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7.1: Greatest common factor and grouping

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In this lesson, we focus on factoring using the greatest common factor, GCF, of a polynomial. When we multiplied polynomials, we multiplied monomials by polynomials by distributing, such as

4x2(2x23x+8)=8x412x3+32x2

We work out the same problem, but backwards. We will start with 8x212x3+32x2 and obtain its factored form.

First, we have to identify the GCF of a polynomial. We introduce the GCF of a polynomial by looking at an example in arithmetic. The method in which we obtained the GCF between numbers in arithmetic is the same method we use to obtain the GCF with polynomials.

Finding the Greatest Common Factor

Definition: Greatest Common Factor

The factored form of a number or expression is the expression written as a product of factors.

The greatest common factor (GCF) of a polynomial is the largest polynomial that is a factor of all terms in the polynomial.

Example 7.1.1

Find the GCF of 15, 24, and 27.

Solution

First we obtain the prime factorization of each number:

15=3524=23327=33

Next, we take only the common factors and if any common factors repeat, we take the factor with the smallest exponent. Recall, the GCF is the largest factor that divides into all numbers. We need to take 3 (we only take 31 because there is only one three in common in all three numbers). Notice, there are no other factors in common with all three numbers. Hence, GCF(15,24,27)=3.

When finding the GCF of a polynomial, we use the same method and the only difference is that we have variables in the expression.

Example 7.1.2

Find the GCF of 24x4y2z, 18x2y4, and 12x3yz5

Solution

First we obtain the prime factorization of each monomial:

24x4y2z=233x4y2z18x2y4=232x2y412x3yz5=223x3yz5

Next, we take only the common factors and if any common factors repeat, we take the factor with the smallest exponent. Recall, the GCF is the largest factor that divides into all terms in the expression. We need to take 23x2yz. Hence, GCF(24x4y2z,18x2y4,12x3yz5)=6x2yz.

Factoring the Greatest Common Factor

Once we obtain the GCF, we can start factoring the GCF from an expression. Eventually, we want to be able to see the GCF quickly and begin factoring right away. Of course, only with practice we obtain better skills. Let’s try an example.

Example 7.1.3

Factor out the GCF: 4x220x+16

Solution

Looking at each term, let’s write the prime factorization of each term:

4x2=22x220x=225x16=24

We need to take 22. Hence, GCF(4x2,20x,16)=22=4. Let’s rewrite each term in the expression as the product of the GCF and the factors left:

4x220x+16Rewrite with the GCF 44x245x+44Rewrite the expression with the GCF and parenthesis4(x25x+4)Factored form

Steps for factoring out the greatest common factor

Step 1. Find the GCF of the expression.

Step 2. Rewrite each term as a product of the GCF and the remaining factors.

Step 3. Rewrite as a product of the GCF and the remaining factors in parenthesis.

Step 4. Verify the factored form by multiplying. The product should be the original expression.

Example 7.1.4

Factor out the GCF: 25x415x3+20x2

Solution

Looking at each term, let’s start by writing the prime factorization of each term.

25x4=52x415x3=35x320x2=225x2

Step 1. We need to take 5x2. Hence, GCF(25x4,15x3,20x2)=5x2.

Step 2. Let’s rewrite each term in the expression as the product of the GCF and the factors left: 25x415x3+20x2Rewrite with the GCF 5x25x25x25x23x+5x24

Step 3. Rewrite the expression with the GCF and the remaining factors in parenthesis: 5x25x25x23x+5x24Rewrite the expression with the GCF and parenthesis5x2(5x23x+4)Factored form

Step 4. Let’s verify the factored form: 5x2(5x23x+4)Distribute the GCF25x415x3+20x2 Original expression

Thus, the factored form is 5x2(5x23x+4).

Example 7.1.5

Factor out the GCF: 3x3y2z+5x4y3z54xy4

Solution

Looking at each term, let’s start by writing the prime factorization of each term.

3x3y2z=3x3y2z5x4y3z5=5x4y3z54xy4=22xy4

Step 1. We need to take xy2. Hence, GCF(3x3y2z,5x4y3z5,4xy4)=xy2.

