How to Factor Quadratic Equations: Step-by-Step Method with Examples

To factor a quadratic equation of the form ax² + bx + c = 0, you rewrite it as a product of two binomials (px + q)(rx + s) = 0, then set each factor equal to zero to find the solutions. For x² + 5x + 6 = 0, the factored form is (x + 2)(x + 3) = 0, giving x = -2 and x = -3.

Below you'll see the full method walked through twice, plus the mistakes that trip students up most often.

A first worked example: x² + 5x + 6 = 0

We want two numbers that multiply to 6 (the constant) and add to 5 (the middle coefficient).

1. List factor pairs of 6: (1, 6) and (2, 3).
2. Check sums: 1 + 6 = 7, and 2 + 3 = 5. ✓
3. Use 2 and 3 to write the factored form: (x + 2)(x + 3) = 0.
4. Apply the zero product property: x + 2 = 0 or x + 3 = 0.
5. Solve each: x = -2 or x = -3.

Check by expanding: (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6. ✓

The general factoring method in 4 steps

1. Put the equation in standard form. Move everything to one side so it reads ax² + bx + c = 0. Example: x² + 7x = -12 becomes x² + 7x + 12 = 0.
2. Factor out any common factor first. If every term shares a factor, pull it out. Example: 2x² + 10x + 12 = 2(x² + 5x + 6).
3. Find two numbers that multiply to a·c and add to b. When a = 1, this is just "multiply to c, add to b." Example: for x² - x - 12, the numbers are -4 and 3.
4. Write the binomials and apply the zero product property. Set each factor equal to zero and solve. Example: (x - 4)(x + 3) = 0 gives x = 4 or x = -3.

When the leading coefficient isn't 1: the AC method

If a ≠ 1, multiply a·c, find two numbers that multiply to that product and add to b, then split the middle term and factor by grouping. This is the same trick the grounding curriculum uses to break down cubics into quadratic factors.

A second worked example: 2x² + 3x - 2 = 0

Here a = 2, b = 3, c = -2, so we need two numbers that multiply to a·c = -4 and add to 3.

1. Factor pairs of -4 whose sum is 3: 4 and -1 (since 4 · (-1) = -4 and 4 + (-1) = 3). ✓
2. Split the middle term: 2x² + 4x - x - 2 = 0.
3. Group and factor: 2x(x + 2) - 1(x + 2) = 0.
4. Pull out the common binomial: (2x - 1)(x + 2) = 0.
5. Solve each factor: 2x - 1 = 0 gives x = 1/2; x + 2 = 0 gives x = -2.

So the solutions are x = 1/2 and x = -2. Expanding (2x - 1)(x + 2) recovers 2x² + 3x - 2. ✓

Common mistakes when factoring quadratics

• Forgetting to set the equation equal to zero first. You can't apply the zero product property to (x + 2)(x + 3) = 6. Move everything to one side before factoring.
• Getting the signs wrong on the two numbers. For x² - 5x + 6, both numbers must be negative (product +, sum -). Students often write (x + 2)(x + 3) instead of (x - 2)(x - 3).
• Skipping the common factor. Factor out GCFs first — otherwise you'll fight harder numbers. 3x² + 15x + 18 is much easier as 3(x² + 5x + 6).
• Only writing one solution. A quadratic has two roots (counting multiplicity). Even if the factors look identical, like (x - 3)² = 0, state x = 3 as a double root.
• Assuming every quadratic factors nicely. If no integer pair works, use the quadratic formula or complete the square instead.

Practice this method with a guided tutor

You now have the recipe: standard form, GCF, find the pair, split and group, solve. The fastest way to lock it in is to work a few problems where a tutor nudges you when you pick the wrong factor pair or forget a sign. Sign up for LernOS to try the guided quadratic-factoring tutor — it walks you through problems like the ones above and catches the mistakes above before they cost you points on a test.