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Study Guides > College Algebra CoRequisite Course

Factoring Basics

Learning Outcomes

  • Identify and factor the greatest common factor of a polynomial.
  • Factor a trinomial with leading coefficient 1.
  • Factor by grouping.
Recall that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For example, [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]. Finding and factoring out a GCF from a polynomial is the first skill involved in factoring polynomials.

Factoring a GCF out of a polynomial

When factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term. To make it less challenging to find this GCF of the polynomial terms, first 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 a polynomial is the largest polynomial that divides evenly into each term of the polynomial.
To factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property "backwards" to rewrite the polynomial in a factored form. Recall that the distributive property of multiplication over addition states that a product of a number and a sum is the same as the sum of the products.

Distributive Property Forward and Backward

Forward: We distribute [latex]a[/latex] over [latex]b+c[/latex].

[latex]a\left(b+c\right)=ab+ac[/latex].

Backward: We factor [latex]a[/latex] out of [latex]ab+ac[/latex].

[latex]ab+ac=a\left(b+c\right)[/latex].

We have seen that we can distribute a factor over a sum or difference. Now we see that we can "undo" the distributive property with factoring.

Example

Factor [latex]25b^{3}+10b^{2}[/latex].

Answer: Find the GCF.

[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]

Analysis

The factored form of the polynomial [latex]25b^{3}+10b^{2}[/latex] is [latex]5b^{2}\left(5b+2\right)[/latex]. You can check this by doing the multiplication. [latex]5b^{2}\left(5b+2\right)=25b^{3}+10b^{2}[/latex]. Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over. For example:

[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 see two more examples of how to find and factor the GCF from binomials. https://youtu.be/25_f_mVab_4

How To: Given a polynomial expression, factor out the greatest common factor

  1. Identify the GCF of the coefficients.
  2. Identify the GCF of the variables.
  3. Combine to find the GCF of the expression.
  4. Determine what the GCF needs to be multiplied by to obtain each term in the expression.
  5. 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 of the form [latex]{x}^{n}[/latex] will always be the lowest exponent.) 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].

The GCF may not always be a monomial. Here is an example of a GCF that is a binomial.

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]

[ohm_question]7888[/ohm_question]
Watch this video to see more examples of how to factor the GCF from a trinomial. https://youtu.be/3f1RFTIw2Ng

Factoring a Trinomial with Leading Coefficient 1

Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [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]. Trinomials of the form [latex]{x}^{2}+bx+c[/latex] can be factored by finding two numbers with a product of [latex]c[/latex] and a sum of [latex]b[/latex]. The trinomial [latex]{x}^{2}+10x+16[/latex], for example, can be factored using the numbers [latex]2[/latex] and [latex]8[/latex] because the product of these numbers is [latex]16[/latex] and their sum is [latex]10[/latex]. The trinomial can be rewritten as the product of [latex]\left(x+2\right)[/latex] and [latex]\left(x+8\right)[/latex].

A General Note: Factoring a Trinomial with Leading Coefficient 1

A trinomial of the form [latex]{x}^{2}+bx+c[/latex] can be written in factored form as [latex]\left(x+p\right)\left(x+q\right)[/latex] where [latex]pq=c[/latex] and [latex]p+q=b[/latex].

Q & A

Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime.

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].

Example: Factoring a Trinomial with Leading Coefficient 1

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] [latex]14[/latex]
[latex]3,-5[/latex] [latex]-2[/latex]
[latex]-3,5[/latex] [latex]2[/latex]
Now that we have identified [latex]p[/latex] and [latex]q[/latex] as [latex]-3[/latex] and [latex]5[/latex], write the factored form as [latex]\left(x - 3\right)\left(x+5\right)[/latex].

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that [latex]\left(x - 3\right)\left(x+5\right)={x}^{2}+2x - 15[/latex].

Q & A

Does the order of the factors matter? No. Multiplication is commutative, so the order of the factors does not matter.

Try It

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

Answer: [latex]\left(x - 6\right)\left(x - 1\right)[/latex]

[ohm_question]7897[/ohm_question]

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: Factoring by Grouping

To factor a trinomial of 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

  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.

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] [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].

Try It

Factor the following.
  1. [latex]2{x}^{2}+9x+9[/latex]
  2. [latex]6{x}^{2}+x - 1[/latex]

Answer:

  1. [latex]\left(2x+3\right)\left(x+3\right)[/latex]
  2. [latex]\left(3x - 1\right)\left(2x+1\right)[/latex]

[ohm_question]7908[/ohm_question]
In the next video we see another example of how to factor a trinomial by grouping. [embed]https://youtu.be/agDaQ_cZnNc[/embed]

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

  • 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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