Chapter 7 · Higher tier

GCSE Maths Algebra Basics (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Use algebraic notation correctly and know the words expression, equation, formula, identity, term, coefficient and factor
Simplify expressions by collecting like terms, and multiply and divide algebraic terms using the laws of indices
Expand single brackets, double brackets (and three brackets at Higher tier)
Factorise common factors, quadratics of the form x² + bx + c and the difference of two squares (and ax² + bx + c at Higher tier)
Tell the difference between an equation and an identity, show that expressions are equivalent and construct algebraic proofs

You should already know

Negative numbers and BIDMAS (Chapter 1)
Factors, HCF and index laws with numbers (Chapter 2)
Fractions and simplifying them (Chapter 3)

Where this leads

Substitution and rearranging formulae (Chapter 8)
Solving linear equations (Chapter 9)
Solving quadratic equations by factorising (Chapter 14)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: A1, A3, A4, A6.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

7.1 The language of algebra: Worksheet (PDF) · Answers (PDF) · lesson page
7.2 Simplifying: collecting like terms: Worksheet (PDF) · Answers (PDF) · lesson page
7.3 Multiplying and dividing terms, and the laws of indices: Worksheet (PDF) · Answers (PDF) · lesson page
7.4 Expanding single brackets: Worksheet (PDF) · Answers (PDF) · lesson page
7.5 Expanding double brackets: Worksheet (PDF) · Answers (PDF) · lesson page
7.6 Factorising: common factors: Worksheet (PDF) · Answers (PDF) · lesson page
7.7 Factorising x² + bx + c and the difference of two squares: Worksheet (PDF) · Answers (PDF) · lesson page
7.8 Factorising ax² + bx + c: Worksheet (PDF) · Answers (PDF) · lesson page
7.9 Identities and algebraic proof: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWork out (a) −3 × 5 (b) −4 × −6 (c) 12 ÷ −3

(a)−3 × 5

(b)−4 × −6

Done

FWrite 3⁴ × 3² as a single power of 3, and then work out its value.

Done

FList the factors of 18 and the factors of 24. Write down their highest common factor (HCF).

Done

FWork out (a) 5 + 3 × 4 (b) 2 × (3 + 4)

(a)5 + 3 × 4

(b)2 × (3 + 4)

Done

FWork out 7 × 15 by splitting 15 into 10 + 5.

Done

7.1

The language of algebra

Edexcel 1MA1: A1, A3
Grade 1-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The language of algebraComing soon

Algebra is a way of writing about numbers when we do not know the number yet, or when the number can change. Instead of a box or a question mark we use a letter. A letter that stands for a number is called a variable (or an unknown). For example, if a pack holds p pencils, then 5 packs hold 5 × p pencils, whatever p is.

Mathematicians are lazy writers, so algebra has short-cuts. Learn these rules first. You will see them on every page of this book.

In words	Written in algebra	Why
a times b	ab	the × sign is left out
y + y + y	3y	and 3y also means 3 × y
a × a	a²	the little 2 is the index (or power)
a × a × a	a³	three copies multiplied
a × a × b	a²b	the 2 belongs to a only
a ÷ b	ab	the division sign becomes a fraction line
half of x	x2 (not 0.5x)	in algebra we write a fraction rather than a decimal
2 times (a + b)	2(a + b)	a number next to a bracket means multiply

The parts of an expression. A term is one part of an expression, with the sign in front of it.

The vocabulary of algebra

Algebra has its own words. Examination questions use them without explaining them, so make sure you can say what each one means.

Word	Meaning	Example
Term	one part of an expression, joined to the others by + or −	4x², −7x and 5 are the terms of 4x² − 7x + 5
Coefficient	the number multiplying the letters in a term	the coefficient of 4x² is 4; the coefficient of x is 1
Constant	a term that is just a number	the 5 in 4x² − 7x + 5
Factor	something that is multiplied; it divides exactly into the term	2 and x are factors of 2x; (x + 1) is a factor of 3(x + 1)
Expression	letters, numbers and operations, with no equals sign	3x + 2
Equation	two expressions joined by =, true for particular values of the letter	3x + 2 = 11 (true only when x = 3)
Formula	a rule linking quantities, with several letters, that you use by substituting	A = lw (area of a rectangle)
Identity	an equality that is true for every value of the letter; written with ≡	2(x + 3) ≡ 2x + 6
Inequality	two expressions compared with <, >, ≤ or ≥	x + 1 > 4

Key ideas

A letter stands for a number. 3y means 3 × y, and ab means a × b.
a² = a × a and a³ = a × a × a. The index only belongs to the letter it sits on.
a ÷ b is written ab. Write fractional coefficients as fractions (x2), not decimals (0.5x).
An expression has no equals sign. An equation has one and can be solved. A formula links quantities. An identity is true for every value.
The sign in front of a term belongs to that term.

Worked example 7.1.1

Write each of these using algebraic notation: (a) p × q (b) 4 × m (c) n × n × n (d) 5 × a × b × b (e) k ÷ 5.

Method. Use the short-cuts: leave out × signs, number first, repeated letters become powers, divide becomes a fraction.

Recall the rules. Every short-cut is just a way of writing the same thing in less space.
ab means a × b; 3y means 3 × y; a² means a × a
Do (a) and (b): just remove the × sign and put the number first. The number goes at the front by convention.
p × q = pq
4 × m = 4m
Do (c): count the repeated letters. Three copies of n multiplied together is n cubed, written with an index 3.
n × n × n = n³
Do (d): the number first, then each letter, grouping the repeated one. There is one a and two bs.
5 × a × b × b = 5ab²
Do (e): a division becomes a fraction. The fraction bar means divide.
k ÷ 5 = k5

Check. Put a = 2 and b = 3 into (d): 5 × 2 × 3 × 3 = 90 and 5ab² = 5 × 2 × 3² = 5 × 2 × 9 = 90. ✓

Watch out. In 5ab² only the b is squared, not the 5 and not the a.

Worked example 7.1.2

Look at the expression 5x² − 3x + 7. (a) How many terms does it have? (b) What is the coefficient of x²? (c) What is the coefficient of x? (d) What is the constant term? (e) Is it an expression, an equation or a formula?

Key words. A term is one part of an expression. The coefficient is the number multiplying the letters. A constant is a term with no letter.

Method. Split the expression into terms at each + or − sign, and keep each sign with the term after it.

Split into terms. A term is one part of the expression. The sign in front belongs to the term.
5x² and −3x and +7
Count them: (a).
There are 3 terms.
Find the coefficients: (b) and (c). The coefficient is the number multiplying the letters, including its sign.
In 5x² the coefficient is 5. In −3x the coefficient is −3 (the minus sign belongs to it).
Find the constant term: (d). The constant is the term with no letter.
The constant term is +7.
Say what kind of thing it is: (e). There is no equals sign, so it cannot be an equation. A formula needs an equals sign too.
It is an expression.

Check. Join the terms: 5x² + (−3x) + 7 = 5x² − 3x + 7. ✓

Watch out. The coefficient of −3x is −3, not 3. And the coefficient of a lone x is 1.

Worked example 7.1.3

Say whether each of these is an expression, an equation, a formula, an identity or an inequality: (a) 4a − 9 (b) 2x + 1 = 15 (c) P = 2l + 2w (d) 3(x + 2) ≡ 3x + 6 (e) y − 2 < 7.

Key words. An expression has no equals sign. An equation is true for particular values. A formula is a rule linking quantities. An identity is true for every value.

Method. Look for the sign in the middle: none, =, ≡ or an inequality sign, and think about what the statement does.

(a) 4a − 9. There is no equals sign and no inequality sign. It is just a collection of terms.
Expression
(b) 2x + 1 = 15. It has an equals sign and is only true for one value of x (x = 7). Something you can solve is an equation.
Equation
(c) P = 2l + 2w. It links several quantities (perimeter, length and width) and you can use it by substituting numbers. A rule like this is a formula.
Formula
(d) 3(x + 2) ≡ 3x + 6. The sign ≡ means "is always equal to". Expanding the bracket gives 3x + 6, so both sides are the same for every x.
Identity
(e) y − 2 < 7. The sign < compares the two sides. It is true for lots of values of y.
Inequality

Check. Test the equation in (b): x = 7 gives 2 × 7 + 1 = 15, but x = 3 gives 7, which is not 15. So it is only true for some values. ✓

Watch out. A formula and an equation both have an equals sign. The difference is the job: an equation is solved to find the unknown, and a formula is a general rule you substitute into.

Worked example 7.1.4

A taxi charges a fixed fee of £3 plus £2 for each mile. (a) Write an expression for the cost, in pounds, of a journey of m miles. (b) Write a formula for the cost C. (c) A different taxi company charges 80p per mile. Write an expression, in pounds, for a journey of m miles.

Method. Break the problem into the fixed part and the part that changes with the number of miles.

Find the part that does not change. The fixed fee is paid whatever the distance.
fixed part = 3
Find the part that changes. Each mile costs £2, so m miles cost 2 × m.
changing part = 2 × m = 2m
(a) Add them. The total is the fixed part plus the changing part.
Expression: 3 + 2m (or 2m + 3)
(b) Make it a formula. A formula has an equals sign and says what the letter C stands for.
C = 3 + 2m
(c) Change pence to pounds first. 80p = £0.80 = £45, and we write the coefficient as a fraction or keep to pence. Here: 80p × m = 80m pence, which is £0.8m.
Expression: 0.8m pounds (this is a price, so a decimal is sensible)

Check. Try m = 4 miles: the first taxi costs 3 + 2 × 4 = £11. Counting it out: £3 + four lots of £2 = 3 + 8 = £11. ✓

Watch out. Keep the units the same. Do not mix pounds and pence in one expression.

Worked example 7.1.5

A two-digit number has tens digit t and units digit u. (a) Write an expression for the value of the number. (b) Write an expression for the number you get by swapping the digits. (c) Write an expression for the sum of the two numbers.

Method. Use place value: the tens digit is worth ten times its face value.

Think of a real example. In 53 the tens digit is 5 and the units digit is 3. The value is 5 × 10 + 3.
53 = 5 × 10 + 3
(a) Do the same with letters. Tens digit t is worth 10t. The units digit u is worth u.
10t + u
(b) Swap the digits. Now u is in the tens place and t is in the units place.
10u + t
(c) Add the two expressions. Collect the like terms: the t terms together and the u terms together.
(10t + u) + (10u + t) = 10t + t + 10u + u = 11t + 11u

Check. Try 53 and 35: 53 + 35 = 88. The expression gives 11 × 5 + 11 × 3 = 55 + 33 = 88. ✓

Watch out. 10t means 10 × t. Do not write t10.

Watch out

3a and a³ are not the same. 3a means a + a + a and a³ means a × a × a. Try a = 2: 3a = 6 but a³ = 8. Also, 4a means 4 × a, not 4 + a.

At Higher tier you are also expected to write expressions for consecutive numbers, even numbers and odd numbers. If n is a whole number, then n + 1 is the next whole number, 2n is always even (it is 2 lots of something) and 2n + 1 is always odd. You will use these in the proofs at the end of this chapter.

7.2

Simplifying: collecting like terms

Edexcel 1MA1: A4
Grade 1-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Simplifying: collecting like termsComing soon

To simplify an expression means to write it in a shorter, neater way without changing its value. The most important way to do that is collecting like terms.

Think about fruit. 3 apples and 2 apples make 5 apples. But 3 apples and 2 pears do not make 5 of anything useful; you just have “3 apples and 2 pears”. Algebra works in exactly the same way. The terms 3x and 2x are the same kind of thing (lots of x), so we can add them: 3x + 2x = 5x. But 3x + 2y cannot be simplified, because x and y are different things.

Algebra tiles: each blue rod is one x and each orange square is 1. Rods can be added to rods, and squares to squares, but a rod cannot be added to a square.

Like terms have exactly the same letters, with exactly the same powers. The number in front (the coefficient) can be different.

