How to Factor Quadratics Step by Step

Start with a quadratic such as x^2 + 7x + 12. You want two numbers that multiply to 12 and add to 7. The pair 3 and 4 works, so the factorization is (x + 3)(x + 4). Expanding the brackets confirms the result: x^2 + 4x + 3x + 12 becomes x^2 + 7x + 12.

For x^2 - x - 12, look for two numbers that multiply to -12 and add to -1. The numbers -4 and 3 work, giving (x - 4)(x + 3). A negative product means the two numbers have opposite signs.

This is the quickest version of factoring quadratics. Find the product from the constant term, find the sum from the coefficient of x, and place the two numbers into brackets with x.

Now consider 2x^2 + 7x + 3. Multiply the first coefficient and the constant: 2 times 3 equals 6. Find two numbers that multiply to 6 and add to 7. Those numbers are 6 and 1.

Split the middle term using that pair: 2x^2 + 6x + x + 3. Group the expression as 2x(x + 3) + 1(x + 3). The matching bracket appears twice, so factor it out to get (2x + 1)(x + 3).

When factoring quadratics, write every stage rather than trying to jump straight to the brackets. The split-middle-term method is slower on paper but reliable, especially when signs or larger coefficients make mental guessing harder.

Always expand your answer. Multiply the first terms, outer terms, inner terms, and last terms, then combine like terms. If you recover the original quadratic exactly, the factorization is correct. This check takes only a few seconds and catches swapped signs or missing coefficients.

Some quadratics do not factor neatly using integers. For example, a quadratic may have irrational or complex roots, so no pair of simple integer brackets will work. In an equation, you may need the quadratic formula or another method instead. The instructions in a school problem usually hint whether simple factoring is expected.

A special pattern can also save time. A difference of two squares, such as x^2 - 25, factors as (x - 5)(x + 5). If every term has a common factor, take that out first. Good how to factor quadratics habits begin with a quick scan for common factors and familiar patterns before using the full product-sum method. Factoring is especially useful when a quadratic expression is set equal to zero. Once you have (x + 3)(x + 4) = 0, the zero-product rule says at least one bracket must equal zero. That gives x = -3 or x = -4. The factorization therefore reveals the roots directly, which is one reason the skill matters beyond simplifying algebraic expressions. If you get stuck, list factor pairs systematically instead of guessing at random. For a positive constant, test pairs with matching signs. For a negative constant, test opposite signs. Then use the middle coefficient to decide which pair has the required sum. This keeps the search small and organised.

Before hunting for factor pairs, you can test whether a quadratic with integer coefficients will factor neatly at all. For ax^2 + bx + c, calculate b^2 - 4ac, called the discriminant. If it is a perfect square, such as 0, 1, 4, 9 or 25, the quadratic factors into brackets with whole-number or fractional coefficients. If it is not a perfect square, no such brackets exist, and you can move straight to the quadratic formula or completing the square instead of searching for pairs that are not there.

Try it on the examples in this guide. For 2x^2 + 7x + 3, the discriminant is 49 - 24 = 25, a perfect square, so factoring works. For x^2 + 5x + 3, it is 25 - 12 = 13, so stop searching. A discriminant of exactly zero signals a perfect square trinomial, where both brackets are the same. For instance, x^2 + 6x + 9 gives 36 - 36 = 0, and it factors as (x + 3)^2. A negative discriminant means there are no real roots, so the expression cannot be split into real linear factors at all.

The zero-product rule only works when one side of the equation is zero. Suppose you need to solve x^2 + x = 12. It is tempting to write x(x + 1) = 12 and conclude that x = 12 or x + 1 = 12, but that reasoning fails, because two numbers can multiply to 12 in endless ways. Instead, rearrange first to x^2 + x - 12 = 0, then factor as (x + 4)(x - 3) = 0. The solutions are x = -4 and x = 3, and substituting either one into the original equation gives 12.

The second trap is dividing both sides by x. Faced with x^2 = 5x, many students divide by x and get x = 5, which is correct but incomplete. Dividing by x silently assumes that x is not zero, and zero is in fact a solution. The safe route is to move everything to one side, giving x^2 - 5x = 0, take out the common factor to get x(x - 5) = 0, and read off both answers, x = 0 and x = 5. As a visual check, the graph of y = x^2 - 5x crosses the x-axis at exactly those two points.

Some expressions are quadratics in disguise. Take x^4 - 13x^2 + 36. It looks like a fourth-power problem, but if you let u stand for x^2, it becomes u^2 - 13u + 36, which factors as (u - 4)(u - 9). Putting x^2 back gives (x^2 - 4)(x^2 - 9), and each bracket is a difference of two squares, so the full answer is (x - 2)(x + 2)(x - 3)(x + 3). The same trick works for 4^x - 6 times 2^x + 8 = 0, since 4^x is (2^x)^2. Factoring gives 2^x = 2 or 2^x = 4, so x = 1 or x = 2.

Factors also let you sketch a graph quickly. For y = x^2 - 2x - 8, factoring gives y = (x - 4)(x + 2), so the curve crosses the x-axis at x = 4 and x = -2. A parabola is symmetrical, so its turning point lies exactly halfway between those crossings, at x = 1. Substituting gives y = (1 - 4)(1 + 2) = -9, so the minimum point is (1, -9). Setting x = 0 shows the curve crosses the y-axis at -8, the constant term. Those few facts are usually enough for an accurate sketch, and they double as a check on your factorisation.

Factoring quadratics means reversing expansion. For a leading coefficient of 1, find two numbers with the right product and sum. For other leading coefficients, multiply the first and last coefficients, split the middle term, and factor by grouping. Expand your final brackets to check the answer.