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Algebra Factoring Tricks: How to Know Which Method to Use

May 9, 2026 · 7 min · factoring · algebra · quadratics · exam technique · maths

Written & checked by Rabail, a student.

Quick answer: To factor fast, run one fixed order: pull out the highest common factor, then check for a difference of two squares, then count the terms. Three terms with a leading coefficient of 1 means find two numbers that multiply to c and add to b. Anything else means the AC method or grouping.

I used to treat factoring like a guessing game. I would stare at 6x^2 - 19x + 10, try a few random brackets, give up after ninety seconds, and run to the quadratic formula. On my IGCSE mock that cost me four minutes across the paper, and on one question the answer entirely, because "factorise fully" does not accept decimals. What fixed it was not sharper arithmetic. It was a fixed order I run every time, so I never have to decide what to try.

The decision order that removes the guessing

Before you write anything, check for a common factor and count the terms. That one habit tells you which method to use in about ten seconds.

  1. Common factor in every term? Pull it out first, always. This step comes before everything else.
  2. Two terms left? Look for a difference of two squares. If both terms are cubes, use the cube identities.
  3. Three terms left? It is a trinomial. If the x^2 coefficient is 1, hunt for a number pair. If it is not 1, use the AC method.
  4. Four terms left? Group them in pairs.
  5. Then ask: can anything factor again? "Fully factorise" means keep going until nothing moves.

Always strip the common factor first

Pulling out the highest common factor shrinks the numbers you then search through, often tenfold.

Take 12x^2 + 30x - 72. Dive straight into the AC method and you multiply 12 by -72, then hunt for two numbers multiplying to -864 and adding to 30. They exist, 48 and -18, but finding them burns real time.

Strip the 6 first and it becomes 6(2x^2 + 5x - 12). Now AC is 2 times -12, which is -24. Two numbers multiplying to -24 and adding to 5 are 8 and -3. You reach 6(2x - 3)(x + 4) in about twenty seconds.

Two extras. If the x^2 term is negative, pull out -1 too: -x^2 + 7x - 12 becomes -(x^2 - 7x + 12), so -(x - 3)(x - 4). And the common factor can be algebraic: 5x^3 - 20x becomes 5x(x^2 - 4), which factors again to 5x(x - 2)(x + 2). Students lose that last mark constantly.

Difference of two squares, including the disguised versions

Anything of the form a^2 - b^2 factors instantly to (a + b)(a - b). The marks are in spotting it when it does not look obvious.

  • 9x^2 - 25 becomes (3x + 5)(3x - 5). Square root each part.
  • 49 - 4y^2 becomes (7 - 2y)(7 + 2y). The squares can come in either order.
  • 2x^2 - 50 needs the common factor first: 2(x^2 - 25), then 2(x - 5)(x + 5).
  • 16x^4 - 81 becomes (4x^2 - 9)(4x^2 + 9), and the first bracket goes again: (2x - 3)(2x + 3)(4x^2 + 9). The classic "fully factorise" trap on Cambridge IGCSE and GCSE higher papers.
  • (x + 3)^2 - 16 becomes (x + 3 - 4)(x + 3 + 4), which tidies to (x - 1)(x + 7).

A sum of two squares never factors, so x^2 + 25 stays as it is. Bonus non-calculator trick: 61^2 - 39^2 is (61 - 39)(61 + 39), which is 22 times 100, so 2200. That shape turns up in WAEC and CBSE arithmetic and takes five seconds.

Worked example: the AC method, start to finish

Factorise 6x^2 - 19x + 10.

  1. Common factor? None.
  2. Multiply A by C: 6 times 10 is 60.
  3. Find two numbers multiplying to 60 and adding to -19. C is positive and B negative, so both are negative: -15 and -4.
  4. Split the middle term using that pair: 6x^2 - 15x - 4x + 10.
  5. Group into two pairs and factor each: 3x(2x - 5) - 2(2x - 5).
  6. The bracket (2x - 5) is now common, so pull it out: (3x - 2)(2x - 5).
  7. Expand to check: 6x^2 - 19x + 10. Correct.

