3 Ways to Factor Trinomials

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3 Ways to Factor Trinomials
3 Ways to Factor Trinomials

A trinomial is an algebraic expression made up of three terms. You will probably learn to factor quadratic trinomials, which are trinomials written in the form ax2 + bx + c. There are several tricks that can be applied to different types of quadratic trinomials, but you'll get better and faster with practice. Polynomials of higher degrees, with terms like3 or x4, cannot always be solved with the same methods, but you can often resort to simple factorization or term substitution to turn them into problems that can be solved with any quadratic formula.


Method 1 of 3: Factoring x2 + bx + c

Trinomials Factor Step 1

Step 1. Learn the distributive property (also known as FOIL in English), to multiply expressions like (x+2)(x+4).

Before you start factoring, it's good to know how this works:

  • multiply the first terms: (x+2)(x+4) = x2 + __
  • Multiply the terms of outside: (x+2)(x+

    Step 4.) = x2+4x + __

  • Multiply the terms of inside: (x+

    Step 2.)(x+4) = x2+4x+2x + __

  • multiply the last terms: (x+

    Step 2.)(x

    Step 4.) = x2+4x+2x

    Step 8.

  • Simplify: x2+4x+2x+8 = x2+6x+8
Trinomials Factor Step 2

Step 2. Understand factorization

When you multiply two binomials together using the distributive one, you end up with a trinomial (a three-term expression) of the form a x2+ b x+ c, where “a”, “b” and “c” are common numbers. If you start with an equation of the same form, you can factor it back into two binomials.

  • If the equation is not written in that order, move the terms into the proper position. For example, rewrite 3x - 10 + x2 like x2 + 3x - 10.
  • As the largest exponent is 2 (x2, this expression is called "quadratic".
Trinomials Factor Step 3

Step 3. Reserve a space for the answer of the method presented

For now, just write (__ __) (__ __) in the space dedicated to the answer. We will fill in these fields shortly.

Don't put + or – signs between blank terms yet, as we don't know which one will be used

Trinomials Factor Step 4

Step 4. Fill in the first terms

In simple problems where the first term of your trinomial is just x2, the terms of the first position will always be x and x. These are the factors of x.2, because x times x = x2.

  • Our example, x2 + 3x - 10, starts with x2, then we can write:
  • (x __)(x __)
  • We'll look at more elaborate problems in the next section, including trinomials that start with a term like 6x2or -x2. For now, follow the example problem.
Trinomials Factor Step 5

Step 5. Use factorization to guess the last terms

If you go back and reread the method used initially, you will see that multiplying the last terms gives 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.

  • In our example, x2 + 3x - 10, the last term is -10.
  • What are the factors of -10? Which two numbers multiplied together make -10?
  • There are a few possibilities: -1 times 10, 1 time -10, -2 times 5, or 2 times -5. Write these pairs down somewhere so you don't forget.
  • Do not change the answer yet. She still looks like this: (x __)(x __).
Trinomials Factor Step 6

Step 6. Test which possibilities work with outside and inside multiplication

We've reduced the last terms to few possibilities. Test each one by multiplying the external and internal terms, then comparing the result to our trinomial. For example:

  • The "x" term of our original problem is "3x", so that's what we want to get in the test.
  • Test -1 and 10: (x-1)(x+10). Outside + inside value = 10x - x = 9x. Not.
  • Test 1 and -10: (x+1)(x-10). -10x + x = -9x. This is not right. In fact, after testing -1 and 10, you know that the answer 1 and -10 will be just the opposite of the above result: -9x, instead of 9x.
  • Test -2 and 5: (x-2)(x+5). 5x - 2x = 3x. This matches the original polynomial, so this is the correct answer: (x-2)(x+5).
  • In simple cases like this, when there's no constant in front of the x2, you can use a shortcut: just add the two factors and put an "x" after (-2+5 → 3x). This won't work with more complicated problems, so it's good to remember the full path described above.

Method 2 of 3: Factoring out more elaborate trinomials

Trinomials Factor Step 7

Step 1. Use simple factorization to facilitate more elaborate problems

Let's say you need to factor 3x2 + 9x - 30. Look for a number that factors all three terms (their "greatest common divisor", or MDC). In this case, it's 3:

  • 3x2 = (3)(x2)
  • 9x = (3)(3x)
  • -30 = (3)(-10)
  • So 3x2 + 9x - 30 = (3)(x2+3x-10). We can factor out the new trinomial using the steps at the beginning of this article. The answer will be (3)(x-2)(x+5).
Trinomials Factor Step 8

Step 2. Look for more elaborate factors

Sometimes the factor may involve variables, or you may need to factor a few times until you find the simplest expression possible. Here are some 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 one more time, using the steps from the beginning. Check your answer and find similar example issues near the end of this article.
Trinomials Factor Step 9

Step 3. Solve problems with a number in front of the x2.

