic grid.

5. Compare the two results.

Part B: Creating a Table

6. Calculate the area when x = 1, 2, 3, 4, and 5.

7. Organise the results in a table.

8. Determine how the area changes as x increases.

9. Explain whether the relationship is linear or nonlinear.

10. Explain why the expanded expression contains x².

Part C: Applying the Model

11. Calculate the garden's area when x = 6.

12. Calculate the garden's area when x = 10.

13. Find the difference between these two areas.

14. Explain how increasing x changes both dimensions.

15. Explain why the area increases faster as x becomes larger.

Part D: Verification

16. Choose one value of x.

17. Calculate the area directly using length × width.

18. Calculate the area using the expanded expression.

19. Compare the results.

20. Explain why the two methods must agree.

Investigation Answer Check

Original Expression

A = (x + 6)(x + 4)

Expansion

A = x² + 4x + 6x + 24

A = x² + 10x + 24

Completed Table

The area increases by:

13, 15, 17, 19 m²

The differences are not constant.

Therefore, the relationship is nonlinear.

When x = 6

A = 6² + 10(6) + 24

= 36 + 60 + 24

A = 120 m²

When x = 10

A = 10² + 10(10) + 24

= 100 + 100 + 24

A = 224 m²

Difference in Area

224 − 120 = 104 m²

Verification

When x = 6:

Length = 12 m

Width = 10 m

Area = 12 × 10

Area = 120 m²

This agrees with the expanded expression.

Final Reflection

Expanding binomials is an essential algebraic skill that connects multiplication, simplification, factorisation, geometry, and quadratic functions.

By applying the distributive property systematically, we can multiply algebraic expressions accurately and simplify the results.

FOIL, algebraic grids, and area models are different ways of organising the same mathematical process.

The most important principle is that every term in one binomial must multiply every term in the other binomial, and all like terms must then be combined correctly.

4. Factoring Quadratic Expressions

Factoring Quadratic Expressions

Introduction

In algebra, we often need to rewrite expressions in different forms to make them easier to understand or solve.

One of the most important algebraic techniques is factorisation, which involves rewriting an expression as a product of simpler expressions.

Consider:

x² + 5x + 6

This quadratic expression can be written as:

(x + 2)(x + 3)

Both expressions are equivalent, but the factored form is especially useful for solving equations, simplifying algebraic fractions, and identifying important features of quadratic functions.

Factorisation is the reverse of expansion.

For example:

Expansion:

(x + 2)(x + 3) = x² + 5x + 6

Factorisation:

x² + 5x + 6 = (x + 2)(x + 3)

Learning to recognise factor pairs and algebraic patterns makes factorisation faster and more accurate.

What Is a Quadratic Expression?

A quadratic expression is a polynomial expression in one variable whose highest power is 2.

The general form is:

ax² + bx + c

where:

• a, b, and c are constants.

• a ≠ 0.

• x is the variable.

Examples include:

• x² + 7x + 10

• x² − 5x + 6

• 2x² + 7x + 3

• 3x² − 12x

• x² − 16

Understanding the Terms

In:

2x² + 5x + 3

• 2x² is the quadratic term.

• 5x is the linear term.

• 3 is the constant term.

Important Observation

Quadratic expressions contain an x² term, but the linear or constant term may be absent.

What Does Factoring Mean?

Factoring means expressing a quantity as a product of factors.

Numerical Example

12 = 3 × 4

The numbers 3 and 4 are factors of 12.

Algebraic Example

x² + 7x + 12 = (x + 3)(x + 4)

The expressions x + 3 and x + 4 are factors of the quadratic expression.

Why Factor?

Factoring is useful because it can help us:

• Solve quadratic equations.

• Simplify rational expressions.

• Find the zeros of quadratic functions.

• Analyse graphs.

• Recognise algebraic patterns.

• Solve practical problems.

Understanding Factor Pairs

A factor pair consists of two numbers whose product equals a given number.

Example: Factor Pairs of 12

Important Observation

Positive numbers can be produced by multiplying two positive numbers or two negative numbers.

Negative numbers are produced by multiplying numbers with opposite signs.

Factoring Quadratics with a Leading Coefficient of 1

The simplest quadratic expressions have the form:

x² + bx + c

We want to rewrite them as:

(x + m)(x + n)

Expanding:

(x + m)(x + n)

= x² + nx + mx + mn

= x² + (m + n)x + mn

Therefore, we need two numbers that:

• Add to b.

• Multiply to c.

General Rule

For x² + bx + c:

m + n = b

mn = c

Then:

x² + bx + c = (x + m)(x + n)

Factoring When Both Signs Are Positive

Worked Example: x² + 7x + 12

Find two numbers that:

• Multiply to 12.

• Add to 7.

Possible positive factor pairs:

1 and 12 → sum = 13

2 and 6 → sum = 8

3 and 4 → sum = 7

The correct pair is 3 and 4.

Therefore:

x² + 7x + 12 = (x + 3)(x + 4)

Verification

Expand:

(x + 3)(x + 4)

= x² + 4x + 3x + 12

= x² + 7x + 12

The factorisation is correct.

Factoring with a Negative Constant

When the constant term is negative, the factors must have opposite signs.

Worked Example: x² + 2x − 15

Find two numbers that:

• Multiply to −15.

• Add to 2.

Possible pairs:

5 and −3 → sum = 2

−5 and 3 → sum = −2

The correct pair is 5 and −3.

Therefore:

x² + 2x − 15 = (x + 5)(x − 3)

Verification

(x + 5)(x − 3)

= x² − 3x + 5x − 15

= x² + 2x − 15

Important Observation

When the constant is negative, the two binomial factors contain opposite signs.

Factoring with a Negative Middle Term

Worked Example: x² − 9x + 20

Find two numbers that:

• Multiply to 20.

• Add to −9.

Since the product is positive and the sum is negative, both numbers must be negative.

Possible pairs:

−1 and −20 → sum = −21

−2 and −10 → sum = −12

−4 and −5 → sum = −9

Therefore:

x² − 9x + 20 = (x − 4)(x − 5)

Verification

(x − 4)(x − 5)

= x² − 5x − 4x + 20

= x² − 9x + 20

Factoring When Both the Middle and Constant Terms Are Negative

Worked Example: x² − 4x − 21

Find two numbers that:

• Multiply to −21.

• Add to −4.

The correct pair is −7 and 3.

