Algebraic Manipulation

Syllabus coverage

Simplify expressions by collecting like terms, e.g. $2 a + 3 b + 5 a - 9 b = 7 a - 6 b$
Expand products of algebraic expressions, including a single bracket such as $3 x (2 x - 4 y)$ and the product of two brackets such as $(2 x + 1) (x - 4)$
Factorise by extracting common factors, and factorise fully, e.g. $9 x^{2} + 15 x y = 3 x (3 x + 5 y)$
Extended only: expand products of more than two brackets, e.g. $(x - 2) (x + 3) (2 x + 1)$
Extended only: factorise expressions of the form $a x + b x + k a y + k b y$ , $a^{2} x^{2} - b^{2} y^{2}$ , $a^{2} + 2 a b + b^{2}$ , $a x^{2} + b x + c$ and $a x^{3} + b x^{2} + c x$
Extended only: complete the square for expressions in the form $a x^{2} + b x + c$
Extended only: manipulate algebraic fractions, including $\frac{x}{3} + \frac{x - 4}{2}$ , $\frac{2 x}{3} - \frac{3 (x - 5)}{2}$ , $\frac{3 a}{4} \times \frac{9 a}{10}$ and $\frac{3 a}{4} \div \frac{9 a}{10}$
Extended only: factorise and simplify rational expressions, e.g. $\frac{x^{2} - 2 x}{x^{2} - 5 x + 6}$ , and write $\frac{1}{x - 2} + \frac{x + 1}{x - 3}$ as a single fraction
Extended only: change the subject of a formula where the subject appears twice, or where there is a power or root of the subject

The factorised form gives the roots.

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

, so the curve crosses the x-axis at

x = - 1

and

x = 3

, crosses the y-axis at

- 3

, and turns at

(1, - 4)

Collecting like terms: what 'simplify' means

A term is a single number, a single letter, or a product of the two. In $5 x^{2} - 3 x + 2$ the terms are $5 x^{2}$ , $- 3 x$ and $2$ . The number in front of a letter is its coefficient: in $- 3 x$ the coefficient is $- 3$ , and the minus sign belongs to the term.

Like terms have exactly the same letters with exactly the same powers. $3 x$ and $- 7 x$ are like terms, so they can be added; $3 x$ and $3 x^{2}$ are not like terms, and neither are $3 a$ and $3 b$ . 'Simplify' means collect the like terms so that the expression has as few terms as it can. The syllabus's own Core example is $2 a + 3 b + 5 a - 9 b = 7 a - 6 b$ : the a terms collect to $7 a$ and the b terms to $- 6 b$ .

The Extended paper collects longer expressions with powers and two letters: $2 a^{2} + 3 a b - 1 + 5 a^{2} - 9 a b + 4 = 7 a^{2} - 6 a b + 3$ . Nothing about the method changes. Deal with one family of like terms at a time and keep each sign attached to its term as you go.

Terms	Like terms?	Why
$3 x$ and $- 7 x$	Yes	same letter, same power, so they collect to $- 4 x$
$3 x$ and $3 x^{2}$	No	the powers are different: x and $x^{2}$ are not the same thing
$5 a b$ and $- 2 a b$	Yes	same product of letters, so they collect to $3 a b$
$3 a$ and $3 b$	No	different letters, so the terms stay apart
the numbers	Yes	constant terms collect with constant terms: $2 - 9 = - 7$

Keep the sign in front of a term when you collect it.

- 9 b

is a term of

- 9 b

, not of

9 b

, and dropping that minus is the commonest way to lose a mark in this topic.

Expanding a single bracket

Expanding means multiplying out, using $a (b + c) = a b + a c$ . The term outside the bracket multiplies every term inside it. So $3 x (2 x - 4 y) = 6 x^{2} - 12 x y$ : the $3 x$ multiplies the $2 x$ and then the $- 4 y$ .

A number outside the bracket behaves in exactly the same way. $5 (2 x - 3) = 10 x - 15$ , and $- 2 (3 x - 5) = - 6 x + 10$ . The negative sign in front of the bracket multiplies both terms, so the second term changes sign as well as the first.

