Algebraic ManipulationAQA GCSE Maths: Revision notes

Section 1

How do you use and interpret algebraic notation?

Algebraic notation is a shorthand way of writing mathematical expressions. Understanding how to read and write it correctly is fundamental to all algebra work.

ab means a × b (no multiplication sign written)
2a means 2 × a (the number in front is called a coefficient)
a² means a × a (the small number above is called a power or exponent)
Brackets group terms together and must be dealt with carefully during manipulation
Terms are individual parts separated by + or − signs (e.g. in 3x + 2y − 5, there are three terms: 3x, 2y, and −5)
Like terms have the same letters and powers (e.g. 3x and 5x are like terms; 3x and 3x² are not)

When substituting numerical values, always follow the correct order of operations (BIDMAS/BODMAS).

Key termsalgebraic notationcoefficientpowerexponentlike terms

Example

Write in algebraic notation: the product of a and b, minus twice c. Answer: ab − 2c. This shows ab (not a × b), the coefficient 2 before c, and the minus sign separating the terms.

Section 2

How do you simplify expressions by collecting like terms?

Collecting like terms (also called combining like terms) is the process of adding and subtracting terms that have identical letter parts.

Step-by-step process:

Identify all the like terms (same letters and powers)
Add or subtract their coefficients
Write the answer with the simplified terms

Examples of like terms:

3x and 7x are like terms → combine to 10x
2xy and −5xy are like terms → combine to −3xy
4a² and a² are like terms → combine to 5a²

Examples of unlike terms (cannot be combined):

3x and 3x² (different powers)
2a and 3b (different letters)
5 and 5x (one has no letter)

When simplifying, it often helps to group like terms together first: 3x + 5y + 2x − y = (3x + 2x) + (5y − y) = 5x + 4y

Key termscollecting like termscombining like termssimplify

Common mistake

Students often try to combine unlike terms, e.g. writing 3x + 2y = 5xy. Remember: you can only combine terms if the letter parts are exactly the same. 3x + 2y stays as 3x + 2y.

Example

Simplify 4a + 3b − 2a + b + 5. Collect like terms: (4a − 2a) + (3b + b) + 5 = 2a + 4b + 5.

Section 3

How do you expand brackets and factorise expressions?

Expanding means removing brackets by multiplying out. Factorising is the opposite process—finding common factors and writing them outside brackets.

Expanding a single term over a bracket:

Multiply the term outside by every term inside
Example: 3(2x + 5) = 6x + 15
Example: −2(a − 3) = −2a + 6 (note the sign change)

Expanding products of two binomials (use FOIL or the grid method):

(x + 3)(x + 2) = x² + 2x + 3x + 6 = x² + 5x + 6
(2x − 1)(x + 4) = 2x² + 8x − x − 4 = 2x² + 7x − 4

Factorising by removing common factors:

Identify the highest common factor (HCF) of all terms
Write it outside the bracket
Example: 6x + 9 = 3(2x + 3)
Example: 12a²b + 8ab = 4ab(3a + 2)

Difference of two squares (HT):

(a + b)(a − b) = a² − b²
This means a² − b² = (a + b)(a − b)
Example: x² − 9 = (x + 3)(x − 3)
Example: 4a² − 25 = (2a + 5)(2a − 5)

Key termsexpandingfactorisingbinomialdifference of two squaresFOIL

Exam tip

When expanding (x + a)(x + b), remember the middle term comes from the cross-multiplication (outer and inner products). For (x + 2)(x + 3), the middle term is 2x + 3x = 5x.

Example

Factorise 15xy + 10x². Find HCF: 5x. Answer: 5x(3y + 2x). Always check: 5x × 3y + 5x × 2x = 15xy + 10x² ✓

Section 4

How do you factorise quadratic expressions?

Quadratic expressions are of the form ax² + bx + c. Factorising depends on the value of a.