Therefore:

x² − 4x − 21 = (x − 7)(x + 3)

Verification

(x − 7)(x + 3)

= x² + 3x − 7x − 21

= x² − 4x − 21

Important Observation

The sign of the sum helps determine which factor has the larger absolute value.

A Systematic Factor-Pair Strategy

Consider:

x² + 11x + 24

Step 1: Identify b and c

b = 11

c = 24

Step 2: List Factor Pairs of 24

Step 3: Choose the Pair with the Correct Sum

3 and 8

Step 4: Write the Factors

(x + 3)(x + 8)

Step 5: Verify

(x + 3)(x + 8)

= x² + 11x + 24

Important Observation

A systematic factor-pair table is particularly helpful when the constant has several possible factor pairs.

Factoring Using an Area Model

Quadratic factorisation can also be represented geometrically.

Consider:

x² + 5x + 6

We can arrange algebra tiles to form a rectangle.

The area can be written as:

x² + 3x + 2x + 6

Combine like terms:

x² + 5x + 6

Therefore:

x² + 5x + 6 = (x + 3)(x + 2)

Important Observation

Factoring can be interpreted as finding the dimensions of a rectangle from its area expression.

Factoring Out a Greatest Common Factor

Before factoring a quadratic, always check whether all terms share a common factor.

Worked Example: 3x² + 12x

Both terms contain 3x.

Factor out 3x:

3x² + 12x = 3x(x + 4)

Verification

3x(x + 4)

= 3x² + 12x

Worked Example: 2x² + 10x + 12

All terms share a factor of 2.

Factor out 2:

2(x² + 5x + 6)

Now factor the quadratic:

x² + 5x + 6 = (x + 2)(x + 3)

Therefore:

2x² + 10x + 12 = 2(x + 2)(x + 3)

Important Observation

Factoring out the greatest common factor first often makes the remaining quadratic easier to factor.

Factoring Quadratics with Leading Coefficients Greater Than 1

Some quadratic expressions have the form:

ax² + bx + c

where a is greater than 1.

These require a slightly different strategy.

Worked Example: 2x² + 7x + 3

We want two binomials whose product gives the original expression.

Step 1: Multiply a and c

a × c = 2 × 3 = 6

Step 2: Find Two Numbers

We need two numbers that:

• Multiply to 6.

• Add to 7.

The numbers are 6 and 1.

Step 3: Split the Middle Term

2x² + 6x + x + 3

Step 4: Factor by Grouping

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

Step 5: Factor the Common Binomial

(2x + 1)(x + 3)

Verification

(2x + 1)(x + 3)

= 2x² + 6x + x + 3

= 2x² + 7x + 3

The AC Method

The previous strategy is often called the AC method.

For:

ax² + bx + c

follow these steps:

1. Multiply a × c.

2. Find two numbers whose product is ac and whose sum is b.

3. Split the middle term using those numbers.

4. Factor by grouping.

5. Factor out the common binomial.

Worked Example

Factor:

3x² + 11x + 6

a × c = 3 × 6 = 18

Find two numbers that multiply to 18 and add to 11.

The numbers are 9 and 2.

Split the middle term:

3x² + 9x + 2x + 6

Group:

3x(x + 3) + 2(x + 3)

Factor:

(3x + 2)(x + 3)

Factoring with Negative Coefficients

Worked Example

Factor:

2x² − 5x − 3

a × c = 2 × (−3) = −6

Find two numbers that:

• Multiply to −6.

• Add to −5.

The numbers are −6 and 1.

Split:

2x² − 6x + x − 3

Group:

2x(x − 3) + 1(x − 3)

Factor:

(2x + 1)(x − 3)

Verification

(2x + 1)(x − 3)

= 2x² − 6x + x − 3

= 2x² − 5x − 3

Special Case: Difference of Two Squares

Some quadratic expressions contain only two terms.

For example:

x² − 25

This is a difference of two squares.

Use the identity:

a² − b² = (a − b)(a + b)

Therefore:

x² − 25 = (x − 5)(x + 5)

Another Example

Factor:

9x² − 16

Recognise:

9x² = (3x)²

16 = 4²

Therefore:

9x² − 16 = (3x − 4)(3x + 4)

Important Observation

Recognising special patterns can make factorisation faster than using a general method.

Special Case: Perfect Square Trinomials

Some quadratic expressions are the squares of binomials.

Example

x² + 10x + 25

Recognise:

x² = (x)²

25 = 5²

Middle term:

2(x)(5) = 10x

Therefore:

x² + 10x + 25 = (x + 5)²

Another Example

4x² − 12x + 9

Recognise:

4x² = (2x)²

9 = 3²

Middle term:

−2(2x)(3) = −12x

Therefore:

4x² − 12x + 9 = (2x − 3)²

Recognising Unfactorable Expressions

Not every quadratic expression can be factored using integer coefficients.

Consider:

x² + 3x + 1

To factor over the integers, we need two integers that:

• Multiply to 1.

• Add to 3.

The integer factor pairs of 1 are:

1 and 1

−1 and −1

Their sums are 2 and −2.

Neither gives 3.

Therefore:

x² + 3x + 1 cannot be factored into binomials with integer coefficients.

However, it can be factored using irrational real numbers.

Important Observation

An expression that cannot be factored using integers may still be factorable over the real or complex numbers.

Using the Discriminant to Recognise Factorability

For a quadratic expression:

ax² + bx + c

the discriminant is:

D = b² − 4ac

The discriminant provides information about the roots of the corresponding quadratic equation.

Interpretation

• D > 0: Two distinct real roots.

• D = 0: One repeated real root.

• D < 0: No real roots.

For quadratic expressions with integer coefficients, a nonnegative perfect-square discriminant indicates that the expression can be factored into linear factors with rational coefficients.

Example

Consider:

x² + 3x + 1

a = 1

b = 3

c = 1

D = 3² − 4(1)(1)

D = 9 − 4

D = 5

Since 5 is not a perfect square, the expression cannot be factored into linear factors with rational coefficients.

Important Observation

The discriminant helps distinguish between expressions that factor neatly over the rational numbers and those that require irrational or complex roots.

Quadratics That Cannot Be Factored Over the Real Numbers

Consider:

x² + 4

The corresponding equation is:

x² + 4 = 0

x² = −4

There is no real number whose square equals −4.

Therefore, x² + 4 cannot be factored into real linear factors.

