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Study Guides > ALGEBRA / TRIG I

Factoring a Trinomial with a Leading Coefficient of 1

Learning Outcomes

  • Factor a trinomial with a leading coefficient of [latex]1[/latex]
  • Use a shortcut to factor trinomials of the form [latex]x^2+bx+c[/latex]
Now that we have shown you how to factor by grouping, We are going to show you how to factor a trinomial whose leading coefficient is [latex]1[/latex].  That is, trinomials of the form [latex]x^2+bx+c[/latex].  Polynomials whose leading coefficients are 1 can be factored using the grouping method that we showed you in the previous section.  We will begin by showing a few examples using the grouping method and then we will show you a shortcut that will make factoring trinomials with a leading coefficient of 1 even easier! The following is a summary of the method, then we will show some examples of how to use it.

Factoring Trinomials in the form [latex]x^{2}+bx+c[/latex]

To factor a trinomial in the form [latex]x^{2}+bx+c[/latex], find two integers, r and s, whose product is c and whose sum is b.

[latex]\begin{array}{l}r\cdot{s}=c\\\text{ and }\\r+s=b\end{array}[/latex]

Rewrite the trinomial as [latex]x^{2}+rx+sx+c[/latex] and then use grouping and the distributive property to factor the polynomial. The resulting factors will be [latex]\left(x+r\right)[/latex] and [latex]\left(x+s\right)[/latex].
Special Note:  Does the summary above look familiar?  It should!  How we factor polynomials of the form [latex]x^{2}+bx+c[/latex] using grouping is exactly the same as for polynomials of the form [latex]ax^{2}+bx+c[/latex].  The only difference is that, previously, we were looking for two integers, r and s, whose sum is b and whose product is ac.  Notice that if our polynomial is of the form  [latex]x^{2}+bx+c[/latex], then [latex]a=1[/latex], making [latex]ac=1\cdot{c}=c[/latex].  As a result, we can skip the step of multiplying a by c! Let’s factor the trinomial [latex]x^{2}+5x+6[/latex]. In this polynomial, the b part of the middle term is  [latex]5[/latex] and the c term is  [latex]6[/latex]. A chart will help us organize possibilities. On the left, list all possible factors of the c term, [latex]6[/latex]; on the right you'll find the sums.
Factors whose product is  [latex]6[/latex] Sum of the factors
[latex]1\cdot6=6[/latex] [latex]1+6=7[/latex]
[latex]2\cdot3=6[/latex] [latex]2+3=5[/latex]
There are only two possible factor combinations, [latex]1[/latex] and [latex]6[/latex], and [latex]2[/latex] and [latex]3[/latex]. You can see that [latex]2+3=5[/latex]. So [latex]2x+3x=5x[/latex], giving us the correct middle term.

Example

Factor [latex]x^{2}+5x+6[/latex].

Answer: Use values from the chart above. Replace [latex]5x[/latex] with [latex]2x+3x[/latex].

[latex]x^{2}+2x+3x+6[/latex]

Group the pairs of terms.

[latex]\left(x^{2}+2x\right)+\left(3x+6\right)[/latex]

Factor x out of the first pair of terms

[latex]x\left(x+2\right)+\left(3x+6\right)[/latex]

Factor [latex]3[/latex] out of the second pair of terms.

[latex]x\left(x+2\right)+3\left(x+2\right)[/latex]

Factor out [latex]\left(x+2\right)[/latex].