Quadratics where a = 1 (form x² + bx + c):

Find two numbers that multiply to give c and add to give b
Example: x² + 5x + 6 = (x + 2)(x + 3) because 2 × 3 = 6 and 2 + 3 = 5
Example: x² − 3x − 10 = (x − 5)(x + 2) because −5 × 2 = −10 and −5 + 2 = −3

Quadratics where a ≠ 1 (form ax² + bx + c) [Higher Tier]:

Method 1: Find factors of ac that add to b, then group and factorise
- Example: 2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)
Method 2: Use the grid/box method or trial and error

Completing the square (HT):

Rearrange x² + bx + c into the form (x + p)² + q
Formula: x² + bx + c = (x + b/2)² − (b/2)² + c
Example: x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4
This is useful for solving quadratics and finding the turning point of a parabola

Key termsquadratic expressionfactorising quadraticscompleting the square

Exam tip

For x² + bx + c, always check your factors: multiply the brackets back out to verify. For 2x² + 7x + 3, check: (x + 3)(2x + 1) = 2x² + x + 6x + 3 = 2x² + 7x + 3 ✓

Example

Complete the square for x² + 8x − 3. Using x² + 8x = (x + 4)² − 16, so x² + 8x − 3 = (x + 4)² − 16 − 3 = (x + 4)² − 19.

Common mistake

When factorising x² + bx + c, students often find the right numbers but write them in the wrong order. For x² + 5x + 6, the factors 2 and 3 give (x + 2)(x + 3), not (x + 5)(x + 6).

Section 5

How do you simplify algebraic fractions and rearrange formulae?

Simplifying algebraic fractions (HT):

Cancel common factors from numerator and denominator
Example: (3x + 6)/(x + 2) = 3(x + 2)/(x + 2) = 3 (when x ≠ −2)
Example: (x² − 4)/(x − 2) = (x + 2)(x − 2)/(x − 2) = x + 2 (when x ≠ 2)
When combining fractions, find a common denominator:
- a/b + c/d = (ad + bc)/(bd)
- Example: 2/x + 3/y = (2y + 3x)/(xy)

Rearranging formulae to change the subject:

The subject is the letter being solved for (usually on its own on one side)
Use inverse operations: if it's +, subtract; if it's ×, divide
Example: If A = lw, rearrange for w: w = A/l
Example: If v = u + at, rearrange for a: v − u = at, so a = (v − u)/t

Cases where the subject appears twice (HT):

Collect terms containing the subject on one side
Example: v² = u² + 2as, rearrange for a: 2as = v² − u², so a = (v² − u²)/(2s)
Example: y = mx + c, rearrange for x: mx = y − c, so x = (y − c)/m
More complex: If p = 2a + 3, and a = p − 5, substitute to find: p = 2(p − 5) + 3 → p = 2p − 10 + 3 → −p = −7 → p = 7

Substituting values into formulae and expressions:

Replace each letter with its numerical value
Follow order of operations (BIDMAS/BODMAS)
Example: If A = πr², and r = 3, then A = π × 3² = 9π ≈ 28.3
Example: If s = ut + ½at², with u = 5, t = 2, a = 3: s = 5(2) + ½(3)(2²) = 10 + 6 = 16

Key termsalgebraic fractionscommon denominatorrearranging formulaesubjectsubstitution

Exam tip

When simplifying algebraic fractions, factorise the numerator and denominator first, then cancel common factors. Always state any restrictions (e.g. x ≠ 0) where division by zero would occur.

Example

Rearrange T = 2π√(L/g) for g. Square both sides: T² = 4π²(L/g). Multiply by g: gT² = 4π²L. Divide by T²: g = 4π²L/T².

Think of it like this

Rearranging a formula is like untangling a knot—use inverse operations to systematically undo each operation until your variable is isolated.

Must Know

Like terms have identical letter parts and powers; only these can be combined by adding/subtracting their coefficients
Expanding uses the distributive property: 3(2x + 5) = 6x + 15; factorising is the reverse, finding common factors
Quadratics x² + bx + c factor into (x + p)(x + q) where p and q multiply to c and add to b; use the method of completing the square to rewrite as (x + b/2)² − (b/2)² + c (HT)
Difference of two squares: a² − b² = (a + b)(a − b) (HT)
Rearranging formulae requires using inverse operations and collecting terms with the subject on one side; when the subject appears twice, rearrange to isolate it (HT)
Algebraic fractions simplify by factorising and cancelling common factors; combine using a common denominator (HT)
Always substitute values carefully following BIDMAS/BODMAS and include units/π where needed

That's the notes covered.

Carry on to the next subtopic.