Edexcel IGCSE

Algebraic manipulation

Algebraic manipulation is rewriting an expression so that it looks different but is worth exactly the same for every value of the letter. It is section 2.2 of the Edexcel International GCSE Mathematics A specification, and almost every question on it is one of two moves: take the brackets out, or put them back in.

Edexcel IGCSE 4MA1Specification 2.269 approved practice questions

Set up your class — free Practise on your own

No card needed. Free for individual students and teachers.

Before you start

What you need to be able to do already

If any of these is shaky, fix it first — every worked example below leans on all three.

Multiply and divide with negative numbers without losing a sign
Work out $- 2 \times - 5$ , and then $- 2 \times 3$ . They are $10$ and $- 6$ . If either took thought, fix directed number first: expanding $- 2 (3 x - 5)$ needs both of those rules in the same line, and a sign error there is invisible afterwards.
Use the index laws on numbers written as powersPowers and roots
Simplify $2^{3} \times 2^{4}$ . It is $2^{7}$ , because multiplying powers of the same base adds the indices. The same law is what turns $x \times x$ into $x^{2}$ when you expand $(x + 4) (x + 7)$ , so if it is not automatic the quadratic work will be guesswork.
Add, multiply and cancel ordinary fractionsFractions
Work out $\frac{1}{4} + \frac{2}{5}$ . It is $\frac{13}{20}$ : a common denominator of $20$ turns it into $\frac{5}{20} + \frac{8}{20}$ . Algebraic fractions are the same procedure with letters in place of some of the numbers, so the Higher-tier objective here is unreachable until this one is effortless.

The idea

What is algebraic manipulation?

An expression is a piece of algebra with no equals sign in it, such as $6 x + 9$ . Manipulating it means writing it a different way without changing its value: expanding removes brackets by multiplying everything inside by whatever is outside, and factorising puts brackets back by pulling out what the terms have in common. Two expressions are equivalent when they give the same number for every value of the letter, which is also the fastest check on your own work: if one value disagrees, something is wrong. Agreement at a single value is encouraging rather than proof, so avoid testing with $0$ or $1$ .

The notation, and how to read it

Every worked solution below is written in these five symbols.

SymbolReads asMeans
$3 x$ three x $3 \times x$ . The multiplication sign is left out between a number and a letter, and the number is always written first.
$x^{2}$ x squared $x \times x$ . Note that $3 x^{2}$ means $3 \times x \times x$ , and is not the same as $(3 x)^{2} = 9 x^{2}$ .
$a (b + c)$ a times the bracket b plus cEverything inside the bracket is multiplied by $a$ , giving $ab + a c$ . This one line is the whole of expanding.
$\equiv$ is identically equal toTrue for every value of the letter, not just some. $3 (2 x + 5) \equiv 6 x + 15$ is an identity; $3 x = 12$ is an equation, true only at $x = 4$ .
$\frac{3 x + 6}{x ^{2} - 4}$ an algebraic fractionA fraction with letters in the top, the bottom, or both. You may cancel only a factor of the whole top and the whole bottom, never a single term.

The one idea underneath all of it

Expanding and factorising are the same law read in opposite directions: $a (b + c) = ab + a c$ . Read left to right it removes brackets, read right to left it creates them, and every objective in this section — collecting like terms, expanding two or three brackets, taking out a common factor, factorising a quadratic, completing the square — is that law applied once or applied repeatedly. Because the two forms are equal for every value of the letter, substituting any number you like into both is always a valid check.

The smallest possible example

Expand $3 (2 x + 5)$ .

$6 x + 15$

The $3$ multiplies each term inside, not just the first: $3 \times 2 x = 6 x$ and $3 \times 5 = 15$ . Check with $x = 1$ : the original is $3 \times 7 = 21$ , and the answer is $6 + 15 = 21$ . Reaching only the first term inside the bracket is the error our question bank records most often on this subtopic.

Key terms

The words a question will use

Exam wording is precise. These five carry the meaning a question depends on.