Like terms (can be added)	Unlike terms (cannot be added)
3x and 7x	3x and 7y (different letters)
5x² and −x²	5x² and 5x (different powers)
4ab and 9ba (the order of letters does not matter)	4ab and 4a (one has a b, one does not)
6 and −2 (plain numbers)	6 and 6x (one has a letter)

Method. (1) Underline or circle each term, with the sign in front of it. (2) Group the like terms together. (3) Add or subtract the coefficients. A letter on its own, like x, has a coefficient of 1.

Key ideas

Like terms have the same letters with the same powers. 3x and 5x are like; 3x and 5x² are not.
Add the coefficients and keep the letters: 3x + 5x = 8x (the x does not change).
The sign in front of a term belongs to the term: in 7x − 3y the terms are +7x and −3y.
x means 1x, so x + 4x = 5x. And −x means −1x.
xy and yx are the same term (multiplication can be done in any order).

Worked example 7.2.1

Simplify 3x + 5x − 2x.

Key words. A term is one part of an expression, together with the sign in front of it (for example 3x or −5). Like terms have exactly the same letters with exactly the same powers (3x and 5x are like terms, 3x and 5x² are not). The coefficient is the number in front of the letters.

Method. Collect like terms: group the terms that have the same letters and powers, then add or subtract their coefficients (the numbers in front).

Write the terms out one by one, each with the sign in front of it. A term is one part of the expression. The sign in front of a term belongs to that term, so in 5x − 2y the terms are +5x and −2y. (A term with no sign in front is positive.)
+3x, +5x, −2x
Sort the terms into groups of like terms. Like terms have the same letters with the same powers, so they are the same kind of thing. We can only add the same kind of thing: 3 apples + 2 apples = 5 apples, but 3 apples + 2 pears cannot be made into one number. Remember that x and x² are different kinds of term, and so are x and xy.
x terms: +3x, +5x, −2x
add the coefficients: 3 + 5 − 2 = 6 so they give 6x
Write the answer. Join the results of the groups. Put the term with the highest power first, and the number last.
6x

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives 18 and the answer gives 18. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. The sign in front of a term goes with it. In 7x − 3x you are taking away 3 lots of x. Also, a letter on its own, such as x, has a coefficient of 1 (and −x has −1).

Worked example 7.2.2

Simplify 4a + 3b − a + 5b − 2 + 7.

Method. Collect like terms: group the terms that have the same letters and powers, then add or subtract their coefficients (the numbers in front).

Write the terms out one by one, each with the sign in front of it. A term is one part of the expression. The sign in front of a term belongs to that term, so in 5x − 2y the terms are +5x and −2y. (A term with no sign in front is positive.)
+4a, +3b, −a, +5b, −2, +7
Sort the terms into groups of like terms. Like terms have the same letters with the same powers, so they are the same kind of thing. We can only add the same kind of thing: 3 apples + 2 apples = 5 apples, but 3 apples + 2 pears cannot be made into one number. Remember that x and x² are different kinds of term, and so are x and xy.
a terms: +4a, −a
   add the coefficients: 4 − 1 = 3 so they give 3a
b terms: +3b, +5b
   add the coefficients: 3 + 5 = 8 so they give 8b
number terms: −2, +7
   add the coefficients: −2 + 7 = 5 so they give 5
Write the answer. Join the results of the groups. Put the term with the highest power first, and the number last.
3a + 8b + 5

Check. Test with numbers. Put a = 3, b = 4 into the original and into your answer. The original gives 46 and the answer gives 46. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. The sign in front of a term goes with it. In 7x − 3x you are taking away 3 lots of x. Also, a letter on its own, such as x, has a coefficient of 1 (and −x has −1).

Worked example 7.2.3

Simplify 3x² + 2x − x² + 5x − 4.

Method. Collect like terms: group the terms that have the same letters and powers, then add or subtract their coefficients (the numbers in front).

Write the terms out one by one, each with the sign in front of it. A term is one part of the expression. The sign in front of a term belongs to that term, so in 5x − 2y the terms are +5x and −2y. (A term with no sign in front is positive.)
+3x², +2x, −x², +5x, −4
Sort the terms into groups of like terms. Like terms have the same letters with the same powers, so they are the same kind of thing. We can only add the same kind of thing: 3 apples + 2 apples = 5 apples, but 3 apples + 2 pears cannot be made into one number. Remember that x and x² are different kinds of term, and so are x and xy.
x² terms: +3x², −x²
add the coefficients: 3 − 1 = 2 so they give 2x²
x terms: +2x, +5x
add the coefficients: 2 + 5 = 7 so they give 7x
number terms: −4 (only one of this type, so it stays as it is)
Write the answer. Join the results of the groups. Put the term with the highest power first, and the number last.
2x² + 7x − 4

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives 35 and the answer gives 35. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. The sign in front of a term goes with it. In 7x − 3x you are taking away 3 lots of x. Also, a letter on its own, such as x, has a coefficient of 1 (and −x has −1).

Worked example 7.2.4

Simplify 4ab − 7ba + 2a² + ab.

Method. Collect like terms: group the terms that have the same letters and powers, then add or subtract their coefficients (the numbers in front).

Write the terms out one by one, each with the sign in front of it. A term is one part of the expression. The sign in front of a term belongs to that term, so in 5x − 2y the terms are +5x and −2y. (A term with no sign in front is positive.)
+4ab, −7ab, +2a², +ab
Sort the terms into groups of like terms. Like terms have the same letters with the same powers, so they are the same kind of thing. We can only add the same kind of thing: 3 apples + 2 apples = 5 apples, but 3 apples + 2 pears cannot be made into one number. Remember that x and x² are different kinds of term, and so are x and xy.
a² terms: +2a² (only one of this type, so it stays as it is)
ab terms: +4ab, −7ab, +ab
add the coefficients: 4 − 7 + 1 = −2 so they give −2ab
Write the answer. Join the results of the groups. Put the term with the highest power first, and the number last.
2a² − 2ab

Check. Test with numbers. Put a = 3, b = 4 into the original and into your answer. The original gives −6 and the answer gives −6. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. The sign in front of a term goes with it. In 7x − 3x you are taking away 3 lots of x. Also, a letter on its own, such as x, has a coefficient of 1 (and −x has −1).

Worked example 7.2.5

Simplify 3x²y + 5xy² − x²y + 2xy² − 4xy.

Method. Collect like terms: group the terms that have the same letters and powers, then add or subtract their coefficients (the numbers in front).

Write the terms out one by one, each with the sign in front of it. A term is one part of the expression. The sign in front of a term belongs to that term, so in 5x − 2y the terms are +5x and −2y. (A term with no sign in front is positive.)
+3x²y, +5xy², −x²y, +2xy², −4xy
Sort the terms into groups of like terms. Like terms have the same letters with the same powers, so they are the same kind of thing. We can only add the same kind of thing: 3 apples + 2 apples = 5 apples, but 3 apples + 2 pears cannot be made into one number. Remember that x and x² are different kinds of term, and so are x and xy.
x²y terms: +3x²y, −x²y
add the coefficients: 3 − 1 = 2 so they give 2x²y
xy² terms: +5xy², +2xy²
add the coefficients: 5 + 2 = 7 so they give 7xy²
xy terms: −4xy (only one of this type, so it stays as it is)
Write the answer. Join the results of the groups. Put the term with the highest power first, and the number last.
2x²y + 7xy² − 4xy

Check. Test with numbers. Put x = 3, y = 4 into the original and into your answer. The original gives 360 and the answer gives 360. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. The sign in front of a term goes with it. In 7x − 3x you are taking away 3 lots of x. Also, a letter on its own, such as x, has a coefficient of 1 (and −x has −1).

Watch out

x and x² are different kinds of term, so 3x + 2x² cannot be simplified to 5x³ or 5x². Also watch the signs: in 5x − 2x + 3 the term being added to 5x is −2x.

7.3

Multiplying and dividing terms, and the laws of indices

Edexcel 1MA1: A1, A4
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Multiplying and dividing terms, and the laws of indicesComing soon

You already know how to multiply numbers. Multiplying algebraic terms follows the same idea: multiply the numbers, then deal with the letters.

What is x × x? Picture a square with sides of length x. Its area is x × x, which we write as x² (“x squared”). The little 2 is called the index or power. It says how many copies of the letter are multiplied together.

A square with side x has area x × x = x².

Multiplying terms

3a × 4b means 3 × a × 4 × b. Multiplication can be done in any order, so we can swap the numbers round: 3 × 4 × a × b = 12ab. When the same letter appears in both terms, count the copies. 5x × 2x = 10 × x × x = 10x².

The laws of indices

Writing the letters out in full shows why these rules work. For example x³ × x² = (x × x × x) × (x × x), which is five xs multiplied together, so it is x⁵.

Law	Example	In words
a^m × aⁿ = a^m+n	x³ × x⁴ = x⁷	multiplying: add the powers
a^m ÷ aⁿ = a^m−n	x⁵ ÷ x² = x³	dividing: subtract the powers
(a^m)ⁿ = a^mn	(x²)³ = x⁶	a power of a power: multiply the powers
a¹ = a	y = y¹	a letter on its own has power 1
a⁰ = 1	x⁴ ÷ x⁴ = x⁰ = 1	anything (except 0) to the power 0 is 1

The number parts of the terms are just multiplied or divided as usual. Only the letter parts use the index laws. And when there is a power outside a bracket, it applies to everything in the bracket: (2x³)² = 2² × (x³)² = 4x⁶.

You have met these laws with numbers in Chapter 2 (for example 2³ × 2⁴ = 2⁷). Here the base is a letter, but nothing else changes.

Dividing terms

To divide terms, divide the numbers and then subtract the powers of each letter. It is the same as cancelling. 12x⁵ ÷ 3x² = 12x⁵3x² = 4x³. You can write a division of terms as a fraction and cancel the numbers and the letters.

Negative and fractional indices

The same laws still work when the indices are negative or fractions. A negative index means “one over”: x⁻² = 1x². A fractional index means a root: x^1/2 = √x, and x^1/3 is the cube root of x. These are covered with numbers in Chapter 6; here we use the laws with letters. Example: x⁻² × x⁵ = x⁻²⁺⁵ = x³.

Key ideas

Multiply terms: multiply the numbers, and add the powers of each letter. 3x² × 4x³ = 12x⁵.
Divide terms: divide the numbers, and subtract the powers. 12x⁵ ÷ 3x² = 4x³.
Power of a power: multiply the powers. (x²)³ = x⁶. A power outside a bracket applies to the number and every letter inside.
a¹ = a and a⁰ = 1.
The laws only work when the letters (the bases) are the same.

Worked example 7.3.1

Simplify 2a²b × 3ab³.

Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x³ = x × x × x. The coefficient is the number in front of the letters.

Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.

Write the question. We are going to simplify one product or quotient of terms.
2a²b × 3ab³
Multiply the terms: first the numbers, then each letter. Multiply the numbers (coefficients) together. For each letter, add the powers, because x² × x³ = (x × x) × (x × x × x) = x⁵ (the law a^m × aⁿ = a^m+n). A letter written on its own has power 1.
Numbers: 2 × 3 = 6
a: add the powers 2 + 1 = 3
b: add the powers 1 + 3 = 4
So the top is 6a³b⁴
Write the answer. Put the number first, then the letters.
6a³b⁴

Check. Test with numbers: put a = 2, b = 3 into the question and into the answer. The question gives 3888 and the answer gives 3888. They are the same. ✓

Watch out. Add the powers when you multiply (x² × x³ = x⁵, not x⁶). Multiply the powers when there is a power of a power. And only the powers change: 3x × 2x = 6x², the 3 and 2 are multiplied, not added.

Worked example 7.3.2

Simplify 5x × 2x.

Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x³ = x × x × x. The coefficient is the number in front of the letters.

Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.

Write the question. We are going to simplify one product or quotient of terms.
5x × 2x
Multiply the terms: first the numbers, then each letter. Multiply the numbers (coefficients) together. For each letter, add the powers, because x² × x³ = (x × x) × (x × x × x) = x⁵ (the law a^m × aⁿ = a^m+n). A letter written on its own has power 1.
Numbers: 5 × 2 = 10
x: add the powers 1 + 1 = 2
So the top is 10x²
Write the answer. Put the number first, then the letters.
10x²

Check. Test with numbers: put x = 2 into the question and into the answer. The question gives 40 and the answer gives 40. They are the same. ✓

Worked example 7.3.3

Simplify 15a³b²5ab.

Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x³ = x × x × x. The coefficient is the number in front of the letters.

Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.

Write the question. We are going to simplify one product or quotient of terms.
15a³b²5ab
Now divide: divide the numbers, and for each letter subtract the powers. Dividing by a term is the same as cancelling. x⁵ ÷ x² = (x × x × x × x × x) ÷ (x × x) = x³ (the law a^m ÷ aⁿ = a^m−n). Divide the numbers as you would divide any numbers.
Numbers: 15 ÷ 5 = 3
a: subtract the powers 3 − 1 = 2
b: subtract the powers 2 − 1 = 1
Write the answer. Put the number first, then the letters.
3a²b

Check. Test with numbers: put a = 2, b = 3 into the question and into the answer. The question gives 36 and the answer gives 36. They are the same. ✓

Worked example 7.3.4

Simplify (3ab²)³.

Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x³ = x × x × x. The coefficient is the number in front of the letters.

Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.

Write the question. We are going to simplify one product or quotient of terms.
(3ab²)³
Deal with any power outside a bracket first. A power outside a bracket applies to everything inside it: the number and every letter. For example (2x³)² means 2x³ × 2x³. The number is raised to the power, and each letter's own power is multiplied by it (the law (a^m)ⁿ = a^mn).
(3ab²)³ → number: 3³ = 27; a: 1 × 3 = 3; b: 2 × 3 = 6 → 27a³b⁶
Write the answer. Put the number first, then the letters.
27a³b⁶

Check. Test with numbers: put a = 2, b = 3 into the question and into the answer. The question gives 157464 and the answer gives 157464. They are the same. ✓

Worked example 7.3.5

Simplify (4x³y)² × 3xy6x²y.

Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x³ = x × x × x. The coefficient is the number in front of the letters.

Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.

Write the question. We are going to simplify one product or quotient of terms.
(4x³y)² × 3xy6x²y
Deal with any power outside a bracket first. A power outside a bracket applies to everything inside it: the number and every letter. For example (2x³)² means 2x³ × 2x³. The number is raised to the power, and each letter's own power is multiplied by it (the law (a^m)ⁿ = a^mn).
(4x³y)² → number: 4² = 16; y: 1 × 2 = 2; x: 3 × 2 = 6 → 16x⁶y²
Multiply the terms: first the numbers, then each letter. Multiply the numbers (coefficients) together. For each letter, add the powers, because x² × x³ = (x × x) × (x × x × x) = x⁵ (the law a^m × aⁿ = a^m+n). A letter written on its own has power 1.
Numbers: 16 × 3 = 48
x: add the powers 6 + 1 = 7
y: add the powers 2 + 1 = 3
So the top is 48x⁷y³
Now divide: divide the numbers, and for each letter subtract the powers. Dividing by a term is the same as cancelling. x⁵ ÷ x² = (x × x × x × x × x) ÷ (x × x) = x³ (the law a^m ÷ aⁿ = a^m−n). Divide the numbers as you would divide any numbers.
Numbers: 48 ÷ 6 = 8
x: subtract the powers 7 − 2 = 5
y: subtract the powers 3 − 1 = 2
Write the answer. Put the number first, then the letters.
8x⁵y²

Check. Test with numbers: put x = 2, y = 3 into the question and into the answer. The question gives 2304 and the answer gives 2304. They are the same. ✓

Worked example 7.3.6

Simplify x⁻² × 3x⁵.

Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x³ = x × x × x. The coefficient is the number in front of the letters.

Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.

Write the question. We are going to simplify one product or quotient of terms.
x⁻² × 3x⁵
Multiply the terms: first the numbers, then each letter. Multiply the numbers (coefficients) together. For each letter, add the powers, because x² × x³ = (x × x) × (x × x × x) = x⁵ (the law a^m × aⁿ = a^m+n). A letter written on its own has power 1.
Numbers: 1 × 3 = 3
x: add the powers −2 + 5 = 3
So the top is 3x³
Write the answer. Put the number first, then the letters.
3x³

Check. Test with numbers: put x = 2 into the question and into the answer. The question gives 24 and the answer gives 24. They are the same. ✓

Worked example 7.3.7

Simplify (27x⁶)^1/3.

Key words. The power or index (plural indices) tells you how many copies of the letter are multiplied together: x³ = x × x × x. The coefficient is the number in front of the letters.

Method. To multiply terms, multiply the numbers and add the powers of each letter. To divide, divide the numbers and subtract the powers. A power outside a bracket multiplies every power inside.

Write the question. We are going to simplify one product or quotient of terms.
(27x⁶)^1/3
Deal with any power outside a bracket first. A power outside a bracket applies to everything inside it: the number and every letter. For example (2x³)² means 2x³ × 2x³. The number is raised to the power, and each letter's own power is multiplied by it (the law (a^m)ⁿ = a^mn).
(27x⁶)^1/3 → number: 27^1/3 = 3; x: 6 × 1/3 = 2 → 3x²
Write the answer. Put the number first, then the letters.
3x²

Check. Test with numbers: put x = 2 into the question and into the answer. The question gives 12 and the answer gives 12. They are the same. ✓

Watch out

Only the powers are added when you multiply. 3x × 2x = 6x²: the numbers are multiplied (3 × 2 = 6), they are not added, and the power of x goes from 1 + 1 = 2. Also x³ × x⁴ is x⁷, not x¹². And x² + x³ cannot be simplified at all: the laws are for multiplying and dividing, not for adding.

7.4

Expanding single brackets

Edexcel 1MA1: A1, A4
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Expanding single bracketsComing soon

A number or letter written directly next to a bracket means multiply. So 3(x + 4) means 3 × (x + 4). Expanding (or multiplying out) the bracket means writing the same thing without the brackets.

Here is why the rule works. Picture a rectangle that is 3 units tall. Its width is made of two pieces: a piece of length x and a piece of length 4. The total area is 3 × (x + 4). But you can also find it by adding the areas of the two parts: 3 × x and 3 × 4.

The whole rectangle has area 3(x + 4). Adding the two parts gives 3x + 12.

So the rule is: multiply every term inside the bracket by the term outside: a(b + c) = ab + ac. It is the same idea you used in the “Do you remember?” questions: 7 × 15 = 7 × (10 + 5) = 70 + 35.

Letters outside the bracket. The rule is the same when the term outside is a letter. x(x + 3) = x × x + x × 3 = x² + 3x. Use the index laws from the last section when you multiply the letters.

The area of this rectangle is x(x + 3) and also x² + 3x.

Negative numbers. Use the sign rules for multiplying. A positive times a negative is negative, and a negative times a negative is positive. So −2(x − 5) = (−2 × x) + (−2 × −5) = −2x + 10. A minus sign in front of a bracket changes the sign of every term inside it.

Expand and simplify. Sometimes there are two brackets. Expand each bracket first, then collect the like terms (see Section 7.2).

Key ideas

a(b + c) = ab + ac. Multiply every term inside by the term outside.
A letter outside multiplies like any other term: x(x + 3) = x² + 3x.
Signs: positive × negative = negative; negative × negative = positive.
−(x − 4) means −1 × (x − 4) = −x + 4.
To expand and simplify: expand every bracket, then collect like terms.

Worked example 7.4.1

Expand 3(x + 4).

Key words. To expand a bracket means to multiply it out so that the brackets disappear. The term outside the bracket is multiplied by each term inside.

Method. To expand a bracket, multiply every term inside the bracket by the term outside it.

Say what the bracket means. A number or letter written next to a bracket means multiply. So 3(x + 4) means 3 multiplied by everything inside the bracket, not just the first term.
3(x + 4) = 3 × x + 3 × 4
Multiply the outside term by each term inside, one at a time. Multiply the numbers first, then the letters (add the powers of any letter that appears in both). All the signs are positive here, so every answer is positive.
3 × x = 3x
3 × 4 = 12
Put the answers together. The results join up as a single expression. Each keeps its own sign.
3x + 12

Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives 18 and the answer gives 18. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. Multiply every term in the bracket, including the last one. And be careful with minus signs: −3(x − 2) = −3x + 6, because −3 × −2 = +6.

Worked example 7.4.2

Expand 5(2a − 3).

Key words. To expand a bracket means to multiply it out so that the brackets disappear. The term outside the bracket is multiplied by each term inside.

Method. To expand a bracket, multiply every term inside the bracket by the term outside it.

Say what the bracket means. A number or letter written next to a bracket means multiply. So 5(2a − 3) means 5 multiplied by everything inside the bracket, not just the first term.
5(2a − 3): 5 is multiplied by each of the terms +2a, −3
Multiply the outside term by each term inside, one at a time. Multiply the numbers first, then the letters (add the powers of any letter that appears in both). Watch the signs: Use the sign rules: a positive times a positive is positive, a negative times a negative is positive, and a positive times a negative (either way round) is negative.
5 × 2a = 10a
5 × (−3) = −15
Put the answers together. The results join up as a single expression. Each keeps its own sign.
10a − 15

Check. Test with numbers. Put a = 2 into the original and into your answer. The original gives 5 and the answer gives 5. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. Multiply every term in the bracket, including the last one. And be careful with minus signs: −3(x − 2) = −3x + 6, because −3 × −2 = +6.

Worked example 7.4.3

Expand 3x(2x − 5).

Key words. To expand a bracket means to multiply it out so that the brackets disappear. The term outside the bracket is multiplied by each term inside.

Method. To expand a bracket, multiply every term inside the bracket by the term outside it.

Say what the bracket means. A number or letter written next to a bracket means multiply. So 3x(2x − 5) means 3x multiplied by everything inside the bracket, not just the first term.
3x(2x − 5): 3x is multiplied by each of the terms +2x, −5
Multiply the outside term by each term inside, one at a time. Multiply the numbers first, then the letters (add the powers of any letter that appears in both). Watch the signs: Use the sign rules: a positive times a positive is positive, a negative times a negative is positive, and a positive times a negative (either way round) is negative.
3x × 2x = 6x²
3x × (−5) = −15x
Put the answers together. The results join up as a single expression. Each keeps its own sign.
6x² − 15x

Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives −6 and the answer gives −6. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. Multiply every term in the bracket, including the last one. And be careful with minus signs: −3(x − 2) = −3x + 6, because −3 × −2 = +6.

Worked example 7.4.4

Expand −2(x − 5).

Key words. To expand a bracket means to multiply it out so that the brackets disappear. The term outside the bracket is multiplied by each term inside.

Method. To expand a bracket, multiply every term inside the bracket by the term outside it.

Say what the bracket means. A number or letter written next to a bracket means multiply. So −2(x − 5) means −2 multiplied by everything inside the bracket, not just the first term.
−2(x − 5): (−2) is multiplied by each of the terms +x, −5
Multiply the outside term by each term inside, one at a time. Multiply the numbers first, then the letters (add the powers of any letter that appears in both). Watch the signs: Use the sign rules: a positive times a positive is positive, a negative times a negative is positive, and a positive times a negative (either way round) is negative.
(−2) × x = −2x
(−2) × (−5) = 10
Put the answers together. The results join up as a single expression. Each keeps its own sign.
−2x + 10

Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives 6 and the answer gives 6. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. Multiply every term in the bracket, including the last one. And be careful with minus signs: −3(x − 2) = −3x + 6, because −3 × −2 = +6.

Worked example 7.4.5

Expand and simplify 3(x + 2) − 2(x − 5).

Method. Expand every bracket first (keeping each sign with its term), then collect the like terms.

