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Factoring Polynomials: Factoring a Monomial from a Polynomial

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

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Summary: 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.

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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 +20x 12 x 2 +20x is the product and one of the factors is 4x 4x . To find the other factor we could set up the problem this way:

4x⋅(               )=12 x 2 +20x 4x⋅(               )=12 x 2 +20x

Since the product 12 x 2 +20x 12 x 2 +20x consists of two terms, the expression multiplying 4x 4x 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.

1st term: 4x(                 ) = 12 x 2 2nd term: 4x(                 ) = 20x (                 ) = 12 x 2 4x (                 ) = 20x 4x (                 ) = 3x (                 ) = 5 1st term: 4x(                 ) = 12 x 2 2nd term: 4x(                 ) = 20x (                 ) = 12 x 2 4x (                 ) = 20x 4x (                 ) = 3x (                 ) = 5

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

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

Sample Set A

Example 1

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

We have the problem: x 4 x 4 times "what expression" yields 3 x 7 −2 x 6 +4 x 5 −3 x 4 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 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 x 4 . The resulting quotient will be the necessary term of the factor.

1st term: 3 x 7 x 4 = 3 x 7−4 =3 x 3 Place 3 x 3  into the 1st position in the( ). 2nd term: −2 x 6 x 4 = −2 x 2 Place −2 x 2  into the 2nd position in the( ). 3rd term: 4 x 5 x 4 = 4x Place 4x into the 3rd position in the( ). 4th term: −3 x 4 x 4 = −3 Place −3 into the 4th position in the( ). 1st term: 3 x 7 x 4 = 3 x 7−4 =3 x 3 Place 3 x 3  into the 1st position in the( ). 2nd term: −2 x 6 x 4 = −2 x 2 Place −2 x 2  into the 2nd position in the( ). 3rd term: 4 x 5 x 4 = 4x Place 4x into the 3rd position in the( ). 4th term: −3 x 4 x 4 = −3 Place −3 into the 4th position in the( ).

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

This result can be checked by applying the distributive property.

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

Example 2

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

1st term: 10 x 3 y 6 5 x 2 y 4 = 2x y 2 Place the 2x y 2  into the 1st position in the( ). 2nd term: 15 x 3 y 4 5 x 2 y 4 = 3x Place the 3x into the 2nd position in the( ). 3rd term: −5 x 2 y 4 5 x 2 y 4 = −1 Place the −1 into the 3rd position in the( ). 1st term: 10 x 3 y 6 5 x 2 y 4 = 2x y 2 Place the 2x y 2  into the 1st position in the( ). 2nd term: 15 x 3 y 4 5 x 2 y 4 = 3x Place the 3x into the 2nd position in the( ). 3rd term: −5 x 2 y 4 5 x 2 y 4 = −1 Place the −1 into the 3rd position in the( ).

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

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

Example 3

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

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

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 .

1st term: −4 a 2 −1 = 4 a 2 Place 4 a 2  into the 1st position inside the( ). 2nd term: − b 3 −1 = b 3 Place  b 3  into the 2nd position inside the( ). 3rd term: 2c −1 = −2c Place −2c into the 3rd position inside the( ). 1st term: −4 a 2 −1 = 4 a 2 Place 4 a 2  into the 1st position inside the( ). 2nd term: − b 3 −1 = b 3 Place  b 3  into the 2nd position inside the( ). 3rd term: 2c −1 = −2c Place −2c into the 3rd position inside the( ).

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

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

Without writing the −1 −1 , we get

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

Example 4

The product is −3 a 2 b 5 −15 a 3 b 2 +9 a 2 b 2 −3 a 2 b 5 −15 a 3 b 2 +9 a 2 b 2 and a factor is −3 a 2 b 2 −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 −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 −3 a 2 b 2 , we get b 3 +5a−3 b 3 +5a−3 as the other factor, and

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

Practice Set A

Exercise 1

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

Solution

x−2 x−2

Exercise 2

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

Solution

y 2 +2y−3 y 2 +2y−3

Exercise 3

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

Solution

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

Exercise 4

The product is -25 a4−35 a2+5 25 a 4 35 a 2 5 and a factor is −5 −5 . Find the other factor.

