12.1.1: Greatest Common Factor

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Dec 15, 2024
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- 12.1: Introduction to Factoring
- 12.2: Factoring Polynomials

The NROC Project

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Learning Objectives

Find the greatest common factor (GCF) of monomials.
Factor polynomials by factoring out the greatest common factor (GCF).
Factor expressions with four terms by grouping.

Introduction

Factors are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: 2 and 10 are factors of 20, as are 4 and 5 and 1 and 20. To factor a number is to rewrite it as a product. $\ 20=4 \cdot 5$ .

Likewise to factor a polynomial, you rewrite it as a product. Just as any integer can be written as the product of factors, so too can any monomial or polynomial be expressed as a product of factors. Factoring is very helpful in simplifying and solving equations using polynomials.

A prime factor is similar to a prime number: it has only itself and 1 as factors. The process of breaking a number down into its prime factors is called prime factorization.

Greatest Common Factor

Let’s first find the greatest common factor (GCF) of two whole numbers. The GCF of two numbers is the greatest number that is a factor of both of the numbers. Take the numbers 50 and 30.

$\ \begin{array}{l} 50=10 \cdot 5 \\ 30=10 \cdot 3 \end{array}$

Their greatest common factor is 10, since 10 is the greatest factor that both numbers have in common.

To find the GCF of greater numbers, you can factor each number to find their prime factors, identify the prime factors they have in common, and then multiply those together.

Example

Find the greatest common factor of 210 and 168.

Solution

$\ \begin{array}{l} 210=\mathbf{2} \cdot \mathbf{3} \cdot 5 \cdot \mathbf{7} \\ 168=\mathbf{2} \cdot 2 \cdot 2 \cdot \mathbf{3} \cdot \mathbf{7} \\ \mathrm{GCF}=\mathbf{2} \cdot \mathbf{3} \cdot \mathbf{7} \end{array}$

$\ \mathrm{GCF}=42$

Because the GCF is the product of the prime factors that these numbers have in common, you know that it is a factor of both numbers. (If you want to test this, go ahead and divide both 210 and 168 by 42: they are both evenly divisible by this number!)

Finding the greatest common factor in a set of monomials is not very different from finding the GCF of two whole numbers. The method remains the same: factor each monomial independently, look for common factors, and then multiply them to get the GCF.

Example

Find the greatest common factor of $\ 25 b^{3}$ and $\ 10 b^{2}$ .

Solution

$\ \begin{array}{l} 25 b^{3}=\mathbf{5} \cdot 5 \cdot \mathbf{b} \cdot \mathbf{b} \cdot b \\ 10 b^{2}=\mathbf{5} \cdot 2 \cdot \mathbf{b} \cdot \mathbf{b} \\ \mathrm{GCF}=\mathbf{5} \cdot \mathbf{b} \cdot \mathbf{b} \end{array}$

$\ \mathrm{GCF}=5 b^{2}$

The monomials have the factors 5, $\ b$ , and $\ b$ in common, which means their greatest common factor is $\ 5 \cdot b \cdot b$ , or simply $\ 5 b^{2}$ .

Example

Find the greatest common factor of $\ 81 c^{3} d$ and $\ 45 c^{2} d^{2}$ .

Solution

$\ \begin{array}{l} 81 c^{3} d=\mathbf{3} \cdot \mathbf{3} \cdot 3 \cdot 3 \cdot \mathbf{c} \cdot \mathbf{c} \cdot c \cdot d \\ 45 c^{2} d^{2}=\mathbf{3} \cdot \mathbf{3} \cdot 5 \cdot \mathbf{c} \cdot \mathbf{c} \cdot \mathbf{d} \cdot d \\ \mathrm{GCF}=\mathbf{3} \cdot \mathbf{3} \cdot \mathbf{c} \cdot \mathbf{c} \cdot \mathbf{d} \end{array}$

$\ \mathrm{GCF}=9 c^{2} d$

Exercise

Find the greatest common factor of $\ 56xy$ and $\ 16 y^{3}$ .

$\ 8$
$\ 8y$
$\ 16y$
$\ 8 x y^{3}$

Answer

Incorrect. 8 is a common factor, but you need to account for the variable terms that the two monomials have in common as well. The correct answer is $\ 8y$ .
Correct. The expression $\ 56 x y$ can be factored as $\ 2 \cdot 2 \cdot 2 \cdot 7 \cdot x \cdot y$ , and $\ 16 y^{3}$ can be factored as $\ 2 \cdot 2 \cdot 2 \cdot 2 \cdot y \cdot y \cdot y$ . They have the factors $\ 2 \cdot 2 \cdot 2$ and $\ y$ in common. Multiplying them together will give you the GCF: $\ 8y$ .
Incorrect. $\ y$ is a common factor, but 16 is not: 56 is not evenly divisible by 16. Think of numbers that are factors of both 56 and 16. The correct answer is $\ 8y$ .
Incorrect. 8 is a common factor, but you need to account for the variable terms that the two monomials have in common as well. The entire term $\ x y^{3}$ is not a factor of either monomial. The correct answer is $\ 8y$ .