Some quadratic trinomials cannot be simplified until you reach the easiest type of problem. Learn how to solve problems like 3x2 + 10x + 8 and then practice yourself with the example problems at the end of this article:

  • Assemble the answer: (__ __)(__ __)
  • The first terms have an "x" each and, when multiplied, result in 3x2. There is only one possible option here: (3x __)(x __).
  • List the factors of 8. Our options are 1 times 8, or 2 times 4.
  • Test them using the terms outside and inside. Note that the order of factors matters, since the outside term is being multiplied by "3x", not by "x". Try all the possibilities until you get a result from outside + within 10x (according to the original problem):
  • (3x+1)(x+8) → 24x+x = 25x Not.
  • (3x+8)(x+1) → 3x+8x = 11x Not.
  • (3x+2)(x+4) → 12x+2x=14x Not.
  • (3x+4)(x+2) → 6x+4x=10x Yes, this is the correct factor.
Trinomials Factor Step 10

Step 4. Use substitution for higher-grade trinomials

Your math textbook might surprise you with a high x exponent equation4, even after having already used simple factorization to ease the problem. Try replacing it with a new variable that turns the equation into something you can solve. For example:

  • x5+13x3+36x
  • =(x)(x4+13x2+36)
  • Let's invent a new variable. We will say that y = x2 and we will make the substitutions:
  • (x)(y2+13y+36)
  • =(x)(y+9)(y+4). Now go back to using the original variable:
  • =(x)(x2+9)(x2+4)
  • =(x)(x±3)(x±2)

Method 3 of 3: Factoring in Special Cases

Trinomials Factor Step 11

Step 1. Look for prime numbers

Check whether the constant in the first or third term of the trinomial is a prime number. A prime number can only be divided equally by itself and by 1, so there is only one possible pair of binomial factors.br>

  • For example, in x2 + 6x + 5, "5" is a prime number, so the binomial should look like this: (__ 5)(__ 1).
  • In 3x problem2+10x+8, 3 is a prime number, so the binomial should look like this: (3x __)(x __).
  • For the 3x problem2+4x+1, both "3" and "1" are prime numbers, so the only possible solution is (3x+1)(x+1). (You should still perform this multiplication to check your calculation, as some expressions cannot be factored – for example, 3x2 + 100x + 1 has no factors).
Trinomials Factor Step 12

Step 2. Check that the trinomial is a perfect square

A perfect square trinomial can be factored into two identical binomials, and the factor is usually written as (x+1)2, instead of (x+1)(x+1). Here are some common ones that tend to show up in trouble:

  • 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
  • In a perfect square trinomial in the shape of a x2 + bx + c, the terms "a" and "c" are always positive perfect squares (such as 1, 4, 9, 16 or 25), and the term b (positive or negative) is always equal to 2(√a * √c).
Trinomials Factor Step 13

Step 3. Check if there is no solution

Not all trinomials can be factored. If you are stuck on a quadratic trinomial (ax2+bx+c), use the quadratic formula to find the result. If the only answers are the square root of a negative number, then there is no real solution, so there are no factors.

For non-quadratic trinomials, use the Eisenstein criterion, which is described in the hints section

Answers and Problem Examples

  1. Answers to the most elaborate factoring problems.

    These are the problems with the part about “more elaborate” trinomials. We've already simplified them, making them an easier problem. Now try to solve them using the steps from the beginning, then check your calculations 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
  2. Try to solve more complex factoring problems.

    These problems have a common factor in each term that needs to be factored in first. Highlight the space after the equals signs to see the answer and check your calculations here:

    • 3x3+3x2-6x = (3x)(x+2)(x-1) ← highlight this space to see your answer
    • -5x3y2+30x2y2-25y2x = (-5xy^2)(x-5)(x-1)
  3. Practice with difficult problems.

    These problems cannot be factored into easier equations, so you will need to craft an answer in the form of (_x + __)(_x + __) by testing:

    • 2x2+3x-5 = (2x+5)(x-1) ← highlight to see the answer
    • 9x2+6x+1 = (3x+1)(3x+1)=(3x+1)2 (Hint: you might need to try more than a couple of factors for 9x).


    • If you don't know how to factor a quadratic trinomial (ax2+bx+c), can use the quadratic formula to find the value of x.
    • Although you don't need to know how to do this, you can use Eisenstein's criterion to quickly determine whether a polynomial is irreducible and cannot be factored. This criterion applies to any polynomial, but it works particularly well with trinomials. If there is a prime number "p" that divides the last two terms equally and satisfies the following conditions, then the polynomial is irreducible:

      • The constant term (no variable) is a multiple of p, but not p.2.
      • The main term (eg "a" in ax2+bx+c) is not a multiple of p.
      • For example, 14x2 + 45x + 51 is irreducible, as there is a prime number (3) that divides 45 and 51 equally, but not 14, and 51 cannot be divided equally by 32.


    • While this is true for quadratic equations, factorable trinomials are not necessarily the product of two binomials. For example: x4 + 105x + 46 = (x2 + 5x + 2)(x2 - 5x + 23).

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