Write the expression. There is more than one bracket. We expand each bracket separately, and then collect like terms.
3(x + 2) − 2(x − 5)
Expand each bracket on its own. Multiply every term in the bracket by the term in front of it, including the sign in front. A minus in front of a bracket means you multiply by a negative number, so every sign inside the bracket flips.
Bracket: 3(x + 2)
3 × x = 3x
3 × 2 = 6

Bracket: −2(x − 5)
(−2) × x = −2x
(−2) × (−5) = 10
Write all the pieces in one line. Now there are no brackets, only terms.
3x + 6 − 2x + 10
Collect the like terms. Group the terms with the same letters and powers and add their coefficients.
x terms: +3x, −2x
add the coefficients: 3 − 2 = 1 so they give x
number terms: +6, +10
add the coefficients: 6 + 10 = 16 so they give 16
Answer: x + 16

Watch out. A minus sign in front of a bracket changes the sign of every term inside it. −2(x − 5) = −2x + 10. Many mistakes are made with the last term.

Worked example 7.4.6

Expand 2xy(3x − 4y).

Key words. To expand a bracket means to multiply it out so that the brackets disappear. The term outside the bracket is multiplied by each term inside.

Method. To expand a bracket, multiply every term inside the bracket by the term outside it.

Say what the bracket means. A number or letter written next to a bracket means multiply. So 2xy(3x − 4y) means 2xy multiplied by everything inside the bracket, not just the first term.
2xy(3x − 4y): 2xy is multiplied by each of the terms +3x, −4y
Multiply the outside term by each term inside, one at a time. Multiply the numbers first, then the letters (add the powers of any letter that appears in both). Watch the signs: Use the sign rules: a positive times a positive is positive, a negative times a negative is positive, and a positive times a negative (either way round) is negative.
2xy × 3x = 6x²y
2xy × (−4y) = −8xy²
Put the answers together. The results join up as a single expression. Each keeps its own sign.
6x²y − 8xy²

Check. Test with numbers. Put x = 2, y = 3 into the original and into your answer. The original gives −72 and the answer gives −72. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. Multiply every term in the bracket, including the last one. And be careful with minus signs: −3(x − 2) = −3x + 6, because −3 × −2 = +6.

Worked example 7.4.7

Expand and simplify 2x(3x − 1) − x(x − 4).

Method. Expand every bracket first (keeping each sign with its term), then collect the like terms.

Write the expression. There is more than one bracket. We expand each bracket separately, and then collect like terms.
2x(3x − 1) − x(x − 4)
Expand each bracket on its own. Multiply every term in the bracket by the term in front of it, including the sign in front. A minus in front of a bracket means you multiply by a negative number, so every sign inside the bracket flips.
Bracket: 2x(3x − 1)
2x × 3x = 6x²
2x × (−1) = −2x

Bracket: −x(x − 4)
(−x) × x = −x²
(−x) × (−4) = 4x
Write all the pieces in one line. Now there are no brackets, only terms.
6x² − 2x − x² + 4x
Collect the like terms. Group the terms with the same letters and powers and add their coefficients.
x² terms: +6x², −x²
add the coefficients: 6 − 1 = 5 so they give 5x²
x terms: −2x, +4x
add the coefficients: −2 + 4 = 2 so they give 2x
Answer: 5x² + 2x

Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives 24 and the answer gives 24. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. A minus sign in front of a bracket changes the sign of every term inside it. −2(x − 5) = −2x + 10. Many mistakes are made with the last term.

Watch out

Multiply every term in the bracket, not just the first. 5(x − 3) is 5x − 15, not 5x − 3. And take care with a minus outside: −3(x − 2) is −3x + 6, because two negatives make a positive.

7.5

Expanding double brackets

Edexcel 1MA1: A4
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Expanding double bracketsComing soon

An expression like (x + 3)(x + 2) is a product of two brackets. We have already met “a number times a bracket”. Now there are two brackets, and the rule gets one step bigger: every term in the first bracket must be multiplied by every term in the second bracket.

The easiest way to see why is a picture. A rectangle has length (x + 3) and width (x + 2). Cut it into four parts. The area of the whole rectangle is (x + 3)(x + 2), and it is also the total of the four parts.

The four parts have areas x², 3x, 2x and 6. Total: x² + 3x + 2x + 6 = x² + 5x + 6.

This is the grid method: put the terms of one bracket along the top and the terms of the other down the side. Each box is the row term times the column term. Then add up all four boxes and collect the like terms. Another way to organise the same four multiplications is F-O-I-L: multiply the First terms, then the Outside terms, then the Inside terms, then the Last terms. Both methods give the same answer. Always count your pieces: two terms times two terms gives four pieces.

Negative terms. Keep the sign with the term, and use the sign rules when multiplying. For (x − 2)(x + 5) the second row of the grid is −2 × x = −2x and −2 × 5 = −10.

Red boxes have a negative area: they are taken away. Total: x² + 5x − 2x − 10 = x² + 3x − 10.

Three special cases

Squared brackets. (x + 3)² means (x + 3)(x + 3), so there are four pieces. The two middle pieces are equal, so you get 2 × 3x: (x + 3)² = x² + 6x + 9. It is not x² + 9.

Difference of two squares. (x + 4)(x − 4) = x² − 4x + 4x − 16 = x² − 16. When the two brackets have the same numbers but opposite signs, the middle terms cancel. You will use this again when you factorise.

Brackets with more terms. The rule does not change. Every term in the first bracket multiplies every term in the second: a bracket with 2 terms times a bracket with 3 terms gives 6 pieces.

Three brackets

To expand three brackets, expand two of them first and collect like terms. Then multiply that answer by the third bracket, again multiplying every term by every term. For example (x + 1)(x + 2)(x + 3) = (x² + 3x + 2)(x + 3) = x³ + 6x² + 11x + 6.

Key ideas

Multiply every term in the first bracket by every term in the second, then collect like terms.
Two terms times two terms makes four pieces (use a grid, or F-O-I-L).
(x + a)² = (x + a)(x + a) = x² + 2ax + a².
(x + a)(x − a) = x² − a² (difference of two squares).
Keep signs with their terms. A negative times a negative is positive.
Higher: for three brackets, expand two, collect, then multiply by the third.

Worked example 7.5.1

Expand and simplify (x + 3)(x + 5).

Key words. A binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.

Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.

Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 2 in the second, so we need 2 × 2 = 4 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. (First, Outer, Inner, Last is a way of remembering the order.) Multiply the numbers and letters, and take care with the signs.
× x 5
x x² 5x
3 3x 15
First: x × x = x²
Outer: x × 5 = 5x
Inner: 3 × x = 3x
Last: 3 × 5 = 15
Write down all the pieces. Add together the results of all the multiplications, each with its own sign.
x² + 5x + 3x + 15
Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.
x² terms: +x² (only one of this type, so it stays as it is)
x terms: +5x, +3x
add the coefficients: 5 + 3 = 8 so they give 8x
number terms: +15 (only one of this type, so it stays as it is)
Answer: x² + 8x + 15

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives 48 and the answer gives 48. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. There should be four pieces before you collect (2 × 2). Count them. A very common error is to multiply only the First and Last terms.

Worked example 7.5.2

Expand and simplify (x − 2)(x + 7).

Key words. A binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.

Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.

Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 2 in the second, so we need 2 × 2 = 4 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. (First, Outer, Inner, Last is a way of remembering the order.) Multiply the numbers and letters, and take care with the signs.
× x 7
x x² 7x
−2 −2x −14
First: x × x = x²
Outer: x × 7 = 7x
Inner: (−2) × x = −2x
Last: (−2) × 7 = −14
Write down all the pieces. Add together the results of all the multiplications, each with its own sign.
x² + 7x − 2x − 14
Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.
x² terms: +x² (only one of this type, so it stays as it is)
x terms: +7x, −2x
add the coefficients: 7 − 2 = 5 so they give 5x
number terms: −14 (only one of this type, so it stays as it is)
Answer: x² + 5x − 14

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives 10 and the answer gives 10. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. There should be four pieces before you collect (2 × 2). Count them. A very common error is to multiply only the First and Last terms.

Worked example 7.5.3

Expand and simplify (x + 4)².

Key words. A binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.

Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.

Write the squared bracket as two brackets. Squaring means multiplying by itself: (a + b)² means (a + b)(a + b). It does not mean you can just square each term.
(x + 4)² = (x + 4)(x + 4)
Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 2 in the second, so we need 2 × 2 = 4 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. (First, Outer, Inner, Last is a way of remembering the order.) Multiply the numbers and letters, and take care with the signs.
× x 4
x x² 4x
4 4x 16
First: x × x = x²
Outer: x × 4 = 4x
Inner: 4 × x = 4x
Last: 4 × 4 = 16
Write down all the pieces. Add together the results of all the multiplications, each with its own sign.
x² + 4x + 4x + 16
Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.
x² terms: +x² (only one of this type, so it stays as it is)
x terms: +4x, +4x
add the coefficients: 4 + 4 = 8 so they give 8x
number terms: +16 (only one of this type, so it stays as it is)
Answer: x² + 8x + 16

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives 49 and the answer gives 49. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. Do not write (x + 3)² = x² + 9. The middle term is missing! (x + 3)(x + 3) has four pieces: x², 3x, 3x and 9.

Worked example 7.5.4

Expand and simplify (2x + 3)(x − 5).

Key words. A binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.

Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.

Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 2 in the second, so we need 2 × 2 = 4 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. (First, Outer, Inner, Last is a way of remembering the order.) Multiply the numbers and letters, and take care with the signs.
× x −5
2x 2x² −10x
3 3x −15
First: 2x × x = 2x²
Outer: 2x × (−5) = −10x
Inner: 3 × x = 3x
Last: 3 × (−5) = −15
Write down all the pieces. Add together the results of all the multiplications, each with its own sign.
2x² − 10x + 3x − 15
Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.
x² terms: +2x² (only one of this type, so it stays as it is)
x terms: −10x, +3x
add the coefficients: −10 + 3 = −7 so they give −7x
number terms: −15 (only one of this type, so it stays as it is)
Answer: 2x² − 7x − 15

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives −18 and the answer gives −18. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. There should be four pieces before you collect (2 × 2). Count them. A very common error is to multiply only the First and Last terms.

Worked example 7.5.5

Expand and simplify (x + 6)(x − 6). What do you notice?

Key words. A binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.

Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.

Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 2 in the second, so we need 2 × 2 = 4 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. (First, Outer, Inner, Last is a way of remembering the order.) Multiply the numbers and letters, and take care with the signs.
× x −6
x x² −6x
6 6x −36
First: x × x = x²
Outer: x × (−6) = −6x
Inner: 6 × x = 6x
Last: 6 × (−6) = −36
Write down all the pieces. Add together the results of all the multiplications, each with its own sign.
x² − 6x + 6x − 36
Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.
x² terms: +x² (only one of this type, so it stays as it is)
x terms: −6x, +6x
add the coefficients: −6 + 6 = 0 so they cancel to give 0
number terms: −36 (only one of this type, so it stays as it is)
Answer: x² − 36

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives −27 and the answer gives −27. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. There should be four pieces before you collect (2 × 2). Count them. A very common error is to multiply only the First and Last terms.

Worked example 7.5.6

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

Method. Expand three brackets in two stages: expand the first two and tidy up, then multiply the result by the third bracket.

Write the question. Three brackets multiplied together. We can only multiply two things at a time, so we do it in two stages.
(x + 1)(x + 2)(x + 3)
Stage 1: expand the first two brackets. Multiply every term of one bracket by every term of the other, exactly as for two brackets.
× x 2
x x² 2x
1 x 2
x × x = x²
x × 2 = 2x
1 × x = x
1 × 2 = 2
Collect the like terms from stage 1. We want a neat answer before the next multiplication. This reduces the amount of work in the next step.
x² terms: +x² (only one of this type, so it stays as it is)
x terms: +2x, +x
add the coefficients: 2 + 1 = 3 so they give 3x
number terms: +2 (only one of this type, so it stays as it is)
So (x + 1)(x + 2) = x² + 3x + 2
Stage 2: multiply the answer from stage 1 by the third bracket. Now we have 3 terms times 2 terms, so 6 multiplications. Again every term must meet every term.
× x 3
x² x³ 3x²
3x 3x² 9x
2 2x 6
x² × x = x³
x² × 3 = 3x²
3x × x = 3x²
3x × 3 = 9x
2 × x = 2x
2 × 3 = 6
Collect the like terms. Add up the terms of the same kind.
x³ + 3x² + 3x² + 9x + 2x + 6
x³ terms: +x³ (only one of this type, so it stays as it is)
x² terms: +3x², +3x²
add the coefficients: 3 + 3 = 6 so they give 6x²
x terms: +9x, +2x
add the coefficients: 9 + 2 = 11 so they give 11x
number terms: +6 (only one of this type, so it stays as it is)
Answer: x³ + 6x² + 11x + 6

Check. Test with numbers. Put x = 2 into the original and into your answer. The original gives 60 and the answer gives 60. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. Do not try to do all three at once. Tidy the answer to stage 1 (collect like terms) before you multiply by the third bracket, or the algebra gets very long.