Using the GCF to Factor Polynomials

When two or more monomials are combined (either added or subtracted), the resulting expression is called a polynomial. If you can find common factors for each term of the polynomial, then you can factor the polynomial.

As you look at the examples of simple polynomials below, try to identify factors that the terms of the polynomial have in common.

Polynomial	Terms	Common Factors
$\ 6 x+9$	$\ 6 x \text { and } 9$	$\ 3 \text { is a factor of } 6 x \text { and } 9$
$\ a^{2}-2 a$	$\ a^{2} \text { and } 2 a$	$\ a \text { is a factor of } a^{2} \text { and } 2 a$
$\ 4 c^{3}+4 c$	$\ 4 c^{3} \text { and } 4 c$	$\ 4 \text { and } c \text { are factors of } 4 c^{3} \text { and } 4 c$

To factor a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property 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.

Product of a number and a sum: $\ a(b+c)=a \cdot b+a \cdot c$ . You can say that “ $\ a$ is being distributed over $\ b+c$ .”)

Sum of the products: $\ a \cdot b+a \cdot c=a(b+c)$ . Here you can say that “ $\ a$ is being factored out.”

In both cases, it is the distributive property that is being used.

Example

Factor $\ 25 b^{3}+10 b^{2}$ .

Solution

$\ \begin{array}{r} 25 b^{3}=\mathbf{5} \cdot 5 \cdot \mathbf{b} \cdot \mathbf{b} \cdot b \\ 10 b^{2}=\mathbf{5} \cdot 2 \cdot \mathbf{b} \cdot \mathbf{b} \end{array}$	Find the GCF. From a previous example, you found the GCF of $\ 25 b^{3}$ and $\ 10 b^{2}$ to be $\ 5 b^{2}$ .
$\ \begin{array}{r} \mathrm{GCF}=\mathbf{5} \cdot \mathbf{b} \cdot \mathbf{b}=5 b^{2} \\ 25 b^{3}=5 b^{2} \cdot 5 b \\ 10 b^{2}=5 b^{2} \cdot 2 \end{array}$	Rewrite each term with the GCF as one factor.
$\ 5 b^{2} \cdot 5 b+5 b^{2} \cdot 2$	Rewrite the polynomial using the factored terms in place of the original terms.
$\ 5 b^{2}(5 b+2)$	Factor out the $\ 5 b^{2}$ .

$\ 5 b^{2}(5 b+2)$

The factored form of the polynomial $\ 25 b^{3}+10 b^{2}$ is $\ 5 b^{2}(5 b+2)$ . You can check this by doing the multiplication. $\ 5 b^{2}(5 b+2)=25 b^{3}+10 b^{2}$ .

Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.

For example:

Factor out 5: $\ 25 b^{3}+10 b^{2}=5\left(5 b^{3}+2 b^{2}\right)$

Then factor out $\ b^{2}$ : $\ 5 b^{2}(5 b+2)$

Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.

Example

Factor $\ 81 c^{3} d+45 c^{2} d^{2}$ .

Solution

$\ \mathbf{3} \cdot \mathbf{3} \cdot 9 \cdot \mathbf{c} \cdot \mathbf{c} \cdot c \cdot \mathbf{d}$	Factor $\ 81 c^{3} d$ .
$\ \mathbf{3} \cdot \mathbf{3} \cdot 5 \cdot \mathbf{c} \cdot \mathbf{c} \cdot \mathbf{d} \cdot d$	Factor $\ 45 c^{2} d^{2}$ .
$\ \mathbf{3} \cdot \mathbf{3} \cdot \mathbf{c} \cdot \mathbf{c} \cdot \mathbf{d}=9 c^{2} d$	Find the GCF.
$\ \begin{aligned} 81 c^{3} d &=9 c^{2} d \cdot 9 c \\ 45 c^{2} d^{2} &=9 c^{2} d \cdot 5 d \end{aligned}$	Rewrite each term as the product of the GCF and the remaining terms.
$\ 9 c^{2} d \cdot 9 c+9 c^{2} d \cdot 5 d$	Rewrite the polynomial expression using the factored terms in place of the original terms.
$\ 9 c^{2} d(9 c+5 d)$	Factor out $\ 9 c^{2} d$ .