Over the complex numbers:

x² + 4 = (x − 2i)(x + 2i)

Important Observation

Whether an expression is factorable depends on the number system being used.

Verifying Factorisation by Expansion

One of the most reliable ways to check factorisation is to multiply the factors.

Worked Example

Verify:

x² + 9x + 20 = (x + 4)(x + 5)

Expand:

(x + 4)(x + 5)

= x² + 5x + 4x + 20

= x² + 9x + 20

The expanded expression matches the original.

Therefore, the factorisation is correct.

Verifying by Substitution

We can also substitute a value for x.

Example

Check:

x² + 7x + 10 = (x + 2)(x + 5)

Choose x = 3.

Original expression:

3² + 7(3) + 10

= 9 + 21 + 10

= 40

Factored expression:

(3 + 2)(3 + 5)

= 5 × 8

= 40

Both expressions give the same result.

Important Observation

Substitution is useful for detecting mistakes, but checking one value does not prove that two expressions are equivalent for all values.

Using Factorisation to Solve Quadratic Equations

Factoring is especially useful when solving equations.

Worked Example

Solve:

x² + 5x + 6 = 0

Factor:

(x + 2)(x + 3) = 0

Apply the zero-product property.

If a product equals zero, at least one factor must equal zero.

Therefore:

x + 2 = 0

or

x + 3 = 0

Solutions:

x = −2 or x = −3

Solving Quadratic Equations with Coefficients

Worked Example

Solve:

2x² + 7x + 3 = 0

Factor:

(2x + 1)(x + 3) = 0

Set each factor equal to zero.

2x + 1 = 0

2x = −1

x = −1/2

Or:

x + 3 = 0

x = −3

Final Answer

x = −1/2 or x = −3

Simplifying Algebraic Fractions

Factorisation is useful when simplifying rational expressions.

Worked Example

Simplify:

(x² + 5x + 6)/(x + 2)

Factor the numerator:

(x + 2)(x + 3)/(x + 2)

Cancel the common factor:

x + 3

However, the original denominator cannot equal zero.

Therefore:

Final Answer: x + 3, where x ≠ −2

Important Observation

When simplifying algebraic fractions, only common factors may be cancelled.

Simplifying More Complex Rational Expressions

Worked Example

Simplify:

(x² − 9)/(x² + 6x + 9)

Factor the numerator:

x² − 9 = (x − 3)(x + 3)

Factor the denominator:

x² + 6x + 9 = (x + 3)²

Therefore:

[(x − 3)(x + 3)]/[(x + 3)(x + 3)]

Cancel one common factor:

(x − 3)/(x + 3), where x ≠ −3

Important Observation

Restrictions from the original denominator must be preserved after cancellation.

Factoring and Quadratic Graphs

Factoring helps identify where a quadratic graph crosses the x-axis.

Consider:

y = x² − 5x + 6

Factor:

y = (x − 2)(x − 3)

The graph crosses the x-axis where y = 0.

Therefore:

x − 2 = 0 or x − 3 = 0

x = 2 or x = 3

Important Observation

Factoring a quadratic function makes its zeros easier to identify.

Applications in Geometry

Quadratic expressions often represent areas of rectangles.

Worked Example

A rectangular garden has an area of:

x² + 9x + 20

Find expressions for its length and width.

Step 1: Factor the Area Expression

Find two numbers that:

• Multiply to 20.

• Add to 9.

The numbers are 4 and 5.

Therefore:

x² + 9x + 20 = (x + 4)(x + 5)

Step 2: Interpret the Factors

Possible dimensions are:

Length = x + 5

Width = x + 4

Important Observation

Factoring an area expression can reveal possible dimensions of a rectangle.

Applications in Number Problems

Factorisation can help solve problems involving unknown numbers.

Worked Example

The product of two consecutive positive integers is 56.

Find the integers.

Let x = smaller integer.

The next integer is x + 1.

Therefore:

x(x + 1) = 56

Expand:

x² + x = 56

Rearrange:

x² + x − 56 = 0

Factor:

(x + 8)(x − 7) = 0

Therefore:

x = −8 or x = 7

Since the integers must be positive:

The numbers are 7 and 8.

Common Mistakes

Choosing factors with the correct product but incorrect sum

Both conditions must be satisfied.

Ignoring negative factor pairs

Negative numbers are often required when factoring expressions with negative coefficients.

Forgetting the greatest common factor

Always check whether a common factor can be removed first.

Confusing factoring with expansion

Factoring rewrites a sum as a product. Expansion rewrites a product as a sum.

Making sign errors

For example:

x² − 5x + 6 = (x − 2)(x − 3)

not (x + 2)(x + 3).

Assuming every quadratic factors over the integers

Some quadratics require irrational or complex roots.

Failing to factor completely

An expression may contain a common factor or another factorable pattern.

Cancelling terms instead of factors

Only common multiplicative factors may be cancelled in rational expressions.

Forgetting restrictions

When simplifying fractions, preserve excluded values from the original denominator.

Not verifying

Expand the factors to confirm that they reproduce the original expression.

Key Terms

• Quadratic Expression: Polynomial expression with highest variable power 2.

• Factorisation: Rewriting an expression as a product of factors.

• Factor: Quantity multiplied by another quantity to form a product.

• Factor Pair: Two numbers whose product equals a specified value.

• Coefficient: Numerical factor multiplying a variable.

• Constant Term: Term without a variable.

• Leading Coefficient: Coefficient of the highest-power term.

• Greatest Common Factor: Largest factor shared by all terms.

• Expansion: Multiplying factors to produce an equivalent expression.

• Distributive Property: Rule that distributes multiplication across addition or subtraction.

• AC Method: Factoring method involving the product of the leading coefficient and constant term.

• Factoring by Grouping: Rewriting and grouping terms to identify common factors.

• Perfect Square Trinomial: Trinomial that is the square of a binomial.

• Difference of Squares: Expression of the form a² − b².

• Zero-Product Property: If a product equals zero, at least one factor must equal zero.

• Quadratic Equation: Equation containing a quadratic expression.

• Root: Value that satisfies an equation.

• Zero of a Function: Input value that makes the function equal zero.

• Discriminant: Quantity b² − 4ac used to analyse quadratic roots.

• Rational Expression: Fraction containing algebraic expressions.

• Excluded Value: Value that makes a denominator equal to zero.

• Verification: Checking that a mathematical result is correct.