The same care is needed when a bracket is subtracted from something. $4 x - 3 (x - 2) = 4 x - 3 x + 6 = x + 6$ : the $+ 6$ appears because $- 3 \times - 2 = + 6$ , and a student who writes $- 6$ has lost the mark.

Expand	Answer	The trap
$3 x (2 x - 4 y)$	$6 x^{2} - 12 x y$	$3 x \times 4 y$ gives $12 x y$ , not $12 x$ or $12 y$
$- 2 (3 x - 5)$	$- 6 x + 10$	the $- 2$ changes the sign of the second term
$4 x - 3 (x - 2)$	$x + 6$	$- 3 \times - 2 = + 6$ , so the constant is positive
$5 (2 x - 3)$	$10 x - 15$	both terms inside the bracket are multiplied by 5

Write the multiplication out for one line even when you can do it in your head. That line is the method mark, and it is also how you spot a sign slip before it reaches the answer.

Expanding two brackets

With two brackets, every term in the first multiplies every term in the second, which gives four products. $(2 x + 1) (x - 4) = 2 x^{2} - 8 x + x - 4 = 2 x^{2} - 7 x - 4$ . Write the four products first, then collect the two middle terms in a second line.

A grid makes the four products visible and stops you leaving one out. The Extended paper also expands two brackets in two letters: $(3 x + y) (x - 4 y) = 3 x^{2} - 12 x y + x y - 4 y^{2} = 3 x^{2} - 11 x y - 4 y^{2}$ . The method is unchanged - the middle terms are like terms because they are both $x y$ .

Two patterns turn up so often that they are worth knowing by sight. $(x + 5) (x - 5) = x^{2} - 25$ is the difference of two squares: the two middle terms are $- 5 x$ and $+ 5 x$ , so they cancel. $(x + 4)^{2} = x^{2} + 8 x + 16$ is a perfect square: the two middle terms are the same, so the middle term doubles.

$(2 x + 1) (x - 4) = 2 x^{2} - 7 x - 4$
$(x + a) (x - a) = x^{2} - a^{2}$ , so $(3 x + 2) (3 x - 2) = 9 x^{2} - 4$
$(x + a)^{2} = x^{2} + 2 a x + a^{2}$ , so $(2 x - 3)^{2} = 4 x^{2} - 12 x + 9$
$(3 x + y) (x - 4 y) = 3 x^{2} - 11 x y - 4 y^{2}$

	$x$	$- 4$
$2 x$	$2 x^{2}$	$- 8 x$
$+ 1$	$x$	$- 4$

Squaring a bracket is not squaring each term.

(x + 3)^{2}

x^{2} + 6 x + 9

, and the missing

6 x

is exactly where the mark goes.

Expanding three brackets (Extended only)

With three brackets you expand in two stages: multiply the first two together, simplify, then multiply that result by the third bracket. $(x - 2) (x + 3) = x^{2} + x - 6$ , and then $(x^{2} + x - 6) (2 x + 1) = 2 x^{3} + x^{2} + 2 x^{2} + x - 12 x - 6 = 2 x^{3} + 3 x^{2} - 11 x - 6$ .

There are six products altogether, which is why doing it all in one line goes wrong. Keep the two stages separate and write both lines down, because the method marks sit in the first stage as much as in the answer.

You can always check a long expansion by substituting a value. Putting $x = 2$ into $(x - 2) (x + 3) (2 x + 1)$ gives $0 \times 5 \times 5 = 0$ , and into $2 x^{3} + 3 x^{2} - 11 x - 6$ gives $16 + 12 - 22 - 6 = 0$ . Both forms give the same value, so the expansion is right.