TermWhat it meansNotation
Term
One piece of an expression, separated from the next by $+$ or $-$ . In $5 x + 3 y - 2 x$ there are three terms, and the sign in front belongs to the term that follows it, so the third term is $- 2 x$ and not $2 x$ . $- 2 x$
Like terms
Terms with exactly the same letters raised to exactly the same powers, so they can be combined into one. $5 x$ and $- 2 x$ are like terms and give $3 x$ ; $x$ and $y$ are not, and $x$ and $x^{2}$ never are. $5 x - 2 x = 3 x$
Common factor
Something that divides exactly into every term. Factorising takes out the highest one: $6 x + 9 = 3 (2 x + 3)$ . Choosing a factor that does not divide all the terms, such as writing $6 x + 9 = 2 (3 x + 4.5)$ , is catalogued as a misconception on this subtopic. $6 x + 9 = 3 (2 x + 3)$
Quadratic expression
An expression whose highest power of the letter is $2$ , such as $x^{2} + 7 x + 12$ . Factorising one means writing it as two brackets multiplied together, which is exactly the reverse of expanding them. $x^{2} + 7 x + 12$
Completing the square
Rewriting a quadratic in the form $(x + p)^{2} + q$ , so the letter appears once instead of twice. Halve the coefficient of $x$ to get $p$ , then correct the constant — squaring it instead of halving it is a recorded misconception here. $(x + 4)^{2} + 2$

The method

How to work out a algebraic manipulation

Five steps, in this order. The worked examples below point back to these numbers.

Read the command word: expand takes brackets out, factorise puts them back.The two are opposite directions of the same law, so doing the wrong one produces correct algebra that scores nothing. ‘Simplify’ usually means expand and then collect; ‘write in the form $(x + p)^{2} + q$ ’ names the finished shape it wants, which tells you the method before you start.
To expand, multiply every term outside the bracket by every term inside it. $3 (2 x + 5)$ is $6 x + 15$ , not $6 x + 5$ . With two brackets there are four products: $(x + 4) (x + 7)$ gives $x^{2} + 7 x + 4 x + 28$ . With three brackets, expand two of them first and multiply the three-term result by what is left.
Collect like terms, carrying the sign in front of each term with it. $4 (2 x + 5) - 3 x$ becomes $8 x + 20 - 3 x$ and then $5 x + 20$ . The $x$ terms combine and the constant does not, because $x$ and a number are never like terms. On a two-mark question the expanded line before collecting is the method mark.
To factorise, take out the highest common factor first; if a quadratic is left, find the pair of numbers that multiply to the constant and add to the coefficient of $x$ . $6 x + 9 = 3 (2 x + 3)$ , because $3$ is the highest number dividing both. For $x^{2} + 7 x + 12$ , $3$ and $4$ multiply to $12$ and add to $7$ , giving $(x + 3) (x + 4)$ . When the coefficient of $x^{2}$ is not $1$ , split the middle term instead: $6 x^{2} - 5 x - 6 = (2 x - 3) (3 x + 2)$ .
Check by substituting one number into the original expression and into your answer.Put $x = 2$ into $4 (2 x + 5) - 3 x$ : that is $4 \times 9 - 6 = 30$ . Put $x = 2$ into $5 x + 20$ : that is $30$ as well. Equivalent expressions must agree at every value, so one disagreement proves an error and takes about ten seconds to find.

Use it when

There is no equals sign to solve. The question says expand, simplify, factorise, or ‘write in the form’, and the answer it wants is another expression rather than a value for the letter.

Do not use it when

There is an equals sign and the command word is solve. $x^{2} + 7 x + 12 = 0$ is a quadratic *equation*: factorising to $(x + 3) (x + 4) = 0$ is a step, but the answer is $x = - 3$ and $x = - 4$ , and that belongs to section 2.7. Beware one move that is never valid in either place: you may cancel only a factor of the whole numerator and the whole denominator, so $\frac{x + 3}{3}$ does not simplify to $x$ , while $\frac{3 ( x + 2 )}{3}$ does simplify to $x + 2$ .

Worked examples

6 IGCSE Maths practice questions on algebraic manipulation, fully worked

One per objective, so the range is covered rather than the same question repeated. Every step is written out, and the answer is shown — this is study material, not a test.

Example 1

Work out the value of 3a + 2c when a = 4 and c = 7. [1 mark]

Evaluate expressions by substituting numerical values for letters

Worked solution

Step 1: Write down the expression: 3a + 2c.
Step 2: Replace a with 4 and c with 7: $3 \times 4 + 2 \times 7$ .
Step 3: Work out each multiplication: $12 + 14$ .
Step 4: Add the results together: $12 + 14 = 26$ .

