Factorising Quadratics

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1. 1

   Find two numbers that multiply to c and add to b

   x² + 7x + 12 → need product 12, sum 7 → 3 and 4

   List the factor pairs of c and test their sums. For 12: 1 and 12 sum to 13, 2 and 6 sum to 8, 3 and 4 sum to 7. Working through the pairs systematically is faster than guessing and guarantees you find the pair if it exists.

2. 2

   Write the brackets

   x² + 7x + 12 = (x + 3)(x + 4)

   The two numbers go straight into the brackets with the sign that made the sum work. Expanding mentally to check takes seconds and confirms both the numbers and their signs at once.

3. 3

   Let the signs of b and c guide the search

   c positive → same signs. c negative → opposite signs

   If c is positive, both numbers share a sign, and b tells you which. If c is negative, one is positive and one negative, and the larger takes b's sign. This halves the search before any pairs are listed.

4. 4

   When a is not 1, split the middle term

   2x² + 7x + 3 → need product 2×3 = 6, sum 7 → 6 and 1 → 2x² + 6x + x + 3 → 2x(x+3) + 1(x+3) → (2x+1)(x+3)

   Multiply a by c, find two numbers with that product and sum b, then split the middle term into those two and factorise in pairs. The common bracket that appears is the confirmation that the split was right.

5. 5

   Recognise the difference of two squares

   x² − 25 = (x + 5)(x − 5)

   With no x term and a subtracted square, the factorisation is immediate. Students who do not recognise the pattern try to find two numbers multiplying to −25 and summing to 0, eventually find 5 and −5, and arrive at the same place more slowly.

Product and sum

Every quadratic factorisation with a leading coefficient of 1 comes down to one search: two numbers whose product is the constant and whose sum is the coefficient of x. That is because expanding (x + p)(x + q) gives x² + (p+q)x + pq, so the middle coefficient is the sum and the constant is the product.

Showing that expansion once makes the method obvious rather than magical. Students who have seen where product-and-sum comes from search far more confidently, and they recover it themselves when they forget which is which.

• x² + bx + c → two numbers with product c, sum b
• c positive → both numbers same sign
• c negative → opposite signs, larger takes b's sign
• a ≠ 1 → product a×c, then split the middle term
• x² − k² → (x + k)(x − k)

When it will not factorise

Not every quadratic factorises over whole numbers, and students can lose several minutes searching for a pair that does not exist. The discriminant b² − 4ac settles it: if it is not a perfect square, no whole-number factorisation exists and the quadratic formula or completing the square is required.

Checking the discriminant first takes ten seconds and prevents the most wasteful failure mode in the topic — a student convinced they have missed something, hunting for factors of a number that has none that work.

What factorising is for

A factorised quadratic gives its solutions immediately: if (x + 3)(x + 4) = 0 then one bracket must be zero, so x is −3 or −4. This is the zero product property, and it is the reason factorising is worth doing at all rather than an end in itself.

It also gives the x-intercepts of the curve directly, which makes sketching straightforward, and it is what allows algebraic fractions to cancel. Students who see those three uses treat the topic as a tool rather than a puzzle.

How we teach factorising

We teach the sign rules before the search, because they cut the work roughly in half and give students a way to check a candidate pair before testing it. Knowing that a negative constant forces opposite signs eliminates most wrong guesses immediately.

We also teach the discriminant check early, well before the quadratic formula is needed for anything else. Knowing when to stop looking is a genuine exam skill, and it is not usually taught as one.