Factoring Polynomials

Introduction

Factoring is the process of rewriting a polynomial as a product of simpler expressions.

It is the reverse of polynomial multiplication.

For example, when we multiply:

(x + 2)(x + 3)

we obtain:

x² + 5x + 6

Factoring reverses this process:

x² + 5x + 6 = (x + 2)(x + 3)

Factoring is an essential algebraic skill used to simplify expressions, solve equations, identify patterns, and model real-world situations.

There are several important factoring techniques:

• Factoring out a greatest common factor.

• Factoring quadratic trinomials.

• Recognising special products.

• Factoring by grouping.

• Combining multiple techniques to factor completely.

A reliable way to verify any factorization is to multiply the factors together and check that the original polynomial is recovered.

What Does Factoring Mean?

A factor is a number or expression that is multiplied by another factor to produce a product.

For example:

3 × 4 = 12

The numbers 3 and 4 are factors of 12.

Similarly:

(x + 2)(x + 5) = x² + 7x + 10

The expressions x + 2 and x + 5 are factors of x² + 7x + 10.

Expanded Form and Factored Form

Consider:

x² + 7x + 10

This is the expanded form.

The factored form is:

(x + 2)(x + 5)

Both expressions are equivalent.

They produce the same result for every value of x.

Why Is Factoring Useful?

Factoring helps us:

• Simplify algebraic expressions.

• Solve polynomial equations.

• Find zeros of polynomial functions.

• Identify the dimensions of geometric shapes.

• Recognise algebraic patterns.

• Simplify rational expressions.

• Understand the structure of polynomials.

Review: Factors and Greatest Common Factors

Before factoring polynomials, it is important to understand factors of numbers.

Consider:

12 and 18

Factors of 12:

1, 2, 3, 4, 6, 12

Factors of 18:

1, 2, 3, 6, 9, 18

The greatest factor shared by both numbers is 6.

Therefore:

GCF(12, 18) = 6

The greatest common factor (GCF) is the largest positive factor shared by two or more integers.

For polynomial terms, the GCF may contain numerical coefficients and variables.

Example

Find the GCF of:

12x³ and 18x²

The greatest common numerical factor is 6.

The greatest common variable factor is x².

Therefore:

GCF = 6x²

Factoring Out a Greatest Common Factor

The simplest factoring method is to identify and remove a factor shared by every term.

This is the reverse of the distributive property.

Recall:

a(b + c) = ab + ac

Factoring reverses this:

ab + ac = a(b + c)

Worked Example

Factor:

6x + 12

The GCF of 6x and 12 is 6.

Divide each term by 6:

6x ÷ 6 = x

12 ÷ 6 = 2

Therefore:

6x + 12 = 6(x + 2)

Verification

Expand:

6(x + 2)

= 6x + 12

The original polynomial is recovered.

Factoring Common Variable Factors

Worked Example

Factor:

8x² + 12x

Find the GCF.

Numerical GCF:

GCF(8, 12) = 4

Variable GCF:

x² and x share x.

Therefore:

GCF = 4x

Divide:

8x² ÷ 4x = 2x

12x ÷ 4x = 3

Thus:

8x² + 12x = 4x(2x + 3)

Another Example

Factor:

15a³b² − 10a²b

Numerical GCF:

5

Variable factors:

a² and b

Therefore:

GCF = 5a²b

Divide each term:

15a³b² ÷ 5a²b = 3ab

−10a²b ÷ 5a²b = −2

Final answer:

15a³b² − 10a²b = 5a²b(3ab − 2)

Factoring a Negative Common Factor

Sometimes it is helpful to factor out a negative number.

Worked Example

Factor:

−6x² + 9x

A common factor is 3x.

However, factoring out −3x makes the leading term inside the parentheses positive.

−6x² ÷ (−3x) = 2x

9x ÷ (−3x) = −3

Therefore:

−6x² + 9x = −3x(2x − 3)

Why Factor Out a Negative?

Factoring out a negative can make the remaining expression easier to work with, particularly when the leading coefficient is negative.

Factoring Common Factors from Three or More Terms

Worked Example

Factor:

12x³ + 18x² − 6x

Find the GCF of all three terms.

