How to factor quadratics: 4 methods, step by step

October 30, 2026 · The QuizeGpt Team · 6 min read

Short answer: to factor a quadratic ax² + bx + c, first pull out any common factor. If a = 1, find two numbers that multiply to c and add to b. For example, x² + 5x + 6 = (x + 2)(x + 3), because 2 × 3 = 6 and 2 + 3 = 5. If a ≠ 1, use the AC method: find two numbers that multiply to a × c and add to b, split the middle term, and factor by grouping. Always check by multiplying back out.

Below are all four methods with worked examples, in the order you should try them.

What “factoring” means

Factoring is multiplying in reverse. You know how to expand:

(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6

Factoring starts with x² + 5x + 6 and works back to (x + 2)(x + 3). It matters because it lets you solve equations: if (x + 2)(x + 3) = 0, one of the factors must be zero, so x = −2 or x = −3.

Step 0: always pull out the GCF first

Before anything else, look for a factor every term shares, called the greatest common factor (GCF). Skipping this step is the most common reason problems get harder than they need to be.

Example: 3x² + 12x − 36

Every term is divisible by 3:

3x² + 12x − 36 = 3(x² + 4x − 12)

Now factor the simpler part. We need two numbers that multiply to −12 and add to 4: 6 and −2.

= 3(x + 6)(x − 2)

Method 1: x² + bx + c (when a = 1)

Find two numbers that multiply to c and add to b. Those are the numbers in your factors.

Example: x² − 5x + 6

Multiply to 6, add to −5. Pairs that multiply to 6: (1, 6), (2, 3), (−1, −6), (−2, −3). Only −2 and −3 add to −5.

x² − 5x + 6 = (x − 2)(x − 3)

Example: x² + 2x − 15

Multiply to −15, add to 2: 5 and −3.

x² + 2x − 15 = (x + 5)(x − 3)

The sign rules (a shortcut)

The signs of b and c tell you the signs of your two numbers before you start:

Method 2: the AC method (when a ≠ 1)

When the x² term has a coefficient other than 1, guessing gets slow. The AC method always works if the quadratic factors at all.

1. Multiply a × c.
2. Find two numbers that multiply to a × c and add to b.
3. Split the middle term using those two numbers.
4. Factor by grouping.

Example: 2x² + 7x + 3

1. a × c = 2 × 3 = 6
2. Multiply to 6, add to 7: 6 and 1
3. Split: 2x² + 6x + x + 3
4. Group: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

Example with negatives: 6x² + 11x − 10

1. a × c = 6 × (−10) = −60
2. Multiply to −60, add to 11: 15 and −4
3. Split: 6x² + 15x − 4x − 10
4. Group: 3x(2x + 5) − 2(2x + 5) = (3x − 2)(2x + 5)

Check: (3x − 2)(2x + 5) = 6x² + 15x − 4x − 10 = 6x² + 11x − 10 ✓

Tip: after grouping, the two brackets must match, like (2x + 5) and (2x + 5). If they don’t, check your signs. When the third term is negative, you usually need to factor out a negative number from the second group.

Method 3: special patterns

Some quadratics follow patterns you can spot right away.

Difference of squares: a² − b² = (a − b)(a + b)

• x² − 49 = (x − 7)(x + 7)
• 4x² − 9 = (2x − 3)(2x + 3)

Perfect square trinomials: a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)²

• x² + 10x + 25 = (x + 5)². Check: the middle term is 2 × x × 5 = 10x ✓
• 9x² − 12x + 4 = (3x − 2)². Check: the middle term is 2 × 3x × 2 = 12x ✓

A sum of squares like x² + 9 does not factor using real numbers.

Method 4: when it won’t factor, use the quadratic formula

Not every quadratic factors nicely. To check quickly, compute the discriminant, b² − 4ac:

• If it’s a perfect square (0, 1, 4, 9, 16, 25…), the quadratic factors over whole numbers.
• If it’s positive but not a perfect square, the solutions are irrational. Use the quadratic formula.
• If it’s negative, there are no real solutions.

Example: x² + x − 1: b² − 4ac = 1 + 4 = 5. That’s not a perfect square, so this won’t factor nicely. The quadratic formula x = (−b ± √(b² − 4ac)) / 2a gives x = (−1 ± √5) / 2.

Example: x² + 4x + 5: b² − 4ac = 16 − 20 = −4. It’s negative, so there are no real solutions.

Using factoring to solve an equation

Once it’s factored, set each factor equal to zero.

Solve x² − 5x + 6 = 0

(x − 2)(x − 3) = 0, so x − 2 = 0 or x − 3 = 0, which gives x = 2 or x = 3.

Check: put x = 2 into the original equation: 4 − 10 + 6 = 0 ✓. Put in x = 3: 9 − 15 + 6 = 0 ✓.

The equation must equal zero before you factor. If you have x² − 5x = −6, add 6 to both sides first.

Quick decision checklist

1. Is there a GCF? Pull it out.
2. Is it a special pattern (difference of squares or perfect square)? Use it.
3. Is a = 1? Find two numbers that multiply to c and add to b.
4. Is a ≠ 1? Use the AC method.
5. Nothing works? Check the discriminant and use the quadratic formula.
6. Multiply back out to check your answer.

How to get faster

Factoring is a skill you build with practice, not something you memorize. Do 10 problems a day for a few days, mixing all the types. Mixed practice trains you to recognize which method to use, and that’s what tests actually check.

Stuck on a homework problem? Our free algebra solver factors it step by step and explains why each step works. Read it, close it, and redo the problem yourself. That last part is where the learning happens. (Moving on to calculus next? Read how to actually understand derivatives.)

FAQ

What is the easiest way to factor quadratics?

When a = 1, find two numbers that multiply to c and add to b. For x² + 7x + 12, that’s 3 and 4, giving (x + 3)(x + 4). Always pull out a common factor first.

How do you factor when a is not 1?

Use the AC method: multiply a × c, find two numbers that multiply to that and add to b, split the middle term into two terms, then factor by grouping.

How do I know if a quadratic can be factored?

Compute the discriminant, b² − 4ac. If it’s a perfect square, the quadratic factors over whole numbers. If not, use the quadratic formula instead.

How do I check my factoring?

Multiply your factors back out (FOIL). If you get the original quadratic, your answer is right.

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