Step 2. Let’s rewrite each term in the expression as the product of the GCF and the factors left: 3x3y2z+5x4y3z54xy4Rewrite with the GCF xy2xy23x2z+xy25x3yz5xy24y2

Step 3. Rewrite the expression with the GCF and the remaining factors in parenthesis: xy23x2z+xy25x3yz5xy24y2Rewrite the expression with the GCF and parenthesisxy2(3x2z+5x2yz54y2)Factored form

Step 4. Let’s verify the factored form: xy2(3x2z+5x2yz54y2)Distribute the GCF3x3y2z+5x4y3z54xy4 Original expression

Thus, the factored form is xy2(3x2z+5x2yz54y2).

Note

The first recorded algorithm for finding the greatest common factor comes from Greek mathematician Euclid around the year 300 BC.

Let’s try an example with eye-balling the GCF and rewriting in factored form.

Example 7.1.6

Factor out the GCF: 21x3+14x2+7x

Solution

Looking at the coefficients, we can see that there is common factor of 7 in each term. Furthermore, we see a factor of x in common in all three terms. Hence, we take 7x as the GCF. Notice we didn’t take a larger exponent on x because only one factor of x is common in all three terms.

Let's rewrite the expression in factored form.

21x3+14x2+7xRewrite with the GCF 7x7x3x2+7x2x+7x1Rewrite the expression with the GCF and parenthesis7x(3x2+2x+1)Factored form

We can always verify the factored form by distributing the 7x and obtaining the original expression.

A Binomial as the Greatest Common Factor

As part of a general strategy for factoring, we always look for a GCF. Sometimes the GCF is a monomial, like in the previous examples, or a binomial. Here we discuss factoring a polynomial where the GCF is a binomial. We usually call this factor by grouping. Consider the below example.

Example 7.1.7

Factor: 3ax7bx

Solution

3ax7bxBoth have x in common, factor it outx(3a7b)Factored form

Let’s make this interesting. Let’s apply this same method, but instead of the GCF being x, let the GCF be the binomial 2a+5b.

Example 7.1.8

Find: 3a(2a+5b)7b(2a+5b)

Solution

3a(2a+5b)7b(2a+5b)Both have (2a+5b) in common, factor it out(2a+5b)(3a7b)Factored form

Factor by Grouping

In Example 7.1.8 , we factored out a GCF of (2a+5b) the same way we factored out an x in Example 7.1.7 . This process can be extended to factor expressions where there isn’t a GCF. We will use a process known as factor by grouping. Factor by grouping is a method used to factor polynomials when there is at least four terms in the expression. Take the next example.

Example 7.1.9

Multiply: (2a+3)(5b+2)

Solution

(2a+3)(5b+2)Distribute (2a+3) into second parenthesis5b(2a+3)+2(2a+3)Distribute10ab+15b+4a+6Product

Notice the product has four terms none of which share a common factor.

Steps for factoring by grouping

To factor by grouping, we first notice the polynomial expression obtains four terms.

Step 1. Group two sets of two terms, e.g., ax+ay+bx+by=(ax+ay)+(bx+by).

Step 2. Factor the GCF from each group, e.g., a(x+y)+b(x+y)

Step 3. Factor the GCF from the expression, e.g., (x+y)(a+b).

Example 7.1.10

Factor: 10ab+15b+4a+6

Solution

Notice we have 4 terms none of which share a common factor. Hence, we use factor by grouping.

Step 1. Group two sets of two terms: 10ab+15b+4a+6Group the first two terms and the last two terms(10ab+15b)+(4a+6)

Step 2. Factor the GCF from each group: (10ab+15b)+(4a+6)Factor 5b from the first group and 2 from the second group5b(2a+3)+2(2a+3)

Step 3. Factor the GCF from the expression: 5b(2a+3)+2(2a+3)Factor the GCF (2a+3)(2a+3)(5b+2)Factored form

Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.

Example 7.1.11

Factor: 6x2+9xy14x21y

Solution

Notice we have 4 terms none of which share a common factor. Hence, we use factor by grouping.

