Algebraic manipulation: simplifying expressions, expanding single and double brackets, factorising (common factors, quadratics and the difference of two squares), and rearranging (changing the subject of) formulae.

What this dot point is asking

Algebraic manipulation is the toolkit that every other algebra topic depends on. Edexcel expects you to simplify expressions, expand single and double brackets, factorise in several ways, and change the subject of a formula. None of these is hard in isolation, but accuracy with signs and indices is what separates full marks from dropped ones, and these skills reappear in equations, graphs and proof throughout both tiers.

Simplifying expressions

Simplifying means combining like terms and tidying powers. Like terms have identical letter parts: $3x$ and $5x$ combine to $8x$, but $3x$ and $3x^2$ do not. When multiplying terms, multiply the numbers and add the indices: $4a \times 3a^2 = 12a^3$. When dividing, subtract the indices: $\dfrac{12y^5}{3y^2} = 4y^3$.

Expanding brackets

Expanding removes brackets by multiplying out.

For a single bracket, multiply every inside term by the outside term: $3(2x - 5) = 6x - 15$, and $-2(4 - x) = -8 + 2x$ (watch the sign change on both terms).

For double brackets, multiply each term in the first bracket by each term in the second. FOIL (First, Outer, Inner, Last) is a useful order. Squaring a bracket is a double bracket too: $(x + 4)^2 = (x+4)(x+4) = x^2 + 8x + 16$, not $x^2 + 16$.

Factorising

Factorising reverses expanding: it writes an expression as a product. There are three main types at GCSE.

To factorise $x^2 + 7x + 12$, find two numbers multiplying to $12$ and adding to $7$: those are $3$ and $4$, so $(x + 3)(x + 4)$. To factorise $x^2 - 25$, recognise the difference of two squares: $(x + 5)(x - 5)$.

Changing the subject of a formula

Rearranging a formula uses the same inverse operations as solving an equation, but the answer is in terms of letters. To make $u$ the subject of $v = u + at$, subtract $at$ from both sides: $u = v - at$. When the required letter appears inside a power or root, undo that last: to make $r$ the subject of $A = \pi r^2$, divide by $\pi$ then square-root, giving $r = \sqrt{\dfrac{A}{\pi}}$.

A harder case has the subject appearing twice. Collect those terms on one side, factorise out the letter, then divide. This pattern is examined at Higher tier and rewards careful, ordered working.

Simplifying algebraic fractions (Higher)

At Higher tier you simplify algebraic fractions by factorising the top and bottom and cancelling common brackets. For $\dfrac{x^2 - 9}{x^2 + 5x + 6}$, factorise both: the top is the difference of two squares $(x + 3)(x - 3)$, and the bottom factorises to $(x + 2)(x + 3)$. The common factor $(x + 3)$ cancels, leaving $\dfrac{x - 3}{x + 2}$. The crucial point is that you may only cancel whole brackets (factors), never individual terms, so you cannot cancel the $x^2$ terms before factorising. This connects directly to the factorising skills above, which is why fluency with the difference of two squares and quadratic factorising pays off.

Try this

Q1. Expand and simplify $3(2x + 1) - 2(x - 4)$. [2 marks]

• Cue. $6x + 3 - 2x + 8 = 4x + 11$.

Q2. Factorise fully $x^2 - 16$. [1 mark]

• Cue. Difference of two squares: $(x + 4)(x - 4)$.