How to Factor Quadratic Trinomials, Step by Step
Factoring turns a sum like x² + 7x + 12 into a product like (x + 3)(x + 4). It is the reverse of multiplying out two binomials, and once you can do it, many quadratic equations can be solved in a few lines.
The Common Core high school standards give the reason to learn it: "Factor a quadratic expression to reveal the zeros of the function it defines." The same standards list factoring as one of several ways to solve quadratic equations, to be used "as appropriate to the initial form of the equation." The cases below are in the order to try them.
1. Step 0: always look for a common factor first
Before anything else, check whether every term shares a factor.
Example: 3x² + 12x + 9. Every term is divisible by 3:
3x² + 12x + 9 = 3(x² + 4x + 3)
Now factor what is inside the parentheses (two numbers that multiply to 3 and add to 4 are 1 and 3):
3(x + 1)(x + 3)
Skipping this step makes the rest harder and often leaves an answer that is not fully factored.
2. Case 1: x² + bx + c (leading coefficient 1)
Find two numbers that multiply to c and add to b.
Example: x² + 7x + 12.
| Pair that multiplies to 12 | Sum |
|---|---|
| 1 and 12 | 13 |
| 2 and 6 | 8 |
| 3 and 4 | 7 |
The pair 3 and 4 adds to 7, so x² + 7x + 12 = (x + 3)(x + 4).
Check by multiplying back: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12. It matches.
3. The sign rules
The signs of b and c tell you the signs of the two numbers before you start searching.
| c is | b is | The two numbers are | Example |
|---|---|---|---|
| positive | positive | both positive | x² + 7x + 12 = (x + 3)(x + 4) |
| positive | negative | both negative | x² - 9x + 20 = (x - 4)(x - 5) |
| negative | either | one positive, one negative | x² - 2x - 15 = (x - 5)(x + 3) |
For x² - 2x - 15: you need a product of -15 and a sum of -2. The pair -5 and 3 works, because -5 x 3 = -15 and -5 + 3 = -2. When c is negative, the larger number (by size) takes the sign of b.
Check: (x - 5)(x + 3) = x² + 3x - 5x - 15 = x² - 2x - 15.
4. Case 2: ax² + bx + c (the AC method)
When the x² term has a coefficient other than 1, use the AC method.
Example: 2x² + 7x + 3. Here a = 2, b = 7, c = 3.
- Multiply a and c: 2 x 3 = 6.
- Find two numbers that multiply to 6 and add to 7: 6 and 1.
- Split the middle term using those numbers: 2x² + 6x + x + 3.
- Group and factor each pair: 2x(x + 3) + 1(x + 3).
- Factor out the shared binomial: (2x + 1)(x + 3).
Check: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3.
If step 4 does not produce the same binomial in both groups, recheck your two numbers or the signs in the split.
5. Case 3: special patterns
Difference of squares: a² - b² = (a - b)(a + b).
Example: x² - 49 = (x - 7)(x + 7). There is no middle term because +7x and -7x cancel when you multiply back.
The standards use a deeper version of this pattern to show what "seeing structure" means: see x⁴ - y⁴ as (x²)² - (y²)², "thus recognizing it as a difference of squares that can be factored as (x² - y²)(x² + y²)." The first factor is itself a difference of squares, so it can be factored once more into (x - y)(x + y).
Perfect square trinomial: a² + 2ab + b² = (a + b)².
Example: x² + 10x + 25 = (x + 5)². The first and last terms are squares (x² and 5²), and the middle term is 2 times x times 5.
Sum of squares: x² + 49 does not factor using real numbers. Do not write (x + 7)(x + 7), which multiplies to x² + 14x + 49.
6. Using factoring to solve an equation
Factoring "reveals the zeros" because of one fact: if a product is 0, at least one factor must be 0.
Solve x² - 2x - 15 = 0.
- Factor: (x - 5)(x + 3) = 0
- Set each factor to zero: x - 5 = 0 or x + 3 = 0
- Solve: x = 5 or x = -3
Check: 5² - 2(5) - 15 = 25 - 10 - 15 = 0. And (-3)² - 2(-3) - 15 = 9 + 6 - 15 = 0. Both work.
The equation must equal zero before you factor. If you have x² - 2x = 15, subtract 15 from both sides first.
7. When factoring is not the right tool
Some trinomials have no pair of whole numbers that works. For x² + 3x + 1, the only whole-number pair with product 1 is 1 and 1, which adds to 2, not 3. That does not mean the equation has no solution. It means another method, such as completing the square or the quadratic formula, is the appropriate choice, which is exactly the judgment the standards ask students to make.
8. Key takeaways
- Always pull out a common factor first.
- For x² + bx + c, find two numbers that multiply to c and add to b, using the signs of b and c to narrow the search.
- For ax² + bx + c, multiply a and c, split the middle term, and factor by grouping.
- Watch for the difference of squares and perfect square patterns.
- Multiply your factors back out to check, and set the equation equal to zero before solving by factoring.