Polynomials, Expanding and Factorising — Edexcel A-level Mathematics (Pure)

SPEC 2.6 • ALGEBRA AND FUNCTIONS

Polynomials: Expanding and Factorising

Master the essential algebraic skills of expanding brackets and factorising expressions

What is a Polynomial?

Definition

A polynomial is an expression consisting of terms with non-negative integer powers of a variable, combined using addition and subtraction.

Key Terminology

Term: Each part separated by + or − (e.g., 3x², 5x, −2)

Coefficient: The number in front of a variable (e.g., 3 in 3x²)

Degree: The highest power of the variable (e.g., degree 2 for x² + 3x − 1)

Constant term: The term with no variable (e.g., −2 in 3x² + 5x − 2)

NameDegreeExample

Linear1

2 x + 5

Quadratic2

3 x^{2} - 4 x + 1

Cubic3

x^{3} + 2 x^{2} - x + 7

Quartic4

2 x^{4} - x^{2} + 3

Expanding Single Brackets

Multiply each term inside the bracket by the term outside:

Formula

a (b + c) = ab + a c

Example 1

3 (x + 4) = 3 x + 12

Example 2

2 x (3 x - 5) = 6 x^{2} - 10 x

Watch Out

When the term outside is negative, signs change inside:

- 2 (x - 3) = - 2 x + 6

✓

NOT

- 2 x - 6

✗

Worked Example01

Expand and simplify: $3 x (2 x - 5) - 2 (x^{2} - 4 x + 1)$

1

Expand first bracket:

3 x (2 x - 5) = 6 x^{2} - 15 x

2

Expand second bracket:

- 2 (x^{2} - 4 x + 1) = - 2 x^{2} + 8 x - 2

3

Combine:

6 x^{2} - 15 x - 2 x^{2} + 8 x - 2

4

Collect like terms:

6 x^{2} - 2 x^{2} = 4 x^{2}

and

- 15 x + 8 x = - 7 x

Answer: $4 x^{2} - 7 x - 2$

Expanding Double Brackets

Use the FOIL method or grid method to expand two brackets:

FOIL Method

(a + b) (c + d) = a c + a d + b c + b d

First • Outer • Inner • Last

Example

Expand

(x + 3) (x + 5)

:

First:

x \times x = x^{2}

Outer:

x \times 5 = 5 x

Inner:

3 \times x = 3 x

Last:

3 \times 5 = 15

Result:

x^{2} + 5 x + 3 x + 15 = x^{2} + 8 x + 15

Worked Example02

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

1

Use FOIL with two variables:

2

First:

2 x \times x = 2 x^{2}

3

Outer:

2 x \times (- 2 y) = - 4 x y

4

Inner:

3 y \times x = 3 x y

5

Last:

3 y \times (- 2 y) = - 6 y^{2}

6

Combine:

2 x^{2} - 4 x y + 3 x y - 6 y^{2} = 2 x^{2} - x y - 6 y^{2}

Answer: $2 x^{2} - x y - 6 y^{2}$

Special Cases: Perfect Squares & DOTS

Perfect Squares

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(a - b)^{2} = a^{2} - 2 ab + b^{2}

Difference of Two Squares

(a + b) (a - b) = a^{2} - b^{2}

Common Error

(x + 3)^{2} \neq = x^{2} + 9

✗

(x + 3)^{2} = x^{2} + 6 x + 9

✓

Don't forget the middle term

2 ab

!

Worked Example03

Expand: $(3 x - 4)^{2}$

1

Use perfect square formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

2

Here

a = 3 x

and

b = 4

3

a^{2} = (3 x)^{2} = 9 x^{2}

4

2 ab = 2 \times 3 x \times 4 = 24 x

5

b^{2} = 4^{2} = 16

Answer: $9 x^{2} - 24 x + 16$

Expanding Triple Brackets

Expand two brackets first, then multiply by the third:

Method

Expand the first two brackets

Simplify and collect like terms

Multiply the result by the third bracket

Simplify and collect like terms again

Example

Expand

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

:

Step 1:

(x + 1) (x + 2) = x^{2} + 2 x + x + 2 = x^{2} + 3 x + 2

Step 2:

(x^{2} + 3 x + 2) (x + 3)

= x^{3} + 3 x^{2} + 3 x^{2} + 9 x + 2 x + 6

Result:

= x^{3} + 6 x^{2} + 11 x + 6

Worked Example04

Expand and simplify: $(x + 2) (x - 1) (x + 3)$

1

First expand

(x + 2) (x - 1)

:

2

= x^{2} - x + 2 x - 2 = x^{2} + x - 2

3

Now multiply by

(x + 3)

:

4

(x^{2} + x - 2) (x + 3)

5

= x^{3} + 3 x^{2} + x^{2} + 3 x - 2 x - 6

6

= x^{3} + 4 x^{2} + x - 6

Answer: $x^{3} + 4 x^{2} + x - 6$

Expanding with Fractional Indices

The same bracket expansion rules apply when working with fractional powers:

Key Rule

x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x^{1} = x

Remember: Add powers when multiplying same bases

Single Bracket

x^{\frac{1}{2}} (x^{\frac{1}{2}} + 2)

= x + 2 x^{\frac{1}{2}}

DOTS with Surds

(x^{\frac{1}{2}} + 3) (x^{\frac{1}{2}} - 3)

= x - 9

Remember

x^{\frac{1}{2}} = x

and

x^{\frac{1}{3}} = 3 x

. Use index laws to simplify products.

Worked Example05

Expand and simplify: $(x^{\frac{1}{2}} + 3) (x^{\frac{1}{2}} - 2)$

1

Use FOIL method:

2

First:

x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x^{\frac{1}{2} + \frac{1}{2}} = x^{1} = x

3

Outer:

x^{\frac{1}{2}} \times (- 2) = - 2 x^{\frac{1}{2}}

4

Inner:

3 \times x^{\frac{1}{2}} = 3 x^{\frac{1}{2}}

5

Last:

3 \times (- 2) = - 6

6

Combine:

x - 2 x^{\frac{1}{2}} + 3 x^{\frac{1}{2}} - 6 = x + x^{\frac{1}{2}} - 6

Answer: $x + x^{\frac{1}{2}} - 6$ or equivalently

x + x - 6

Factorising: Common Factors

Factorising is the reverse of expanding - we write an expression as a product of factors.

Method

Identify the Highest Common Factor (HCF) of all terms

Write the HCF outside a bracket

Divide each term by the HCF to find what goes inside

Check by expanding!

Example 1

6 x + 15 = 3 (2 x + 5)

HCF = 3

Example 2

4 x^{2} - 10 x = 2 x (2 x - 5)

HCF = 2x

Factorise Completely

6 x^{3} - 15 x^{2} + 9 x = 3 x (2 x^{2} - 5 x + 3)

✓

NOT just

3 (2 x^{3} - 5 x^{2} + 3 x)

- you can take out more!

Worked Example06

Factorise completely: $6 x^{\frac{3}{2}} - 15 x^{\frac{1}{2}}$

1

Find HCF of coefficients: HCF(6, 15) = 3

2

Find HCF of powers: HCF(

x^{\frac{3}{2}}, x^{\frac{1}{2}}

) =

x^{\frac{1}{2}}

(take smallest power)

3

Overall HCF =

3 x^{\frac{1}{2}}

4

Factor out:

6 x^{\frac{3}{2}} \div 3 x^{\frac{1}{2}} = 2 x^{\frac{3}{2} - \frac{1}{2}} = 2 x

5

Factor out:

- 15 x^{\frac{1}{2}} \div 3 x^{\frac{1}{2}} = - 5

Answer: $3 x^{\frac{1}{2}} (2 x - 5)$ or equivalently

3 x (2 x - 5)

Factorising Quadratics (Monic)

A monic quadratic has coefficient 1 in front of x²:

Method for x² + bx + c

Find two numbers

p

and

q

where:

p \times q = c

and

p + q = b

Then:

x^{2} + b x + c = (x + p) (x + q)

Example

Factorise

x^{2} + 7 x + 12

:

Need:

p \times q = 12

and

p + q = 7

Factor pairs of 12: (1,12), (2,6), (3,4)

3 + 4 = 7

✓

Answer:

x^{2} + 7 x + 12 = (x + 3) (x + 4)

Worked Example07

Factorise: $x^{2} + 5 x y + 6 y^{2}$

1

This is like

x^{2} + 5 x + 6

but with y as a "multiplier"

2

Need two numbers: product = 6, sum = 5

3

Factor pairs of 6: (1,6), (2,3)

4

2 + 3 = 5

✓

5

Write as

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

- note the y in each bracket

6

Check:

(x + 2 y) (x + 3 y) = x^{2} + 3 x y + 2 x y + 6 y^{2} = x^{2} + 5 x y + 6 y^{2}

✓

Answer: $(x + 2 y) (x + 3 y)$

Factorising Quadratics (Non-Monic)

For

a x^{2} + b x + c

where

a \neq = 1

, use the AC method:

AC Method

Calculate

a c

Find two numbers that multiply to

a c

and add to

b

Split the middle term using these numbers

Factorise by grouping

Example

Factorise

2 x^{2} + 7 x + 3

:

a c = 2 \times 3 = 6

Need two numbers: product = 6, sum = 7 → 6 and 1

2 x^{2} + 6 x + 1 x + 3

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

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

Worked Example08

Factorise: $6 x^{2} + 7 x - 3$

1

Use AC method:

a c = 6 \times (- 3) = - 18

2

Need two numbers: product = −18, sum = 7

3

Try: 9 and −2 → 9 + (−2) = 7 ✓ and 9 × (−2) = −18 ✓

4

Split middle term:

6 x^{2} + 9 x - 2 x - 3

5

Group:

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

Answer: $(3 x - 1) (2 x + 3)$

Difference of Two Squares

Formula

a^{2} - b^{2} = (a + b) (a - b)

Recognise expressions that can be written as "something² minus something²":

Example 1

x^{2} - 25

= x^{2} - 5^{2}

= (x + 5) (x - 5)

Example 2

4 x^{2} - 9

= (2 x)^{2} - 3^{2}

= (2 x + 3) (2 x - 3)

Remember

a^{2} + b^{2}

CANNOT be factorised using real numbers!

Worked Example09

Factorise: $4 x^{2} - 25 y^{2}$

1

Recognise this as difference of two squares

2

4 x^{2} = (2 x)^{2}

3

25 y^{2} = (5 y)^{2}

4

So

4 x^{2} - 25 y^{2} = (2 x)^{2} - (5 y)^{2}

5

Apply DOTS:

a^{2} - b^{2} = (a + b) (a - b)

where

a = 2 x

and

b = 5 y

Answer: $(2 x + 5 y) (2 x - 5 y)$

Practice

Question 1: Expand and simplify:

(2 x + 3) (x - 4) (x + 1)

Step 1: Expand

(2 x + 3) (x - 4)

:

= 2 x^{2} - 8 x + 3 x - 12 = 2 x^{2} - 5 x - 12

Step 2: Multiply by

(x + 1)

:

(2 x^{2} - 5 x - 12) (x + 1)

= 2 x^{3} + 2 x^{2} - 5 x^{2} - 5 x - 12 x - 12

= 2 x^{3} - 3 x^{2} - 17 x - 12

Answer: $2 x^{3} - 3 x^{2} - 17 x - 12$

Practice

Question 2: Expand and simplify:

(x^{\frac{1}{2}} - 2)^{2}

Step 1: Use perfect square formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

Step 2: Here

a = x^{\frac{1}{2}}

and

b = 2

a^{2} = (x^{\frac{1}{2}})^{2} = x

2 ab = 2 \times x^{\frac{1}{2}} \times 2 = 4 x^{\frac{1}{2}}

b^{2} = 4

Answer: $x - 4 x^{\frac{1}{2}} + 4$ or equivalently

x - 4 x + 4

Practice

Question 3: Factorise:

x^{2} - 4 x y - 12 y^{2}

Step 1: This is like

x^{2} - 4 x - 12

with y as a "multiplier"

Step 2: Need two numbers: product = −12, sum = −4

Factor pairs of 12: (1,12), (2,6), (3,4)

We need −6 and +2 since −6 × 2 = −12 and −6 + 2 = −4 ✓

Step 3: Write with y in each factor

Answer: $(x - 6 y) (x + 2 y)$

Summary

Key Formulas

TypeFormula

Single bracket

a (b + c) = ab + a c

Double bracket

(a + b) (c + d) = a c + a d + b c + b d

Perfect square

(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}

Difference of squares

a^{2} - b^{2} = (a + b) (a - b)

Key Exam Skills

Fractional indices follow the same rules: add powers when multiplying

With two variables, the second variable appears in each factor

DOTS works with surds:

x - 9 = (x + 3) (x - 3)

"Factorise completely" means check for further factorisation