• Unfactorable over the Integers: Expression that cannot be written as a product of lower-degree polynomials with integer coefficients.

• Parabola: Graph of a quadratic function.

• x-Intercept: Point where a graph crosses or touches the x-axis.

Key Takeaways

• Factoring is the reverse of expansion.

• Quadratic expressions have the general form ax² + bx + c.

• For x² + bx + c, find two numbers that add to b and multiply to c.

• Positive constants may require two positive or two negative factors.

• Negative constants require factors with opposite signs.

• The greatest common factor should be removed first.

• The AC method can factor quadratics with leading coefficients greater than 1.

• Special patterns include perfect square trinomials and differences of squares.

• Not every quadratic factors over the integers.

• The discriminant helps identify the nature of quadratic roots.

• Factorisation can be verified by expansion.

• Factoring is useful for solving quadratic equations.

• Factoring can simplify algebraic fractions.

• Factored form helps identify x-intercepts of quadratic graphs.

• Factoring is useful in geometry, number problems, and other real-world applications.

Check Your Understanding

Understanding Factorisation

1. What does it mean to factor a quadratic expression?

2. What is the general form of a quadratic expression?

3. What two conditions must factor pairs satisfy when factoring x² + bx + c?

4. Why should the greatest common factor be checked first?

5. How can factorisation be verified?

Basic Factorisation

6. Factor x² + 5x + 6.

7. Factor x² + 9x + 20.

8. Factor x² − 7x + 12.

9. Factor x² + x − 12.

10. Factor x² − 2x − 15.

More Advanced Factorisation

11. Factor 2x² + 7x + 3.

12. Factor 3x² + 10x + 3.

13. Factor 2x² − 5x − 3.

14. Factor completely: 2x² + 14x + 24.

15. Factor x² − 49.

Applications

16. Solve x² + 7x + 12 = 0.

17. Solve x² − 5x + 6 = 0.

18. Simplify (x² + 5x + 6)/(x + 2), stating restrictions.

19. Explain why x² + 3x + 1 cannot be factored using integer coefficients.

20. Explain how factorisation helps identify the x-intercepts of a quadratic graph.

Answer Key

Extended Challenge

1. Factor completely: 4x² + 20x + 24.

2. Factor: 6x² + 11x + 3.

3. Factor: 8x² − 2x − 3.

4. Factor completely: 3x² − 27.

5. Factor: 9x² − 24x + 16.

6. Solve: 2x² + 9x + 7 = 0.

7. Solve: 3x² − 13x + 4 = 0.

8. Simplify (x² − 16)/(x² + 8x + 16), stating restrictions.

9. Determine whether x² + 4x + 7 can be factored into real linear factors.

10. A rectangle has area x² + 13x + 42. Find possible expressions for its dimensions.

Extended Challenge Answers

Final Investigation: Designing a Rectangular Garden

A school is designing a rectangular garden.

The area is represented by:

A = x² + 11x + 30

The school wants to determine possible dimensions of the garden.

Part A: Understanding the Expression

1. Identify the quadratic, linear, and constant terms.

2. Identify the coefficient of x².

3. Identify the coefficient of x.

4. Identify the constant term.

5. Explain why this is a quadratic expression.

Part B: Finding Factor Pairs

6. List the positive factor pairs of 30.

7. Calculate the sum of each pair.

8. Identify the pair that adds to 11.

9. Factor the quadratic expression.

10. Explain how the factors represent possible dimensions.

Part C: Verifying

11. Expand the factored expression.

12. Combine like terms.

13. Compare the result with the original expression.

14. Explain why the factorisation is correct.

Part D: Applying the Model

15. Calculate the garden's dimensions when x = 2.

16. Calculate the area directly using length × width.

17. Calculate the area using the quadratic expression.

18. Compare the results.

19. Explain why both methods agree.

20. Describe how the area changes as x increases.

Investigation Answer Check

Original Expression

A = x² + 11x + 30

Finding Factor Pairs

We need two numbers that:

• Multiply to 30.

• Add to 11.

The numbers are 5 and 6.

Therefore:

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

Dimensions

Possible dimensions are:

Length = x + 6

Width = x + 5

Verification

(x + 5)(x + 6)

= x² + 6x + 5x + 30

= x² + 11x + 30

The factorisation is correct.

When x = 2

Length = 2 + 6 = 8 m

Width = 2 + 5 = 7 m

Area = 8 × 7

Area = 56 m²

Using the quadratic expression:

A = 2² + 11(2) + 30

= 4 + 22 + 30

A = 56 m²

Both methods agree.

Final Reflection

Factoring quadratic expressions is an important algebraic skill that connects multiplication, polynomial structure, equation solving, and graph analysis.

By identifying suitable factor pairs, checking for common factors, recognising special patterns, and verifying results through expansion, we can factor quadratic expressions accurately and efficiently.

Factorisation also helps us solve equations, simplify rational expressions, and interpret real-world problems involving areas and unknown quantities.

The most important principle is that factoring rewrites a quadratic expression as a product of simpler factors while preserving its mathematical value.

5. Algebraic Patterns and Identities

Introduction

Algebra is full of patterns. When we recognise these patterns, we can often simplify expressions, perform calculations, and solve problems much more efficiently.

For example, consider:

(x + 5)²

We could multiply the two binomials using the distributive property. However, recognising a familiar algebraic pattern allows us to expand immediately:

(x + 5)² = x² + 10x + 25

Similarly, consider:

(x + 7)(x − 7)

Rather than multiplying every term, we can recognise a difference-of-squares pattern:

(x + 7)(x − 7) = x² − 49

These relationships are examples of algebraic identities.

An algebraic identity is an equation that remains true for every permitted value of its variables.

Understanding identities is important because they help us:

• Expand algebraic expressions efficiently.

• Factor polynomials.

• Simplify complicated expressions.

• Recognise mathematical patterns.

• Solve equations.

• Perform mental arithmetic.

• Understand geometric relationships.

• Apply algebra in science, engineering, and other fields.

What Is an Algebraic Identity?

An algebraic identity is an equation that is true for every value of its variables for which both sides are defined.

Consider:

2(x + 3) = 2x + 6

This equation is true regardless of the value of x.