$\ 9 c^{2} d(9 c+5 d)$

Exercise

Factor $\ 8 a^{6}-11 a^{5}$ .

$\ 88\left(a^{6}-a^{5}\right)$
$\ 8 a\left(a^{5}-3\right)$
$\ a^{5}(a-1)$
$\ a^{5}(8 a-11)$

Answer

Incorrect. 88 is the least common multiple, not the greatest common factor, of 11 and 8. If $\ 88\left(a^{6}-a^{5}\right)$ were expanded it would become $\ 88 a^{6}-88 a^{5}$ , not $\ 8 a^{6}-11 a^{5}$ . The correct answer is $\ a^{5}(8 a-11)$ .
Incorrect. 8 is not a common factor of 8 and 11. If $\ 8 a\left(a^{5}-3\right)$ were expanded, it would become $\ 8 a^{6}-24 a$ , not $\ 8 a^{6}-11 a^{5}$ . The correct answer is $\ a^{5}(8 a-11)$ .
Incorrect. $\ a^{5}$ is a common factor, but the values 8 and 11 have been left out of this factorization. If $\ a^{5}(a-1)$ were expanded it would become $\ a^{6}-a^{5}$ , not $\ 8 a^{6}-11 a^{5}$ . The correct answer is $\ a^{5}(8 a-11)$ .
Correct. The values 8 and 11 share no common factors, but the GCF of $\ a^{6}$ and $\ a^{5}$ is $\ a^{5}$ . So you can factor out $\ a^{5}$ and rewrite the polynomial as $\ a^{5}(8 a-11)$ .

Factoring by Grouping

The distributive property allows you to factor out common factors. However, what do you do if the terms within the polynomial do not share any common factors?

If there is no common factor for all of the terms in the polynomial, another technique needs to be used to see if the polynomial can be factored. It involves organizing the polynomial in groups.

Example

Factor $\ 4 a b+12 a+3 b+9$

Solution

$\ (4 a b+12 a)+(3 b+9)$	Group terms into pairs.
$\ \begin{array}{r} 4 a b=\mathbf{2} \cdot \mathbf{2} \cdot \mathbf{a} \cdot b \\ 12 a=3 \cdot \mathbf{2} \cdot \mathbf{2} \cdot \mathbf{a} \\ \quad \text { GCF }=4 a \end{array}$	Find the GCF of the first pair of terms.
$\ \begin{array}{r} (4 a \cdot b+4 a \cdot 3)+(3 b+9) \\ 4 a(b+3)+(3 b+9) \end{array}$	Factor the GCF, $\ 4a$ , out of the first group.
$\ \begin{array}{r} 3 b=\mathbf{3} \cdot b \\ 9=\mathbf{3} \cdot 3 \\ \text { GCF }=3 \end{array}$	Find the GCF of the second pair of terms.
$\ \begin{array}{r} 4 a(b+3)+(3 \cdot b+3 \cdot 3) \\ 4 a(b+3)+3(b+3) \end{array}$	Factor 3 out of the second group.
$\ 4 a(b+3)+3(b+3)$	Notice that the two terms have a common factor, $\ (b+3)$ .
$\ (b+3)(4 a+3)$	Factor out the common factor $\ (b+3)$ from the two terms.

$\ (b+3)(4 a+3)$

Notice that when you factor two terms, the result is a monomial times a polynomial. But the factored form of a four-term polynomial is the product of two binomials.

This process is called the grouping technique. Broken down into individual steps, here's how to do it (you can also follow this process in the example below).

Group the terms of the polynomial into pairs.
Factor each pair of terms (find the greatest common factor and then use the distributive property to pull out the GCF).
Look for common factors between the factored forms of the paired terms.
Factor the common polynomial out of the groups.

Let’s try factoring a few more four-term polynomials. Notice that in the example below, the first term is $\ x^{2}$ , and $\ x$ is the only variable present.

Example

Factor $\ x^{2}+2 x+5 x+10$ .

Solution

$\ \left(x^{2}+2 x\right)+(5 x+10)$	Group terms of the polynomial into pairs.
$\ x(x+2)+(5 x+10)$	Factor out the like factor, $\ x$ , from the first group.
$\ x(x+2)+5(x+2)$	Factor out the like factor, $\ 5$ , from the second group.
$\ (x+2)(x+5)$	Look for common factors between the factored forms of the paired terms. Here, the common factor is $\ (x+2)$ . Factor out the common factor, $\ (x+2)$ , from both terms. The polynomial is now factored.

$\ (x+2)(x+5)$

This method of factoring only works in some cases. Notice that both factors here contain the term.

