Factor Trinomials
9.1 Learning Objectives
- Greatest common factor
- Identify the difference between a factor and the act of factoring
- Identify the greatest common factor of a polynomial
- Factor the greatest common factor out of a polynomial
- Factor a Trinomial with Leading Coefficient = 1
- Identify a trinomial
- Identify the leading coefficient of a trinomial
- Use a method to factor a trinomial with a leading coefficient of 1
- Factor by Grouping
- Factor a trinomial with leading coefficient other than 1 using grouping
- Recognize when a trinomial cannot be factored
Greatest Common Factor
The greatest common factor (GCF) of a group of given polynomials is the largest polynomial that divides evenly into the polynomials.Example 9.1.A
Find the greatest common factor of [latex]25b^{3}[/latex] and [latex]10b^{2}[/latex].Answer:
[latex]\begin{array}{l}\,\,25b^{3}=5\cdot5\cdot{b}\cdot{b}\cdot{b}\\\,\,10b^{2}=5\cdot2\cdot{b}\cdot{b}\\\text{GCF}=5\cdot{b}\cdot{b}\end{array}[/latex]
Answer
[latex-display]\text{GCF}=5b^{2}[/latex-display]Example 9.1.B
Find the greatest common factor of [latex]81c^{3}d[/latex] and [latex]45c^{2}d^{2}[/latex].Answer:
[latex]\begin{array}{l}\,\,\,81c^{3}d=3\cdot3\cdot3\cdot3\cdot{c}\cdot{c}\cdot{c}\cdot{d}\\45c^{2}d^{2}=3\cdot3\cdot5\cdot{c}\cdot{c}\cdot{d}\cdot{d}\\\,\,\,\,\text{GCF}=3\cdot3\cdot{c}\cdot{c}\cdot{d}\end{array}[/latex]
Answer
[latex-display]\text{GCF}=9c^{2}d[/latex-display]Distributive Property Forward and Backward
Forward: Product of a number and a sum: [latex]a\left(b+c\right)=a\cdot{b}+a\cdot{c}[/latex]. You can say that “[latex]a[/latex] is being distributed over [latex]b+c[/latex].” Backward: Sum of the products: [latex]a\cdot{b}+a\cdot{c}=a\left(b+c\right)[/latex]. Here you can say that “a is being factored out.” We first learned that we could distribute a factor over a sum or difference, now we are learning that we can "undo" the distributive property with factoring.Example 9.1.C
Factor [latex]25b^{3}+10b^{2}[/latex].Answer: Find the GCF. From a previous example, you found the GCF of [latex]25b^{3}[/latex] and [latex]10b^{2}[/latex] to be [latex]5b^{2}[/latex].
[latex]\begin{array}{l}\,\,25b^{3}=5\cdot5\cdot{b}\cdot{b}\cdot{b}\\\,\,10b^{2}=5\cdot2\cdot{b}\cdot{b}\\\text{GCF}=5\cdot{b}\cdot{b}=5b^{2}\end{array}[/latex]
Rewrite each term with the GCF as one factor.[latex]\begin{array}{l}25b^{3} = 5b^{2}\cdot5b\\10b^{2}=5b^{2}\cdot2\end{array}[/latex]
Rewrite the polynomial using the factored terms in place of the original terms.[latex]5b^{2}\left(5b\right)+5b^{2}\left(2\right)[/latex]
Factor out the [latex]5b^{2}[/latex].[latex]5b^{2}\left(5b+2\right)[/latex]
Answer
[latex-display]5b^{2}\left(5b+2\right)[/latex-display][latex]\begin{array}{l}25b^{3}+10b^{2}=5\left(5b^{3}+2b^{2}\right)\,\,\,\,\,\,\,\,\,\,\,\text{Factor out }5.\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=5b^{2}\left(5b+2\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Factor out }b^{2}.\end{array}[/latex]
Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually. In the following video we show two more examples of how to find and factor the GCF from binomials. https://youtu.be/25_f_mVab_4 We will show one last example of finding the GCF of a polynomial with several terms and two variables. No matter how large the polynomial, you can use the same technique described below to factor out it's GCF.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 9.1.d
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.
