Quadratics

🔢 Maths Years 9–11 Foundation

Factorise and solve quadratic equations of the form x² + bx + c = 0.

Open this lesson, with practice and a printable worksheet →

A quadratic is an expression with an x² term in it, like x² + 7x + 12. They describe the shape of a thrown ball, the path of water from a fountain, and the area of a square whose side you don't yet know — anything where a quantity depends on the square of something. At GCSE the key skill is factorising them into two brackets.

Why this matters: factorising turns a hard-looking equation into an easy one. Once a quadratic is a product of two brackets, you can read off exactly where its graph crosses the x-axis — which is how you solve real problems about maximum area, when a ball lands, or where two paths meet.
A quadratic graph is a U-shaped curve called a parabola

What "factorising" undoes

x 3 x 4 x² 3x 4x 12
Area model: (x+3)(x+4) = x² + 3x + 4x + 12 = x² + 7x + 12.

Expanding (x + 3)(x + 4) with the grid or FOIL method gives x² + 4x + 3x + 12 = x² + 7x + 12. Factorising is that process in reverse — you start with x² + 7x + 12 and work back to the two brackets. Notice where the numbers came from: the 12 is 3 × 4, and the 7 is 3 + 4. That is the whole secret.

Factorising x² + bx + c

Find two numbers that multiply to give c (the last number) and add to give b (the middle number). Those two numbers go into the brackets.

x² + 7x + 12: we need two numbers that multiply to 12 and add to 7 → that's 3 and 4. So it factorises to (x + 3)(x + 4).
3 4 × = 12 + = 7
Find two numbers that multiply to c and add to b.

Signs are a helpful clue. Think of the pair of numbers you need:

Sign of cSign of bThe two numbers are…Example
++both positivex² + 7x + 12 → (x+3)(x+4)
+−both negativex² − 7x + 12 → (x−3)(x−4)
−eitherone positive, one negativex² + x − 12 → (x+4)(x−3)

When c is negative, the two numbers have opposite signs, and the bigger one takes the sign of b. That single rule clears up most sign muddles.

Solving a quadratic equation

To solve "= 0", factorise first, then use this powerful idea:

If two things multiply to make 0, then one of them must be 0. So if (x + 5)(x − 3) = 0, then x + 5 = 0 or x − 3 = 0.

Analogy: if you multiply two numbers and get zero, at least one of them was zero — there's no other way. Nothing times nothing-but-zero gives zero. That's why each bracket gets its own turn at being zero.

x² + 2x − 15 = 0: factorises to (x + 5)(x − 3) = 0, so x + 5 = 0 gives x = −5, and x − 3 = 0 gives x = 3. So x = −5 or x = 3. (Quadratics usually have two answers — the graph crosses the x-axis in two places.)
−53
The roots x = −5 and x = 3 are where the curve crosses the x-axis.

Difference of two squares — the special case

x 3 x² − 9 → (x+3) (x−3)
Difference of two squares: x² − 9 = (x + 3)(x − 3).

Sometimes there is no middle term, just a square number being subtracted, like x² − 9. Here the two numbers multiply to −9 and add to 0, so they must be +3 and −3.

Difference of two squares: x² − 9 factorises to (x + 3)(x − 3). Spot these whenever a square number is being subtracted (x² − 25, x² − 36, x² − 49). Also watch for a common factor first: x² + 3x = x(x + 3), which solves to x = 0 or x = −3.
In a nutshell: factorise by finding two numbers that multiply to c and add to b; solve by setting each bracket to 0 — and expect two solutions. Multiply your brackets back out to check you land on the original quadratic.
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