Notation, vocabulary and manipulation

Substitution is replacing letters with numbers and then evaluating, following the order of operations from Structure and calculation. Put the value in brackets when it is negative — otherwise −3² is easily read as −9 rather than (−3)² = 9, which is the commonest substitution error there is.

Collecting like terms means adding terms with exactly the same letters and powers. 3x and 5x are like terms; 3x and 3x² are not, and neither are 3x and 3y. So 4x + 3y − x + 2y simplifies to 3x + 5y and stops there.

Expanding a single bracket multiplies every term inside by the term outside. Watch the signs: 3(2x − 5) = 6x − 15, and −2(x − 4) = −2x + 8, because a negative times a negative is positive.

Expanding two brackets multiplies every term in the first by every term in the second — four products from two binomials — then collect like terms:

(x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15

At Higher tier this extends to three brackets. Expand two of them first, then multiply the result by the third, taking each term in turn.

Factorising is expanding in reverse, and there are three kinds to recognise.

Common factors. Take out the largest thing every term shares, numbers and letters both: 6x² + 9x = 3x(2x + 3). Check by expanding.

Quadratics of the form x² + bx + c. Find two numbers that multiply to give c and add to give b, then write them in brackets. For x² − 2x − 15 the numbers are −5 and 3, so it factorises to (x − 5)(x + 3).

The difference of two squares. Whenever an expression is one square subtracted from another, it factorises immediately:

a² − b² = (a + b)(a − b)

So x² − 49 = (x + 7)(x − 7), and 9x² − 25 = (3x + 5)(3x − 5). Spotting this saves a great deal of time, and it appears in surd and fraction questions as well as in factorising ones.

Algebraic fractions follow exactly the same rules as numerical ones — the difficulty is that simplifying usually requires factorising first.

Simplifying. Factorise the top and the bottom, then cancel any bracket that appears in both:

(x² − 9) ⁄ (x² + 7x + 12) = ((x + 3)(x − 3)) ⁄ ((x + 3)(x + 4)) = (x − 3) ⁄ (x + 4)

Only whole factors cancel. In (x + 3) ⁄ (x + 4) nothing cancels at all, because x is a term rather than a factor.

Multiplying. Multiply the tops and the bottoms, then cancel — or cancel first, which keeps the numbers small.

Dividing. Turn the second fraction upside down and multiply.

Adding and subtracting. Find a common denominator exactly as with numbers, usually by multiplying the two denominators together, then combine the numerators and simplify.

Rearranging a formula to change the subject uses the same inverse operations as solving an equation — whatever is done to one side is done to the other. The order is the reverse of the order of operations: undo addition and subtraction first, then multiplication and division, then powers and roots.

At Higher tier the subject often appears more than once. Then the method is: collect every term containing it on one side, factorise it out, and divide by the bracket.

Proof means showing something is true for every value, not checking a few examples. Testing three numbers is not a proof, and gains no marks.

The method is to write the general case algebraically and manipulate it. The standard forms are worth having ready: any integer is n, an even number is 2n, an odd number is 2n + 1, and consecutive integers are n, n + 1, n + 2.

To show that the sum of two consecutive odd numbers is a multiple of 4: take 2n + 1 and 2n + 3. Their sum is 4n + 4 = 4(n + 1), which is 4 times an integer, so it is a multiple of 4 whatever n is. Finishing with a sentence saying what has been shown is part of the mark.

Showing two expressions are equivalent works the same way — expand and simplify one side until it matches the other, writing ≡ rather than = because the statement holds for every value.

A function is a rule with inputs and outputs, written f(x). So f(x) = 3x − 1 means f(4) = 3(4) − 1 = 11: substitute the input for x.

An inverse function, written f⁻¹(x), reverses the process — it takes the output back to the input. Find it by writing y = f(x), rearranging to make x the subject, then swapping the letters. For f(x) = 3x − 1: y = 3x − 1, so x = (y + 1) ⁄ 3, giving f⁻¹(x) = (x + 1) ⁄ 3.

A composite function applies one function then another. fg(x) means do g first, then f — read it from the inside out, right to left, which is the opposite of how it looks. If f(x) = 3x − 1 and g(x) = x², then fg(2) = f(g(2)) = f(4) = 11, while gf(2) = g(5) = 25. The order matters.