How to Factor a Quadratic Equation

The Quick Test Before You Start

Compute b² − 4ac. If it is a negative number, the quadratic does not factor over the rationals and you should go straight to the quadratic formula. If it is a perfect square, factorising will give clean answers. This check costs two operations and saves the trial and error of hunting for factors.

The AC Method Step by Step

Multiply a by c to get ac. Find two numbers that multiply to ac and add to b. Split the middle term using those two numbers, then factor by grouping. For x² + 5x + 6, ac is 6 and the numbers are 2 and 3, giving x² + 2x + 3x + 6 which factors as x(x + 2) + 3(x + 2) = (x + 3)(x + 2).

Factoring With a Leading Coefficient

When a is not 1, factor out the common factor first if there is one. For 2x² + 8x + 6, dividing everything by 2 gives x² + 4x + 3, which factors to (x + 1)(x + 3). Restoring the 2 gives the original coefficients.

The Difference of Two Squares

A perfect square trinomial of the form x² ± 2kx + k² factors as (x ± k)². For x² − 6x + 9 this is (x − 3)², giving a repeated root of 3 and a discriminant of zero. Recognising the pattern saves a long search for two numbers.

When No Two Numbers Work

If no pair of numbers multiplies to ac and adds to b, the quadratic will not factor over the integers. That is not a failure. Switch to the quadratic formula, which handles every case, and remember that an irrational factorisation is rarely worth chasing.

Checking Your Factorisation

Expand your brackets and confirm you recover the original coefficients, especially the middle term. Then confirm the roots sum to −b / a and multiply to c / a. For x² + 5x + 6 the roots −2 and −3 sum to −5, matching −b / a = −5.