For x = 1:

Left side = 2(1 + 3) = 8

Right side = 2(1) + 6 = 8

For x = 5:

Left side = 2(5 + 3) = 16

Right side = 2(5) + 6 = 16

For x = −2:

Left side = 2(−2 + 3) = 2

Right side = 2(−2) + 6 = 2

The two expressions always produce the same value.

This is because the distributive property guarantees their equivalence.

Identity Versus Equation

Not every equation is an identity.

Consider:

x + 3 = 7

This equation is true only when x = 4.

It is not an identity.

Compare:

Important Observation

An identity describes a relationship that is always true, while an ordinary equation may be true only for particular values.

Why Are Algebraic Identities Useful?

Algebraic identities allow us to recognise the structure of expressions.

For example:

(a + b)² = a² + 2ab + b²

Once we know this identity, we can apply it to many different expressions.

Examples:

(x + 3)² = x² + 6x + 9

(2x + 5)² = 4x² + 20x + 25

(3a + 4b)² = 9a² + 24ab + 16b²

The variables a and b in the identity can represent numbers, variables, or entire algebraic expressions.

Important Observation

Learning an identity means understanding a general relationship, not memorising a separate rule for every example.

The Distributive Identity

One of the most fundamental algebraic identities is the distributive property.

a(b + c) = ab + ac

Similarly:

a(b − c) = ab − ac

This identity allows multiplication to be distributed across addition or subtraction.

Worked Example

Expand:

4(x + 7)

Apply the distributive property:

4(x + 7) = 4x + 28

Worked Example

Expand:

−3(2x − 5)

Distribute −3:

−3(2x) + (−3)(−5)

= −6x + 15

Important Observation

The distributive property is the foundation of many more advanced algebraic identities.

The Square of a Sum Identity

One of the most frequently used identities is:

(a + b)² = a² + 2ab + b²

This identity describes the square of a binomial containing addition.

Worked Example

Expand:

(x + 6)²

Identify:

a = x

b = 6

Apply the identity:

(x + 6)² = x² + 2(x)(6) + 6²

Simplify:

(x + 6)² = x² + 12x + 36

Another Example

Expand:

(3x + 2)²

Identify:

a = 3x

b = 2

Substitute:

(3x + 2)² = (3x)² + 2(3x)(2) + 2²

= 9x² + 12x + 4

Important Observation

The middle term is twice the product of the two terms in the original binomial.

Understanding the Square of a Sum Geometrically

Consider a square with side length a + b.

Its area is:

A = (a + b)²

The square can be divided into four smaller regions.

The total area is:

a² + ab + ab + b²

Combine like terms:

a² + 2ab + b²

Therefore:

(a + b)² = a² + 2ab + b²

Important Observation

The identity works because the four smaller regions exactly fill the larger square without gaps or overlaps.

The Square of a Difference Identity

The second important identity is:

(a − b)² = a² − 2ab + b²

This identity describes the square of a binomial containing subtraction.

Worked Example

Expand:

(x − 7)²

Identify:

a = x

b = 7

Apply the identity:

(x − 7)² = x² − 2(x)(7) + 7²

Simplify:

(x − 7)² = x² − 14x + 49

Another Example

Expand:

(4x − 3)²

Identify:

a = 4x

b = 3

Substitute:

(4x − 3)² = (4x)² − 2(4x)(3) + 3²

= 16x² − 24x + 9

Important Observation

The final term is positive because squaring a negative quantity produces a nonnegative result.

Why the Square of a Difference Works

We can prove the identity using multiplication.

Start with:

(a − b)²

Rewrite:

(a − b)(a − b)

Apply the distributive property:

= a² − ab − ab + b²

Combine the middle terms:

= a² − 2ab + b²

Therefore:

(a − b)² = a² − 2ab + b²

This proof works for all real values of a and b.

Important Observation

Algebraic identities can be proved using fundamental operations rather than relying only on examples.

Comparing the Two Square Identities

The square of a sum and square of a difference are closely related.

Example

(x + 5)² = x² + 10x + 25

(x − 5)² = x² − 10x + 25

Only the sign of the middle term changes.

Important Observation

Recognising the sign pattern helps prevent mistakes when expanding squared binomials.

The Difference of Two Squares Identity

Another extremely useful identity is:

a² − b² = (a − b)(a + b)

This identity allows us to factor the difference between two squared quantities.

Worked Example

Factor:

x² − 49

Recognise:

x² = (x)²

49 = 7²

Apply the identity:

x² − 49 = (x − 7)(x + 7)

Worked Example

Factor:

25x² − 16

Recognise:

25x² = (5x)²

16 = 4²

Therefore:

25x² − 16 = (5x − 4)(5x + 4)

Important Observation

The difference-of-squares identity applies when two squared expressions are separated by subtraction.

Why the Difference of Squares Identity Works

Start with:

(a − b)(a + b)

Expand:

= a² + ab − ab − b²

The middle terms cancel:

+ab − ab = 0

Therefore:

(a − b)(a + b) = a² − b²

The identity is true because the cross-products are equal in magnitude and opposite in sign.

Geometric Interpretation

Imagine a large square of side length a and a smaller square of side length b.

When a ≥ b ≥ 0, the difference in their areas is:

a² − b²

The remaining area can be rearranged into a rectangle with dimensions:

a − b and a + b

Therefore:

a² − b² = (a − b)(a + b)

The Product of Two Binomials with a Common Term

Another useful identity is:

(x + a)(x + b) = x² + (a + b)x + ab

This identity is especially useful when expanding or factoring quadratic expressions.

Worked Example

Expand:

(x + 3)(x + 8)

Apply the identity:

= x² + (3 + 8)x + (3)(8)

= x² + 11x + 24

Another Example

Expand:

(x − 4)(x + 7)

Treat a = −4 and b = 7.

= x² + (−4 + 7)x + (−4)(7)

= x² + 3x − 28

Important Observation

The coefficient of x is the sum of the two constants, while the constant term is their product.

The Cube of a Sum

Algebraic identities also apply to powers greater than two.

One important identity is:

(a + b)³ = a³ + 3a²b + 3ab² + b³

Worked Example

Expand:

(x + 2)³

Identify:

a = x

b = 2

Apply the identity:

(x + 2)³ = x³ + 3(x²)(2) + 3(x)(2²) + 2³

Simplify:

= x³ + 6x² + 12x + 8

Important Observation

The coefficients 1, 3, 3, 1 form a pattern that also appears in Pascal's triangle.