Answer26

Example 2

Factorise $x^{2} + 7 x + 12$ fully. [2 marks]

Factorising quadratic expressions

Worked solution

Step 1: We need two numbers that multiply to give 12 and add to give 7.
Step 2: List factor pairs of 12: (1, 12), (2, 6), (3, 4).
Step 3: Check which pair adds to 7: $1 + 12 = 13$ (too big), $2 + 6 = 8$ (too big), $3 + 4 = 7$ (perfect).
Step 4: So the numbers are 3 and 4.
Step 5: Write the factorised form as (x + 3)(x + 4).

Answer(x + 3)(x + 4)

Example 3

Simplify the expression 5a + 3b - 2a + 4b. [2 marks]

Collect like terms

Worked solution

Step 1: Identify the like terms: terms with 'a' are 5a and -2a.
Step 2: Combine the 'a' terms: 5a - 2a = 3a.
Step 3: Identify the 'b' terms: 3b and 4b.
Step 4: Combine the 'b' terms: 3b + 4b = 7b.
Step 5: Write the simplified expression: 3a + 7b.

Answer3a + 7b

Show 3 more worked examples

Example 4

Work out 4(3x + 7). Give your answer in the form ax + b, e.g. 2x + 3. [2 marks]

Multiply a single term over a bracket

Worked solution

Step 1: The expression means 4 multiplied by the bracket (3x + 7).
Step 2: Multiply the 4 by the first term inside: $4 \times 3$ x = 12x.
Step 3: Multiply the 4 by the second term inside: $4 \times 7 = 28$ .
Step 4: Add the two results together: 12x + 28.

Answer12x + 28

Example 5

Factorise 6x + 9 by taking out the common factor. Give your answer in fully factorised form, e.g. 2(3x + 5). [2 marks]

Take out common factors

Worked solution

Step 1: Look at the two numbers: 6 and 9.
Step 2: Find the largest number that divides both 6 and 9. The factors of 6 are 1, 2, 3, and 6. The factors of 9 are 1, 3, and 9. The largest common factor is 3.
Step 3: This largest common factor, 3, will go outside the bracket.
Step 4: Now, work out what must go inside the bracket. Divide 6x by 3 to get 2x. Divide 9 by 3 to get 3.
Step 5: So inside the bracket we write 2x + 3.
Step 6: The fully factorised expression is 3(2x + 3).

Answer3(2x + 3)

Example 6

Simplify fully $\frac{x ^{2} + 5 x + 6}{x ^{2} - 4}$ . Show your working. [3 marks]

Simplifying algebraic fractions

Worked solution

Step 1: Factorise the numerator: x² + 5x + 6 = (x + 2)(x + 3).
Step 2: Factorise the denominator: x² - 4 = (x - 2)(x + 2).
Step 3: Cancel the common factor (x + 2) from top and bottom.

Answer

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

Common mistakes

Where marks actually get lost

Learnly tags 15 distinct misconceptions on this subtopic. These are errors that show up in real student work — not generic revision advice.

Examiner traps

Only multiplying the first term
What goes wrong: Students multiply the term outside the bracket only by the first term inside, ignoring the second term.
What to do instead: Remember to multiply the outside term by every term inside the bracket. For 2(3x+5), multiply 2 by 3x AND 2 by 5.
Adding instead of multiplying
What goes wrong: Students add the outside term to each term inside the bracket instead of multiplying.
What to do instead: When expanding brackets, you multiply the outside term with each inside term, not add. For 2(3x+5), do $2 \times 3$ x and $2 \times 5$ .
Sign errors with negative numbers
What goes wrong: Students make mistakes with signs when the term outside or inside the bracket is negative, e.g., -2(3x-5) gives -6x-10 instead of -6x+10.
What to do instead: Be careful with signs. Multiply the sign as well: negative times positive gives negative, negative times negative gives positive.
Forgetting to multiply the constant
What goes wrong: Students multiply the variable term correctly but forget to multiply the constant term inside the bracket.
What to do instead: Don't forget the constant term! For 2(3x+5), you must multiply 2 by 5 as well to get 6x+10.
Adding instead of factoring
What goes wrong: Students add the common factor to each term instead of multiplying it out, e.g., writing 6x + $9 = 3 + 2$ x + 3 instead of 3(2x + 3).
What to do instead: When factorising, think of the common factor as something you multiply out, not add. Check by expanding your answer.
Only factoring one term
What goes wrong: Students factor only the first term, e.g., writing 6x + 9 = 3(2x) + 9, leaving the second term unchanged.
What to do instead: The common factor must be taken out of every term. Divide each term by the common factor and put the results inside brackets.