Step 1. Group two sets of two terms: 6x2+9xy14x21yGroup the first two terms and the last two terms(6x2+9xy)+(14x21y)

Step 2. Factor the GCF from each group. (6x2+9xy)+(14x21y)Factor 3x from the first group and 7from the second group3x(2x+3y)7(2x+3y)

Step 3. Factor the GCF from the expression: 3x(2x+3y)7(2x+3y)Factor the GCF (2x+3y)(2x+3y)(3x7)Factored form

Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.

Note

Notice after Step 2., we want the binomial’s GCFto be identical so that we can factor it out in Step 3. Be sure these binomials are identical. A common error is when the binomials aren’t identical (sometimes by a negative) and students factor anyways.

Example 7.1.12

Factor: 5xy8x10y+16

Solution

Notice we have 4 terms none of which share a common factor. Hence, we use factor by grouping.

Step 1. Group two sets of two terms: 5xy8x10y+16Group the first two terms and the last two terms(5xy8x)+(10y+16)

Step 2. Factor the GCF from each group: (5xy8x)+(10y+16)Factor x from the first group and 2from the second groupx(5y8)2(5y8)Both binomials are identical

Step 3. Factor the GCF from the expression: x(5y8)2(5y8)Factor the GCF (5y8)(5y8)(x2)Factored form

Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.

Example 7.1.13

Factor: 12ab14a6b+7

Solution

Notice we have 4 terms none of which share a common factor. Hence, we use factor by grouping.

Step 1. Group two sets of two terms: 12ab14a6b+7Group the first two terms and the last two terms(12ab14a)+(6b+7)

Step 2. Factor the GCF from each group: (12ab14a)+(6b+7)Factor 2a from the first group and 1from the second group2a(6b7)1(6b7)Both binomials are identical

Step 3. Factor the GCF from the expression: 2a(6b7)1(6b7)Factor the GCF (6b7)(6b7)(2a1)Factored form

Careful in these types of expressions, where we factor the entire second binomial and are left with the term 1. This occurs sometimes with factoring and it’s important to always write the 1 in Step 2. so that we do not forget it is there. Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.

Factor by Grouping by Rearranging Terms

Sometimes after completing Step 2., the binomials aren’t identical (by more than a negative sign). At this point we must return to the original problem and rearrange the terms so that when we factor by grouping, we obtain identical binomials in Step 2.

Example 7.1.14

Factor: 4a221b3+6ab14ab2

Solution

Notice we have 4 terms none of which share a common factor. Hence, we use factor by grouping.

Step 1. Group two sets of two terms: 4a221b3+6ab14ab2Group the first two terms and the last two terms(4a221b3)+(6ab14ab2)

Step 2. Factor the GCF from each group: (4a221b3)+(6ab14ab2)Factor 2ab from the second group(4a221b3)+2a(3b7b2)Binomials are NOT identical

Since these binomials aren’t identical, we return to the original expression and rearrange the terms. Let’s try moving 6ab to the first group and 21b3 to the second group.

Step 1. Group two sets of two terms: 4a2+6ab21b314ab2Group the first two terms and the last two terms(4a2+6ab)+(21b314ab2)

Step 2. Factor the GCF from each group: (4a2+6ab)+(21b314ab2)Factor 2a from the first group and 7b2from the second group2a(2a+3b)7b2(3b+2a)Rewrite so the binomials are identical2a(2a+3b)7b2(2a+3b)Binomials are identical

Step 3. Factor the GCF from the expression: 2a(2a+3b)7b2(2a+3b)Factor the GCF (2a+3b)(2a+3b)(2a7b2)Factored form

Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.

Example 7.1.15

Factor: 8xy12y+1510x

Solution

Notice we have 4 terms none of which share a common factor. Hence, we use factor by grouping.

Step 1. Group two sets of two terms: 8xy12y+1510xGroup the first two terms and the last two terms(8xy12y)+(1510x)

Step 2. Factor the GCF from each group: (8xy12y)+(1510x)Factor 4y from the first group and 5from the second group4y(2x3)+5(32x)Binomials are NOT identical, but VERY close

Since these binomials aren’t identical but close to it, we can think about it some more. These binomials would be identical if only the 3 and 2x in the second binomial were switched. Let’s factor a 1 out of the second binomial: 4y(2x3)+5(32x)Factor a 1 from the second binomial4y(2x3)+51(3+2x)Rewrite so the binomials are identical4y(2x3)5(2x3)Binomials are identical

Step 3. Factor the GCF from the expression: 4y(2x3)5(2x3)Factor the GCF (2x3)(2x3)(4y5)Factored form

Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.