Numerical GCF = 6

Variable GCF = x

Therefore:

GCF = 6x

Divide:

12x³ ÷ 6x = 2x²

18x² ÷ 6x = 3x

−6x ÷ 6x = −1

Final answer:

12x³ + 18x² − 6x = 6x(2x² + 3x − 1)

The trinomial inside the parentheses cannot be factored further using integer coefficients.

This is completely factored over the integers.

Factoring Quadratic Trinomials

A quadratic trinomial has three terms and a highest exponent of 2.

A common form is:

x² + bx + c

We want to rewrite it as:

(x + m)(x + n)

where:

m + n = b

and:

mn = c

This means we need two numbers that:

• Add to the coefficient of x.

• Multiply to the constant term.

Worked Example

Factor:

x² + 7x + 12

Find two numbers that:

Add to 7.

Multiply to 12.

The numbers are 3 and 4.

Because:

3 + 4 = 7

3 × 4 = 12

Therefore:

x² + 7x + 12 = (x + 3)(x + 4)

Verification

Expand:

(x + 3)(x + 4)

= x² + 4x + 3x + 12

= x² + 7x + 12

Correct.

Factoring Trinomials with Positive Constants

Worked Example

Factor:

x² + 9x + 20

We need two numbers that add to 9 and multiply to 20.

Possible factor pairs of 20:

The correct pair is 4 and 5.

Therefore:

x² + 9x + 20 = (x + 4)(x + 5)

Another Example

Factor:

x² + 11x + 24

Find two numbers that:

Add to 11.

Multiply to 24.

The numbers are 3 and 8.

x² + 11x + 24 = (x + 3)(x + 8)

Factoring Trinomials with Negative Constants

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

Worked Example

Factor:

x² + 2x − 15

We need two numbers that:

Add to 2.

Multiply to −15.

The numbers are 5 and −3.

Because:

5 + (−3) = 2

5 × (−3) = −15

Therefore:

x² + 2x − 15 = (x + 5)(x − 3)

Another Example

Factor:

x² − 4x − 21

We need two numbers that:

Add to −4.

Multiply to −21.

The numbers are −7 and 3.

Therefore:

x² − 4x − 21 = (x − 7)(x + 3)

Factoring Trinomials with Negative Middle Terms

When the constant is positive and the middle coefficient is negative, both numbers must be negative.

Worked Example

Factor:

x² − 9x + 20

We need two numbers that:

Add to −9.

Multiply to 20.

The numbers are −4 and −5.

Therefore:

x² − 9x + 20 = (x − 4)(x − 5)

Sign Pattern Summary

This table applies to monic trinomials of the form x² + bx + c that factor into integer binomials.

Factoring Trinomials with Leading Coefficients Greater Than One

Some trinomials have the form:

ax² + bx + c

where a is not 1.

These expressions require additional steps.

One useful approach is the AC method, also called factoring by splitting the middle term.

Worked Example

Factor:

2x² + 7x + 3

Step 1: Multiply a and c

a = 2

c = 3

ac = 6

Step 2: Find Two Numbers

Find two numbers that:

Multiply to 6.

Add to 7.

The numbers are 6 and 1.

Step 3: Split the Middle Term

Rewrite:

2x² + 6x + x + 3

Step 4: Group Terms

(2x² + 6x) + (x + 3)

Step 5: Factor Each Group

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

Step 6: Factor the Common Binomial

(2x + 1)(x + 3)

Verification

(2x + 1)(x + 3)

= 2x² + 6x + x + 3

= 2x² + 7x + 3

Correct.

Another Example Using the AC Method

Factor:

6x² + 11x + 3

Multiply:

a × c = 6 × 3 = 18

Find two numbers that:

Multiply to 18.

Add to 11.

The numbers are 9 and 2.

Split:

6x² + 9x + 2x + 3

Group:

(6x² + 9x) + (2x + 3)

Factor:

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

Factor the common binomial:

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

Factoring Trinomials with Negative Coefficients

Worked Example

Factor:

3x² − 10x + 8

Multiply:

a × c = 3 × 8 = 24

Find two numbers that:

Multiply to 24.