See 9 more examiner traps

Using wrong common factor
What goes wrong: Students choose a factor that doesn't divide all terms, e.g., factoring 6x + 9 as 2(3x + 4.5) or 6(x + 1.5) instead of the highest common factor 3.
What to do instead: Find the largest number that divides into all coefficients. For 6 and 9, the highest common factor is 3, not 2 or 6.
Sign error in brackets
What goes wrong: Students incorrectly handle signs when taking out a negative common factor, e.g., writing -6x - 9 = -3(2x + 3) instead of -3(2x + 3) is correct, but they might write -3(2x - 3).
What to do instead: When taking out a negative factor, every sign inside the brackets changes. Check by expanding: -3(2x+3) = -6x-9.
Forgetting to halve the coefficient
What goes wrong: When completing the square, students often forget to halve the coefficient of x before squaring. For example, for x^ $2 + 6$ x, they write (x+6)^2 instead of (x+3)^2.
What to do instead: Always halve the coefficient of x before squaring. For x^2 + bx, use (x + b/2)^2.
Incorrect sign in the bracket
What goes wrong: Students may put the wrong sign inside the bracket, e.g., for x^ $2 - 4$ x, they write (x-2)^2 correctly but then add the constant incorrectly, or they write (x+2)^2.
What to do instead: The sign inside the bracket is the same as the sign of the coefficient of x. For x^ $2 - 6$ x, use (x - 3)^2.
Not factoring out negative coefficient
What goes wrong: When the coefficient of x^2 is negative, students fail to factor it out first, leading to errors. For example, for -x^ $2 + 4$ x, they try to complete the square without factoring -1.
What to do instead: If the x^2 coefficient is negative, factor it out first. For -x^ $2 + 4$ x, rewrite as -(x^ $2 - 4$ x) then complete the square.
Adding constant incorrectly
What goes wrong: Students forget to adjust the constant term when completing the square, e.g., for x^ $2 + 6$ x + 5, they write (x+3)^ $2 + 5$ instead of (x+3)^ $2 - 4$ .
What to do instead: After forming the square, subtract the square of half the coefficient and add the original constant. For x^ $2 + 6$ x + 5: (x+3)^ $2 - 9 + 5$ = (x+3)^ $2 - 4$ .
Wrong sign in the bracket
What goes wrong: Students may write (x - b/2)² instead of (x + b/2)² when b is positive, or vice versa.
What to do instead: The sign in the bracket is the same as the sign of the coefficient of x. For x² + bx, use (x + b/2)².
Incorrect adjustment for negative constant
What goes wrong: When the constant term is negative, students forget to subtract the squared term from the original constant, leading to an error.
What to do instead: After adding the squared term inside the bracket, subtract it outside. For x² + bx + c, write (x + b/2)² - (b/2)² + c.
Ignoring coefficient of x²
What goes wrong: Students attempt to complete the square without factoring out the coefficient of x² when it is not 1.
What to do instead: If the coefficient of x² is not 1, factor it out first. For ax² + bx + c, rewrite as a(x² + (b/a)x) + c.

Exam technique

How the marks are actually awarded

Method marks are given for working a marker can follow. This is what to put on the page.

Where the marks are: Questions in this section run from $1$ to $4$ marks. Expanding a single bracket is usually $1$ ; ‘expand and simplify’ with two brackets is $2$ , one for the four correct products and one for the collected answer. Factorising a quadratic is $2$ , and a harder one such as $6 x^{2} - 5 x - 6$ is $2$ or $3$ , with the marks spread across the split middle term and the two brackets. Completing the square is $3$ . On a ‘show that’ question the working *is* the answer: you are being asked to get from one line to another, so a bare final line scores nothing however correct it is. Everywhere else, write the expanded line down anyway — it is the method mark, and it survives a slip in the collecting that comes after it.
Command words to expect: ExpandExpand and simplifySimplifyFactoriseFactorise completelyMultiply out the bracketsWrite in the formShow thatWork out the value ofGive your answer in its simplest form
Accuracy and rounding: Nothing rounds here — algebra is exact — so the marks go on form instead. ‘Factorise completely’ means nothing is left to take out: $4 x + 8 = 4 (x + 2)$ , and $2 (2 x + 4)$ is not finished. Write a quadratic with the highest power first, $x^{2} + 11 x + 28$ . Keep fractions as fractions: completing the square on $x^{2} + 5 x$ gives $(x + \frac{5}{2})^{2} - \frac{25}{4}$ , never $2.5$ and $6.25$ , unless the question asks for decimals. And when a question names a form, hand back that form — an answer of $5 (x - 2)^{2} - 7$ answers ‘in the form $a (x + b)^{2} + c$ ’, while the expanded version does not.
Calculator: A calculator is allowed in every paper of this qualification, on both Foundation and Higher tiers, and it cannot expand or factorise anything for you. Its one honest use on this topic is the substitution check in step 5: put the same number into the original expression and into your answer and confirm the two agree.