[latex]\left(x+2\right)\left(x+3\right)[/latex]

Answer

[latex-display]\left(x+2\right)\left(x+3\right)[/latex-display]

Note that if you wrote [latex]x^{2}+5x+6[/latex] as [latex]x^{2}+3x+2x+6[/latex] and grouped the pairs as [latex]\left(x^{2}+3x\right)+\left(2x+6\right)[/latex]; then factored, [latex]x\left(x+3\right)+2\left(x+3\right)[/latex], and factored out [latex]x+3[/latex], the answer would be [latex]\left(x+3\right)\left(x+2\right)[/latex]. Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well; they are equivalent answers. In the following video, we present another example of how to use grouping to factor a quadratic polynomial. https://youtu.be/_Rtp7nSxf6c   Finally, let’s take a look at the trinomial [latex]x^{2}+x–12[/latex]. In this trinomial, the c term is [latex]−12[/latex]. So look at all of the combinations of factors whose product is [latex]−12[/latex]. Then see which of these combinations will give you the correct middle term, where b is [latex]1[/latex].
Factors whose product is [latex]−12[/latex] Sum of the factors
[latex]1\cdot−12=−12[/latex] [latex]1+−12=−11[/latex]
[latex]2\cdot−6=−12[/latex] [latex]2+−6=−4[/latex]
[latex]3\cdot−4=−12[/latex] [latex]3+−4=−1[/latex]
[latex]4\cdot−3=−12[/latex] [latex]4+−3=1[/latex]
[latex]6\cdot−2=−12[/latex] [latex]6+−2=4[/latex]
[latex]12\cdot−1=−12[/latex] [latex]12+−1=11[/latex]
There is only one combination where the product is [latex]−12[/latex] and the sum is [latex]1[/latex], and that is when [latex]r=4[/latex], and [latex]s=−3[/latex]. Let’s use these to factor our original trinomial.

Example

Factor [latex]x^{2}+x–12[/latex].

Answer: Rewrite the trinomial using the values from the chart above. Use values [latex]r=4[/latex] and [latex]s=−3[/latex].

[latex]x^{2}+4x+−3x–12[/latex]

Group pairs of terms.

[latex]\left(x^{2}+4x\right)+\left(−3x–12\right)[/latex]

Factor [latex]x[/latex] out of the first group.

[latex]x\left(x+4\right)+\left(-3x-12\right)[/latex]

Factor [latex]−3[/latex]  out of the second group.

[latex]x\left(x+4\right)–3\left(x+4\right)[/latex]

Factor out [latex]\left(x+4\right)[/latex].

[latex]\left(x+4\right)\left(x-3\right)[/latex]

Answer

[latex-display]\left(x+4\right)\left(x-3\right)[/latex-display]

In the above example, you could also rewrite [latex]x^{2}+x-12[/latex] as [latex]x^{2}– 3x+4x–12[/latex] first. Then factor [latex]x\left(x – 3\right)+4\left(x–3\right)[/latex], and factor out [latex]\left(x–3\right)[/latex] getting [latex]\left(x–3\right)\left(x+4\right)[/latex]. Since multiplication is commutative, this is the same answer.

Factoring Tips

Factoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored. While there is no foolproof way to find the right combination on the first guess, there are some tips that can ease the way.

Tips for Finding Values that Work

When factoring a trinomial in the form [latex]x^{2}+bx+c[/latex], consider the following tips. Look at the c term first.
  • If the c term is a positive number, then the factors of c will both be positive or both be negative. In other words, r and s will have the same sign.
  • If the c term is a negative number, then one factor of c will be positive, and one factor of c will be negative. Either r or s will be negative, but not both.
Look at the b term second.
  • If the c term is positive and the b term is positive, then both r and s are positive.
  • If the c term is positive and the b term is negative, then both r and s are negative.
  • If the c term is negative and the b term is positive, then the factor that is positive will have the greater absolute value. That is, if [latex]|r|>|s|[/latex], then r is positive and s is negative.
  • If the c term is negative and the b term is negative, then the factor that is negative will have the greater absolute value. That is, if [latex]|r|>|s|[/latex], then r is negative and s is positive.
After you have factored a number of trinomials in the form [latex]x^{2}+bx+c[/latex], you may notice that the numbers you identify for r and s end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews three examples.
Trinomial [latex]x^{2}+7x+10[/latex] [latex]x^{2}+5x+6[/latex] [latex]x^{2}+x-12[/latex]
r and s values [latex]r=+5,s=+2[/latex] [latex]r=+2,s=+3[/latex] [latex]r=+4,s=–3[/latex]
Factored form [latex]\left(x+5\right)\left(x+2\right)[/latex] [latex]\left(x+2\right)\left(x+3\right)[/latex] [latex](x+4)(x–3)[/latex]

