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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student's understanding of the factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be placed inside the parentheses.The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with leading coefficients different from 1. Objectives of this module: be able to factor a monomial from a polynomial.

Overview

  • The Factorization Process

The factorization process

We introduce the process of factoring a monomial from a polynomial by examining a problem: Suppose that 12 x 2 + 20 x is the product and one of the factors is 4 x . To find the other factor we could set up the problem this way:

4 x ( ) = 12 x 2 + 20 x

Since the product 12 x 2 + 20 x consists of two terms, the expression multiplying 4 x must consist of two terms, since, by the distributive property

Four x multiplied by an empty set of parentheses with curved arrows pointing from four x to the inside of the empty parentheses is equal to twelve x squared plus twenty x.

Now we see that this problem is simply an extension of finding the factors of a monomial.

1 s t t e r m : 4 x ( ) = 12 x 2 2 n d t e r m : 4 x ( ) = 20 x ( ) = 12 x 2 4 x ( ) = 20 x 4 x ( ) = 3 x ( ) = 5

Thus, 4 x ( 3 x + 5 ) = 12 x 2 + 20 x .

Usually, these divisions can be done mentally and the terms of the factor filled in directly.

Sample set a

The product is 3 x 7 2 x 6 + 4 x 5 3 x 4 and one factor is x 4 . Find the other factor.

We have the problem: x 4 times "what expression" yields 3 x 7 2 x 6 + 4 x 5 3 x 4 ? Mathematically,

x 4 ( ) = 3 x 7 2 x 6 + 4 x 5 3 x 4

Since there are four terms in the product, there must be four terms inside the parentheses. To find each of the four terms, we’ll divide (mentally) each term of the product by x 4 . The resulting quotient will be the necessary term of the factor.

1 s t t e r m : 3 x 7 x 4 = 3 x 7 4 = 3 x 3 Place 3 x 3 into the 1 st position in the ( ) . 2 n d t e r m : 2 x 6 x 4 = 2 x 2 Place 2 x 2 into the 2 nd position in the ( ) . 3 r d t e r m : 4 x 5 x 4 = 4 x Place 4 x into the 3 rd position in the ( ) . 4 t h t e r m : 3 x 4 x 4 = 3 Place 3 into the 4 th position in the ( ) .

Therefore, the other factor is 3 x 3 2 x 2 + 4 x 3 .

This result can be checked by applying the distributive property.

x 4 ( 3 x 3 2 x 2 + 4 x 3 ) = 3 x 7 2 x 6 + 4 x 5 3 x 4 Is this correct?

An equation showing the product of a and the sum of b and c equal to ab plus ac. The product on the left are identified as factors and the expression on the right of the equal sign is identified as the product.

3 x 4 + 3 2 x 4 + 2 + 4 x 4 + 1 = 3 x 7 2 x 6 + 4 x 5 3 x 4 Is this correct? 3 x 7 2 x 6 + 4 x 5 3 x 4 = 3 x 7 2 x 6 + 4 x 5 3 x 4 Yes, this is correct .

Thus,

x 4 ( 3 x 3 2 x 2 + 4 x 3 ) = 3 x 7 2 x 6 + 4 x 5 3 x 4

Again, if the divisions can be performed mentally, the process can proceed very quickly.

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The product is 10 x 3 y 6 + 15 x 3 y 4 5 x 2 y 4 and a factor is 5 x 2 y 4 . Find the other factor.

5 x 2 y 4 ( ) = 10 x 3 y 6 + 15 x 3 y 4 5 x 2 y 4

Since there are three terms in the product, there must be three terms inside the parentheses. To find each of these three terms, we’ll divide each term of the product by 5 x 2 y 4 .

1 s t t e r m : 10 x 3 y 6 5 x 2 y 4 = 2 x y 2 Place the 2 x y 2 into the 1st position in the ( ) . 2 n d t e r m : 15 x 3 y 4 5 x 2 y 4 = 3 x Place the 3 x into the 2nd position in the ( ) . 3 r d t e r m : 5 x 2 y 4 5 x 2 y 4 = 1 Place the 1 into the 3rd position in the ( ) .

The other factor is 2 x y 2 + 3 x 1 , and

5 x 2 y 4 ( 2 x y 2 + 3 x 1 ) = 10 x 3 y 6 + 15 x 3 y 4 5 x 2 y 4

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The product is 4 a 2 b 3 + 2 c and a factor is 1 . Find the other factor.

1 ( ) = 4 a 2 b 3 + 2 c

Since there are three terms in the product, there must be three terms inside the parentheses. We will divide (mentally) each term of the product by 1 .

1 s t t e r m : 4 a 2 1 = 4 a 2 Place 4 a 2 into the 1st position inside the ( ) . 2 n d t e r m : b 3 1 = b 3 Place b 3 into the 2nd position inside the ( ) . 3 r d t e r m : 2 c 1 = 2 c Place 2 c into the 3rd position inside the ( ) .

