Factoring Trinomials Calculator is a free online tool that identifies the factors of a given trinomial. STUDYQUERIES’s online calculator for factoring trinomials makes the calculation faster, and it displays the factors of a trinomial in a matter of seconds.

How to Use the Factoring Trinomials Calculator?

To use the factoring trinomials calculator, follow these steps:

  • Step 1: Enter the trinomial function into the input field
  • Step 2: To get an answer, click “FACTOR”
  • Step 3: In the new window, you will see the factors of the trinomial

Factoring Trinomials Calculator

How to Factoring Trinomials?: 3 Easy Methods

Trinomials are algebraic expressions made up of three terms. Most likely, you’ll learn how to factor quadratic trinomials, which are trinomials written as \(ax^2 + bx + c\). Using these tricks will improve your ability to factor quadratic trinomials, but it will take some practice.

Factoring Trinomials Calculator
Factoring Trinomials Calculator

Polynomials with terms like \(x^3\) or \(x^4\) are not always solvable by the same methods, but you can factor or substitute them into quadratic formulas just like any other polynomial.

Factoring Trinomials Method 1: Factoring x² + bx + c

Become familiar with FOIL multiplication.

Using the FOIL method, or “First, Outside, Inside, Last,” you can multiply expressions like \((x+2)(x+4)\). Before we factor, it is useful to understand how this strategy works:

Multiply the \(\color{red}{First}\) two terms:

$$(\color{red}{x} + 2)(\color{red}{x} + 4)=\color{red}{x^2} + {\_}$$

The \(\color{red}{Outside}\) terms are multiplied as follows:

$$(\color{red}{x}+2)(x+\color{red}{4}) = x^2+\color{red}{4x} + {\_}$$

Multiply the \(\color{red}{Inside}\) terms:

$$(x+\color{red}{2})(\color{red}{x}+4) = x^2+4x+\color{red}{2x} +{\_}$$

Multiply the \(\color{red}{Last}\) terms:

$$(x+\color{red}{2})(x+\color{red}{4}) = x^2+4x+2x+\color{red}{8}$$

Simply write \(x^2+\color{red}{4x}+\color{red}{2x}+8\ as\ x^2+6x+8\)

Understand factoring.

Multiplying two binomials together in \(FOIL\) produces a trinomial (an expression with three terms) in the form \(ax^2+bx+c\), where \(a, b,\ and\ c\) are ordinary numbers. You can factor back an equation in the same form into two binomials if you start with the same form.

  • If the terms aren’t written in this order, arrange them so they are. For example, rewrite \(3x – 10 + x^2 as x^2 + 3x – 10\).
  • As the highest exponent is \(2\), this type of expression is \(“quadratic.”\)

Write a space for the answer in FOIL form.

For now, just write \((\_)(\_)\) in the space where the answer is going to be written. We’ll fill this in as we go.

Please don’t write \(+\) or \(-\) between the blank terms yet, since we don’t know which one it will be.

Fill out the First terms.

Where the first term of your trinomial is just \(x^2\), the terms in the First position will always be \(x\) and \(x\). These are the factors of the term \(x^2\), since \(x \times x) = x^2\).

The example \(x^2 + 3x – 10\) starts with \(x^2\), so we can write: \((x\_)(x\_)\).

In the next section, we’ll look at more complicated problems, including trinomials that start with a term like \(6x^2\) or \(-x^2\). For now, just follow the example problem.

Use factoring to guess at the Last terms.

When you read the \(FOIL\ method\) step again, you’ll see that multiplying the Last terms together gives you the final term in the polynomial (the one with no x). To the factor, we need to find two numbers that multiply to create the last term.

In our example \(x^2 + 3x – 10\), the last term is \(-10\).

What are the factors of \(-10\)? What two numbers multiplied together equal \(-10\)?

A few possibilities are: \(-1 \times 10\), \(1 \times -10\), \(-2 \times 5\), or \(2 \times -5\). Write these pairs down somewhere to remember them.

We won’t change our answer yet. It still looks this way: \((x\_)(x\_)\).

Test which possibilities work with Outside and Inside multiplication.

The Last terms have been narrowed down to a few options. Test each possibility by multiplying the \(Outside\) and \(Inside\) terms and comparing the result to our trinomial. For example:

Within this test, we want to end up with an \(“x”\) term of \(3x\), as this is what we had in our original problem.