The Cube of a Difference

The corresponding identity is:

(a − b)³ = a³ − 3a²b + 3ab² − b³

Worked Example

Expand:

(x − 3)³

Substitute:

= x³ − 3(x²)(3) + 3(x)(3²) − 3³

Simplify:

x³ − 9x² + 27x − 27

Important Observation

The signs alternate in the cube-of-a-difference identity.

The Sum and Difference of Cubes

Two further identities are useful for factoring cubic expressions.

Sum of Cubes

a³ + b³ = (a + b)(a² − ab + b²)

Difference of Cubes

a³ − b³ = (a − b)(a² + ab + b²)

Worked Example: Sum of Cubes

Factor:

x³ + 8

Recognise:

8 = 2³

Apply the identity:

x³ + 8 = (x + 2)(x² − 2x + 4)

Worked Example: Difference of Cubes

Factor:

x³ − 27

Recognise:

27 = 3³

Apply the identity:

x³ − 27 = (x − 3)(x² + 3x + 9)

Important Observation

Unlike the sum of two squares, the sum of two cubes can be factored using a standard identity.

Choosing the Correct Algebraic Identity

A major skill in algebra is recognising which identity is appropriate.

The best choice depends on the structure of the expression.

Example

Consider:

(2x + 5)²

The entire binomial is squared.

Choose the square-of-a-sum identity.

Example

Consider:

9x² − 25

There are two squared terms separated by subtraction.

Choose the difference-of-squares identity.

Example

Consider:

x³ + 64

There are two cubes separated by addition.

Choose the sum-of-cubes identity.

Important Observation

Identifying the structure of an expression is often more important than immediately beginning calculations.

A Practical Strategy for Recognising Identities

When examining an algebraic expression, ask yourself:

Does the expression contain brackets raised to a power?

If yes, check whether it matches a square or cube identity.

Does the expression contain two terms?

Check for a difference of squares or a sum/difference of cubes.

Does the expression contain three terms?

Check whether it is a perfect square trinomial.

Does the expression contain two multiplied binomials?

Check for conjugate factors, repeated factors, or a common variable term.

Do all terms have a common factor?

Factor out the greatest common factor before applying another identity.

Worked Example

Simplify:

3x² − 75

First, factor out the common factor:

3(x² − 25)

Recognise the difference of squares:

x² − 25 = (x − 5)(x + 5)

Therefore:

3x² − 75 = 3(x − 5)(x + 5)

Important Observation

Some expressions require more than one algebraic technique.

Common Mistakes in Recognising Identities

Confusing a sum of squares with a difference of squares

x² + 25 does not factor using the difference-of-squares identity over the real numbers.

Forgetting the middle term when squaring

Incorrect:

(x + 4)² = x² + 16

Correct:

(x + 4)² = x² + 8x + 16

Using the wrong sign in a cube identity

The signs in (a − b)³ alternate.

Confusing squares and cubes

x² − 8 is not a difference of cubes.

Applying an identity before removing a common factor

Expressions may become easier to recognise after factoring out a common factor.

Memorising formulas without understanding them

Understanding how identities are derived helps prevent errors and makes them easier to remember.

Part 1 Summary

The most important algebraic identities introduced in this part are:

• Distributive property: a(b + c) = ab + ac

• Square of a sum: (a + b)² = a² + 2ab + b²

• Square of a difference: (a − b)² = a² − 2ab + b²

• Difference of squares: a² − b² = (a − b)(a + b)

• Product of binomials: (x + a)(x + b) = x² + (a + b)x + ab

• Cube of a sum: (a + b)³ = a³ + 3a²b + 3ab² + b³

• Cube of a difference: (a − b)³ = a³ − 3a²b + 3ab² − b³

• Sum of cubes: a³ + b³ = (a + b)(a² − ab + b²)

• Difference of cubes: a³ − b³ = (a − b)(a² + ab + b²)

These identities are useful because they represent general mathematical relationships that can be applied to many different expressions.

Simplifying Expressions Using Algebraic Identities

Algebraic simplification means rewriting an expression in an equivalent, often more useful form.

The goal is not always to make an expression shorter. Sometimes we want to make it easier to factor, evaluate, or solve.

Worked Example: Simplifying Two Squared Binomials

Simplify:

(x + 4)² − (x − 4)²

We could expand both expressions separately.

First:

(x + 4)² = x² + 8x + 16

Second:

(x − 4)² = x² − 8x + 16

Subtract:

(x² + 8x + 16) − (x² − 8x + 16)

Remove brackets carefully:

x² + 8x + 16 − x² + 8x − 16

Combine like terms:

16x

A More Efficient Method

Recognise that the entire expression is a difference of two squares.

Let:

a = x + 4

b = x − 4

Apply:

a² − b² = (a − b)(a + b)

Therefore:

[(x + 4) − (x − 4)][(x + 4) + (x − 4)]

Simplify each bracket:

(8)(2x)

= 16x

Both methods give the same answer, but recognising the identity reduces the amount of work.

Important Observation

The most efficient method is often the one that recognises a larger algebraic pattern before expanding individual terms.

Recognising Hidden Difference-of-Squares Patterns

Sometimes the two squared quantities are themselves algebraic expressions.

Worked Example

Simplify:

(3x + 2)² − (3x − 2)²

Let:

a = 3x + 2

b = 3x − 2

Apply the difference-of-squares identity:

(a − b)(a + b)

First bracket:

(3x + 2) − (3x − 2) = 4

Second bracket:

(3x + 2) + (3x − 2) = 6x

Multiply:

4 × 6x = 24x

Final Answer: 24x

Another Example

Simplify:

(a + b)² − (a − b)²

Apply the same method:

[(a + b) − (a − b)][(a + b) + (a − b)]

= (2b)(2a)

= 4ab

This produces another useful identity:

(a + b)² − (a − b)² = 4ab

Using Identities to Simplify Algebraic Fractions

Algebraic identities are especially useful when simplifying fractions containing polynomials.

The key strategy is to factor the numerator and denominator before cancelling common factors.

Worked Example

Simplify:

(x² − 36)/(x + 6)

Recognise the numerator as a difference of squares:

x² − 36 = (x − 6)(x + 6)

Substitute:

[(x − 6)(x + 6)]/(x + 6)

Cancel the common factor:

x − 6

However, the original denominator cannot equal zero.

x + 6 ≠ 0

Therefore:

Final Answer: x − 6, where x ≠ −6

Important Observation

Cancelling a common factor does not remove restrictions from the original expression.