Solution

5 a 4 +7 a 2 −1 5 a 4 +7 a 2 −1

Exercise 5

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

Solution

a 2 − b 2 a 2 − b 2

Exercises

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

Exercise 6

4x+10,     2 4x+10,     2

Solution

2x+5 2x+5

Exercise 7

6y+18,     3 6y+18,     3

Exercise 8

5x+25,     5 5x+25,     5

Solution

x+5 x+5

Exercise 9

16a+64,     8 16a+64,     8

Exercise 10

3 a 2 +9a,     3a 3 a 2 +9a,     3a

Solution

a+3 a+3

Exercise 11

14 b 2 +16b,     2b 14 b 2 +16b,     2b

Exercise 12

21 x 2 +28x,     7x 21 x 2 +28x,     7x

Solution

3x+4 3x+4

Exercise 13

45 y 2 +50y,     5y 45 y 2 +50y,     5y

Exercise 14

18 a 2 −4a,     2a 18 a 2 −4a,     2a

Solution

9a−2 9a−2

Exercise 15

20 a 2 −12a,     4a 20 a 2 −12a,     4a

Exercise 16

7 x 2 −14x,     7x 7 x 2 −14x,     7x

Solution

x−2 x−2

Exercise 17

6 y 2 −24y,     6y 6 y 2 −24y,     6y

Exercise 18

8 a 2 +4a,     4a 8 a 2 +4a,     4a

Solution

2a+1 2a+1

Exercise 19

26 b 2 +13b,     13b 26 b 2 +13b,     13b

Exercise 20

9 x 2 +6x+18,     6 9 x 2 +6x+18,     6

Solution

3 2 x 2 +x+3 3 2 x 2 +x+3

Exercise 21

12 b 2 +16b+20,     4 12 b 2 +16b+20,     4

Exercise 22

21 x 2 +7x−14,     7 21 x 2 +7x−14,     7

Solution

3 x 2 +x−2 3 x 2 +x−2

Exercise 23

35 x 2 +40x−5,     5 35 x 2 +40x−5,     5

Exercise 24

14 y 2 −28y+14,     14 14 y 2 −28y+14,     14

Solution

y 2 −2y+1 y 2 −2y+1

Exercise 25

36 a 2 −16a+12,     4 36 a 2 −16a+12,     4

Exercise 26

4 y 2 −10y−12,     2 4 y 2 −10y−12,     2

Solution

2 y 2 −5y−6 2 y 2 −5y−6

Exercise 27

6 b 2 −6b−3,     3 6 b 2 −6b−3,     3

Exercise 28

18 x 3 +20x,     2x 18 x 3 +20x,     2x

Solution

9 x 2 +10 9 x 2 +10

Exercise 29

40 y 3 +24y,     4y 40 y 3 +24y,     4y

Exercise 30

16 x 3 −12 x 2 ,     4 x 2 16 x 3 −12 x 2 ,     4 x 2

Solution

4x−3 4x−3

Exercise 31

11 x 3 −11x+11,     11 11 x 3 −11x+11,     11

Exercise 32

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

Solution

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

Exercise 33

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

Exercise 34

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

Solution

2 a 3 − a 2 −3a+4 2 a 3 − a 2 −3a+4

Exercise 35

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

Exercise 36

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

Solution

x 5 +4 x 3 −4 x 5 +4 x 3 −4

Exercise 37

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

Exercise 38

10 x 3 −35 x 2 ,     5 x 2 10 x 3 −35 x 2 ,     5 x 2

Solution

2x−7 2x−7

Exercise 39

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

Exercise 40

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

Solution

5a+2b 5a+2b

Exercise 41

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 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

Exercise 42

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

Solution

4 x 4 y 4 +x+1 4 x 4 y 4 +x+1

Exercise 43

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

Exercise 44

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 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

Solution

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

Exercise 45

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

Exercise 46

5a+10,     −5 5a+10,     −5

Solution

−a−2 −a−2

Exercise 47

6b+8,     −2 6b+8,     −2

Exercise 48

8 x 2 +12x,     −4x 8 x 2 +12x,     −4x

Solution

−2x−3 −2x−3

Exercise 49

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

Exercise 50

a+b,     −1 a+b,     −1

Solution

−a−b −a−b

Exercise 51

x+y,     −1 x+y,     −1

Exercise 52

a−b+c,     −1 a−b+c,     −1

Solution

−a+b−c −a+b−c

Exercise 53

2x+4y−z,     −1 2x+4y−z,     −1

Exercise 54

−a−b−c,     −1 −a−b−c,     −1

Solution

a+b+c a+b+c

Exercise 55

x 2 −x+1,     −1 x 2 −x+1,     −1

Exercises for Review

Exercise 56

((Reference)) How many 4 y 2 4 y 2 ’s are there in 24 x2 y3 24 x 2 y 3 ?

Solution

6 x 2 y 6 x 2 y

Exercise 57

((Reference)) Find the product. (2y−3) 2 (2y−3) 2 .

Exercise 58

((Reference)) Solve 2(2a−1)−a=7 2(2a−1)−a=7 .

Solution

a=3 a=3

Exercise 59

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

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