The Shortcut

Notice that in each of the examples above, the r and s values are repeated in the factored form of the trinomial. So what does this mean? It means that in trinomials of the form [latex]x^{2}+bx+c[/latex] (where the coefficient in front of [latex]x^{2}[/latex] is [latex]1[/latex]), if you can identify the correct r and s values, you can effectively skip the grouping steps and go right to the factored form. For those of you that like shortcuts, let's look at some examples where we use this idea.  
Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way. Shortcut This Way
In the next two examples, we will show how you can skip the step of factoring by grouping and move directly to the factored form of a product of two binomials with the r and s values that you find. The idea is that you can build factors for a trinomial in this form: [latex]x^2+bx+c[/latex] by finding r and s, then placing them in two binomial factors like this:

[latex]\left(x+r\right)\left(x+s\right)\text{ OR }\left(x+s\right)\left(x+r\right)[/latex]

Example

Factor: [latex]y^2+6y-27[/latex]

Answer: Find r and s:

Factors whose product is -27 Sum of the factors
[latex]1\cdot{-27}=-27[/latex] [latex]1-27=-26[/latex]
[latex]3\cdot{-9}=-27[/latex] [latex]3-9=-6[/latex]
[latex]-3\cdot{9}=-27[/latex]  [latex]-3+9=6[/latex]
Instead of rewriting the middle term, we will use the values of r and s that give the product and sum that we need. In this case:

[latex]\begin{array}{l}r=-3\\s=9\end{array}[/latex]

It helps to start by writing two empty sets of parentheses:

[latex]\left(\,\,\,\,\,\,\,\,\,\,\,\,\right)\left(\,\,\,\,\,\,\,\,\,\,\,\,\right)[/latex]

The squared term is y, so we will place a y in each set of parentheses:

[latex]\left(y\,\,\,\,\,\,\,\,\right)\left(y\,\,\,\,\,\,\,\,\right)[/latex]

Now we can fill in the rest of each binomial with the values we found for r and s.

[latex]\left(y-3\right)\left(y+9\right)[/latex]

Note how we kept the sign on each of the values.  The nice thing about factoring is you can check your work. Multiply the binomials together to see if you did it correctly.

[latex]\begin{array}{l}\left(y-3\right)\left(y+9\right)\\=y^2+9y-3y-27\\=y^2+6y-27\end{array}[/latex]

Answer

[latex-display]\left(y-3\right)\left(y+9\right)[/latex-display]

We will show one more example so you can gain more experience.

Example

Factor: [latex]-m^2+16m-48[/latex]

Answer: There is a negative in front of the squared term, so we will factor out a negative one from the whole trinomial first. Remember, this boils down to changing the sign of all the terms:

[latex]-m^2+16m-48=-1\left(m^2-16m+48\right)[/latex]

Now we can factor [latex]\left(m^2-16m+48\right)[/latex] by finding r and s. Note that b is negative, and c is positive so we are probably looking for two negative numbers:

Factors whose product is 48 Sum of the factors
[latex]-1\cdot{-48}=48[/latex] [latex]-1-48=-49[/latex]
[latex]-2\cdot{-12}=48[/latex] [latex]-2-12=-14[/latex]
[latex]-3\cdot{-16}=-48[/latex] [latex]-3-16=-19[/latex]
[latex]-4\cdot{-12}=-48[/latex] [latex]-4-12=-16[/latex]

There are more factors whose product is [latex]48[/latex], but we have found the ones that sum to  [latex]-16[/latex], so we can stop.