Simplifying Fractions with Perfect Square Trinomials

Worked Example

Simplify:

(x² + 10x + 25)/(x² − 25)

First, factor the numerator.

x² + 10x + 25 = (x + 5)²

Next, factor the denominator.

x² − 25 = (x − 5)(x + 5)

Substitute:

[(x + 5)(x + 5)]/[(x − 5)(x + 5)]

Cancel one common factor:

(x + 5)/(x − 5)

The original denominator is zero when x = 5 or x = −5.

Therefore:

Final Answer: (x + 5)/(x − 5), where x ≠ 5 and x ≠ −5

Important Observation

The expression may look simpler after cancellation, but all restrictions from the original denominator remain.

Simplifying Fractions Using Cubic Identities

Worked Example

Simplify:

(x³ − 8)/(x − 2)

Recognise the numerator as a difference of cubes.

x³ − 8 = x³ − 2³

Apply:

a³ − b³ = (a − b)(a² + ab + b²)

Therefore:

x³ − 8 = (x − 2)(x² + 2x + 4)

Substitute:

[(x − 2)(x² + 2x + 4)]/(x − 2)

Cancel:

x² + 2x + 4

Restriction:

x ≠ 2

Final Answer

x² + 2x + 4, where x ≠ 2

Important Observation

Recognising cubic identities can simplify expressions that would otherwise require polynomial division.

Proving Algebraic Identities

An algebraic identity must be true for every permitted value of its variables.

Testing several numerical values can provide evidence, but it does not constitute a general proof.

To prove an identity, we can transform one side algebraically until it matches the other.

Worked Example

Prove:

(a + b)² − (a − b)² = 4ab

Start with the left-hand side:

(a + b)² − (a − b)²

Expand both squares:

(a² + 2ab + b²) − (a² − 2ab + b²)

Remove the second set of brackets:

a² + 2ab + b² − a² + 2ab − b²

Combine like terms:

4ab

This matches the right-hand side.

Therefore, the identity is proved.

Important Observation

A proof demonstrates why an identity works for all permitted values, not merely for selected examples.

Proving an Identity Using Two Different Methods

Sometimes the same expression can be simplified in two ways.

Consider:

(a + b)² + (a − b)²

Method A: Expansion

Expand:

(a² + 2ab + b²) + (a² − 2ab + b²)

Combine like terms:

2a² + 2b²

Factor out 2:

2(a² + b²)

Therefore:

(a + b)² + (a − b)² = 2(a² + b²)

Method B: Geometric Interpretation

For real a and b, the quantities (a + b)² and (a − b)² represent the areas of squares with side lengths |a + b| and |a − b|.

Their combined area equals twice the sum of the squares of a and b.

This gives the same relationship:

(a + b)² + (a − b)² = 2(a² + b²)

Important Observation

Algebraic identities often have geometric interpretations that provide additional insight into why they work.

Proving the Difference of Cubes Identity

We can prove:

a³ − b³ = (a − b)(a² + ab + b²)

Start with the right-hand side:

(a − b)(a² + ab + b²)

Distribute a:

a³ + a²b + ab²

Distribute −b:

−a²b − ab² − b³

Combine:

a³ + a²b + ab² − a²b − ab² − b³

The middle terms cancel:

a³ − b³

Therefore, the identity is proved.

Important Observation

The signs in the second factor are arranged so that the unwanted terms cancel during expansion.

Recognising Patterns in Consecutive Numbers

Algebraic identities can reveal useful relationships between integers.

Worked Example

Find the difference between the squares of two consecutive integers.

Let the smaller integer be n.

The next integer is n + 1.

The difference between their squares is:

(n + 1)² − n²

Expand:

n² + 2n + 1 − n²

Simplify:

2n + 1

Therefore, the difference between the squares of consecutive integers is always odd.

Numerical Examples

Every result is odd.

Important Observation

Algebraic identities can be used to prove number patterns rather than merely observing them.

Difference of Squares and Consecutive Odd Numbers

Consider two consecutive odd integers.

Let the smaller integer be:

2n + 1

The next odd integer is:

2n + 3

Their squares differ by:

(2n + 3)² − (2n + 1)²

Use the difference-of-squares identity:

[(2n + 3) − (2n + 1)][(2n + 3) + (2n + 1)]

Simplify:

(2)(4n + 4)

= 8n + 8

Factor:

8(n + 1)

Therefore, the difference between the squares of consecutive odd integers is always divisible by 8.

Example

7² − 5² = 49 − 25 = 24

24 is divisible by 8.

Important Observation

Algebraic identities can reveal divisibility properties that are not immediately obvious.

Efficient Mental Arithmetic Using Squares

Perfect square identities can simplify calculations involving numbers close to convenient values.

Worked Example: 104²

Rewrite:

104 = 100 + 4

Apply:

(a + b)² = a² + 2ab + b²

Therefore:

104² = (100 + 4)²

= 100² + 2(100)(4) + 4²

= 10000 + 800 + 16

104² = 10816

Worked Example: 97²

Rewrite:

97 = 100 − 3

Apply:

(a − b)² = a² − 2ab + b²

Therefore:

97² = (100 − 3)²

= 10000 − 600 + 9

97² = 9409

Important Observation

Numbers close to powers of 10 are often good candidates for perfect square identities.

Efficient Multiplication Using Conjugate Numbers

Consider:

58 × 62

These numbers are equally spaced around 60.

Rewrite:

58 = 60 − 2

62 = 60 + 2

Apply the difference-of-squares identity:

(60 − 2)(60 + 2)

= 60² − 2²

= 3600 − 4

58 × 62 = 3596

Another Example: 997 × 1003

997 = 1000 − 3

1003 = 1000 + 3

Therefore:

997 × 1003 = 1000² − 3²

= 1000000 − 9

= 999991

Important Observation

The difference-of-squares identity is especially efficient when multiplying numbers that are equally distant from a convenient central value.

Solving Quadratic Equations Using Identities

Algebraic identities can transform equations into forms that are easier to solve.

Worked Example

Solve:

x² − 49 = 0

Recognise a difference of squares:

(x − 7)(x + 7) = 0

Apply the zero-product property:

x − 7 = 0

or

x + 7 = 0

Therefore:

x = 7 or x = −7

Verification

For x = 7:

7² − 49 = 0

For x = −7:

(−7)² − 49 = 0

Both solutions are correct.