$(x - 2) (x + 3) = x^{2} + x - 6$ , the first stage
$(x - 2) (x + 3) (2 x + 1) = 2 x^{3} + 3 x^{2} - 11 x - 6$ , the full expansion
check with $x = 2$ : $0 \times 5 \times 5 = 0$ and $16 + 12 - 22 - 6 = 0$

Factorising by taking out the common factor

Factorising is expanding backwards: you write the expression as a product instead of a sum. The first thing to look for is a common factor, meaning a number or a letter that divides every term exactly. The syllabus example is $9 x^{2} + 15 x y = 3 x (3 x + 5 y)$ : the highest common factor of $9 x^{2}$ and $15 x y$ is $3 x$ , so the $3$ and the $x$ both come outside the bracket.

Factorise fully means take out the highest common factor, not just any factor. $12 x^{2} + 18 x = 6 x (2 x + 3)$ , whereas $2 x (6 x + 9)$ is not fully factorised because the bracket still has a factor of 3 inside it. The check is always the same: expand your answer back and see whether you recover the expression you started with.

Extended candidates also meet the form $a x^{3} + b x^{2} + c x$ , where the common factor is $x$ and possibly a number as well: $4 x^{3} - 6 x^{2} + 2 x = 2 x (2 x^{2} - 3 x + 1)$ .

$9 x^{2} + 15 x y = 3 x (3 x + 5 y)$
$12 x^{2} + 18 x = 6 x (2 x + 3)$
$4 x^{3} - 6 x^{2} + 2 x = 2 x (2 x^{2} - 3 x + 1)$
check: $2 x (2 x^{2} - 3 x + 1) = 4 x^{3} - 6 x^{2} + 2 x$

The common factor has to divide every term. In

3 x^{2} + 6 x

the

3 x

comes out and the answer is

3 x (x + 2)

. In

3 x^{2} + 6

only the 3 is common to both terms, so the answer is

3 (x^{2} + 2)

Factorising quadratics (Extended only)

A quadratic expression has the form $a x^{2} + b x + c$ with $a \neq 0$ . When $a = 1$ the factorisation is two brackets whose numbers multiply to $c$ and add to $b$ : $x^{2} + 7 x + 12 = (x + 3) (x + 4)$ because $3 \times 4 = 12$ and $3 + 4 = 7$ . The signs follow from that pair of numbers: $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ , and $x^{2} + 2 x - 15 = (x + 5) (x - 3)$ , where one number is negative because $c$ is negative.

When $a$ is not 1, look for two numbers that multiply to $a c$ and add to $b$ , use them to split the middle term, then factorise in pairs. For $2 x^{2} + 7 x + 3$ : $a c = 6$ and $6 + 1 = 7$ , so $2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + 1 (x + 3) = (2 x + 1) (x + 3)$ .

Four special forms are named in the syllabus. The difference of two squares $a^{2} x^{2} - b^{2} y^{2} = (a x + b y) (a x - b y)$ , the perfect square $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ , the four-term form $a x + b x + k a y + k b y$ which is factorised by grouping in pairs, and the common factor form $a x^{2} + b x + c$ where the bracket itself needs factorising.

A factorised quadratic also tells you where its curve crosses the x-axis. $x^{2} - 2 x - 3 = (x - 3) (x + 1)$ , so $y = x^{2} - 2 x - 3$ is zero at $x = 3$ and $x = - 1$ , which is what the diagram below shows. Completing the square is the third way of rewriting the same expression, and the syllabus lists it here: $x^{2} + 6 x + 1 = (x + 3)^{2} - 9 + 1 = (x + 3)^{2} - 8$ , because $(x + 3)^{2} = x^{2} + 6 x + 9$ and the extra 9 has to be taken away again.

