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Study Guides > Intermediate Algebra

Read: Factor by Grouping

Learning Objectives

  • Factor a trinomial with leading coefficient other than [latex]1[/latex] using grouping
Trinomials with leading coefficients other than [latex]1[/latex] are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[/latex] can be rewritten as [latex]\left(2x+3\right)\left(x+1\right)[/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[/latex] and then factor each portion of the expression to obtain [latex]2x\left(x+1\right)+3\left(x+1\right)[/latex]. We then pull out the GCF of [latex]\left(x+1\right)[/latex] to find the factored expression. The first step in this process is to figure out what two numbers to use to re-write the x term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of [latex]6[/latex] and a sum of [latex]5[/latex]
Factors of [latex]2\cdot3=6[/latex] Sum of Factors
[latex]1,6[/latex] [latex]7[/latex]
[latex]-1,-6[/latex] [latex]-7[/latex]
[latex]2,3[/latex] [latex]5[/latex]
[latex]-2,-3[/latex] [latex]-5[/latex]
  The pair [latex]p=2,\text{ and }q=3[/latex] will give the correct x term, so we will rewrite it using the new factors:

[latex]2{x}^{2}+5x+3=2x^2+2x+3x+3[/latex]

Now we can group the polynomial into two binomials.

[latex]2x^2+2x+3x+3=(2x^2+2x)+(3x+3)[/latex]

Identify the GCF of each binomial.

[latex]2x[/latex] is the GCF of [latex](2x^2+2x)[/latex] and [latex]3[/latex] is the GCF of [latex](3x+3)[/latex], use this to rewrite the polynomial:

[latex](2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)[/latex]

Note how we leave the signs in the binomials and the addition that joins them, be careful with signs when you factor out the GCF. The GCF of our new polynomial is [latex](x+1)[/latex], we factor this out as well:

[latex]2x(x+1)+3(x+1)=(x+1)(2x+3)[/latex].

Sometimes it helps visually to write the polynomial this way [latex](x+1)2x+(x+1)3[/latex] before you factor out the GCF. This is purely a matter of preference, multiplication is commutative, so order doesn't matter.

 A General Note: Factor by Grouping

To factor a trinomial in the form [latex]a{x}^{2}+bx+c[/latex] by grouping, we find two numbers with a product of [latex]ac[/latex] and a sum of [latex]b[/latex]. We use these numbers to divide the [latex]x[/latex] term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.

Example

Factor [latex]5{x}^{2}+7x - 6[/latex] by grouping.

Answer: We have a trinomial with [latex]a=5,b=7[/latex], and [latex]c=-6[/latex]. First, determine [latex]ac=-30[/latex]. We need to find two numbers with a product of [latex]-30[/latex] and a sum of [latex]7[/latex]. In the table, we list factors until we find a pair with the desired sum.

Factors of [latex]-30[/latex] Sum of Factors
[latex]1,-30[/latex] [latex]-29[/latex]
[latex]-1,30[/latex] [latex]29[/latex]
[latex]2,-15[/latex] [latex]-13[/latex]
[latex]-2,15[/latex] [latex]13[/latex]
[latex]3,-10[/latex] [latex]-7[/latex]
[latex]-3,10[/latex] [latex]7[/latex]
So [latex]p=-3[/latex] and [latex]q=10[/latex].
[latex]\begin{array}{cc}5{x}^{2}-3x+10x - 6 \hfill & \text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\hfill \\ x\left(5x - 3\right)+2\left(5x - 3\right)\hfill & \text{Factor out the GCF of each part}.\hfill \\ \left(5x - 3\right)\left(x+2\right)\hfill & \text{Factor out the GCF}\text{ }\text{ of the expression}.\hfill \end{array}[/latex]

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that [latex]\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6[/latex].
We can summarize our process in the following way:

Given a trinomial in the form [latex]a{x}^{2}+bx+c[/latex], factor by grouping.

  1. List factors of [latex]ac[/latex].
  2. Find [latex]p[/latex] and [latex]q[/latex], a pair of factors of [latex]ac[/latex] with a sum of [latex]b[/latex].
  3. Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[/latex].
  4. Pull out the GCF of [latex]a{x}^{2}+px[/latex].
  5. Pull out the GCF of [latex]qx+c[/latex].
  6. Factor out the GCF of the expression.
In the following video we present one more example of factoring a trinomial whose leading coefficient is not 1 using the grouping method. https://youtu.be/agDaQ_cZnNc Factoring trinomials whose leading coefficient is not [latex]1[/latex] becomes quick and kind of fun once you get the idea.  Give the next example a try on your own before you look at the solution.

We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.

Example

Factor [latex]2{x}^{2}+9x+9[/latex].

Answer: Find two numbers p, q such that [latex]p\cdot{q}=18[/latex], and [latex]p + q = 9[/latex]. [latex]9[/latex] and [latex]18[/latex] are both positive, so we will only consider positive factors.

Factors of [latex]2\cdot9=18[/latex] Sum of Factors
[latex]1, 18[/latex] [latex]19[/latex]
[latex]3,6[/latex] [latex]9[/latex]
We can stop because we have found our factors. Rewrite the original expression, and group.

[latex]2x^2+3x+6x+9=(2x^2+3x)+(6x+9)[/latex]

Factor out the GCF of each binomial, and write as a product of two binomials:

[latex](2x^2+3x)+(6x+9)=x(2x+3)+3(2x+3)=(x+3)(2x+3)[/latex]

[latex-display]2{x}^{2}+9x+9=(x+3)(2x+3)[/latex-display]

Here is an example where the x term is positive and c is negative.

Example

Factor [latex]6{x}^{2}+x - 1[/latex].

Answer:  

Factors of [latex]6\cdot-1=-6[/latex] Sum of Factors
[latex]-1,6[/latex] [latex]5[/latex]
[latex]1,-6[/latex] [latex]-5[/latex]
[latex]-2,3[/latex] [latex]1[/latex]
We can stop because we have found our factors. Rewrite the original expression, and group.

[latex]6{x}^{2}+x - 1=6x^2-2x+3x-1[/latex]

Factor out the GCF of each binomial, and write as a product of two binomials:

[latex](6x^2-2x)+(3x-1)=2x(3x-1)+1(3x-1)=(2x+1)(3x-1)[/latex]

[latex-display]6{x}^{2}+x - 1=(2x+1)(3x-1)[/latex-display]

In the following video example, we will factor a trinomial whose leading term is negative. https://youtu.be/zDAMjdBfkDs For our last example, you will see that sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.

Example

Factor [latex]7x^{2}-16x–5[/latex].

Answer: Find [latex]p, q[/latex] such that [latex]p\cdot{q}=-35\text{ and }p+q=-16[/latex]

Factors of [latex]7\cdot{-5}=-35[/latex] Sum of Factors
[latex]-1, 35[/latex] [latex]34[/latex]
[latex]1, -35[/latex] [latex]-34[/latex]
[latex]-5, 7[/latex] [latex]2[/latex]
[latex]-7,5[/latex] [latex]-2[/latex]
 

Answer

Cannot be factored. None of the factors add up to [latex]-16[/latex]

Licenses & Attributions

CC licensed content, Original

  • Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Factor a Trinomial in the Form -ax^2+bx+c Using the Grouping Technique. 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.