Algebraic Expressions

(a + b)(c + d + e) = ac + ad + ae + bc + bd + be

Each term of the binomial multiplies every term of the trinomial — six products in total.

Write your answer in standard form

Arrange the final answer with the highest power first (x³, then x², then x, then the constant). This makes like terms easy to spot and answers easy to check.

5a(x − 2) + 3(x − 2) = (x − 2)(5a + 3)

Both terms contain the identical bracket (x − 2), so the whole bracket comes out as the common factor. Treat it exactly as you would treat a single letter. This pattern is the last step of factorising by grouping, so getting comfortable with it here makes that section much easier.

−4x³ + 8x² = −4x²(x − 2)

When the leading term is negative it is usually tidier to take the negative out too. Every sign inside the bracket then flips: −4x³ ÷ (−4x²) = x, and +8x² ÷ (−4x²) = −2. Expanding always tells you whether you got the signs right.

Two mistakes that cost marks

Losing a term. If a term is the common factor, dividing it by itself leaves 1, not nothing: 6x² + 6x = 6x(x + 1), never 6x(x).

Taking out too much. The common factor must divide into every term. In 4x² + 6x + 3, nothing divides into all three, so there is no common factor at all — move on to another method.

a² − b² = (a + b)(a − b)

The brackets are identical except for the sign. It does not matter which one you write first — (a + b)(a − b) and (a − b)(a + b) are the same answer.

Factorise fully — sometimes it works twice

x⁴ − 16 = (x² + 4)(x² − 4), and the second bracket is itself a difference of two squares, giving (x² + 4)(x + 2)(x − 2). Always look at each new bracket and ask whether it factorises again. The first bracket, x² + 4, is a sum of squares and stops there.

Two mistakes that cost marks

Trying it on a sum. x² + 9 does not factorise. Only the difference works.

Squaring only the letter. In 9x² the square root is 3x, not 9x or 3x². Take the root of the number and the variable.

Always check by expanding

After factorising, expand your answer to verify it matches the original expression. This takes seconds and guarantees your answer is correct.

Grouping order matters

Both groups must produce the same binomial factor after factoring. If they do not, try rearranging the terms into a different grouping before concluding the expression cannot be factored this way.

Remembering the signs — SOAP

Reading the three signs in order: Same (the binomial matches the question), Opposite (the middle term of the trinomial is the other sign), Always Positive (the last term of the trinomial is always +). Both identities follow SOAP, so you only have to remember one shape.

Two mistakes that cost marks

Writing 2ab in the middle. The trinomial is a² − ab + b², not a² − 2ab + b². That doubled version is the expansion of (a − b)², a completely different thing — and if you use it, your check by expanding will not come back to the question.

Trying to factorise the trinomial again. a² ± ab + b² never factorises further at this level. Once you have the two brackets, you are finished.

Step 1 — common factor, every single time

Before anything else, ask whether a number or variable divides into every term. Skipping this is the most common reason a factorisation “does not work”: 2x² − 50 refuses to fit the difference-of-squares pattern until you write it as 2(x² − 25).

Step 3 — look at every bracket again

“Factorise fully” means keep going until nothing else moves. After grouping, one bracket is often a difference of squares. After a common factor, the bracket is often a trinomial. x⁴ − 16 needs the difference of squares twice. Only stop when every bracket is checked.

a² + b² — a sum of two squares never factorises. (A sum of two cubes does; do not mix them up.)

a² ± ab + b² — the trinomial left behind by the cubes identities stops there.

A trinomial with no whole-number pair — if nothing multiplies to ac and adds to b, it does not factorise over the integers, and you would need the quadratic formula instead.

State restrictions on the variable

The denominator of a fraction can never equal zero. Always state restrictions such as x ≠ 0 when the variable appears in the denominator. This is a required part of a complete answer in Grade 10.

Cancel factors, never terms

x + 3x does not simplify to 31. The x on top is part of a sum, so it is a term, not a factor.

x(x + 3)x does simplify to x + 3, because there the x is multiplying the whole bracket — it is a factor.

The (a − b) and (b − a) trick

Brackets that are reversed differ only by a factor of −1: (3 − x) = −(x − 3). So x − 33 − x = x − 3−(x − 3) = −1. Whenever two brackets look like mirror images, pull a −1 out of one of them and they will cancel. Examiners use this almost every year.

Restrictions come from the original denominator

Work out what makes the denominator zero before you cancel. In x² − 4x² − 5x + 6 the denominator factorises to (x − 2)(x − 3), so x ≠ 2 and x ≠ 3 — even though the (x − 2) disappears when you cancel. Forgetting the restriction that cancels away is the single most common lost mark in this topic.

Where marks get lost

Cancelling the x³ with the x. In x³ + 8x + 2 you cannot strike out x³ against x — they are terms inside sums. Factorise first, always.

Using a² − 2ab + b². The cubes trinomial has a single ab, not 2ab. If your bracket does not cancel cleanly, that doubled middle term is usually why.

Denominators x² − 9 and x + 3 factorise to (x + 3)(x − 3) and (x + 3).

The LCD contains every distinct bracket, at its highest power — so here the LCD is (x + 3)(x − 3), not the product of the two denominators.

Multiplying the denominators together always works, but it produces bigger numbers and an answer you then have to cancel back down. Building the LCD bracket by bracket keeps the work small.

The minus-sign trap

When you subtract a fraction, the minus applies to the whole numerator, not just its first term. Put the converted numerator in brackets before you subtract:

Wrong: 5x − 2x + 6    Right: 5x − (2x + 6) = 5x − 2x − 6 = 3x − 6.