Form	Method	Example
$x^{2} + b x + c$	two numbers multiplying to $c$ and adding to $b$	$x^{2} - x - 12 = (x - 4) (x + 3)$
$a x^{2} + b x + c$	two numbers multiplying to $a c$ and adding to $b$ , then split the middle term	$2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3)$
$a^{2} x^{2} - b^{2} y^{2}$	difference of two squares	$9 x^{2} - 25 = (3 x + 5) (3 x - 5)$
$a^{2} + 2 a b + b^{2}$	perfect square	$x^{2} + 10 x + 25 = (x + 5)^{2}$
$a x + b x + k a y + k b y$	group in pairs and take out the repeated bracket	$3 x + 3 y + a x + a y = (x + y) (3 + a)$
$a x^{3} + b x^{2} + c x$	common factor	$4 x^{3} - 6 x^{2} + 2 x = 2 x (2 x^{2} - 3 x + 1)$

Always expand your answer back as a check. If

(2 x + 1) (x + 3)

gives

2 x^{2} + 7 x + 3

again, the factorisation is right, and the check costs you fifteen seconds.

Algebraic fractions (Extended only)

An algebraic fraction has a letter in the numerator or the denominator. The rules are the ones you already use for numerical fractions, with one extra step: factorise first, because you can only cancel factors, never terms that are added or subtracted.

Multiplying and dividing is the straightforward case. $\frac{3 a}{4} \times \frac{9 a}{10} = \frac{27 a^{2}}{40}$ , and to divide you turn the second fraction upside down and multiply: $\frac{3 a}{4} \div \frac{9 a}{10} = \frac{3 a}{4} \times \frac{10}{9 a} = \frac{5}{6}$ . Notice that the $a$ cancels in the division because it is a factor of the top and of the bottom.

Simplifying needs both parts factorised: $\frac{x^{2} - 2 x}{x^{2} - 5 x + 6} = \frac{x (x - 2)}{(x - 2) (x - 3)} = \frac{x}{x - 3}$ . The $(x - 2)$ cancels because it is a factor of the numerator and of the denominator. Nothing else does: the $x$ on the top is multiplied by $(x - 2)$ but the $x$ on the bottom is part of the sum $x - 3$ .

Adding and subtracting needs a common denominator, exactly as with numbers. $\frac{1}{x - 2} + \frac{x + 1}{x - 3} = \frac{(x - 3) + (x + 1) (x - 2)}{(x - 2) (x - 3)} = \frac{x^{2} - 5}{(x - 2) (x - 3)}$ . When a whole fraction is subtracted, put its numerator in a bracket before you multiply: $\frac{2 x}{3} - \frac{3 (x - 5)}{2} = \frac{4 x - 9 (x - 5)}{6} = \frac{45 - 5 x}{6}$ .

$\frac{3 a}{4} \times \frac{9 a}{10} = \frac{27 a^{2}}{40}$
$\frac{3 a}{4} \div \frac{9 a}{10} = \frac{5}{6}$
$\frac{x^{2} - 2 x}{x^{2} - 5 x + 6} = \frac{x}{x - 3}$
$\frac{2 x}{3} - \frac{3 (x - 5)}{2} = \frac{45 - 5 x}{6}$

Cancelling only works on factors.

\frac{x + 3}{x + 5}

does not simplify to

\frac{3}{5}

: the x's are terms of a sum, not factors of a product, so that fraction is already in its simplest form.

Rearranging a formula: changing the subject

The subject of a formula is the letter that stands on its own, usually on the left. Rearranging means doing the same thing to both sides until the letter you want is alone. To make $x$ the subject of $y = 3 x + 7$ , subtract 7 from both sides and then divide both sides by 3: $x = \frac{y - 7}{3}$ . Undo the operations in the reverse order to the one that built the formula.

If the subject is trapped in a fraction or a bracket, clear that first. $y = \frac{x + 5}{4}$ becomes $4 y = x + 5$ , so $x = 4 y - 5$ . Where the subject is squared or under a root, undo it with the opposite operation: $y = \sqrt{x - 1}$ gives $x = y^{2} + 1$ , and $y = x^{2} + 3$ gives $x = \pm \sqrt{y - 3}$ .

Extended candidates also have to deal with a subject that appears twice. For $a x + b = c x + d$ : collect the $x$ terms on one side, $a x - c x = d - b$ ; take out the common factor, $x (a - c) = d - b$ ; then divide, $x = \frac{d - b}{a - c}$ . The three steps are always the same, and the factorising in the middle step is what most students forget.