Note

If the binomials are (ab) and (ba), we will factor 1 out of one of the binomials to obtain identical binomials.

(ba)Factor out 11(b+a)Apply the commutative property to the addition1(ab)Now the binomial is written as subtraction

Note

Sofia Kovalevskaya of Russia was the first woman on the editorial staff of a mathematical journal in the late 19th century. She also researched the planet Saturn’s rotating rings.

Greatest Common Factor and Grouping Homework

Factor the greatest common factor.

Exercise 7.1.1

9+8b2

Exercise 7.1.2

45x225

Exercise 7.1.3

5635p

Exercise 7.1.4

7ab35a2b

Exercise 7.1.5

3a2b+6a3b2

Exercise 7.1.6

5x25x315x4

Exercise 7.1.7

20x430x+30

Exercise 7.1.8

28m4+40m3+8

Exercise 7.1.9

30b9+5ab15a2

Exercise 7.1.10

48a2b256a3b56a5b

Exercise 7.1.11

20x8y2z2+15x5y2z+35x3y3z

Exercise 7.1.12

50x2y+10y2+70xz2

Exercise 7.1.13

30qpr5qp+5q

Exercise 7.1.14

18n5+3n321n+3

Exercise 7.1.15

40x1120x12+50x1350x14

Exercise 7.1.16

32mn8+4m6n+12mn4+16mn

Exercise 7.1.17

x5

Exercise 7.1.18

1+2n2

Exercise 7.1.19

50x80y

Exercise 7.1.20

27x2y572x3y2

Exercise 7.1.21

8x3y2+4x3

Exercise 7.1.22

32n9+32n6+40n5

Exercise 7.1.23

21p6+30p2+27

Exercise 7.1.24

10x4+20x2+12x

Exercise 7.1.25

27y7+12y2x+9y2

Exercise 7.1.26

30m6+12mn225

Exercise 7.1.27

3p+12q15q2r2

Exercise 7.1.28

30y4z3x5+50y4z510y4z3x

Exercise 7.1.29

28b+14b2+35b3+7b5

Exercise 7.1.30

30a8+6a5+27a3+21a2

Exercise 7.1.31

24x64x4+12x3+4x2

Exercise 7.1.32

10y7+6y104y10x8y8x

Factor each completely.

Exercise 7.1.33

40r38r225r+5

Exercise 7.1.34

3n32n29n+6

Exercise 7.1.35

15b3+21b235b49

Exercise 7.1.36

3x3+15x2+2x+10

Exercise 7.1.37

35x328x220x+16

Exercise 7.1.38

7xy49x+5y35

Exercise 7.1.39

32xy+40x2+12y+15x

Exercise 7.1.40

16xy56x+2y7

Exercise 7.1.41

2xy8x2+7y328y2x

Exercise 7.1.42

40xy+35x8y27y

Exercise 7.1.43

32uv20u+24v15

Exercise 7.1.44

10xy+30+25x+12y

Exercise 7.1.45

3uv+14u6u27v

Exercise 7.1.46

16xy3x6x2+8y

Exercise 7.1.47

35x310x256x+16

Exercise 7.1.48

14v3+10v27v5

Exercise 7.1.49

6x348x2+5x40

Exercise 7.1.50

28p3+21p2+20p+15

Exercise 7.1.51

7n3+21n25n15

Exercise 7.1.52

42r349r2+18r21

Exercise 7.1.53

15ab6a+5b32b2

Exercise 7.1.54

3mn8m+15n40

Exercise 7.1.55

5mn+2m25n10

Exercise 7.1.56

8xy+56xy7

Exercise 7.1.57

4uv+14u2+12v+42u

Exercise 7.1.58

24xy+25y220x30y3

Exercise 7.1.59

56ab+1449a16b


This page titled 7.1: Greatest common factor and grouping is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz (ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts platform.

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