Add to −10.

The numbers are −6 and −4.

Split:

3x² − 6x − 4x + 8

Group:

(3x² − 6x) + (−4x + 8)

Factor:

3x(x − 2) − 4(x − 2)

Therefore:

3x² − 10x + 8 = (3x − 4)(x − 2)

Factoring by Grouping

Factoring by grouping is useful for polynomials with four or more terms.

The goal is to create common factors within groups.

Worked Example

Factor:

ax + ay + bx + by

Group:

(ax + ay) + (bx + by)

Factor each group:

a(x + y) + b(x + y)

Now factor the common binomial:

(a + b)(x + y)

Numerical Example

Factor:

3x² + 6x + 2x + 4

Group:

(3x² + 6x) + (2x + 4)

Factor:

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

Factor the common binomial:

(3x + 2)(x + 2)

Important Observation

The grouping method is closely related to the AC method.

Both involve creating groups that share a common binomial factor.

Special Products: Difference of Squares

Some polynomials follow patterns that allow us to factor them quickly.

One of the most important is the difference of squares.

Recall:

(a + b)(a − b) = a² − b²

Reversing this identity gives:

a² − b² = (a + b)(a − b)

A difference of squares contains:

• Two terms.

• A subtraction sign between them.

• Two expressions that are perfect squares.

Worked Example

Factor:

x² − 25

Recognise:

x² = (x)²

25 = 5²

Therefore:

x² − 25 = x² − 5²

Apply the identity:

x² − 25 = (x + 5)(x − 5)

Verification

(x + 5)(x − 5)

= x² − 5x + 5x − 25

= x² − 25

The middle terms cancel.

Factoring More Complicated Differences of Squares

Worked Example

Factor:

9x² − 16

Recognise:

9x² = (3x)²

16 = 4²

Therefore:

9x² − 16 = (3x)² − 4²

Apply the identity:

9x² − 16 = (3x + 4)(3x − 4)

Another Example

Factor:

25a² − 49b²

Recognise:

25a² = (5a)²

49b² = (7b)²

Therefore:

25a² − 49b² = (5a + 7b)(5a − 7b)

Important Reminder

The sum of two squares does not factor using the difference-of-squares identity.

For example:

x² + 25

cannot be rewritten as (x + 5)(x − 5).

That product equals x² − 25, not x² + 25.

The expression x² + 25 is irreducible over the real numbers.

Perfect Square Trinomials

Another important pattern is the perfect square trinomial.

Recall:

(a + b)² = a² + 2ab + b²

and:

(a − b)² = a² − 2ab + b²

Factoring reverses these identities.

a² + 2ab + b² = (a + b)²

a² − 2ab + b² = (a − b)²

How to Recognise a Perfect Square Trinomial

Look for three characteristics:

1. The first term is a perfect square.

2. The last term is a perfect square.

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

Factoring a Perfect Square Trinomial

Worked Example

Factor:

x² + 10x + 25

The first term is:

x² = (x)²

The last term is:

25 = 5²

Check the middle term:

2 × x × 5 = 10x

The pattern matches.

Therefore:

x² + 10x + 25 = (x + 5)²

Verification

(x + 5)²

= (x + 5)(x + 5)

= x² + 5x + 5x + 25

= x² + 10x + 25

Correct.

Another Example

Factor:

x² − 12x + 36

Recognise:

x² = (x)²

36 = 6²

Middle term:

−2(x)(6) = −12x

Therefore:

x² − 12x + 36 = (x − 6)²

Perfect Square Trinomials with Coefficients

Worked Example

Factor:

4x² + 20x + 25

The first term is:

4x² = (2x)²

The last term is:

25 = 5²

Check:

2(2x)(5) = 20x

Therefore:

4x² + 20x + 25 = (2x + 5)²

Another Example

Factor:

9x² − 24x + 16

Recognise:

9x² = (3x)²

16 = 4²

Middle term:

−2(3x)(4) = −24x

Therefore:

9x² − 24x + 16 = (3x − 4)²

Difference of Cubes

Some higher-degree polynomials can also be factored using special identities.