Worked example 7.5.7

Expand and simplify (x + 2)(x² − 3x + 4).

Key words. A binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.

Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.

Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 3 in the second, so we need 2 × 3 = 6 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. Multiply the numbers and letters, and take care with the signs.
× x² −3x 4
x x³ −3x² 4x
2 2x² −6x 8
x × x² = x³
x × (−3x) = −3x²
x × 4 = 4x
2 × x² = 2x²
2 × (−3x) = −6x
2 × 4 = 8
Write down all the pieces. Add together the results of all the multiplications, each with its own sign.
x³ − 3x² + 4x + 2x² − 6x + 8
Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.
x³ terms: +x³ (only one of this type, so it stays as it is)
x² terms: −3x², +2x²
add the coefficients: −3 + 2 = −1 so they give −x²
x terms: +4x, −6x
add the coefficients: 4 − 6 = −2 so they give −2x
number terms: +8 (only one of this type, so it stays as it is)
Answer: x³ − x² − 2x + 8

Check. Test with numbers. Put x = 3 into the original and into your answer. The original gives 20 and the answer gives 20. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. There should be four pieces before you collect (2 × 2). Count them. A very common error is to multiply only the First and Last terms.

Worked example 7.5.8

Expand and simplify (2a − 3b)².

Key words. A binomial is an expression with two terms, like x + 3. Expanding two brackets means multiplying every term of one by every term of the other.

Method. To expand two brackets, multiply every term in the first bracket by every term in the second, then collect the like terms.

Write the squared bracket as two brackets. Squaring means multiplying by itself: (a + b)² means (a + b)(a + b). It does not mean you can just square each term.
(2a − 3b)² = (2a − 3b)(2a − 3b)
Multiply every term in the first bracket by every term in the second bracket. There are 2 terms in the first bracket and 2 in the second, so we need 2 × 2 = 4 multiplications. A table (the grid method) keeps them organised: each box is the row term times the column term. (First, Outer, Inner, Last is a way of remembering the order.) Multiply the numbers and letters, and take care with the signs.
× 2a −3b
2a 4a² −6ab
−3b −6ab 9b²
First: 2a × 2a = 4a²
Outer: 2a × (−3b) = −6ab
Inner: (−3b) × 2a = −6ab
Last: (−3b) × (−3b) = 9b²
Write down all the pieces. Add together the results of all the multiplications, each with its own sign.
4a² − 6ab − 6ab + 9b²
Collect the like terms. The two middle pieces (the Outer and the Inner) are usually like terms. Add them. Pieces with different letters or powers stay separate.
a² terms: +4a² (only one of this type, so it stays as it is)
ab terms: −6ab, −6ab
add the coefficients: −6 − 6 = −12 so they give −12ab
b² terms: +9b² (only one of this type, so it stays as it is)
Answer: 4a² − 12ab + 9b²

Check. Test with numbers. Put a = 3, b = 4 into the original and into your answer. The original gives 36 and the answer gives 36. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)

Watch out. Do not write (x + 3)² = x² + 9. The middle term is missing! (x + 3)(x + 3) has four pieces: x², 3x, 3x and 9.

Watch out

(x + 5)² is not x² + 25. Squaring a bracket does not square each term. Write it as two brackets, (x + 5)(x + 5), and expand to get x² + 10x + 25. Also do not forget the middle term in the answer!

7.6

Factorising: common factors

Edexcel 1MA1: A4
Grade 2-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Factorising: common factorsComing soon

Factorising is the opposite of expanding. Expanding turns 3(x + 4) into 3x + 12. Factorising turns 3x + 12 back into 3(x + 4). We write the expression as a product (things multiplied together), with brackets.

The idea is to look for a common factor: a number or letter that divides exactly into every term. Think about 6x + 9. The 6x and the 9 can both be divided by 3. So we pull a 3 out in front, and put what is left over inside a bracket: 6x + 9 = 3(2x + 3).

The rectangle has area 6x + 9. Its height is 3, so its length must be 2x + 3. Factorising finds the side lengths of a rectangle when you know its area.

Taking out the highest common factor (HCF)

Step 1. Find the highest number that divides into all the numbers. (For 12x and 18 it is 6.)
Step 2. Look at each letter. If a letter is in every term, take out the lowest power of it. (For x² and x, take out x.)
Step 3. Write the HCF in front of a bracket. Inside the bracket write each term divided by the HCF.
Step 4. Check by expanding. You should get the original expression back.

You must take out the highest common factor. If you take out only a part of it, the bracket still has a common factor and the expression is not fully factorised. For example 12x + 18 = 2(6x + 9) is correct but not finished, because 6x + 9 can be factorised again. The full answer is 6(2x + 3).

If a term is exactly equal to the common factor, put a 1 in the bracket: 3x + 3 = 3(x + 1). A term cannot just disappear.

Common brackets

Sometimes the common factor is a whole bracket. In x(x + 2) + 3(x + 2) both parts contain (x + 2). Take (x + 2) out as the common factor: (x + 2)(x + 3).

Key ideas

Factorising is the reverse of expanding: 3x + 12 = 3(x + 4).
The common factor must divide into every term. Use the highest number and the lowest power of each common letter.
Write the HCF outside the bracket. Inside, put each term divided by the HCF. A term equal to the HCF leaves a 1.
Always check by expanding.
"Factorise fully" means the bracket cannot be factorised any further.

Worked example 7.6.1

Factorise 6x + 15.

Key words. A factor of a term divides into it exactly. A common factor of several terms is a factor of all of them. To factorise an expression means to write it as a product, with brackets. It is the reverse of expanding.

Method. Factorising with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
The terms are: 6x, 15
Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
Factors of 6: 1, 2, 3, 6
Factors of 15: 1, 3, 5, 15
The biggest number that is in every list is 3.
Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
x: it is missing from at least one term (a missing letter has power 0), so it is not common.
Write the highest common factor. Multiply the number part and the common letters together.
HCF = 3
Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
6x ÷ 3 = 2x
15 ÷ 3 = 5
So we can write 3(2x + 5)

Check. Expand the answer to see if you get the original back: 3 × 2x = 6x; 3 × 5 = 15. Together: 6x + 15. ✓

Watch out. Take out the highest common factor. If you take out only 2 from 6x + 12, the bracket (3x + 6) still has a common factor, so it is not fully factorised. And if a term is the common factor itself, the bracket gets a 1, not a blank: 3x + 3 = 3(x + 1).

Worked example 7.6.2

Factorise x² + 5x.

Method. Factorising with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
The terms are: x², 5x
Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
Factors of 1: 1
Factors of 5: 1, 5
The biggest number that is in every list is 1.
Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
x: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so x is common.
Write the highest common factor. Multiply the number part and the common letters together.
HCF = x
Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
x² ÷ x = x
5x ÷ x = 5
So we can write x(x + 5)

Check. Expand the answer to see if you get the original back: x × x = x²; x × 5 = 5x. Together: x² + 5x. ✓

Worked example 7.6.3

Factorise fully 6x² + 9x.

Method. Factorising with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
The terms are: 6x², 9x
Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
Factors of 6: 1, 2, 3, 6
Factors of 9: 1, 3, 9
The biggest number that is in every list is 3.
Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
x: the powers in the terms are 2, 1. It is in every term and the lowest power is 1, so x is common.
Write the highest common factor. Multiply the number part and the common letters together.
HCF = 3x
Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
6x² ÷ 3x = 2x
9x ÷ 3x = 3
So we can write 3x(2x + 3)

Check. Expand the answer to see if you get the original back: 3x × 2x = 6x²; 3x × 3 = 9x. Together: 6x² + 9x. ✓

Worked example 7.6.4

Factorise fully 8xy + 12y.

Method. Factorising with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
The terms are: 8xy, 12y
Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
Factors of 8: 1, 2, 4, 8
Factors of 12: 1, 2, 3, 4, 6, 12
The biggest number that is in every list is 4.
Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
x: it is missing from at least one term (a missing letter has power 0), so it is not common.
y: the powers in the terms are 1, 1. It is in every term and the lowest power is 1, so y is common.
Write the highest common factor. Multiply the number part and the common letters together.
HCF = 4y
Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
8xy ÷ 4y = 2x
12y ÷ 4y = 3
So we can write 4y(2x + 3)

Check. Expand the answer to see if you get the original back: 4y × 2x = 8xy; 4y × 3 = 12y. Together: 8xy + 12y. ✓

Worked example 7.6.5

Factorise fully 12a²b + 18ab² − 6ab.

Method. Factorising with a common factor: find the highest common factor (HCF) of all the terms, write it in front of a bracket, and divide each term by it to get what is inside.

Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
The terms are: 12a²b, 18ab², −6ab
Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 6: 1, 2, 3, 6
The biggest number that is in every list is 6.
Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
a: the powers in the terms are 2, 1, 1. It is in every term and the lowest power is 1, so a is common.
b: the powers in the terms are 1, 2, 1. It is in every term and the lowest power is 1, so b is common.
Write the highest common factor. Multiply the number part and the common letters together.
HCF = 6ab
Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
12a²b ÷ 6ab = 2a
18ab² ÷ 6ab = 3b
−6ab ÷ 6ab = −1
So we can write 6ab(2a + 3b − 1)

Check. Expand the answer to see if you get the original back: 6ab × 2a = 12a²b; 6ab × 3b = 18ab²; 6ab × (−1) = −6ab. Together: 12a²b + 18ab² − 6ab. ✓

Worked example 7.6.6

Factorise x(x + 2) + 3(x + 2).

Method. Look for a common factor that is a whole bracket.

Spot the repeated bracket. Both parts of the expression contain (x + 2).
x(x + 2) + 3(x + 2)
Take the bracket out as a common factor. It is like factorising xB + 3B = B(x + 3), where B stands for (x + 2).
What is left from each part is x and 3.
Write the answer. The common bracket goes in one bracket and what is left goes in the other.
(x + 2)(x + 3)

Check. Expand the answer: (x + 2)(x + 3) = x² + 5x + 6. And expanding the question: x² + 2x + 3x + 6 = x² + 5x + 6. ✓

Watch out. Do not expand the brackets first: that makes the question harder.

Watch out

Do not forget the 1. 4x + 4 = 4(x + 1), not 4(x). Check by expanding: 4 × 1 = 4. And check that you have taken out the highest common factor: 8x + 12 = 4(2x + 3), not 2(4x + 6).

7.7

Factorising x² + bx + c and the difference of two squares

Edexcel 1MA1: A4
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Factorising x² + bx + c and the difference of two squaresComing soon

An expression such as x² + 5x + 6 is called a quadratic expression, because the highest power of x is 2. In Section 7.5 you expanded (x + 2)(x + 3) and got x² + 5x + 6. Now we go backwards.

Look at how the answer was built. When you expand (x + p)(x + q) you get x² + (p + q)x + pq. So the middle number is the sum of p and q, and the last number is their product.

The rectangle with area x² + 5x + 6 has sides (x + 2) and (x + 3). The two numbers 2 and 3 multiply to give 6 and add to give 5.

The method for x² + bx + c

Step 1. Find two numbers that multiply to c (the last number) and add to b (the middle number). List the factor pairs of c to find them.
Step 2. Write them in two brackets: (x + first number)(x + second number).
Step 3. Check by expanding.