Example

Factor $\ 2 x^{2}-3 x+8 x-12$ .

Solution

$\ \left(2 x^{2}-3 x\right)+(8 x-12)$	Group terms into pairs.
$\ x(2 x-3)+4(2 x-3)$	Factor the common factor, $\ x$ , out of the first group and the common factor, 4, out of the second group.
$\ (x+4)(2 x-3)$	Factor out the common factor, $\ (2 x-3)$ , from both terms.

$\ (x+4)(2 x-3)$

Example

Factor $\ 3 x^{2}+3 x-2 x-2$ .

Solution

$\ \left(3 x^{2}+3 x\right)+(-2 x-2)$	Group terms into pairs. Since subtraction is the same as addition of the opposite, you can write $\ \color{green}-2 x-2 \text{ as } +(-2x-2)$
$\ 3 x(x+1)+(-2 x-2)$	Factor the common factor $\ 3x$ out of first group.
$\ 3 x(x+1)-2(x+1)$	Factor the common factor of -2 out of the second group. Notice what happens to the signs within the parentheses once -2 is factored out.
$\ (x+1)(3 x-2)$	Factor out the common factor, $\ (x+1)$ , from both terms.

$\ (x+1)(3 x-2)$

Exercise

Factor $\ 10 a b+5 b+8 a+4$ .

$\ (2 a+1)(5 b+4)$
$\ (5 b+2 a)(4+1)$
$\ 5(2 a b+b+8 a+4)$
$\ (4+2 a)(5 b+1)$

Answer

Correct. The polynomial $\ 10 a b+5 b+8 a+4$ can be grouped as $\ (10 a b+5 b)+(8 a+4)$ . Pulling out common factors, you find: $\ 5 b(2 a+1)+4(2 a+1)$ . Since $\ (2 a+1)$ is a common factor, the factored form is $\ (2 a+1)(5 b+4)$ .
Incorrect. When factoring $\ 5 b$ out of $\ 10 a b$ and $\ 5 b$ , the remaining $\ 2 a$ and 1 must still be added and multiplied by the common factor $\ 5 b$ : $\ 5 b(2 a+1)$ . Similarly, factoring out the 4 from $\ 8 a+4$ leaves $\ 4(2 a+1)$ . Then you can factor $\ (2 a+1)$ from the sum of those expressions to get the correct factorization, $\ (2 a+1)(5 b+4)$ .
Incorrect. The 5 is a common factor only for $\ 10 a b+5 b$ , giving $\ 5 b(2 a+1)$ . The other pair, $\ 8 a+4$ , has a common factor of 4. Factoring them gives $\ 4(2 a+1)$ . Since both expressions have a common factor of $\ 2 a+1$ , you can factor again to get $\ (2 a+1)(5 b+4)$ .
Incorrect. You correctly identified $\ 5 b$ as a factor of one pair, leaving $\ 2 a$ and 1, and 4 as the factor of the other pair, also leaving $\ 2a$ and 1. However, this gives $\ 5 b(2 a+1)+4(2 a+1)$ . If you had $\ 5 b x+4 x$ , you could factor out the $\ x$ to get $\ x(5 b+4)$ , so factoring out the $\ (2 a+1)$ gives $\ (2 a+1)(5 b+4)$ .

Sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.

Example

Factor $\ 7 x^{2}-21 x+5 x-5$ .

Solution

$\ \left(7 x^{2}-21 x\right)+(5 x-5)$	Group terms into pairs.
$\ 7 x(x-3)+(5 x-5)$	Factor the common factor $\ 7x$ out of the first group.
$\ 7 x(x-3)+5(x-1)$	Factor the common factor 5 out of the second group.
$\ 7 x(x-3)+5(x-1)$	The two groups $\ 7 x(x-3)$ and $\ 5(x-1)$ do not have any common factors, so this polynomial cannot be factored any further.

Cannot be factored

In the example above, each pair can be factored, but then there is no common factor between the pairs!

Summary

A whole number, monomial, or polynomial can be expressed as a product of factors. You can use some of the same logic that you apply to factoring integers to factoring polynomials. To factor a polynomial, first identify the greatest common factor of the terms, and then apply the distributive property to rewrite the expression. Once a polynomial in $\ a \cdot b+a \cdot c$ form has been rewritten as $\ a(b+c)$ , where $\ a$ is the GCF, the polynomial is in factored form.

When factoring a four-term polynomial using grouping, find the common factor of pairs of terms rather than the whole polynomial. Use the distributive property to rewrite the grouped terms as the common factor times a binomial. Finally, pull any common binomials out of the factored groups. The fully factored polynomial will be the product of two binomials.

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