A difference of cubes follows:

a³ − b³ = (a − b)(a² + ab + b²)

Worked Example

Factor:

x³ − 8

Recognise:

x³ = (x)³

8 = 2³

Apply the identity:

x³ − 2³

= (x − 2)(x² + 2x + 4)

Therefore:

x³ − 8 = (x − 2)(x² + 2x + 4)

Verification

(x − 2)(x² + 2x + 4)

= x³ + 2x² + 4x − 2x² − 4x − 8

= x³ − 8

Correct.

Sum of Cubes

The corresponding sum-of-cubes identity is:

a³ + b³ = (a + b)(a² − ab + b²)

Worked Example

Factor:

x³ + 27

Recognise:

27 = 3³

Therefore:

x³ + 3³

= (x + 3)(x² − 3x + 9)

Final answer: (x + 3)(x² − 3x + 9)

Another Example

Factor:

8x³ + 125

Recognise:

8x³ = (2x)³

125 = 5³

Apply the identity:

8x³ + 125 = (2x + 5)(4x² − 10x + 25)

Recognising Special Factoring Patterns

Recognising these patterns can make factoring much faster than using a general method.

Factoring Completely

A polynomial is completely factored when it is written as a product of factors that cannot be factored further within the chosen number system.

In most introductory algebra problems, this means factoring completely over the integers, unless otherwise stated.

A common mistake is to stop after removing the greatest common factor.

Worked Example

Factor completely:

2x² + 10x + 12

Step 1: Find the GCF

All coefficients are divisible by 2.

Factor out 2:

2(x² + 5x + 6)

Step 2: Factor the Trinomial

Find two numbers that:

Add to 5.

Multiply to 6.

The numbers are 2 and 3.

Therefore:

x² + 5x + 6 = (x + 2)(x + 3)

Final Answer

2x² + 10x + 12 = 2(x + 2)(x + 3)

Verification

2(x + 2)(x + 3)

= 2(x² + 5x + 6)

= 2x² + 10x + 12

Correct.

Factoring Completely with Special Products

Worked Example

Factor completely:

3x² − 75

First find the GCF:

3x² − 75 = 3(x² − 25)

Recognise the difference of squares:

x² − 25 = (x + 5)(x − 5)

Therefore:

3x² − 75 = 3(x + 5)(x − 5)

Another Example

Factor completely:

4x³ − 36x

First find the GCF:

4x³ − 36x = 4x(x² − 9)

Recognise:

x² − 9 = (x + 3)(x − 3)

Therefore:

4x³ − 36x = 4x(x + 3)(x − 3)

Factoring Repeated Differences of Squares

Sometimes a polynomial can be factored more than once using the same identity.

Worked Example

Factor completely:

x⁴ − 16

Recognise:

x⁴ = (x²)²

16 = 4²

Apply the difference-of-squares identity:

x⁴ − 16 = (x² + 4)(x² − 4)

The factor x² − 4 is also a difference of squares:

x² − 4 = (x + 2)(x − 2)

Therefore:

x⁴ − 16 = (x² + 4)(x + 2)(x − 2)

The factor x² + 4 cannot be factored further over the real numbers.

Factoring Higher-Degree Polynomials by Grouping

Worked Example

Factor:

x³ + 3x² + 2x + 6

Group the first two and last two terms:

(x³ + 3x²) + (2x + 6)

Factor each group:

x²(x + 3) + 2(x + 3)

Now factor the common binomial:

(x + 3)(x² + 2)

The second factor cannot be factored further over the real numbers.

Another Example

Factor:

2x³ − 6x² + 5x − 15

Group:

(2x³ − 6x²) + (5x − 15)

Factor:

2x²(x − 3) + 5(x − 3)

Therefore:

(x − 3)(2x² + 5)

Factoring Expressions with Multiple Variables

The same techniques apply to expressions containing more than one variable.

Worked Example: Greatest Common Factor

Factor:

12a²b + 18ab²

The numerical GCF is 6.

Both terms contain a and b.

