How to Factor Trinomials

Reviewed by Grace Imson, MA

Last Updated: March 16, 2023 Fact Checked

A trinomial is an algebraic expression made up of three terms. Most likely, you'll start learning how to factor quadratic trinomials, meaning trinomials written in the form ax² + bx + c. There are several tricks to learn that apply to different types of quadratic trinomial, but you'll get better and faster at using them with practice. Higher degree polynomials, with terms like x³ or x⁴, are not always solvable by the same methods, but you can often use simple factoring or substitution to turn them into problems that can be solved like any quadratic formula.

Method 1

Method 1 of 3:

Factoring x² + bx + c

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1
Learn FOIL multiplication. You might have already learned the FOIL method, or "First, Outside, Inside, Last," to multiply expressions like (x+2)(x+4). It's useful to know how this strategy works before we get to factoring:^{[1]
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- Multiply the First terms: (x+2)(x+4) = x² + __
- Multiply the Outside terms: (x+2)(x+4) = x²+4x + __
- Multiply the Inside terms: (x+2)(x+4) = x²+4x+2x + __
- Multiply the Last terms: (x+2)(x+4) = x²+4x+2x+8
- Simplify: x²+4x+2x+8 = x²+6x+8
2
Understand factoring. When you multiply two binomials together in the FOIL method, you end up with a trinomial (an expression with three terms) in the form ax²+bx+c, where a, b, and c are ordinary numbers. If you start with an equation in the same form, you can factor it back into two binomials.^{[2]
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- If the equation isn't written in this order, move the terms around so they are. For example, rewrite 3x - 10 + x² as x² + 3x - 10.
- Because the highest exponent is 2 (x²), this type of expression is "quadratic."
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3
Write a space for the answer in FOIL form. For now, just write (__ __)(__ __) in the space where you'll write the answer. We'll fill this out as we go.
- Don't write + or - between the blank terms yet, since we don't know which it will be.
4
Fill out the First terms. For simple problems, where the first term of your trinomial is just x², the terms in the First position will always be x and x. These are the factors of the term x², since x times x = x².^{[3]
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- Our example x² + 3x - 10 just begins with x², so we can write:
- (x __)(x __)
- We'll cover more complicated problems in the next section, including trinomials that begin with a term like 6x² or -x². For now, follow the example problem.
5
Use factoring to guess at the Last terms. If you go back and reread the FOIL method step, you'll see that multiplying the Last terms together gives you the final term in the polynomial (the one with no x). So to factor, we need to find two numbers that multiply to form the last term.^{[4]
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- In our example x² + 3x - 10, the last term is -10.
- What are the factors of -10? What two numbers multiplied together equal -10?
- There are a few possibilities: -1 times 10, 1 times -10, -2 times 5, or 2 times -5. Write these pairs down somewhere to remember them.
- Don't change our answer yet. It still looks like this: (x __)(x __).
6
Test which possibilities work with Outside and Inside multiplication. We've narrowed the Last terms down to a few possibilities. Use trial and error to test each possibility, multiplying the Outside and Inside terms, and comparing the result to our trinomial. For example:
- Our original problem has an "x" term of 3x, so that's what we want to end up with in this test.
- Test -1 and 10: (x-1)(x+10). The Outside + Inside = 10x - x = 9x. Nope.
- Test 1 and -10: (x+1)(x-10). -10x + x = -9x. That's not right. In fact, once you test -1 and 10, you know that 1 and -10 will just be the opposite of the answer above: -9x instead of 9x.
- Test -2 and 5: (x-2)(x+5). 5x - 2x = 3x. That matches the original polynomial, so this is the correct answer: (x-2)(x+5).
- In simple cases like this, when you don't have a constant in front of the x² term, you can use a shortcut: just add the two factors together and put an "x" after it (-2+5 → 3x). This won't work for more complicated problems, though, so it's good to remember the "long way" described above.

Method 2

Method 2 of 3:

Factoring More Complicated Trinomials

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1
Use simple factoring to make more complicated problems easier. Let's say you need to factor 3x² + 9x - 30. Look for something that factors into each of the three terms (the "greatest common factor", or GCF).^{[5]
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Research source} In this case, it's 3:
- 3x² = (3)(x²)
- 9x = (3)(3x)
- -30 = (3)(-10)
- Therefore, 3x² + 9x - 30 = (3)(x²+3x-10). We can factor out the new trinomial using the steps in the section above. Our final answer will be (3)(x-2)(x+5).
2
Look for trickier factors. Sometimes, the factor might involve variables, or you might need to factor a couple times to find the simplest possible expression. Here are a few examples:
- 2x²y + 14xy + 24y = (2y)(x² + 7x + 12)
- x⁴ + 11x³ - 26x² = (x²)(x² + 11x - 26)
- -x² + 6x - 9 = (-1)(x² - 6x + 9)
- Don't forget to factor the new trinomial further, using the steps in method 1. Check your work and find similar example problems in the example problems near the bottom of this page.
3
Solve problems with a number in front of the x². Some quadratic trinomials can't be simplified down to the easiest type of problem. Learn how to solve problems like 3x² + 10x + 8, then practice on your own with the example problems at the bottom of the page:
- Set up our answer: (__ __)(__ __)
- Our "First" terms will each have an x, and will multiply together to make 3x². There's only one possible option here: (3x __)(x __).
- List factors of 8. Our options are 1 times 8, or 2 times 4.
- Test these using the Outside and Inside terms. Note that the order of the factors matter, since the Outside term is being multiplied by 3x instead of x. Try out every possibility until you get an Outside+Inside result of 10x (from the original problem):
- (3x+1)(x+8) → 24x+x = 25x no
- (3x+8)(x+1) → 3x+8x = 11x no
- (3x+2)(x+4) → 12x+2x=14x no
- (3x+4)(x+2) → 6x+4x=10x yes This is the correct factor.
4
Use substitution for higher-degree trinomials. Your math book might surprise you with an equation with a high exponent, such as x⁴, even after you've used simple factoring to make the problem easier. Try substituting a new variable that turns it into a problem you know how to solve. For example:
- x⁵+13x³+36x
- =(x)(x⁴+13x²+36)
- Let's invent a new variable. We'll say y = x², and plug it in:
- (x)(y²+13y+36)
- =(x)(y+9)(y+4). Now switch back to using the original variable:
- =(x)(x²+9)(x²+4)
- =(x)(x±3)(x±2)

