How to Factor Trinomials (a = 1 and a ≠ 1)

What does it mean to factor a trinomial?

Factoring a trinomial means rewriting a three-term expression like x² + 7x + 12 as a product, here (x + 3)(x + 4). It's multiplication in reverse: you start with the answer to a FOIL problem and find the two binomials.

Learning how to factor trinomials is mostly about a quick search for the right pair of numbers. The search is easy when the x² coefficient is 1 and takes one extra step when it isn't.

Factored form is worth the effort because it shows the zeros. Once x² + 7x + 12 is written as (x + 3)(x + 4), you can read off that the expression equals 0 at x = −3 and x = −4. That's the main reason factoring shows up when you solve equations and sketch parabolas.

Why should you factor out the GCF first?

Taking out the greatest common factor makes the numbers smaller and often turns a hard trinomial into an easy one. It's also required for factoring completely.

For 2x² + 10x + 12, every term is divisible by 2. Factor it out to get 2(x² + 5x + 6). Now you only need two numbers that multiply to 6 and add to 5, which are 2 and 3. The full answer is 2(x + 2)(x + 3).

How do you factor trinomials when a = 1?

Find two numbers that multiply to c and add to b. Those numbers go into the binomials: x² + bx + c = (x + m)(x + n).

1. Start with x² + 7x + 12. You need a product of 12 and a sum of 7.
2. List factor pairs of 12: 1 and 12, 2 and 6, 3 and 4.
3. Only 3 + 4 = 7.
4. Write the answer: (x + 3)(x + 4).

How do you factor trinomials when a ≠ 1?

Use the ac method. Multiply a by c, find two numbers that multiply to ac and add to b, split the middle term with them, then factor by grouping. Here is 6x² + 7x − 3.

1. ac = 6 × (−3) = −18. You need a product of −18 and a sum of 7.
2. The pair is 9 and −2, since 9 × (−2) = −18 and 9 + (−2) = 7.
3. Split the middle term: 6x² + 9x − 2x − 3.
4. Group and factor each pair: 3x(2x + 3) − 1(2x + 3).
5. Pull out the shared binomial: (2x + 3)(3x − 1).

Is there a faster way than the ac method?

If a is a prime number, guess and check is often quicker. The first terms of the binomials must be ax and x, so you only need to place the factors of c.

For 3x² + 10x + 8, the binomials start as (3x + ?)(x + ?). The factor pairs of 8 are 1 and 8, and 2 and 4. Try 4 and 2 in the blanks: (3x + 4)(x + 2) = 3x² + 6x + 4x + 8 = 3x² + 10x + 8. That works on the first try.

If a has several factor pairs, like 12 or 24, the guesses multiply quickly. That's when the ac method saves time, because it gives you the split directly.

How do you factor a trinomial with a negative leading term?

Factor out −1 first so the x² term is positive, then factor what is left. Keep the negative sign in front of the final answer.

For −x² + x + 12, pulling out −1 gives −(x² − x − 12). Two numbers that multiply to −12 and add to −1 are −4 and 3, so the answer is −(x − 4)(x + 3).

What special cases should you recognize?

Perfect square trinomials factor as a binomial squared: a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)². For x² − 10x + 25, the first and last terms are x² and 5², and the middle is 2 × x × 5, so it's (x − 5)².

A related shortcut is the difference of squares, a² − b² = (a + b)(a − b). You can think of it as a trinomial with b = 0, such as 4x² − 9 = (2x + 3)(2x − 3). Our factoring formulas and patterns article lists the rest.

How do you know if a trinomial can't be factored?

Check the discriminant b² − 4ac. With integer coefficients, the trinomial factors over the integers only if that number is a perfect square.

For x² + 3x − 2, b² − 4ac = 9 + 8 = 17, which isn't a perfect square. No integer pair will multiply to −2 and add to 3, so it is prime over the integers. If it's part of an equation, use the quadratic formula instead.

How do you solve quadratic equations by factoring?

Move everything to one side so it equals 0, factor, and set each factor equal to 0. This works because of the zero product property: if a product is 0, at least one factor must be 0.

What are common mistakes when factoring trinomials?

Most mistakes are about signs or stopping too early. Check these before you move on.

1. Skipping the GCF, so the answer isn't factored completely.
2. Choosing numbers that multiply to c when a ≠ 1. With the ac method, they must multiply to ac.
3. Getting the sign wrong in the second group, such as writing −1(2x − 3) when you need −1(2x + 3).
4. Trying to solve before the equation equals 0. In x² − x = 12, you must subtract 12 first.
5. Not checking by multiplying the factors back out.