Faster Ways to Factorise Quadratics Under Exam Pressure

Factorising quadratics is one of the few skills that appears in almost every mathematics examination, from Grade 9 through to calculus. It is also where a surprising amount of exam time disappears — not because students cannot do it, but because they approach every quadratic the same way, by trial and error, instead of recognising which of a small number of situations they are in.

What follows is a decision order that removes most of the guesswork, with worked examples. Everything here is standard mathematics; the value is in the sequence and in knowing which method to reach for.

Check in This Order, Every Time

Before factorising anything, run through four checks in this fixed order. The order matters, because each step makes the next one easier:

1. Is there a common factor? Take it out first, always.
2. Is it a special form? Difference of two squares, or a perfect square trinomial.
3. Is the coefficient of x² equal to 1? If so, factorise by inspection.
4. Otherwise, use the AC method — a systematic procedure, not guesswork.

Students who skip step 1 and start hunting for factor pairs on a quadratic with a common factor of 3 make the problem three times harder than it needs to be.

Step 1: Take Out the Common Factor

Look at every term, including the constant, and extract the highest common factor of the coefficients along with any shared variable.

Example: 6x² − 15x − 9.

All three coefficients share a factor of 3, so this is 3(2x² − 5x − 3), and the bracket then factorises as (2x + 1)(x − 3). Final answer: 3(2x + 1)(x − 3).

If the leading coefficient is negative, take out the negative as well. For −2x² + 10x − 12, write −2(x² − 5x + 6) = −2(x − 2)(x − 3). Handling the sign once at the start prevents a whole class of sign errors later.

Step 2: Recognise the Special Forms

Two patterns are worth recognising instantly, because they factorise in one line with no searching at all.

Difference of two squares

a² − b² = (a + b)(a − b).

The trigger is two terms, both perfect squares, separated by a minus sign. So 9x² − 25 = (3x + 5)(3x − 5), and 49 − 4y² = (7 + 2y)(7 − 2y).

Two cautions. A sum of two squares such as x² + 16 does not factorise over the real numbers — this is a very common error. And the squares can themselves be expressions: x⁴ − 81 = (x² + 9)(x² − 9) = (x² + 9)(x + 3)(x − 3), where stopping after the first step leaves marks on the table.

Perfect square trinomial

a² ± 2ab + b² = (a ± b)².

The test is quick: are the first and last terms perfect squares, and is the middle term exactly twice the product of their square roots? For 4x² + 12x + 9, the square roots are 2x and 3, twice their product is 12x, and the middle term matches — so it is (2x + 3)².

Step 3: Monic Quadratics by Inspection

When the coefficient of x² is 1, factorising x² + bx + c means finding two numbers that multiply to c and add to b. Those two numbers go straight into the brackets.

The speed comes from reading the signs first, which narrows the search before you start:

• c positive, b positive → both numbers positive. x² + 7x + 12 = (x + 3)(x + 4).
• c positive, b negative → both numbers negative. x² − 7x + 12 = (x − 3)(x − 4).
• c negative → one positive, one negative, and the larger in size takes the sign of b. x² − 2x − 15 = (x − 5)(x + 3).

Also list factor pairs of c starting from the middle rather than from 1. For c = 36 and b = 13, starting at 6 × 6 and working outward reaches 4 × 9 in two steps, where starting at 1 × 36 takes four.

Step 4: Non-Monic Quadratics — the AC Method

For ax² + bx + c with a ≠ 1, trial and error becomes slow and unreliable. The AC method is systematic and takes the same number of steps every time.

Take 6x² + 11x − 10.

1. Compute ac. Here 6 × (−10) = −60.
2. Find two numbers that multiply to ac and add to b. Multiply to −60, add to 11: that is 15 and −4.
3. Split the middle term using them. 6x² + 15x − 4x − 10.
4. Factorise in pairs. 3x(2x + 5) − 2(2x + 5).
5. Extract the common bracket. (2x + 5)(3x − 2).

Expanding mentally confirms the middle term: 2x × (−2) + 5 × 3x = −4x + 15x = 11x. That five-second check catches almost every error this method can produce.

The one thing to watch is step 4: if the two brackets do not come out identical, the sign split in step 3 was wrong. With 6x² − 11x − 10, for instance, the pair is −15 and 4, giving 3x(2x − 5) + 2(2x − 5) = (2x − 5)(3x + 2).

When Factorising Is the Wrong Move

Factorising is fast when it works, but it is not always the right tool:

• The discriminant is not a perfect square. Use the quadratic formula. For 2x² + 3x − 4, the discriminant is 9 + 32 = 41, which is not a perfect square, so no integer factorisation exists.
• The question asks for the vertex, maximum, minimum or line of symmetry. Complete the square instead — that form gives the turning point directly.
• The question says "give your answer to 2 decimal places" or "in surd form". That phrasing is a strong signal that the roots are irrational and the formula is intended.
• The coefficients are large or awkward. If a and c are large, ac has many factor pairs, and the formula is simply faster.

Recognising these cases early is worth as much as any factorising shortcut, because the most expensive exam mistake is persisting with a method that cannot work.

The Mistakes That Cost the Most Marks

• Not factorising completely. Leaving 2x² − 8 as 2(x² − 4) instead of 2(x + 2)(x − 2).
• Trying to factorise a sum of two squares. x² + 9 has no real factorisation.
• Losing the common factor at the end. Taking out a 3 at the start and forgetting to write it in the final answer.
• Confusing factorising with solving. "Factorise" wants brackets; "solve" wants values of x. Answering the wrong one loses the mark even with correct work.
• Forgetting the zero-product step when solving. From (x − 3)(x + 5) = 0 the roots are x = 3 and x = −5 — note the sign reverses from the bracket.
• Never checking by expansion. Expanding back takes a few seconds and catches nearly every error.

Speed in factorising is a by-product of recognition, not of rushing. Working through the four-step order deliberately for a week or two makes the recognition automatic, after which the shortcuts apply themselves. If algebra is where your marks are going, a free assessment will identify the specific habits behind it — and our algebra tutoring page sets out how we rebuild that fluency.