Factoring by Grouping
Trinomials with leading coefficients other than 1 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.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.How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[/latex], factor by grouping.
- List factors of [latex]ac[/latex].
- Find [latex]p[/latex] and [latex]q[/latex], a pair of factors of [latex]ac[/latex] with a sum of [latex]b[/latex].
- Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[/latex].
- Pull out the GCF of [latex]a{x}^{2}+px[/latex].
- Pull out the GCF of [latex]qx+c[/latex].
- Factor out the GCF of the expression.
Example 3: Factoring a Trinomial by Grouping
Factor [latex]5{x}^{2}+7x - 6[/latex] by grouping.Solution
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] | 29 |
| [latex]2,-15[/latex] | [latex]-13[/latex] |
| [latex]-2,15[/latex] | 13 |
| [latex]3,-10[/latex] | [latex]-7[/latex] |
| [latex]-3,10[/latex] | 7 |
[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].Try It 3
Factor the following.a.[latex]2{x}^{2}+9x+9[/latex] b. [latex]6{x}^{2}+x - 1[/latex]
SolutionLicenses & Attributions
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- College Algebra. Provided by: OpenStax Authored by: OpenStax College Algebra. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