Test -1 and 10:

The \(Outside\) + \(Inside\) = \(10x – x = 9x\). Nope.

Test 1 and -10:

\(-10x + x = -9x\). This isn’t right. Once you test \(-1\) and \(10\), you know that \(1\) and \(-10\) will be just the opposite of the answer above: \(-9x\) instead of \(9x\).

Test -2 and 5:

\(5x – 2x = 3x\). That matches the original polynomial, so the correct answer is \((x-2)(x+5)\).

If you don’t have a constant in front of the \(x^2\) term, you can simply add the two factors together and then put an \(“x”\) after it \((-2+5 \longrightarrow 3x)\). It won’t work for more complex problems, so it’s a good idea to remember the “long way” described above.

Factoring Trinomials Method 2: Factoring More Complicated Trinomials

Use simple factoring to make more complicated problems easier.

Let’s say that you need to factor \(3x^2 + 9x – 30\). Find the most common element (the “greatest common factor” or GCF).  In this case, it’s 3:

$$3x^2 = (3)(x^2)$$

$$9x = (3)(3x)$$

$$-30 = (3)(-10)$$

Therefore, \(3x^2 + 9x – 30 = (3)(x^2+3x-10)\). We can factor out the new trinomial by following the steps in the previous section. The final answer will be \((3)(x-2)(x+5)\).

Be on the lookout for trickier factors.

Factoring may require variables, or you may need to factor several times to find the simplest expression. Here are some examples:

$$2x^2y + 14xy + 24y = (2y)(x^2 + 7x + 12)$$

$$x^4 + 11x^3 – 26x^2 = (x^2)(x^2 + 11x – 26)$$

$$-x^2 + 6x – 9 = (-1)(x^2 – 6x + 9)$$

Don’t forget to factor the new trinomial further using method 1. Please check your work and find similar examples near the bottom of this page.

Solve problems with a number in front of the x².

In some cases, quadratic trinomials cannot be simplified down to the simplest form. Learn how to solve problems like \(3x^2 + 10x + 8\), then practice on your own with the examples at the bottom of the page:

Let’s set up our answer: $${(\_\ \_)}{(\_\ \_)}$$

“First” terms will have an \(x\) and will multiply together to make \(3x^2\). The only option here is \({(3x\ \_\ \_)}{(x\ \_\ \_)}\).

List the factors of \(8\). Our options are \(1 \times 8\), or \(2 \times 4\).

Use the Outside and Inside terms to test them. Since the Outside term is being multiplied by \(3x\) instead of \(x\), the order of the factors matters. You need to try every possible solution until you get an \(Outside+Inside\) result of \(10x\) (from the original problem):

$$(3x+1)(x+8)\longrightarrow 24x+x = 25x\longrightarrow no$$

$$(3x+8)(x+1)\longrightarrow 3x+8x = 11x\longrightarrow no$$

$$(3x+2)(x+4)\longrightarrow 12x+2x=14x\longrightarrow no$$

$$(3x+4)(x+2)\longrightarrow 6x+4x=10x\longrightarrow Yes,\ this\ is\ correct$$

Use substitution for higher-degree trinomials.

Despite using simple factoring to make the problem easier, your math book might present an equation with a high exponent, such as \(x^4\). Try substituting a new variable that turns it into a problem you can solve. Using this example:

$$x^5+13x^3+36x=(x)(x^4+13x^2+36)$$

Let’s invent a new variable. We’ll say \(y = x^2\), and plug it in:

$$(x)(y^2+13y+36)$$

$$=(x)(y+9)(y+4)$$ Now switch back to using the original variable:

$$=(x)(x^2+9)(x^2+4)$$

$$=(x)(x\pm 3)(x\pm 2)$$

Factoring Trinomials Method 3: Factoring Special Cases

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.

For example, in \(x^2 + 6x + 5\), “\(5\) is a prime number, so the binomial must be in the form \({(\_\ \_ 5)}{(\_\ \_ 1)}\)

In the problem \(3x^2+10x+8\), \(3\) is a prime number, so the binomial must be in the form \({(3x\ \_\ \_)}{(x\ \_\ \_)}\)

For the problem \(3x^2+4x+1\), the only possible solution is \((3x+1)(x+1)\). 