Signs of the expression	What the two numbers are	Example
b and c both positive	both positive	x² + 7x + 12 = (x + 3)(x + 4)
b negative, c positive	both negative	x² − 7x + 12 = (x − 3)(x − 4)
c negative	one positive and one negative; the bigger one has the sign of b	x² + x − 12 = (x − 3)(x + 4) x² − x − 12 = (x + 3)(x − 4)

Always check first for a common factor (Section 7.6): 2x² + 10x + 12 = 2(x² + 5x + 6) = 2(x + 2)(x + 3).

A perfect square happens when the two numbers are the same: x² + 6x + 9 = (x + 3)(x + 3) = (x + 3)².

The difference of two squares

You saw in Section 7.5 that (x + 4)(x − 4) = x² − 16: the middle terms cancel. So going backwards, an expression of the form a² − b² (a square, minus a square) factorises as (a − b)(a + b).

A picture of the rule: the big square with a small square removed has area x² − a². Cut off the orange strip and move it to the side to make a rectangle with sides (x − a) and (x + a).

To use it: (1) check there are two terms with a minus between them, (2) check both terms are squares, (3) square root each one, (4) write (a − b)(a + b). For example 4x² − 25: the square root of 4x² is 2x and the square root of 25 is 5, so the answer is (2x − 5)(2x + 5).

It is also a quick way to do arithmetic without a calculator: 51² − 49² = (51 − 49)(51 + 49) = 2 × 100 = 200.

Key ideas

Factorising x² + bx + c: find two numbers that multiply to c and add to b. Then (x + number)(x + number).
Check for a common factor first, and check your answer by expanding.
Difference of two squares: a² − b² = (a − b)(a + b). It needs a minus and two squares.
x² + 9 (a plus) cannot be factorised with this rule.
Square numbers to know: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225.

Worked example 7.7.1

Factorise x² + 7x + 12.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of x² + 7x + 12 have no common factor other than 1, so there is nothing to take out.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 12.
We need two numbers that multiply to 12 (the last number) and add to 7 (the middle number). List the pairs that multiply to 12:
1 × 12 = 12, add: 1 + 12 = 13 ✗
−1 × −12 = 12, add: −1 + −12 = −13 ✗
2 × 6 = 12, add: 2 + 6 = 8 ✗
−2 × −6 = 12, add: −2 + −6 = −8 ✗
3 × 4 = 12, add: 3 + 4 = 7 ✓ this pair works
Put the pair into two brackets: (x + 3)(x + 4).
Check by expanding: x² + 4x + 3x + 12 = x² + 7x + 12. ✓
Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
x² + 7x + 12 = (x + 3)(x + 4)

Check. Expand the answer: (x + 3)(x + 4) gives x² + 7x + 12 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Worked example 7.7.2

Factorise x² − 9x + 20.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of x² − 9x + 20 have no common factor other than 1, so there is nothing to take out.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −9 and c = 20.
We need two numbers that multiply to 20 (the last number) and add to −9 (the middle number). List the pairs that multiply to 20:
1 × 20 = 20, add: 1 + 20 = 21 ✗
−1 × −20 = 20, add: −1 + −20 = −21 ✗
2 × 10 = 20, add: 2 + 10 = 12 ✗
−2 × −10 = 20, add: −2 + −10 = −12 ✗
4 × 5 = 20, add: 4 + 5 = 9 ✗
−4 × −5 = 20, add: −4 + −5 = −9 ✓ this pair works
Put the pair into two brackets: (x − 4)(x − 5).
Check by expanding: x² − 5x − 4x + 20 = x² − 9x + 20. ✓
Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
x² − 9x + 20 = (x − 5)(x − 4)

Check. Expand the answer: (x − 5)(x − 4) gives x² − 9x + 20 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Worked example 7.7.3

Factorise x² + 2x − 15.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of x² + 2x − 15 have no common factor other than 1, so there is nothing to take out.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 2 and c = −15.
We need two numbers that multiply to −15 (the last number) and add to 2 (the middle number). List the pairs that multiply to −15:
1 × −15 = −15, add: 1 + −15 = −14 ✗
−1 × 15 = −15, add: −1 + 15 = 14 ✗
3 × −5 = −15, add: 3 + −5 = −2 ✗
−3 × 5 = −15, add: −3 + 5 = 2 ✓ this pair works
Put the pair into two brackets: (x − 3)(x + 5).
Check by expanding: x² + 5x − 3x − 15 = x² + 2x − 15. ✓
Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
x² + 2x − 15 = (x − 3)(x + 5)

Check. Expand the answer: (x − 3)(x + 5) gives x² + 2x − 15 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Worked example 7.7.4

Factorise x² − 3x − 28.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of x² − 3x − 28 have no common factor other than 1, so there is nothing to take out.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = −3 and c = −28.
We need two numbers that multiply to −28 (the last number) and add to −3 (the middle number). List the pairs that multiply to −28:
1 × −28 = −28, add: 1 + −28 = −27 ✗
−1 × 28 = −28, add: −1 + 28 = 27 ✗
2 × −14 = −28, add: 2 + −14 = −12 ✗
−2 × 14 = −28, add: −2 + 14 = 12 ✗
4 × −7 = −28, add: 4 + −7 = −3 ✓ this pair works
Put the pair into two brackets: (x + 4)(x − 7).
Check by expanding: x² − 7x + 4x − 28 = x² − 3x − 28. ✓
Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
x² − 3x − 28 = (x − 7)(x + 4)

Check. Expand the answer: (x − 7)(x + 4) gives x² − 3x − 28 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Worked example 7.7.5

Factorise x² + 6x + 9.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of x² + 6x + 9 have no common factor other than 1, so there is nothing to take out.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 6 and c = 9.
We need two numbers that multiply to 9 (the last number) and add to 6 (the middle number). List the pairs that multiply to 9:
1 × 9 = 9, add: 1 + 9 = 10 ✗
−1 × −9 = 9, add: −1 + −9 = −10 ✗
3 × 3 = 9, add: 3 + 3 = 6 ✓ this pair works
Put the pair into two brackets: (x + 3)(x + 3).
Check by expanding: x² + 3x + 3x + 9 = x² + 6x + 9. ✓
Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
x² + 6x + 9 = (x + 3)²

Check. Expand the answer: (x + 3)² gives x² + 6x + 9 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Worked example 7.7.6

Factorise x² − 49.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of x² − 49 have no common factor other than 1, so there is nothing to take out.
Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
x² − 49: x² is a square and 49 is a square, with a minus between them. ✓
Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
a² − b² = (a − b)(a + b)
Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
x² = (x)², so a = x
49 = (7)², so b = 7
Put a and b into the rule. One bracket has a minus, the other has a plus.
(x − 7)(x + 7)

Check. Expand to check. (x − 7)(x + 7): x × x = x²; x × 7 = 7x and (−7) × x = −7x (these two cancel); (−7) × 7 = −49. Total: x² − 49. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Worked example 7.7.7

Factorise 4x² − 25.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of 4x² − 25 have no common factor other than 1, so there is nothing to take out.
Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
4x² − 25: 4x² is a square and 25 is a square, with a minus between them. ✓
Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
a² − b² = (a − b)(a + b)
Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
4x² = (2x)², so a = 2x
25 = (5)², so b = 5
Put a and b into the rule. One bracket has a minus, the other has a plus.
(2x − 5)(2x + 5)

Check. Expand to check. (2x − 5)(2x + 5): 2x × 2x = 4x²; 2x × 5 = 10x and (−5) × 2x = −10x (these two cancel); (−5) × 5 = −25. Total: 4x² − 25. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Worked example 7.7.8

Factorise fully 2x² − 32.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
The terms are: 2x², −32
Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
Factors of 2: 1, 2
Factors of 32: 1, 2, 4, 8, 16, 32
The biggest number that is in every list is 2.
Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
x: it is missing from at least one term (a missing letter has power 0), so it is not common.
Write the highest common factor. Multiply the number part and the common letters together.
HCF = 2
Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
2x² ÷ 2 = x²
−32 ÷ 2 = −16
So we can write 2(x² − 16)
Look at the bracket. Can it be factorised again? After taking out the HCF, the bracket might be a difference of two squares or a quadratic that factorises. A question that says "factorise fully" needs every possible factor taken out.
The bracket is (x² − 16) = x² − 16.
Look at the expression. Count the terms and check for squares. A difference of two squares has exactly two terms, both are perfect squares (something multiplied by itself), and there is a minus between them.
x² − 16: x² is a square and 16 is a square, with a minus between them. ✓
Remember the rule. It comes from expanding (a − b)(a + b) = a² + ab − ab − b² = a² − b². The two middle terms cancel. Going backwards:
a² − b² = (a − b)(a + b)
Find a and b by square rooting each term. The square root undoes the square. Take the square root of the number, and halve the power of each letter.
x² = (x)², so a = x
16 = (4)², so b = 4
Put a and b into the rule. One bracket has a minus, the other has a plus.
(x − 4)(x + 4)
Write the full answer. Do not forget the common factor you took out at the start. It stays in front of the brackets.
2x² − 32 = 2(x − 4)(x + 4)

Check. Expand to check. (x − 4)(x + 4): x × x = x²; x × 4 = 4x and (−4) × x = −4x (these two cancel); (−4) × 4 = −16. Total: x² − 16. ✓ Then multiply by 2: 2x² − 32. ✓

Watch out. It only works with a minus between two squares. x² + 9 cannot be factorised like this. Square root the number and the letter: 25x² gives 5x, not 25x.

Worked example 7.7.9

Work out 51² − 49² without a calculator.

Method. Use the difference of two squares, a² − b² = (a − b)(a + b), to avoid awkward squaring.

Spot the pattern. Two square numbers with a minus between them: this is a difference of two squares, a² − b². Using the rule saves us from squaring big or awkward numbers.
51² − 49², with a = 51 and b = 49
Use the rule a² − b² = (a − b)(a + b). The first bracket is the difference of the two numbers and the second is their sum.
(a − b) = 51 − 49 = 2
(a + b) = 51 + 49 = 100
Multiply the two answers. These are easy numbers, which is why the rule helps.
2 × 100 = 200

Check. Calculator check: 51² = 2601 and 49² = 2401, and 2601 − 2401 = 200. ✓

Watch out. Subtract and add the two numbers (not their squares): (a − b) and (a + b).

Watch out

The numbers must multiply to the last number and add to the middle number. Many students find a pair that multiplies correctly but forget to check the sum: 12 = 2 × 6 but 2 + 6 = 8, not 7. And with a minus in front of the last number, the two numbers have different signs. For the difference of two squares, the answer is two brackets with the same numbers but opposite signs; (x − 4)(x − 4) is wrong for x² − 16.

7.8

Factorising ax² + bx + c

Edexcel 1MA1: A4
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Factorising ax² + bx + cComing soon

So far the number in front of x² was 1. In this section it is not: for example 2x² + 7x + 3. Expand (2x + 1)(x + 3) to see where an expression like this comes from.

The four areas are 2x², x, 6x and 3. The two middle areas add to 7x.

Notice that the middle term 7x came from x + 6x. We can use this idea backwards. The method is called splitting the middle term (or the “ac method”).

Step 1. Check for a common factor and take it out.
Step 2. Multiply the number in front of x² (call it a) by the last number (c). Here 2 × 3 = 6.
Step 3. Find two numbers that multiply to ac and add to the middle number b. Here 1 and 6: 1 × 6 = 6 and 1 + 6 = 7.
Step 4. Split the middle term using these two numbers: 2x² + x + 6x + 3.
Step 5. Factorise the first pair and the second pair separately: 2x² + x = x(2x + 1) and 6x + 3 = 3(2x + 1).
Step 6. The two brackets are the same. Take the common bracket out: (2x + 1)(x + 3).
Step 7. Check by expanding.

If the two brackets in step 6 are not the same, a mistake has been made. Go back and check the two numbers. Another way is trial and improvement: write down pairs of brackets such as (2x )(x ) and try the factor pairs of the last number, expanding each time to check the middle term. The splitting method is more reliable.

Key ideas

For ax² + bx + c: find two numbers that multiply to ac and add to b.
Use them to split the middle term, factorise in pairs, then take out the common bracket.
Take out any common factor first: 12x² + 10x − 8 = 2(6x² + 5x − 4).
Always check by expanding.

