Algebraic Identities and Factorization

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