Method 3

Method 3 of 3:

Factoring Special Cases

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1
Check for prime numbers. Check to see if the constant in either the first or third term of the trinomial is a prime number. A prime number can be divided evenly only by itself and 1, so there is only one possible pair of binomial factors.^{[6]
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- For example, in x² + 6x + 5, 5 is a prime number, so the binomial must be in the form (__ 5)(__ 1).
- In the problem 3x²+10x+8, 3 is a prime number, so the binomial must be in the form (3x __)(x __).
- For the problem 3x²+4x+1, the only possible solution is (3x+1)(x+1). (You should still multiply this out to check your work, since some expressions can't be factored at all – for example, 3x²+100x+1 has no factors.)
2
Check to see if the trinomial is a perfect square. A perfect square trinomial can be factored into two identical binomials, and the factor is usually written (x+1)² instead of (x+1)(x+1). Here a few common ones that tend to show up in problems:^{[7]
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- x²+2x+1=(x+1)², and x²-2x+1=(x-1)²
- x²+4x+4=(x+2)², and x²-4x+4=(x-2)²
- x²+6x+9=(x+3)², and x²-6x+9=(x-3)²
- A perfect square trinomial in the form ax² + bx + c always has a and c terms that are positive perfect squares (such as 1, 4, 9, 16, or 25), and a b term (positive or negative) that equals 2(√a * √c).^{[8]
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3
Check whether no solution exists. Not all trinomials can be factored. If you're stuck on a quadratic trinomials (ax²+bx+c), use the quadratic formula to find the answer. If the only answers are the square root of a negative number, no real solutions exist, so there are no factors.
- For non-quadratic trinomials, use Eisenstein's Criterion, described in the Tips section.

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Question

When the factors of a trinomial are (x-p) and (x-q), then the constant term of the trinomial is what?

Donagan

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The constant is +pq.

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Have you heard the term "slip" used to solve a trinomial?

Donagan

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Yes, the term is actually "slip and slide." The explanation is too lengthy to treat here, but you can websearch it.

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Does (x-4)(x-5) mean the same as (x-5)(x-4)?

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Yes. Since you are multiplying the exact same trionimals the answer will be the same.

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Answers and Example Problems

Answers to "tricky factoring" problems. These are the problems from the step about "trickier factors." We already simplified them down to an easier problem, so try to solve them using the steps in method 1, then check your work here:
- (2y)(x² + 7x + 12) = (x+3)(x+4)
- (x²)(x² + 11x - 26) = (x+13)(x-2)
- (-1)(x² - 6x + 9) = (x-3)(x-3) = (x-3)²
Try more tricky factoring problems. These problems have a common factor in each term that needs to be factored out first. Highlight the space after the equal signs to see the answer so you can check your work:
- 3x³+3x²-6x = (3x)(x+2)(x-1) ← highlight that space to see the answer
- -5x³y²+30x²y²-25y²x = (-5xy^2)(x-5)(x-1)
Practice tough problems. These problems can't be factored into easier equations, so you'll need to work out an answer in the form of (_x + __)(_x + __) by trial and error:
- 2x²+3x-5 = (2x+5)(x-1) ← highlight to see answer
- 9x²+6x+1 = (3x+1)(3x+1)=(3x+1)² (Hint: You might need to try more than one pair of factors for 9x.)

Tips

If you can't figure out how to factor a quadratic trinomial (ax²+bx+c), you can instead use the quadratic formula to solve for x.

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Although you don't need to know how to do this, you can use Eisenstein's Criteria to quickly determine if a polynomial is irreducible and cannot be factored. This criteria works for any polynomial but particularly well for trinomials. If there is a prime number p which evenly divides the last two terms and meets the following conditions, then the polynomial is irreducible:
- The constant term (with no variable) is a multiple of p but not of p².
- The leading term (for instance, a in ax²+bx+c) is not a multiple of p.
- For example, 14x² + 45x + 51 is irreducible because there is a prime number (3) that evenly divides both 45 and 51 but not 14, and 51 cannot be evenly divided by 3².
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Warnings

Although true for quadratics, factorable trinomials are not necessarily the product of two binomials. For instance, x⁴ + 105x + 46 = (x² + 5x + 2)(x² - 5x + 23).

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