Step 5 is where people panic. If the two brackets do not match, you split with the wrong pair or made a sign slip. The split order does not matter: 6x^2 - 4x - 15x + 10 gives 2x(3x - 2) - 5(3x - 2), the same answer.

Four-term grouping is the identical move without step 2. For 2x^3 + 6x^2 - 5x - 15 you group straight away: 2x^2(x + 3) - 5(x + 3), giving (x + 3)(2x^2 - 5).

Two sign rules halve your search:

  • If C is positive, both numbers carry the sign of B.
  • If C is negative, the signs differ, and the bigger number carries the sign of B.

The factor-pair search that stops you guessing

List factor pairs from 1 upwards and stop at the square root. That turns a fishing trip into a short finite list.

For AC = 60: 1 and 60, 2 and 30, 3 and 20, 4 and 15, 5 and 12, 6 and 10. Six pairs, and past the square root of 60 there are no more. Scan for the pair whose sum or difference is 19, find 4 and 15, then apply signs. Magnitudes first, signs second, beats trying signed pairs at random.

If no integer pair exists, stop. The fast test is the discriminant, b^2 - 4ac. For 2x^2 + 5x + 1 that is 25 - 8, which is 17, not a perfect square, so no integer factorisation exists. Ten seconds of checking saves two minutes of failure. For a second pair of eyes, the math solver shows the split-the-middle line rather than just an answer, and explain re-words the reasoning.

Where the marks actually get lost

Most lost factoring marks are not arithmetic errors, they are instruction errors.

  • "Factorise" wants brackets. "Solve" wants x values. If it says solve, add the final line: set each bracket to zero, so (3x - 2)(2x - 5) = 0 gives x = 2/3 or x = 5/2. Stopping at brackets loses the answer mark on AQA, Edexcel, OCR and Cambridge papers alike.
  • "Fully factorise" means check again after every success, especially inside new brackets.
  • Show the split-middle-term line. If your final brackets are wrong, that line usually still earns a method mark.
  • At A-Level and AP you also need the cubes: a^3 - b^3 is (a - b)(a^2 + ab + b^2), and a^3 + b^3 is (a + b)(a^2 - ab + b^2). The sign in the long bracket is always opposite the short one, and the middle term is never doubled.

Short bursts beat long sessions. I do ten mixed expressions from the quiz tool with the type unlabelled, because knowing which method to reach for is the whole skill. More practice sits in the maths hub.

Test yourself

  1. Fully factorise 3x^2 - 27.
  2. Factorise 4x^2 + 4x - 15.
  3. Factorise by grouping: 2x^3 + 6x^2 - 5x - 15.

Answers: 3(x - 3)(x + 3); (2x - 3)(2x + 5); (x + 3)(2x^2 - 5).

FAQ

Is factoring actually faster than the quadratic formula?

When it works, yes, roughly three times faster and with far less chance of a slip. Give yourself a hard thirty-second limit. If no integer pair has appeared, check the discriminant and switch method.

How do I know if a quadratic will not factor at all?

Work out b^2 - 4ac. If it is negative there are no real roots. If it is positive but not a perfect square, the roots are irrational, so no integer brackets exist. Perfect squares are 0, 1, 4, 9, 16, 25 and so on.

Do I have to memorise the sum and difference of cubes?

For GCSE and Cambridge IGCSE, usually not. For A-Level, AP and further WASSCE maths topics, yes, and they repay the two minutes. Write both lines from memory once a week and they stick.

How do I check a factorisation without expanding everything?

Substitute a number. Put x = 1 into the original and into your brackets. For 6x^2 - 19x + 10 that gives -3, and (3x - 2)(2x - 5) gives 1 times -3, also -3. Not a proof, but it catches nearly every sign error in five seconds.

In short: do not decide what to try, run the order. Common factor, difference of two squares, then count the terms. List factor pairs systematically instead of guessing, settle magnitudes before signs, and check the discriminant before sinking two minutes into a quadratic that was never going to factor.