$y = 3 x + 7$ rearranges to $x = \frac{y - 7}{3}$
$y = \frac{x + 5}{4}$ rearranges to $x = 4 y - 5$
$a x + b = c x + d$ rearranges to $x = \frac{d - b}{a - c}$
$y = \sqrt{x - 1}$ rearranges to $x = y^{2} + 1$

Check a rearrangement by substituting numbers into the original formula and into your answer. If the two do not agree, one of them is wrong - and the check costs less than the mark you would otherwise lose.

Key terms you must be able to define

Term: A single number, a single letter, or a product of numbers and letters. In $5 x^{2} - 3 x + 2$ the terms are $5 x^{2}$ , $- 3 x$ and $2$ .

Coefficient: The number multiplying a letter in a term. In $- 3 x$ the coefficient is $- 3$ .

Constant term: A term with no letter in it. In $3 x^{2} - 5 x + 2$ the constant term is 2.

Like terms: Terms with exactly the same letters and the same powers, so they can be added or subtracted: $3 x$ and $- 7 x$ .

Simplify: Collect like terms until the expression has as few terms as possible. $2 a + 3 b + 5 a - 9 b$ simplifies to $7 a - 6 b$ .

Expand: Multiply out the brackets and write the expression as a sum of terms. $3 x (2 x - 4 y) = 6 x^{2} - 12 x y$ .

Factorise: Write an expression as a product of factors - the reverse of expanding. $9 x^{2} + 15 x y = 3 x (3 x + 5 y)$ .

Factorise fully: Take out the highest common factor of every term, so that no bracket has a common factor left inside it. $12 x^{2} + 18 x = 6 x (2 x + 3)$ .

Common factor: A number or letter that divides every term of the expression exactly. In $9 x^{2} + 15 x y$ the common factor is $3 x$ .

Quadratic expression: An expression of the form $a x^{2} + b x + c$ with $a \neq 0$ , such as $2 x^{2} + 7 x + 3$ .

Difference of two squares: $a^{2} - b^{2} = (a + b) (a - b)$ . The two middle terms cancel when you expand, e.g. $9 x^{2} - 25 = (3 x + 5) (3 x - 5)$ .

Perfect square: $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ , e.g. $x^{2} + 10 x + 25 = (x + 5)^{2}$ .

Completing the square: Writing $a x^{2} + b x + c$ in the form $(x + p)^{2} + q$ , e.g. $x^{2} + 6 x + 1 = (x + 3)^{2} - 8$ .

Algebraic fraction: A fraction with a letter in the numerator, the denominator or both, such as $\frac{x + 1}{x - 3}$ .

Cancel: Divide the numerator and the denominator of a fraction by a common factor. Only factors can be cancelled, never terms.

Numerator: The number or expression on the top of a fraction.

Denominator: The number or expression on the bottom of a fraction. Fractions are added by writing them over a common denominator.

Subject of a formula: The letter that stands alone in a formula, usually on the left-hand side, such as $A$ in $A = \frac{1}{2} (a + b) h$ .

Change the subject: Rearrange a formula so that a different letter is on its own. $y = 3 x + 7$ becomes $x = \frac{y - 7}{3}$ .

In context

Cambridge sets 0580 in real situations - a recipe, a bank statement, a map, a garden, a graph of a car's journey. These show how the mathematics appears in a question, then you write your own working for the numbers you are given.

Collecting like terms

Simplify $5 x^{2} - 3 x + 2 + 4 x^{2} + 7 x - 9$ . Collect the $x^{2}$ terms: $5 x^{2} + 4 x^{2} = 9 x^{2}$ . Collect the $x$ terms: $- 3 x + 7 x = 4 x$ . Collect the numbers: $2 - 9 = - 7$ . The answer is $9 x^{2} + 4 x - 7$ , and nothing else can be collected because those three families are not like terms.