Therefore:

GCF = 6ab

Factor:

12a²b + 18ab² = 6ab(2a + 3b)

Worked Example: Difference of Squares

Factor:

25x² − 36y²

Recognise:

25x² = (5x)²

36y² = (6y)²

Therefore:

25x² − 36y² = (5x + 6y)(5x − 6y)

Worked Example: Perfect Square Trinomial

Factor:

x² + 8xy + 16y²

Recognise:

x² = (x)²

16y² = (4y)²

Middle term:

2(x)(4y) = 8xy

Therefore:

x² + 8xy + 16y² = (x + 4y)²

Factoring and the Zero Product Property

Factoring is especially useful when solving polynomial equations.

The zero product property states:

If:

ab = 0

then:

a = 0 or b = 0

This works because the product of two numbers is zero only when at least one factor is zero.

Worked Example

Solve:

x² + 5x + 6 = 0

Factor:

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

Apply the zero product property:

x + 2 = 0

or:

x + 3 = 0

Solve:

x = −2

or:

x = −3

Final Answer

x = −2 or x = −3

These values are called the roots or zeros of the polynomial.

Factoring and Graphs of Quadratic Functions

Consider:

y = x² − 5x + 6

Factor:

y = (x − 2)(x − 3)

The x-intercepts occur when y = 0.

Therefore:

(x − 2)(x − 3) = 0

So:

x = 2 or x = 3

The graph crosses the x-axis at:

(2, 0) and (3, 0)

Factored form makes the zeros of a polynomial easy to identify.

Repeated Factors

Consider:

y = (x − 3)²

The only zero is:

x = 3

Because the factor is repeated, the parabola touches the x-axis at (3, 0) rather than crossing it.

Factoring to Simplify Rational Expressions

Factoring is also useful for simplifying algebraic fractions.

Worked Example

Simplify:

(x² + 5x + 6)/(x + 2)

Factor the numerator:

x² + 5x + 6 = (x + 2)(x + 3)

Substitute:

[(x + 2)(x + 3)]/(x + 2)

Cancel the common nonzero factor:

= x + 3

However, the original expression is undefined when x = −2.

Therefore, the simplified expression is equivalent to the original only for:

x ≠ −2

Important Reminder

We may cancel common factors, not arbitrary terms.

For example:

(x + 3)/(x + 5)

cannot be simplified by cancelling x.

Choosing the Correct Factoring Method

When factoring a polynomial, examine its structure before beginning.

A Reliable Factoring Strategy

1. Arrange the polynomial in standard form.

2. Look for a greatest common factor.

3. Count the terms.

4. Check for a special product.

5. If it is a trinomial, attempt trinomial factoring.

6. If there are four terms, consider grouping.

7. Check whether any resulting factors can be factored further.

8. Expand the factors to verify the result.

Verifying Factorizations by Expansion

Every factorization should be checked to ensure it produces the original polynomial.

Worked Example

Verify:

6x² + 11x + 3 = (3x + 1)(2x + 3)

Expand the right side:

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

Multiply:

3x × 2x = 6x²

3x × 3 = 9x

1 × 2x = 2x

1 × 3 = 3

Combine:

6x² + 9x + 2x + 3

= 6x² + 11x + 3

The original polynomial is recovered.

The factorization is correct.

Worked Example: Finding an Error

A student writes:

x² − 7x + 12 = (x − 2)(x − 6)

Check by expansion:

(x − 2)(x − 6)

= x² − 6x − 2x + 12

= x² − 8x + 12

This is not the original polynomial.

The correct factors must add to −7 and multiply to 12.

The numbers are −3 and −4.

Therefore:

x² − 7x + 12 = (x − 3)(x − 4)

Why Verification Matters

Expansion can reveal:

• Incorrect signs.

• Incorrect factor pairs.

• Missing coefficients.

• Errors when splitting the middle term.

• Incomplete calculations.

Real-World Application: Finding Rectangle Dimensions

Suppose the area of a rectangular garden is:

A = x² + 9x + 20

The area of a rectangle is:

A = Length × Width

Factor the polynomial:

x² + 9x + 20

Find two numbers that add to 9 and multiply to 20.

The numbers are 4 and 5.