Check yourself

You should now be able to:

Substitute given numbers for letters and work out the value of an expression, keeping to the order of operations.
Collect like terms, carrying the sign in front of each term with it.
Multiply a single term over a bracket, reaching every term inside and not just the first.
Take out the highest common factor of every term and write the expression as a product.
Expand the product of two linear brackets and collect the four terms into three.
Factorise a quadratic $x^{2} + b x + c$ using the pair of numbers that multiply to $c$ and add to $b$ .
Expand the product of three linear brackets, simplifying in two stages (Higher).
Factorise a harder quadratic such as $6 x^{2} - 5 x - 6$ , where the coefficient of $x^{2}$ is not $1$ (Higher).
Add, subtract, multiply, divide and simplify algebraic fractions, cancelling only whole factors (Higher).
Complete the square, writing $x^{2} + b x + c$ as $(x + \frac{b}{2})^{2} + c - \frac{b ^{2}}{4}$ (Higher).
Use algebra to construct a proof, or to show that one expression is identically equal to another (Higher).

Specification coverage

The 11 things 4MA1 asks you to do

The assessable objectives for equations, formulae and identities — algebraic manipulation, in the order the specification lists them.

Factorising quadratic expressions
Evaluate expressions by substituting numerical values for letters
Collect like terms
Multiply a single term over a bracket
Take out common factors
Simplifying algebraic fractions

Show 5 more objectives

Expand the product of two simple linear expressions
Expand the product of two or more linear expressions
Factorise harder quadratic expressions (e.g. 6x^2 - 5x - 6)
Complete the square for a given quadratic expression
Use algebra to support and construct proofs

See who is stuck on algebraic manipulation before you teach it.

Create a class, share the 8-character code, and set algebraic manipulation as practice. Your students work through all 69 approved practice questions — not just the 6 on this page — and you see which objectives the class has not got yet. Free for individual students and teachers.

How Learnly works for teachers

Set up your class — free Practise on your own

What you need to be able to do already

Multiply and divide with negative numbers without losing a sign

Use the index laws on numbers written as powersPowers and roots

Add, multiply and cancel ordinary fractionsFractions

What is algebraic manipulation?

The notation, and how to read it

The words a question will use

Term

Like terms

Common factor

Quadratic expression

Completing the square

How to work out a algebraic manipulation

6 IGCSE Maths practice questions on algebraic manipulation, fully worked

Work out the value of 3a + 2c when a = 4 and c = 7. [1 mark]

Factorise x2+7x+12 fully. [2 marks]

Simplify the expression 5a + 3b - 2a + 4b. [2 marks]

Work out 4(3x + 7). Give your answer in the form ax + b, e.g. 2x + 3. [2 marks]

Factorise 6x + 9 by taking out the common factor. Give your answer in fully factorised form, e.g. 2(3x + 5). [2 marks]

Simplify fully x2−4x2+5x+6​. Show your working. [3 marks]

Where marks actually get lost

Only multiplying the first term

Adding instead of multiplying

Sign errors with negative numbers

Forgetting to multiply the constant

Adding instead of factoring

Only factoring one term

Using wrong common factor

Sign error in brackets

Forgetting to halve the coefficient

Incorrect sign in the bracket

Not factoring out negative coefficient

Adding constant incorrectly

Wrong sign in the bracket

Incorrect adjustment for negative constant

Ignoring coefficient of x²

How the marks are actually awarded

You should now be able to:

The 11 things 4MA1 asks you to do

See who is stuck on algebraic manipulation before you teach it.

Factorise $x^{2} + 7 x + 12$ fully. [2 marks]

Simplify fully $\frac{x ^{2} + 5 x + 6}{x ^{2} - 4}$ . Show your working. [3 marks]