Expanding two brackets

Expand and simplify $(3 x - 2) (x + 5)$ . Four products: $3 x \times x = 3 x^{2}$ , $3 x \times 5 = 15 x$ , $- 2 \times x = - 2 x$ , $- 2 \times 5 = - 10$ . The middle two are like terms, $15 x - 2 x = 13 x$ , so the answer is $3 x^{2} + 13 x - 10$ .

Difference of two squares

Expand $(4 x + 3) (4 x - 3)$ . The four products are $16 x^{2} - 12 x + 12 x - 9$ . The two middle terms are equal and opposite, so they cancel and the answer is $16 x^{2} - 9$ . Once you recognise the pattern you can square each term and subtract: $(4 x)^{2} = 16 x^{2}$ and $3^{2} = 9$ .

Factorising a quadratic

Factorise $x^{2} - x - 12$ . You need two numbers that multiply to $- 12$ and add to $- 1$ . They are $- 4$ and $3$ , so $x^{2} - x - 12 = (x - 4) (x + 3)$ . Check by expanding: $x^{2} + 3 x - 4 x - 12 = x^{2} - x - 12$ .

Rearranging a formula

The area of a trapezium is $A = \frac{1}{2} (a + b) h$ . To make $a$ the subject, multiply both sides by 2: $2 A = (a + b) h$ . Divide both sides by $h$ : $\frac{2 A}{h} = a + b$ . Subtract $b$ : $a = \frac{2 A}{h} - b$ . The same three steps work whatever the letters are called.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What are like terms?

Terms with exactly the same letters and powers.

3 x

and

- 7 x

are like terms;

3 x

and

3 x^{2}

are not.

Q2Expand

5 (2 x - 3)

10 x - 15

. The 5 multiplies both terms inside the bracket.

Q3Expand and simplify

(x + 4) (x - 7)

x^{2} - 3 x - 28

, from the four products

x^{2} - 7 x + 4 x - 28

Q4Expand

(x + 6) (x - 6)

x^{2} - 36

. The two middle terms cancel: this is the difference of two squares.

Q5Expand

(2 x - 5)^{2}

4 x^{2} - 20 x + 25

. The middle term doubles, so it is not

4 x^{2} + 25

Q6Expand

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

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

. Expand the first pair to

x^{2} + x - 6

, then multiply by

(2 x + 1)

Q7Factorise fully

12 x^{2} + 18 x

6 x (2 x + 3)

. The highest common factor is

6 x

Q8Factorise

x^{2} - 7 x + 12

(x - 3) (x - 4)

, because

3 \times 4 = 12

and

3 + 4 = 7

Q9Factorise

x^{2} + 2 x - 15

(x + 5) (x - 3)

, because

5 \times - 3 = - 15

and

5 - 3 = 2

Q10Factorise

9 x^{2} - 25

(3 x + 5) (3 x - 5)

. A difference of two squares:

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

Q11Factorise

2 x^{2} + 7 x + 3

(2 x + 1) (x + 3)

. Split the middle term using

6 + 1 = 7

Q12What does 'factorise fully' mean?

Take out the highest common factor of every term, so that no bracket has a common factor left inside it.

Q13Simplify

\frac{x^{2} - 2 x}{x^{2} - 5 x + 6}

\frac{x}{x - 3}

. Factorise to

\frac{x (x - 2)}{(x - 2) (x - 3)}

and cancel the

(x - 2)

Q14Write

\frac{1}{x - 2} + \frac{x + 1}{x - 3}

as a single fraction.

\frac{x^{2} - 5}{(x - 2) (x - 3)}

, using the common denominator

(x - 2) (x - 3)

Q15Why can you not cancel the x's in

\frac{x + 3}{x + 5}

Because

x + 3

and

x + 5

are sums. Cancelling only removes common factors, so that fraction is already in its simplest form.

Q16Make

x

the subject of

a x + b = c x + d

x = \frac{d - b}{a - c}

: collect the

x

terms, factorise to

x (a - c) = d - b

, then divide.

