Factoring Basics
Learning Objectives
- Identify and factor the GCF of a polynomial
- Factor a Trinomial with Leading Coefficient 1
- Factor by Grouping
When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, [latex]4[/latex] is the GCF of [latex]16[/latex] and [latex]20[/latex] because it is the largest number that divides evenly into both [latex]16[/latex] and [latex]20[/latex] The GCF of polynomials works the same way: [latex]4x[/latex] is the GCF of [latex]16x[/latex] and [latex]20{x}^{2}[/latex] because it is the largest polynomial that divides evenly into both [latex]16x[/latex] and [latex]20{x}^{2}[/latex].
When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.
A General Note: Greatest Common Factor
The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.How To: Given a polynomial expression, factor out the greatest common factor.
- Identify the GCF of the coefficients.
- Identify the GCF of the variables.
- Combine to find the GCF of the expression.
- Determine what the GCF needs to be multiplied by to obtain each term in the expression.
- Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.
Example: Factoring the Greatest Common Factor
Factor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[/latex].Answer: First, find the GCF of the expression. The GCF of [latex]6,45[/latex], and [latex]21[/latex] is [latex]3[/latex]. The GCF of [latex]{x}^{3},{x}^{2}[/latex], and [latex]x[/latex] is [latex]x[/latex]. (Note that the GCF of a set of expressions in the form [latex]{x}^{n}[/latex] will always be the exponent of lowest degree.) And the GCF of [latex]{y}^{3},{y}^{2}[/latex], and [latex]y[/latex] is [latex]y[/latex]. Combine these to find the GCF of the polynomial, [latex]3xy[/latex]. Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that [latex]3xy\left(2{x}^{2}{y}^{2}\right)=6{x}^{3}{y}^{3},3xy\left(15xy\right)=45{x}^{2}{y}^{2}[/latex], and [latex]3xy\left(7\right)=21xy[/latex]. Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.
[latex]\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)[/latex]
Analysis of the Solution
After factoring, we can check our work by multiplying. Use the distributive property to confirm that [latex]\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[/latex].Try It
Factor [latex]x\left({b}^{2}-a\right)+6\left({b}^{2}-a\right)[/latex] by pulling out the GCF.Answer: [latex]\left({b}^{2}-a\right)\left(x+6\right)[/latex]
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: Factoring a Trinomial by Grouping
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] | 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
Factor the following.- [latex]2{x}^{2}+9x+9[/latex]
- [latex]6{x}^{2}+x - 1[/latex]
Answer:
- [latex]\left(2x+3\right)\left(x+3\right)[/latex]
- [latex]\left(3x - 1\right)\left(2x+1\right)[/latex]
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- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
- Example: Greatest Common Factor. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Factoring Trinomials by Grouping. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Question ID 7886, 7897, 7908. Authored by: Tyler Wallace. License: CC BY: Attribution. License terms: IMathAS Community License CC- BY + GPL.
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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.