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:

$$x^2+2x+1=(x+1)^2,\ and\ x^2-2x+1=(x-1)^2$$

$$x^2+4x+4=(x+2)^2,\ and\ x^2-4x+4=(x-2)^2$$

$$x^2+6x+9=(x+3)^2,\ and\ x^2-6x+9=(x-3)^2$$

A perfect square trinomial in the form \(ax^2 + bx + c\) 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 \(2(\sqrt{a} \times \sqrt{c})\).

Check whether no solution exists.

Factoring trinomials is not possible for all trinomials. When you’re stuck on quadratic trinomials \((ax^2+bx+c)\), use the quadratic formula. When all the answers are the square root of a negative number, there are no real solutions, so there are no factors.

Use Eisenstein’s Criterion, which is described in the Tips section, for non-quadratic trinomials.

Important Tips To Factoring Trinomials

Rather than factoring a quadratic trinomial \((ax^2+bx+c)\), you can solve for \(x\) using the quadratic formula.

Eisenstein’s Criteria can be used to determine quickly if a polynomial is irreducible and cannot be factored, even if you don’t know how to do it. For trinomials, this criteria works well, but it is particularly useful for polynomials. The polynomial is irreducible if there is a prime number \(p\) that evenly divides the last two terms and meets the following conditions:

The constant term (with no variable) is a multiple of \(p\) but not of \(p^2\).
The leading term (for instance, \(a\) in \(ax^2+bx+c\)) is not a multiple of \(p\).

For example, \(14x^2 + 45x + 51\) 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 \(3^2\).

Conclusion

Trinomials of the form \(x^2 + bx + c\) can be factored by finding two integers \(r\) and \(s\), whose sum is \(b\) and whose product is \(c\). Rewrite the trinomial as \(x^2 + rx + sx + c\) and then factor the polynomial using the distributive property.

Look first for common factors for all three terms in a trinomial that takes the form \(ax^2 + bx + c\), where \(a\) is a coefficient other than \(1\). The first factor out the common factor, then factor the simpler trinomials. In the case where the remaining trinomial is still of the form \(ax^2 + bx + c\), find two integers, \(r\), and \(s\), whose sum is \(b\) and whose product is \(ac\). Rewrite the trinomial as \(ax^2 + rx + sx + c\), and factor the polynomial using grouping and the distributive property.

When \(ax^2\) is negative, you can factor \(−1\) out of the whole trinomial before continuing.

FAQs

How do you factor Trinomials step by step?

  • Step 1: Identify the values for \(b\) and \(c\). In this example, \(b=6\) and \(c=8\).
  • Step 2: Find two numbers that \(ADD\) to \(b\) and \(MULTIPLY\) to \(c\). This step can take a little bit of trial and error.
  • Step 3: Use the numbers you picked to write out the factors and check.

What is trinomial and example?

A trinomial is an algebraic expression with three terms that are not zero. \(x + y + z\) is a trinomial in three variables \(x, y,\ and\ z\).

What is a basic trinomial?

A polynomial with three terms is called a trinomial. Trinomials often (but not always!) have the form \(x^2 + bx + c\).

How do you write a trinomial in factored form?

Factoring Trinomials of the Form \(x^2 + bx + c\) and \(x^2 – bx + c\)

  • Write out all the pairs of numbers which can be multiplied to produce \(c\).
  • Add each pair of numbers to find a pair that produce \(b\) when added.
  • If \(b > 0\), then the factored form of the trinomial is \((x + d )(x + e)\).

What is a cubic trinomial example?

Cubic Trinomials of the Form \(Ax^3 + Bx+^2 + Cx\). For example, the greatest common factor of the trinomial \(3x^3 – 6x^2 – 9x\) is \(3x\), so the polynomial is equal to \(3x\) times the trinomial \(x^2 – 2x -3\), or \(3x(x^2 – 2x – 3)\).

How do you break down a trinomial?

To the factor, a trinomial is to decompose an equation into the product of two or more binomials. This means that we will rewrite the trinomial in the form \((x + m) (x + n)\). Your task is to determine the value of m and n. In other words, we can say that factoring a trinomial is the reverse process of the foil method.