Q17What is the difference between

3 x + 2 x

and

3 x \times 2 x

5 x

and

6 x^{2}

. Adding collects like terms; multiplying multiplies the coefficients and adds the powers.

Q18Complete the square for

x^{2} + 6 x + 1

(x + 3)^{2} - 8

, because

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

and the extra 9 is taken back off.

0 of 18 marked as known

Exam technique for this topic

Write the expansion on its own line before you collect anything. The method mark is for the products, so a student who jumps straight to the final line can lose the whole question when the arithmetic slips.
Take the sign with the term when a bracket is multiplied by a negative number or subtracted: $- 3 (2 x - 5) = - 6 x + 15$ and $5 - 2 (x + 3) = 5 - 2 x - 6 = - 1 - 2 x$ . Sign errors here cost more marks in this topic than any other single mistake.
Check every factorisation by expanding it back, and every expansion by substituting a number such as $x = 2$ into both forms. They must give the same value, and the check takes seconds.
Factorise fully: if the bracket you have written still has a common factor, take it out. $4 x^{2} + 8 x = 4 x (x + 2)$ , not $2 (2 x^{2} + 4 x)$ . The word 'fully' is doing work in the question.
In an algebraic fraction, factorise the numerator and the denominator before cancelling anything, and only cancel factors. In a 'change the subject' question where the subject appears twice, collect its terms on one side, factorise, then divide - and show all three steps.

Why students lose marks here

A student wrote

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

Why it loses marksAdding like terms adds the coefficients, not the powers.

3 x

and

2 x

are like terms, so they collect to

5 x

The answer that scores

3 x + 2 x = 5 x

. Only multiplication produces

x^{2}

3 x \times 2 x = 6 x^{2}

A student wrote

3 (x - 4) = 3 x - 4

Why it loses marksThe 3 outside the bracket multiplies every term inside it, not just the first.

The answer that scores

3 (x - 4) = 3 x - 12

, because

3 \times - 4 = - 12

A student wrote

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

Why it loses marksA squared bracket is the bracket multiplied by itself, so the two middle terms

3 x + 3 x = 6 x

cannot be left out.

The answer that scores

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

A student wrote

9 x^{2} + 15 x y

factorises to

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

Why it loses marksThe

x

divides both terms as well as the 3, so taking out only the 3 leaves a common factor behind and the expression is not fully factorised.

The answer that scores

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

. Check:

3 x \times 3 x = 9 x^{2}

and

3 x \times 5 y = 15 x y

A student wrote

\frac{x + 3}{x + 5} = \frac{3}{5}

Why it loses marksThe x's here are terms of a sum, not factors of a product, so they cannot be cancelled. Cancelling removes a common factor from the whole of the top and the whole of the bottom.

The answer that scores

\frac{x + 3}{x + 5}

is already in its simplest form. Cancelling needs factors:

\frac{x (x - 2)}{(x - 2) (x - 3)} = \frac{x}{x - 3}

A student wrote

y = 3 x + 7

gives

x = y - \frac{7}{3}

Why it loses marksThe division by 3 has to apply to the whole of

y - 7

, not just to the 7. Undo the operations in reverse order: subtract 7 first, then divide by 3.

The answer that scores

x = \frac{y - 7}{3}

. Check with

y = 13

x = \frac{13 - 7}{3} = 2

, and

3 \times 2 + 7 = 13

, which agrees with the original formula.

Where this fits in IGCSE Maths

This page is one topic of Cambridge IGCSE Maths 0580. IGCSE Maths is the course home: every area and every topic in Cambridge's own order, the papers you sit and what each one tests, and the exam-practice tools. The full topic list, with each topic linked to its revision page, is on the IGCSE Maths syllabus page.

Syllabus coverage

Collecting like terms: what 'simplify' means

Expanding a single bracket

Expanding two brackets

Expanding three brackets (Extended only)

Factorising by taking out the common factor

Factorising quadratics (Extended only)

Algebraic fractions (Extended only)

Rearranging a formula: changing the subject

Key terms you must be able to define

In context

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

Where this fits in IGCSE Maths