Worked example 7.8.1

Factorise 2x² + 7x + 3.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of 2x² + 7x + 3 have no common factor other than 1, so there is nothing to take out.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. A quadratic where the number in front of x² is 2, not 1. We use the split the middle term method.
Multiply the first and last numbers: 2 × (3) = 6. We need two numbers that multiply to 6 and add to 7 (the middle number).
1 × 6 = 6, add: 7 ✓ this pair works
Split the middle term (7x) into 1x and 6x: 2x² + 1x + 6x + 3.
Factorise each pair separately. First pair: 2x² + x = x(2x + 1). Second pair: 6x + 3 = 3(2x + 1).
The two brackets (2x + 1) are the same. That is how we know it has worked. Write: x(2x + 1) + 3(2x + 1).
Take the common bracket out: (2x + 1) multiplied by (x + 3).
Result: (x + 3)(2x + 1).
Check by expanding: (x + 3)(2x + 1) = 2x² + 7x + 3. ✓
Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
2x² + 7x + 3 = (x + 3)(2x + 1)

Check. Expand the answer: (x + 3)(2x + 1) gives 2x² + 7x + 3 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Worked example 7.8.2

Factorise 3x² − 10x + 8.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of 3x² − 10x + 8 have no common factor other than 1, so there is nothing to take out.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. A quadratic where the number in front of x² is 3, not 1. We use the split the middle term method.
Multiply the first and last numbers: 3 × (8) = 24. We need two numbers that multiply to 24 and add to −10 (the middle number).
1 × 24 = 24, add: 25 ✗
−1 × −24 = 24, add: −25 ✗
2 × 12 = 24, add: 14 ✗
−2 × −12 = 24, add: −14 ✗
3 × 8 = 24, add: 11 ✗
−3 × −8 = 24, add: −11 ✗
4 × 6 = 24, add: 10 ✗
−4 × −6 = 24, add: −10 ✓ this pair works
Split the middle term (−10x) into −4x and −6x: 3x² − 4x − 6x + 8.
Factorise each pair separately. First pair: 3x² − 4x = x(3x − 4). Second pair: −6x + 8 = 2(−3x + 4).
The two brackets (3x − 4) are the same. That is how we know it has worked. Write: x(3x − 4) + 2(−3x + 4).
Take the common bracket out: (3x − 4) multiplied by (x + 2).
Result: (x − 2)(3x − 4).
Check by expanding: (x − 2)(3x − 4) = 3x² − 10x + 8. ✓
Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
3x² − 10x + 8 = (x − 2)(3x − 4)

Check. Expand the answer: (x − 2)(3x − 4) gives 3x² − 10x + 8 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Worked example 7.8.3

Factorise 6x² + x − 2.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Check for a common factor first. Always check this first. A common factor is a number or letter that divides into every term.
The terms of 6x² + x − 2 have no common factor other than 1, so there is nothing to take out.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. A quadratic where the number in front of x² is 6, not 1. We use the split the middle term method.
Multiply the first and last numbers: 6 × (−2) = −12. We need two numbers that multiply to −12 and add to 1 (the middle number).
1 × −12 = −12, add: −11 ✗
−1 × 12 = −12, add: 11 ✗
2 × −6 = −12, add: −4 ✗
−2 × 6 = −12, add: 4 ✗
3 × −4 = −12, add: −1 ✗
−3 × 4 = −12, add: 1 ✓ this pair works
Split the middle term (1x) into −3x and 4x: 6x² − 3x + 4x − 2.
Factorise each pair separately. First pair: 6x² − 3x = 3x(2x − 1). Second pair: 4x − 2 = 2(2x − 1).
The two brackets (2x − 1) are the same. That is how we know it has worked. Write: 3x(2x − 1) + 2(2x − 1).
Take the common bracket out: (2x − 1) multiplied by (3x + 2).
Result: (2x − 1)(3x + 2).
Check by expanding: (2x − 1)(3x + 2) = 6x² + x − 2. ✓
Write the answer. The brackets are the factors of the quadratic. We have checked by expanding that they multiply back to the original.
6x² + x − 2 = (2x − 1)(3x + 2)

Check. Expand the answer: (2x − 1)(3x + 2) gives 6x² + x − 2 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Worked example 7.8.4

Factorise fully 12x² + 10x − 8.

Method. Factorising: (1) take out any common factor, (2) look at what is left: a difference of two squares or a quadratic with three terms, (3) check by expanding.

Write the terms and think: what could divide into every one? Factorising is the opposite of expanding: we take a common factor outside a bracket. A common factor is a number or letter that divides exactly into every term. We want the highest common factor, so that nothing is left to take out.
The terms are: 12x², 10x, −8
Find the highest common factor of the numbers. List the factors of each number (all the numbers that divide into it exactly). The biggest number that is in every list is the number part of the common factor.
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 10: 1, 2, 5, 10
Factors of 8: 1, 2, 4, 8
The biggest number that is in every list is 2.
Find the letters that are in every term. A letter is part of the common factor only if it appears in every term. Then use the lowest power of it, because that is the most that divides into all of them (x divides into x² and into x, but x² does not divide into x).
x: it is missing from at least one term (a missing letter has power 0), so it is not common.
Write the highest common factor. Multiply the number part and the common letters together.
HCF = 2
Write the HCF outside a bracket. Inside the bracket put each term divided by the HCF. This is the reverse of expanding. Each division tells us what must be left inside the bracket.
12x² ÷ 2 = 6x²
10x ÷ 2 = 5x
−8 ÷ 2 = −4
So we can write 2(6x² + 5x − 4)
Look at the bracket. Can it be factorised again? After taking out the HCF, the bracket might be a difference of two squares or a quadratic that factorises. A question that says "factorise fully" needs every possible factor taken out.
The bracket is (6x² + 5x − 4) = 6x² + 5x − 4.
Factorise the quadratic. It has three terms, so we use the "two numbers" method. Follow every line carefully.
Spot the pattern. A quadratic where the number in front of x² is 6, not 1. We use the split the middle term method.
Multiply the first and last numbers: 6 × (−4) = −24. We need two numbers that multiply to −24 and add to 5 (the middle number).
1 × −24 = −24, add: −23 ✗
−1 × 24 = −24, add: 23 ✗
2 × −12 = −24, add: −10 ✗
−2 × 12 = −24, add: 10 ✗
3 × −8 = −24, add: −5 ✗
−3 × 8 = −24, add: 5 ✓ this pair works
Split the middle term (5x) into −3x and 8x: 6x² − 3x + 8x − 4.
Factorise each pair separately. First pair: 6x² − 3x = 3x(2x − 1). Second pair: 8x − 4 = 4(2x − 1).
The two brackets (2x − 1) are the same. That is how we know it has worked. Write: 3x(2x − 1) + 4(2x − 1).
Take the common bracket out: (2x − 1) multiplied by (3x + 4).
Result: (2x − 1)(3x + 4).
Check by expanding: (2x − 1)(3x + 4) = 6x² + 5x − 4. ✓
Write the full answer. Do not forget the common factor you took out at the start. It stays in front of the brackets.
12x² + 10x − 8 = 2(2x − 1)(3x + 4)

Check. Expand the answer: 2(2x − 1)(3x + 4) gives 12x² + 10x − 8 again. ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong. And check for a common factor first: it makes the numbers smaller.

Watch out

The numbers must multiply to ac (the first number times the last number), not just to the last number c. And they must add to the middle number b. If the two brackets are not identical after you factorise the pairs, the split was wrong. Also check for a common factor first: 12x² + 10x − 8 is much easier after taking out 2.

7.9

Identities and algebraic proof

Edexcel 1MA1: A3, A6
Grade 4-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Identities and algebraic proofComing soon

In Section 7.1 you met the words equation and identity. Now we use them properly.

An equation such as 2x + 1 = 9 is true only for particular values of x. (Here only x = 4 works.) You solve an equation to find those values.

An identity such as 2(x + 3) ≡ 2x + 6 is true for every value of x. Try x = 1: 8 and 8. Try x = 10: 26 and 26. The sign ≡ means “is identically equal to”. You do not solve an identity. Instead you show that both sides are the same by expanding, simplifying or factorising.

How do you tell the difference? Simplify both sides. If they turn into exactly the same expression it is an identity. If they do not, it is an equation. Trying a few numbers is not enough to prove an identity: 3x = 12 is true for x = 4 but it is an equation, not an identity. Numbers can only disprove an identity: one value where the two sides differ is enough.

Showing that two expressions are equivalent

In a “show that” question you are told the answer and you must explain how to get there. Pick the side that looks more complicated. Expand brackets, collect like terms or factorise until you reach the other side. Write every step and finish with a sentence.

Using algebra to support an argument

Algebra lets us prove statements about all numbers at once. Testing 3 + 5 = 8 and 7 + 9 = 16 shows that the sum of two odd numbers seems to be even, but it does not prove it. To prove it, use letters:

Idea	Algebra
any integer (whole number)	n
consecutive integers	n, n + 1, n + 2, …
an even number	2n
an odd number	2n + 1 (or 2n − 1)
consecutive even numbers	2n, 2n + 2, 2n + 4, …
consecutive odd numbers	2n + 1, 2n + 3, …
a multiple of 3	3 × (an integer), like 3k

If two different numbers are needed, use two different letters (m and n), because the two numbers do not have to be related. A proof always follows the same plan: (1) write the numbers using letters, (2) write the expression, (3) simplify it, (4) show it has the required form, (5) write a conclusion in words.

A picture proof that (n + 1)² − n² = 2n + 1: the L-shape is made of n + n + 1 small squares.

Disproving with a counter-example

To show that a statement is false, you only need one example where it fails. This is called a counter-example. For instance “the sum of two odd numbers is odd” is false: 3 + 5 = 8, which is even.

Key ideas

An equation is true for particular values; an identity (≡) is true for all values.
To show an identity, expand or factorise until both sides match. Testing numbers is not a proof.
Even: 2n. Odd: 2n + 1. Consecutive integers: n, n + 1, n + 2. Use different letters for different numbers.
To prove a multiple of k, show the expression can be written as k × (an integer).
One counter-example is enough to prove a statement false.

Worked example 7.9.1

Is 3(x + 2) = 3x + 6 an equation or an identity?

Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.

Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.

Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.
Left side: 3(x + 2) = 3x + 6
Right side: 3x + 6 = 3x + 6
Compare the two sides. They are identical, so the statement is true for every value of x. That is what an identity is. (It is often written with the sign ≡.)
3x + 6 and 3x + 6 are the same. ✓
State the answer. Both sides are identical, so the statement is true for every value of x.
It is an identity.

Check. Test two values: x = 2: left = 12, right = 12; x = 5: left = 21, right = 21. They match each time (this supports the answer, but the expanding above is what proves it). ✓

Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.

Worked example 7.9.2

Is (x + 1)² = x² + 1 an equation or an identity?

Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.

Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.

Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.
Left side: (x + 1)² = x² + 2x + 1
Right side: x² + 1 = x² + 1
Compare the two sides. They are different expressions, so they cannot be equal for every value of x.
x² + 2x + 1 and x² + 1 are not the same.
Find a value that shows they are not always equal. One value that makes the two sides different is enough to show it is not an identity.
Try x = 1: left side = 4, right side = 2. These are different.
Conclude. A statement that is true only for particular values is an equation.
It is an equation: it is true only when x = 0.

Check. Put x = 0 into both sides: left = 1 and right = 1, which are equal. ✓

Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.

Worked example 7.9.3

Show that (x + 4)² − 16 ≡ x(x + 8).

Method. Work on the more complicated side (the left). Expand, collect like terms, and factorise, to reach the right side.

Start with the left side. It has a squared bracket, so it is the complicated side.
(x + 4)² − 16
Write the squared bracket as two brackets. (x + 4)² means (x + 4)(x + 4).
(x + 4)² − 16 = (x + 4)(x + 4) − 16
Expand the brackets. Multiply every term by every term: x², 4x, 4x and 16.
(x + 4)(x + 4) − 16 = x² + 4x + 4x + 16 − 16
Collect like terms. 4x + 4x = 8x and 16 − 16 = 0.
x² + 4x + 4x + 16 − 16 = x² + 8x
Factorise. x is a common factor of x² and 8x.
x² + 8x = x(x + 8)
Write the conclusion. The last line must be the right side of the question.
The left side simplifies to x(x + 8), which is the right side. So (x + 4)² − 16 ≡ x(x + 8). ✓

Check. Test x = 2: left side (2 + 4)² − 16 = 36 − 16 = 20 and right side 2 × (2 + 8) = 20. ✓ (Testing supports the answer but it is the algebra above that proves it.)