Therefore:

A = (x + 4)(x + 5)

One possible pair of dimensions is:

Length = x + 5 metres

Width = x + 4 metres

Checking the Dimensions

Multiply:

(x + 4)(x + 5)

= x² + 5x + 4x + 20

= x² + 9x + 20

The dimensions produce the required area.

Numerical Example

If x = 3:

Length = 8 m

Width = 7 m

Area = 8 × 7

Area = 56 m²

Check using the original polynomial:

A = 3² + 9(3) + 20

= 9 + 27 + 20

= 56 m²

Both methods agree.

Real-World Application: Projectile Motion

A ball is launched upward from ground level.

Its height after t seconds is modelled by:

h(t) = −5t² + 20t

where height is measured in metres.

We want to determine when the ball returns to the ground.

At ground level:

h(t) = 0

Therefore:

−5t² + 20t = 0

Factor out the greatest common factor:

−5t(t − 4) = 0

Apply the zero product property:

−5t = 0

or:

t − 4 = 0

Therefore:

t = 0 or t = 4

Interpretation

At t = 0, the ball is launched.

At t = 4 seconds, the ball returns to the ground.

The flight lasts 4 seconds.

Real-World Application: Manufacturing

A manufacturer produces rectangular panels.

The area of a panel is represented by:

A = 6x² + 17x + 12

To identify possible expressions for its dimensions, factor the polynomial.

Multiply:

a × c = 6 × 12 = 72

Find two numbers that add to 17 and multiply to 72.

The numbers are 8 and 9.

Split the middle term:

6x² + 8x + 9x + 12

Group:

(6x² + 8x) + (9x + 12)

Factor:

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

Therefore:

A = (2x + 3)(3x + 4)

Possible panel dimensions are:

2x + 3

and:

3x + 4

This demonstrates how factoring can help recover dimensions from an area expression.

Common Mistakes

Forgetting to factor out the GCF first

For example:

2x² + 10x + 12

should be factored completely as:

2(x + 2)(x + 3)

Choosing numbers that multiply correctly but do not add correctly

For x² + 7x + 12, the correct pair is 3 and 4, not 2 and 6.

Ignoring negative signs

For x² − 5x + 6, both numbers must be negative.

The correct factorization is:

(x − 2)(x − 3)

Confusing a difference of squares with a sum of squares

x² − 16 = (x + 4)(x − 4)

But x² + 16 does not factor using this identity.

Forgetting the middle term in a perfect square trinomial

(x + 5)² = x² + 10x + 25

not x² + 25.

Stopping before the expression is completely factored

For example:

x⁴ − 16 = (x² + 4)(x² − 4)

is not completely factored over the integers.

Continue:

(x² + 4)(x + 2)(x − 2)

Incorrect grouping

Both groups must produce the same common binomial factor.

Assuming every quadratic factors into integer binomials

For example:

x² + x + 1

does not factor into real linear factors because its discriminant is negative.

Failing to verify

Always expand the factors to check the original polynomial.

Key Terms

• Factor: Number or expression multiplied by another to form a product.

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

• Expanded Form: Polynomial written as a sum of terms.

• Factored Form: Polynomial written as a product of expressions.

• Greatest Common Factor (GCF): Largest common factor shared by terms.

• Coefficient: Numerical factor multiplying a variable.

• Quadratic Trinomial: Three-term polynomial with degree 2.

• Monic Polynomial: Polynomial whose leading coefficient is 1.

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

• AC Method: Technique for factoring trinomials by splitting the middle term.

• Factoring by Grouping: Method of creating common factors within groups of terms.

• Common Binomial Factor: Binomial expression shared by two or more terms.

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

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

• Sum of Cubes: Expression of the form a³ + b³.

• Difference of Cubes: Expression of the form a³ − b³.

• Completely Factored: Written as a product of factors that cannot be factored further in the chosen number system.

• Irreducible Polynomial: Polynomial that cannot be factored into lower-degree polynomials over the specified number system.

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

• Root or Zero: Input value that makes a polynomial equal zero.

• Expansion: Multiplying factors to obtain a polynomial expression.

• Verification: Checking a factorization by expanding the factors.

Key Takeaways

• Factoring is the reverse of polynomial multiplication.

• The greatest common factor should be checked first.