[latex]\begin{array}{l}r=-4\\s=-12\end{array}[/latex]

Now we can fill in each binomial with the values we found for r and s, make sure to use the correct variable!

[latex]\left(m-4\right)\left(m-12\right)[/latex]

We are not done yet, remember that we factored out a negative sign in the first step. We need to remember to include that.

 [latex]-1\left(m-4\right)\left(m-12\right)[/latex]

Answer

[latex-display]-m^2+16m-48=-1\left(m-4\right)\left(m-12\right)[/latex-display]

In the following video, we present two more examples of factoring a trinomial with a leading coefficient of 1. https://youtu.be/-SVBVVYVNTM

Try It

[ohm_question]7900[/ohm_question]
To summarize our process, consider the following steps:

How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[/latex], factor it

  1. List factors of [latex]c[/latex].
  2. Find [latex]p[/latex] and [latex]q[/latex], a pair of factors of [latex]c[/latex] with a sum of [latex]b[/latex].
  3. Write the factored expression [latex]\left(x+p\right)\left(x+q\right)[/latex].
We will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.  We will show that when c is negative, either p or q will be negative.

Example

Factor [latex]x^{2}+x–12[/latex].

Answer: Consider all the combinations of numbers whose product is [latex]-12[/latex] and list their sum.

Factors whose product is [latex]−12[/latex] Sum of the factors
[latex]1\cdot−12=−12[/latex] [latex]1+−12=−11[/latex]
[latex]2\cdot−6=−12[/latex] [latex]2+−6=−4[/latex]
[latex]3\cdot−4=−12[/latex] [latex]3+−4=−1[/latex]
[latex]4\cdot−3=−12[/latex] [latex]4+−3=1[/latex]
[latex]6\cdot−2=−12[/latex] [latex]6+−2=4[/latex]
[latex]12\cdot−1=−12[/latex] [latex]12+−1=11[/latex]
Choose the values whose sum is [latex]+1[/latex]:  [latex]p=4[/latex] and [latex]q=−3[/latex], and place them into a product of binomials.  

[latex]\left(x+4\right)\left(x-3\right)[/latex]

Think About It

Which property of multiplication can be used to describe why [latex]\left(x+4\right)\left(x-3\right) =\left(x-3\right)\left(x+4\right)[/latex]. Use the textbox below to write down your ideas before you look at the answer. [practice-area rows="2"][/practice-area]

Answer: The commutative property of multiplication states that factors may be multiplied in any order without affecting the product.

[latex]a\cdot b=b\cdot a[/latex]

In our last example, we will show how to factor a trinomial whose b term is negative.

Example

Factor [latex]{x}^{2}-7x+6[/latex].

Answer: List the factors of [latex]6[/latex]. Note that the b term is negative, so we will need to consider negative numbers in our list.

Factors of [latex]6[/latex] Sum of Factors
[latex]1,6[/latex] [latex]7[/latex]
[latex]2, 3[/latex] [latex]5[/latex]
[latex]-1, -6[/latex] [latex]-7[/latex]
[latex]-2, -3[/latex] [latex]-5[/latex]
Choose the pair that sum to [latex]-7[/latex], which is [latex]-1, -6[/latex] Write the pair as constant terms in a product of binomials. [latex-display]\left(x-1\right)\left(x-6\right)[/latex-display]

In the last example, the b term was negative and the c term was positive. This will always mean that if it can be factored, p and q will both be negative.

Think About It

Can every trinomial be factored as a product of binomials? Mathematicians often use a counterexample to prove or disprove a question. A counterexample means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient [latex]1[/latex] that cannot be factored as a product of binomials? Use the textbox below to write your ideas. [practice-area rows="2"][/practice-area]

Answer: Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime. A counterexample would be: [latex]x^2+3x+7[/latex]

Try It

[ohm_question]7907[/ohm_question]

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Licenses & Attributions

CC licensed content, Original

  • Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.

CC licensed content, Shared previously

  • Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.