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]. In the following video you will see two more example of how to find and factor our the greatest common factor of a polynomial. https://youtu.be/3f1RFTIw2Ng9.1.1 Factor a Trinomial with Leading Coefficient = 1
Trinomials are polynomials with three terms. We are going to show you a method for factoring a trinomial whose leading coefficient is 1. Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that trinomials can be factored. The trinomial [latex]{x}^{2}+5x+6[/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\left(x+2\right)[/latex] and [latex]\left(x+3\right)[/latex]. Recall how to use the distributive property to multiply two binomials:[latex]\left(x+2\right)\left(x+3\right) = x^2+3x+2x+6=x^2+5x+6[/latex]
We can reverse the distributive property and return [latex]x^2+5x+6\text{ to }\left(x+2\right)\left(x+3\right) [/latex] by finding two numbers with a product of [latex]6[/latex] and a sum of [latex]5[/latex].
Factoring a Trinomial with Leading Coefficient 1
In general, for a trinomial of the form[latex]{x}^{2}+bx+c[/latex] you can factor a trinomial with leading coefficient 1 by finding two numbers,[latex]p[/latex] and [latex]q[/latex] whose product is c, and whose sum is b.Example 9.1.e
Factor [latex]{x}^{2}+2x - 15[/latex].Answer: We have a trinomial with leading coefficient [latex]1,b=2[/latex], and [latex]c=-15[/latex]. We need to find two numbers with a product of [latex]-15[/latex] and a sum of [latex]2[/latex]. In the table, we list factors until we find a pair with the desired sum.
Factors of [latex]-15[/latex] | Sum of Factors |
---|---|
[latex]1,-15[/latex] | [latex]-14[/latex] |
[latex]-1,15[/latex] | 14 |
[latex]3,-5[/latex] | [latex]-2[/latex] |
[latex]-3,5[/latex] | 2 |
How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[/latex], factor it.
- List factors of [latex]c[/latex].
- Find [latex]p[/latex] and [latex]q[/latex], a pair of factors of [latex]c[/latex] with a sum of [latex]b[/latex].
- Write the factored expression [latex]\left(x+p\right)\left(x+q\right)[/latex].
Example 9.1.f
Factor [latex]x^{2}+x–12[/latex].Answer: Consider all the combinations of numbers whose product is -12, 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] |
[latex]\left(x+4\right)\left(x-3\right)[/latex]
Answer
[latex-display]\left(x+4\right)\left(x-3\right)[/latex-display]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 numbers may be multiplied in any order without affecting the product.
Example 9.1.G
Factor [latex]{x}^{2}-7x+6[/latex].Answer: List the factors of 6. 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] |
Analysis of the solution
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 counter example to prove or disprove a question. A counter example means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient 1 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 counter-example would be: [latex]x^2+3x+7[/latex]
9.1.2 Factor 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. 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]2\cdot3=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] |
[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+2)+(3x+3)[/latex]
Identify the GCF of each binomial.
2x is the GCF of [latex](2x^2+2)[/latex] and 3 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 9.1.H
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 |
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].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.
We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.
Example 9.1.I
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]. 9 and 18 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] |
[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]Example 9.1.J
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] |
[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]Example 9.1.K
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 factoredLicenses & Attributions
CC licensed content, Original
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
- 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.
CC licensed content, Shared previously
- Ex 1: Identify GCF and Factor a Binomial. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- 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.
- Ex 2: Identify GCF and Factor a Trinomial. Authored by: James Sousa (Mathispower4u.com) . 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.
- Factor a Trinomial in the Form -ax^2+bx+c Using the Grouping Technique. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.