The other factor is 4 a 2 + b 3 2 c , and

1 ( 4 a 2 + b 3 2 c ) = 4 a 2 b 3 + 2 c

Without writing the 1 , we get

( 4 a 2 + b 3 2 c ) = 4 a 2 b 3 + 2 c

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The product is 3 a 2 b 5 15 a 3 b 2 + 9 a 2 b 2 and a factor is 3 a 2 b 2 . Find the other factor.

3 a 2 b 2 ( ) = 3 a 2 b 5 15 a 3 b 2 + 9 a 2 b 2

Mentally dividing each term of the original trinomial by 3 a 2 b 2 , we get b 3 + 5 a 3 as the other factor, and

3 a 2 b 2 ( b 3 + 5 a 3 ) = 3 a 2 b 5 15 a 3 b 2 + 9 a 2 b 2

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Practice set a

The product is 3 x 2 6 x and a factor is 3 x . Find the other factor.

x 2

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The product is 5 y 4 + 10 y 3 15 y 2 and a factor is 5 y 2 . Find the other factor.

y 2 + 2 y 3

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The product is 4 x 5 y 3 8 x 4 y 4 + 16 x 3 y 5 + 24 x y 7 and a factor is 4 x y 3 . Find the other factor.

x 4 2 x 3 y + 4 x 2 y 2 + 6 y 4

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The product is 25 a 4 35 a 2 5 and a factor is 5 . Find the other factor.

5 a 4 + 7 a 2 1

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The product is a 2 + b 2 and a factor is 1 . Find the other factor.

a 2 b 2

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Exercises

For the following problems, the first quantity represents the product and the second quantity a factor. Find the other factor.

4 x + 10 , 2

2 x + 5

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6 y + 18 , 3

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5 x + 25 , 5

x + 5

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16 a + 64 , 8

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3 a 2 + 9 a , 3 a

a + 3

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14 b 2 + 16 b , 2 b

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21 x 2 + 28 x , 7 x

3 x + 4

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45 y 2 + 50 y , 5 y

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18 a 2 4 a , 2 a

9 a 2

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20 a 2 12 a , 4 a

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7 x 2 14 x , 7 x

x 2

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6 y 2 24 y , 6 y

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8 a 2 + 4 a , 4 a

2 a + 1

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26 b 2 + 13 b , 13 b

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9 x 2 + 6 x + 18 , 6

3 2 x 2 + x + 3

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12 b 2 + 16 b + 20 , 4

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21 x 2 + 7 x 14 , 7

3 x 2 + x 2

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35 x 2 + 40 x 5 , 5

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14 y 2 28 y + 14 , 14

y 2 2 y + 1

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36 a 2 16 a + 12 , 4

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4 y 2 10 y 12 , 2

2 y 2 5 y 6

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6 b 2 6 b 3 , 3

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18 x 3 + 20 x , 2 x

9 x 2 + 10

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40 y 3 + 24 y , 4 y

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16 x 3 12 x 2 , 4 x 2

4 x 3

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11 x 3 11 x + 11 , 11

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10 a 3 + 12 a 2 + 16 a + 8 , 2

5 a 3 + 6 a 2 + 8 a + 4

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14 b 3 + 16 b 2 + 26 b + 30 , 2

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8 a 3 4 a 2 12 a + 16 , 4

2 a 3 a 2 3 a + 4

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25 x 3 30 x 2 + 15 x 10 , 5

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4 x 6 + 16 x 4 16 x , 4 x

x 5 + 4 x 3 4

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9 a 5 + 6 a 5 18 a 4 + 24 a 2 , 3 a 2

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10 x 3 35 x 2 , 5 x 2

2 x 7

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12 x 3 y 5 + 20 x 3 y 2 , 4 x 3 y 2

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10 a 4 b 3 + 4 a 3 b 4 , 2 a 3 b 3

5 a + 2 b

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8 a 3 b 6 c 8 + 12 a 2 b 5 c 6 16 a 2 b 7 c 5 , 4 a 2 b 5 c 5

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4 x 5 y 4 + x 2 + x , x

4 x 4 y 4 + x + 1

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14 a 5 b 2 3 a 4 b 4 + 7 a 3 , a 3

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64 a 5 b 3 c 11 + 56 a 4 b 4 c 10 48 a 3 b 5 c 9 8 a 3 b 2 c 5 , 8 a 3 b 2 c 5

8 a 2 b c 6 + 7 a b 2 c 5 6 b 3 c 4 1

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3 h 3 b 2 2 h 6 b 3 9 h 2 b + h b , h b

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5 a + 10 , 5

a 2

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6 b + 8 , 2

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8 x 2 + 12 x , 4 x

2 x 3

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20 a 2 b 2 10 a 2 , 10 a 2

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a + b , 1

a b

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x + y , 1

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a b + c , 1

a + b c

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2 x + 4 y z , 1

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a b c , 1

a + b + c

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x 2 x + 1 , 1

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Exercises for review

( [link] ) How many 4 y 2 ’s are there in 24 x 2 y 3 ?

6 x 2 y

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( [link] ) Find the product. ( 2 y 3 ) 2 .

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( [link] ) Solve 2 ( 2 a 1 ) a = 7 .

a = 3

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( [link] ) Given that 3 m 2 n is a factor of 12 m 3 n 4 , find the other factor.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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