Collecting like terms, expanding single and double brackets, factorising into brackets, and simplifying algebraic fractions at Higher tier.

What this dot point is asking

AQA wants you to manipulate algebraic expressions confidently: collect like terms, expand single and double brackets, factorise back into brackets, and at Higher tier simplify algebraic fractions. These are the foundational skills that every later algebra topic depends on, from solving equations to sketching curves, so AQA tests them both as standalone short questions and as the first step inside larger problems.

Collecting like terms

Like terms have exactly the same combination of letters raised to the same powers. You can add or subtract their coefficients but you cannot combine unlike terms. For instance $5a + 3b - 2a + b = 3a + 4b$, where $5a$ and $-2a$ combine and $3b$ and $b$ combine, but $a$ terms and $b$ terms stay separate. Note that $x^2$ and $x$ are unlike terms because the powers differ, so $4x^2 + 3x - x^2 = 3x^2 + 3x$ and no further simplification is possible.

Expanding brackets

For a single bracket, multiply the outside term by each inside term: $3(2x - 5) = 6x - 15$. Watch signs when the outside term is negative: $-2(x - 4) = -2x + 8$.

For double brackets, multiply each term in the first bracket by each term in the second (often remembered as FOIL: First, Outer, Inner, Last), then collect like terms. The special difference of two squares result is worth memorising: $(x + a)(x - a) = x^2 - a^2$. So $(x + 7)(x - 7) = x^2 - 49$ instantly, with no middle term.

Factorising into brackets

Factorising reverses expansion. First always look for a common factor: $6x^2 + 9x = 3x(2x + 3)$, taking out the highest common factor $3x$. For a quadratic $x^2 + bx + c$, find two numbers multiplying to $c$ and adding to $b$: $x^2 + 7x + 12 = (x + 3)(x + 4)$. The difference of two squares factorises in reverse: $x^2 - 25 = (x + 5)(x - 5)$, and $9x^2 - 16 = (3x + 4)(3x - 4)$.

When the leading coefficient is not $1$, split the middle term. For $2x^2 + 7x + 3$, multiply $2 \times 3 = 6$, find two numbers multiplying to $6$ and adding to $7$ (those are $6$ and $1$), rewrite as $2x^2 + 6x + x + 3$, then group: $2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$.

Simplifying algebraic fractions at Higher tier

At Higher tier you simplify, add, subtract and multiply algebraic fractions. The golden rule is that you can only cancel whole factors, never individual terms. To simplify, factorise top and bottom fully and cancel common brackets. To add or subtract, find a common denominator exactly as with numerical fractions: $\dfrac{2}{x} + \dfrac{3}{x + 1} = \dfrac{2(x + 1) + 3x}{x(x + 1)} = \dfrac{5x + 2}{x(x + 1)}$.

Expanding three brackets and squaring binomials

Higher questions sometimes need three brackets multiplied. Expand two of them first, then multiply the result by the third, collecting terms at each stage: $(x + 1)(x + 2)(x + 3)$ becomes $(x^2 + 3x + 2)(x + 3) = x^3 + 6x^2 + 11x + 6$. Squaring a binomial is worth knowing as a pattern: $(a + b)^2 = a^2 + 2ab + b^2$, so $(2x + 3)^2 = 4x^2 + 12x + 9$. The middle term, twice the product of the two terms, is the one most often forgotten, turning a three-term answer into a wrong two-term one.

Algebraic proof

Proof questions ask you to show a statement is always true, using algebra rather than examples. Represent the quantities generally: an even number is $2n$, an odd number is $2n + 1$, and consecutive integers are $n$ and $n + 1$. To prove the sum of two consecutive integers is odd, write $n + (n + 1) = 2n + 1$, which is one more than the even number $2n$, hence odd. The key is to use a general term, manipulate it, and finish with a clear statement that the result has the required form, since a few worked examples never count as a proof.