Watch out. Do not start with both sides and "do the same to both". Start with one side and change it into the other.

Worked example 7.9.4

Prove that the sum of three consecutive integers is a multiple of 3.

Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.

Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.

Use a letter for the first integer. Let n be any integer. The next two integers are 1 more and 2 more.
The three integers are n, n + 1, n + 2.
Write the sum. Add the three numbers.
n + (n + 1) + (n + 2)
Simplify by collecting like terms. There are three ns and the numbers 1 and 2.
n + (n + 1) + (n + 2) = 3n + 3
Factorise to show a multiple of 3. A multiple of 3 is 3 times a whole number, so take out the common factor 3.
3n + 3 = 3(n + 1)
Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.
n is an integer, so n + 1 is also an integer. Therefore 3(n + 1) is 3 multiplied by an integer, so it is a multiple of 3. The sum of any three consecutive integers is a multiple of 3. ✓

Check. Test with 7, 8, 9: the sum is 24 = 3 × 8. The formula 3(n + 1) with n = 7 is 3 × 8 = 24. ✓

Watch out. "Let n be any integer" is the start of nearly every proof. Do not choose a particular number.

Worked example 7.9.5

Prove that the sum of any two odd numbers is even.

Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.

Write each odd number using a letter. An even number is 2 × an integer. An odd number is one more than an even number. The two odd numbers do not have to be linked, so use different letters m and n.
The odd numbers are 2m + 1 and 2n + 1, where m and n are integers.
Write the sum.
(2m + 1) + (2n + 1)
Simplify. Collect like terms: 2m + 2n + 1 + 1.
(2m + 1) + (2n + 1) = 2m + 2n + 2
Factorise to show it is even. An even number is 2 × an integer.
2m + 2n + 2 = 2(m + n + 1)
Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.
m + n + 1 is an integer, so 2(m + n + 1) is 2 multiplied by an integer. So it is even. The sum of any two odd numbers is even. ✓

Check. Test with 3 and 9: 3 + 9 = 12, which is 2 × 6. With m = 1, n = 4: 2(1 + 4 + 1) = 12. ✓

Watch out. Using the same letter for both numbers would only prove the result for equal odd numbers.

Worked example 7.9.6

Prove that the difference between the squares of two consecutive integers is equal to the sum of the two integers.

Method. Write the two numbers with a letter, work out the difference of their squares, and compare it with their sum.

Write the two integers. Let n be any integer. The next integer is n + 1.
The integers are n and n + 1.
Write the difference of the squares. The bigger one minus the smaller one.
(n + 1)² − n²
Expand (n + 1)². (n + 1)(n + 1) = n² + n + n + 1.
(n + 1)² − n² = n² + 2n + 1 − n²
Simplify. The n² terms cancel.
n² + 2n + 1 − n² = 2n + 1
Compare with the sum of the two integers. The sum is n + (n + 1).
n + (n + 1) = 2n + 1
Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.
Both the difference of the squares and the sum of the two integers equal 2n + 1. So the difference between the squares of two consecutive integers equals their sum. ✓

Check. Test with 5 and 6: 6² − 5² = 36 − 25 = 11 and 5 + 6 = 11. ✓

Worked example 7.9.7

Prove that the difference between the squares of two consecutive odd numbers is a multiple of 8.

Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.

Write the two consecutive odd numbers. Let n be any integer. Then 2n + 1 is odd, and the next odd number is 2 more.
The odd numbers are 2n + 1 and 2n + 3.
Write the difference of the squares. The bigger square minus the smaller square.
(2n + 3)² − (2n + 1)²
Expand each squared bracket separately. Four pieces each: (2n + 3)(2n + 3) and (2n + 1)(2n + 1).
(2n + 3)² = 4n² + 6n + 6n + 9 = 4n² + 12n + 9
(2n + 1)² = 4n² + 2n + 2n + 1 = 4n² + 4n + 1
Subtract, using a bracket for the second answer. The minus sign changes every sign in the second bracket.
(2n + 3)² − (2n + 1)² = (4n² + 12n + 9) − (4n² + 4n + 1) = 4n² + 12n + 9 − 4n² − 4n − 1 = 8n + 8
Factorise to show a multiple of 8. Take out the common factor 8.
8n + 8 = 8(n + 1)
Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.
n + 1 is an integer, so 8(n + 1) is 8 multiplied by an integer. So the difference between the squares of any two consecutive odd numbers is a multiple of 8. ✓

Check. Test with 5 and 7: 49 − 25 = 24 = 8 × 3. Here 2n + 1 = 5 so n = 2 and 8(2 + 1) = 24. ✓

Watch out. Expand each bracket separately, then subtract. Never subtract inside the bracket expansion.

Worked example 7.9.8

Edward says: “n² + n + 41 is a prime number for every whole number n from 0 upwards.” Show that Edward is wrong.

Method. To show that a statement is false you need one counter-example: one value that makes it fail.

Try some small values. Substitute n = 0, 1, 2, … into n² + n + 41 and see if the answers are prime.
n = 0: 41 (prime)
n = 1: 43 (prime)
n = 2: 47 (prime)
n = 3: 53 (prime)
Think about when it might fail. The statement is true for the first 40 values (n = 0 to 39) but a clever choice of n can break it. If n = 41 then every term has a factor of 41. If n = 40, then n + 1 = 41.
n = 40: 40² + 40 + 41 = 1600 + 40 + 41 = 1681
Check whether 1681 is prime. 1681 = 41 × 41, because 41² = 1681.
41 × 41 = 1681, so 1681 is not prime.
Conclude. One value that fails is enough.
n = 40 is a counter-example, so the statement is false.

Check. Recalculate 41 × 41: 40 × 41 = 1640 and 1640 + 41 = 1681. ✓

Watch out. A counter-example must be a specific value that you have checked. Showing that the statement works for many values is never enough to prove it true.

Watch out

Checking a few numbers is not a proof. 2x = x + 4 works for x = 4, but it is an equation. In a proof, start with letters (“let n be any integer”), never with particular numbers. And make sure the last line says what has been shown, in words: “so this is a multiple of 3”.

Chapter summary

Skill	Method
Notation	3y = 3 × y; ab = a × b; a² = a × a; a ÷ b = ab; write fractions, not decimals, as coefficients
Vocabulary	Expression (no =), equation (true for some values), formula (a rule), identity (true for all values, ≡), term, coefficient, constant, factor
Collect like terms	Group terms with the same letters and powers and add the coefficients. 4x + 3y − x = 3x + 3y
Index laws	a^m × aⁿ = a^m+n; a^m ÷ aⁿ = a^m−n; (a^m)ⁿ = a^mn; a⁰ = 1
Multiply and divide terms	Multiply or divide the numbers, then add or subtract the powers of each letter
Expand a single bracket	a(b + c) = ab + ac. Every term inside is multiplied; watch the signs
Expand two brackets	Every term times every term (grid or F-O-I-L), then collect like terms. (x + a)(x + b) = x² + (a + b)x + ab
Expand three brackets (Higher)	Expand two, collect, then multiply by the third
Common factor	Take out the highest common factor of all terms: 6x² + 9x = 3x(2x + 3)
x² + bx + c	Two numbers that multiply to c and add to b: x² + 5x + 6 = (x + 2)(x + 3)
Difference of two squares	a² − b² = (a − b)(a + b)
ax² + bx + c (Higher)	Multiply a × c, find two numbers that multiply to it and add to b, split the middle term, factorise in pairs
Show that and proof	Work from one side (expand, simplify, factorise) to the other; use 2n (even), 2n + 1 (odd), n, n + 1 (consecutive); finish with a sentence
Counter-example	One example that makes a statement false

Can you…?

I can write algebra using the correct notation (ab, 3y, a², fractions).
I know the meaning of expression, equation, formula, identity, term, coefficient and factor.
I can simplify an expression by collecting like terms.
I can multiply and divide algebraic terms and use the laws of indices.
I can expand single brackets, including with a negative or a letter outside.
I can expand and simplify two brackets, including squares.
I can take out a common factor and factorise fully.
I can factorise x² + bx + c and the difference of two squares.
I can tell the difference between an equation and an identity.
I can show that two expressions are equal, and use algebra to prove simple statements about numbers.
(Higher) I can expand three brackets, factorise ax² + bx + c, and prove harder statements about integers.

Key terms

Term: One part of an expression, with the sign in front of it, for example 3x or −5.
Coefficient: The number multiplying the letters in a term. The coefficient of 4x² is 4.
Constant: A term that is just a number, such as +7.
Variable: A letter that stands for a number that can change or is not yet known.
Expression: Numbers, letters and operations with no equals sign, such as 3x + 2.
Equation: Two expressions joined by =, true only for particular values of the letter.
Formula: A rule that links quantities, for example A = lw.
Identity: An equality that is true for every value of the letters. Written with ≡.
Inequality: A statement comparing two expressions with <, >, ≤ or ≥.
Factor: Something that is multiplied; a number or expression that divides exactly into another.
Common factor: A factor that divides exactly into every term of an expression.
Highest common factor (HCF): The biggest common factor of a set of terms.
Index (power): The small raised number that says how many copies of a letter are multiplied: x³.
Like terms: Terms with the same letters raised to the same powers. They can be added or subtracted.
Expand: Multiply out brackets so that none are left.
Factorise: Write an expression as a product, using brackets.
Quadratic: An expression whose highest power is 2, such as x² + 5x + 6.
Binomial: An expression with two terms, such as x + 3.
Difference of two squares: An expression a² − b², which factorises as (a − b)(a + b).
Integer: A whole number: positive, negative or zero.
Consecutive: Following one after another, such as 7, 8, 9 or 2n, 2n + 2.
Proof: A logical argument that shows a statement is always true, using algebra rather than examples.
Counter-example: One example that shows a statement is false.

Total 42 marks. Allow about 42 minutes.

← Chapter 6: Standard form, indices and surds Chapter 8: Substitution and formulae →

×	x	−5
2x	2x²	−10x
3	3x	−15

×	x	−6
x	x²	−6x
6	6x	−36

×	2a	−3b
2a	4a²	−6ab
−3b	−6ab	9b²

×	y	2
y	y²	2y
7	7y	14

GCSE Maths Algebra Basics (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

The language of algebra

The vocabulary of algebra

Key ideas

Watch out

Exercise 7A Notation and vocabulary

Simplifying: collecting like terms

Key ideas

Watch out

Exercise 7B Collecting like terms

Multiplying and dividing terms, and the laws of indices

Multiplying terms

The laws of indices

Dividing terms

Negative and fractional indices

Key ideas

Watch out

Exercise 7C Multiplying, dividing and indices

Expanding single brackets

Key ideas

Watch out

Exercise 7D Expanding single brackets

Expanding double brackets

Three special cases

Three brackets

Key ideas

Watch out

Exercise 7E Expanding double brackets

Factorising: common factors

Taking out the highest common factor (HCF)

Common brackets

Key ideas

Watch out

Exercise 7F Factorising with common factors

Factorising x² + bx + c and the difference of two squares

The method for x2 + bx + c

The difference of two squares

Key ideas

Watch out

Exercise 7G Factorising x² + bx + c

Exercise 7H Difference of two squares and mixed factorising

Factorising ax² + bx + c

Key ideas

Watch out

Exercise 7I Factorising ax² + bx + c

Identities and algebraic proof

Showing that two expressions are equivalent

Using algebra to support an argument

Disproving with a counter-example

Key ideas

Watch out

Exercise 7J Identities and proof

Chapter summary

Can you…?

Key terms

Exercise 7Z Check-up (non-calculator)

Exercise 7X Exam-style questions

The method for x² + bx + c