Solving Quadratic Equations by Factorising

Find two numbers that multiply to give the constant and add to give the number in front of x, then write them in two brackets. For x² + 7x + 12, list the factor pairs of 12: 1 and 12, 2 and 6, 3 and 4. Only 3 and 4 add to 7, so x² + 7x + 12 = (x + 3)(x + 4). Expand the brackets to check.

Factorise, then set each bracket equal to zero, because two things only multiply to zero if one of them is zero. (x + 3)(x + 4) = 0 gives x + 3 = 0 or x + 4 = 0, so x = −3 or x = −4. The factorised form is the same equation written differently, so it has the same solutions.

The roots are the solutions: the values of x that make the equation equal to zero. x² + 5x + 6 = 0 factorises to (x + 2)(x + 3) = 0, so its roots are −2 and −3. On the graph of y = x² + 5x + 6, the roots are where the curve crosses the x-axis. Factorising is the quickest way to find them when it works.

Keep the sign when you list the factor pairs: one number will be positive and the other negative. For x² + 3x − 18, the pairs for −18 include 3 and −6, −3 and 6, 2 and −9. Only −3 and 6 add to 3, so the equation becomes (x − 3)(x + 6) = 0 and the solutions are x = 3 and x = −6.

Divide out a common factor first: 5x² + 30x − 80 = 0 becomes x² + 6x − 16 = 0, which factorises as usual: x = 2 or x = −8. With no common factor, as in 2x² + 7x + 3, find two numbers that multiply to a × c = 6 and add to 7: 6 and 1. Split the middle term into 6x + x and factorise in pairs: (2x + 1)(x + 3).

No. It works when the numbers are whole and a pair that fits exists, which is why it is worth trying first; otherwise use completing the square or the quadratic formula. Check by expanding: (x + 2)(x + 3) gives x² + 5x + 6, so it matches. Factorising with a = 1 is around grade 4 to 5 at GCSE; with a ≠ 1 it is grade 6 to 7.