Solving Perfect Square Equations

Worked Example

Solve:

x² + 12x + 36 = 0

Recognise a perfect square trinomial:

(x + 6)² = 0

Therefore:

x + 6 = 0

x = −6

Important Observation

A perfect square quadratic equation set equal to zero has a repeated root.

Solving Equations with a Nonzero Right-Hand Side

Worked Example

Solve:

(x + 3)² = 25

Take square roots:

x + 3 = ±5

Case 1:

x + 3 = 5

x = 2

Case 2:

x + 3 = −5

x = −8

Final Answer

x = 2 or x = −8

Verification

For x = 2:

(2 + 3)² = 25

For x = −8:

(−8 + 3)² = 25

Both solutions are correct.

Important Observation

When taking the square root of a positive number, remember both the positive and negative possibilities.

Solving Cubic Equations Using Identities

Worked Example

Solve:

x³ − 27 = 0

Recognise:

x³ − 27 = x³ − 3³

Factor using the difference-of-cubes identity:

(x − 3)(x² + 3x + 9) = 0

The first factor gives:

x − 3 = 0

x = 3

The quadratic factor has discriminant:

D = 3² − 4(1)(9)

= 9 − 36

= −27

Since D is negative, the quadratic factor has no real roots.

Therefore, the only real solution is:

x = 3

Important Observation

Factoring a cubic expression can reduce a higher-degree equation to simpler equations.

Applying Identities to Expressions with Multiple Variables

Algebraic identities work with expressions containing more than one variable.

Worked Example

Expand:

(2a + 3b)²

Apply:

(a + b)² = a² + 2ab + b²

Treat the first term as 2a and the second as 3b.

Therefore:

(2a + 3b)² = (2a)² + 2(2a)(3b) + (3b)²

= 4a² + 12ab + 9b²

Worked Example

Factor:

16x² − 81y²

Recognise:

16x² = (4x)²

81y² = (9y)²

Therefore:

16x² − 81y² = (4x − 9y)(4x + 9y)

Important Observation

The symbols a and b in an identity can represent entire algebraic expressions, not just individual variables.

Applying Identities to Higher Powers

Some expressions contain powers greater than two but still follow familiar identities.

Worked Example

Factor:

x⁴ − 81

Recognise:

x⁴ = (x²)²

81 = 9²

Apply the difference-of-squares identity:

x⁴ − 81 = (x² − 9)(x² + 9)

The first factor is another difference of squares:

x² − 9 = (x − 3)(x + 3)

Therefore:

x⁴ − 81 = (x − 3)(x + 3)(x² + 9)

Important Observation

Some expressions can be factored repeatedly using the same identity.

Combining Multiple Identities

More advanced problems sometimes require two or more identities.

Worked Example

Simplify:

(x + 2)² − (x − 2)² + (x + 3)(x − 3)

First, simplify the difference of squared binomials:

(x + 2)² − (x − 2)² = 8x

Next, simplify the conjugate product:

(x + 3)(x − 3) = x² − 9

Combine:

8x + x² − 9

Write in descending powers:

x² + 8x − 9

Important Observation

Identities can be applied to individual parts of a larger expression before combining the results.

Choosing Between Expansion and Factorisation

A major part of efficient algebra is knowing when to expand and when to factor.

Expansion Is Often Useful When:

• Combining algebraic expressions.

• Removing brackets.

• Writing expressions in standard polynomial form.

• Comparing coefficients.

• Simplifying sums and differences.

Factorisation Is Often Useful When:

• Solving equations set equal to zero.

• Simplifying rational expressions.

• Finding common factors.

• Identifying roots.

• Recognising geometric relationships.

Example

Consider:

(x + 4)(x − 4)

If the goal is to write a polynomial in standard form, expand:

x² − 16

If the goal is to solve:

x² − 16 = 0

factor:

(x − 4)(x + 4) = 0

Then:

x = ±4

Important Observation

The best algebraic form depends on what we want to accomplish.

Verifying Identities Using Numerical Substitution

Numerical substitution is useful for checking calculations and detecting mistakes.

Worked Example

Check:

(a + b)² = a² + 2ab + b²

Choose:

a = 4

b = 3

Left-hand side:

(4 + 3)² = 49

Right-hand side:

4² + 2(4)(3) + 3²

= 16 + 24 + 9

= 49

Both sides agree.

However, this checks only one pair of values.

Important Observation

Numerical examples can support an identity, but algebraic proof is required to establish that it always works.

Detecting Incorrect Identities

Sometimes an expression looks like a familiar identity but contains an error.

Consider the incorrect statement:

(a + b)² = a² + b²

Choose:

a = 2

b = 3

Left-hand side:

(2 + 3)² = 25

Right-hand side:

2² + 3² = 13

Since 25 ≠ 13, the statement is false.

A single counterexample is enough to disprove a proposed identity.

Important Observation

To prove an identity, we need a general argument. To disprove a claimed identity, one valid counterexample is sufficient.

Using Identities to Compare Expressions

Sometimes we want to determine whether two expressions are equivalent.

Worked Example

Determine whether:

(x + 5)² − 10x

is equivalent to:

x² + 25

Expand:

(x + 5)² − 10x

= x² + 10x + 25 − 10x

Combine like terms:

= x² + 25

Therefore:

(x + 5)² − 10x = x² + 25

The expressions are equivalent for all real x.

Important Observation

Two expressions may look different but represent exactly the same mathematical relationship.

Part 2 Practice

Try these questions before moving to Part 3.

Simplification

1. Simplify (x + 6)² − (x − 6)².

2. Simplify (2x + 3)(2x − 3).

3. Simplify (a + b)² + (a − b)².

4. Factor completely: 2x² − 50.

5. Simplify (x³ − 64)/(x − 4), stating restrictions.

Proof and Reasoning

6. Prove that (a + b)² − (a − b)² = 4ab.

7. Explain why the difference between consecutive square numbers is always odd.

8. Give a counterexample to the incorrect statement (a − b)² = a² − b².

9. Explain why checking three numerical values does not prove an identity.

10. Describe when factoring is more useful than expanding.

Applications

11. Calculate 106² using an identity.

12. Calculate 49 × 51 using an identity.

13. Solve x² − 121 = 0.

14. Solve (x − 4)² = 36.

15. Factor completely: x⁴ − 16.

Answers