• Trinomials of the form x² + bx + c can be factored by finding two numbers whose sum is b and product is c.

• Trinomials with leading coefficients other than 1 can often be factored using the AC method.

• Factoring by grouping is useful for polynomials with four terms.

• A difference of squares factors into conjugate binomials.

• Perfect square trinomials factor into squared binomials.

• Sum-of-cubes and difference-of-cubes identities can factor certain cubic expressions.

• Some polynomials require more than one factoring technique.

• Factoring completely means continuing until no factor can be reduced further over the specified number system.

• Factoring can be used to solve polynomial equations using the zero product property.

• Factored form helps identify zeros and x-intercepts.

• Every factorization can be verified by expansion.

• Factoring is useful in geometry, physics, engineering, and algebraic modelling.

Check Your Understanding

Factor each expression completely over the integers where possible.

Greatest Common Factors

1. 6x + 18

2. 8x² + 12x

3. 15a³ − 10a²

4. 12x³ − 18x² + 6x

Factoring Trinomials

5. x² + 7x + 12

6. x² + 9x + 20

7. x² − 8x + 15

8. x² + 2x − 24

9. 2x² + 7x + 3

10. 3x² + 10x + 8

11. 4x² − 12x + 9

12. 6x² − 7x − 3

Special Products

13. x² − 49

14. 9x² − 25

15. x² + 12x + 36

16. 4x² − 20x + 25

17. x³ − 27

18. x³ + 64

Factoring Completely

19. 2x² + 14x + 24

20. 3x² − 27

21. 5x³ − 45x

22. x⁴ − 81

23. x³ + 4x² + 3x + 12

24. 4x² − 16x + 16

Answer Key

Extended Challenge

Factor each polynomial completely. Show the steps used and verify selected answers by expansion.

Common Factors

1. 18x⁴ − 24x³ + 6x²

2. 20a³b² − 30a²b³ + 10ab

3. −12x³ + 36x² − 24x

4. 16x⁵ − 81x

5. 8a³ − 32a

Quadratic Trinomials

6. 2x² + 9x + 10

7. 3x² − 14x + 8

8. 6x² + x − 12

9. 8x² − 14x + 3

10. 12x² − 11x − 5

Special Products

11. 16x² − 81

12. 25a² − 36b²

13. 9x² + 30x + 25

14. 49x² − 70xy + 25y²

15. 8x³ − 125

16. 27x³ + 64

17. x⁴ − 625

Grouping and Multiple Techniques

18. x³ + 5x² + 2x + 10

19. 3x³ − 6x² + 5x − 10

20. 2x³ + 8x² + 6x + 24

21. x³ − 4x² − 9x + 36

22. 3x³ + 12x² − 27x − 108

23. x⁴ − 5x² + 4

24. 2x⁴ − 18x²

25. x⁶ − 64

Final Challenge: Factoring Investigation

A rectangular garden has an area represented by:

A = 2x² + 13x + 15

The garden designer wants to determine possible expressions for its length and width.

Complete the investigation.

1. Identify the coefficients a, b, and c.

2. Calculate a × c.

3. Find two numbers whose product equals ac and whose sum equals b.

4. Rewrite the middle term using these two numbers.

5. Factor the polynomial by grouping.

6. Identify possible expressions for the garden's length and width.

7. Verify the factorization by expansion.

8. Calculate the garden's area when x = 5.

9. Calculate the dimensions when x = 5.

10. Verify the area using length × width.

11. Explain why factoring can be useful when determining dimensions.

12. Describe how the area model relates to polynomial multiplication and factoring.

Further Investigation: Comparing Factoring Methods

Factor the following expressions:

A = x² − 16

B = x² + 8x + 16

C = 2x² + 10x + 12

D = x³ + 3x² + 4x + 12

E = x⁴ − 81

For each polynomial:

1. Identify the most appropriate factoring method.

2. Explain why you selected that method.

3. Show all factoring steps.

4. Determine whether further factoring is possible.

5. Write the completely factored form over the integers.

6. Expand your answer to verify it.

7. Compare the methods and identify any similarities.

8. Explain why recognising polynomial structure makes factoring more efficient.