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Factoring Polynomials: The Greatest Common Factor

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: understand more clearly the factorization process, be able to determine the greatest common factor of two or more terms.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

Overview

  • Factoring Method
  • Greatest Common Factor

Factoring Method

In the last two types of problems (Sections (Reference) and (Reference)), we knew one of the factors and were able to determine the other factor through division. Suppose, now, we’re given the product without any factors. Our problem is to find the factors, if possible. This procedure and the previous two procedures are based on the distributive property.

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.

We will use the distributive property in reverse.

ab+ac ︸ product = a(b+c) ︸ factors ab+ac ︸ product = a(b+c) ︸ factors

We notice that in the product, a a is common to both terms. (In fact, a a is a common factor of both terms.) Since a a is common to both terms, we will factor it out and write

a(               ) a(               )

Now we need to determine what to place inside the parentheses. This is the procedure of the previous section. Divide each term of the product by the known factor a. a.

ab a =b and ac a =c ab a =b and ac a =c

Thus, b b and c c are the required terms of the other factor. Hence,

ab+ac=a(b+c) ab+ac=a(b+c)

When factoring a monomial from a polynomial, we seek out factors that are not only common to each term of the polynomial, but factors that have these properties:

  1. The numerical coefficients are the largest common numerical coefficients.
  2. The variables possess the largest exponents common to all the variables.

Greatest Common Factor

A monomial factor that meets the above two requirements is called the greatest common factor of the polynomial.

Sample Set A

Example 1

Factor 3x−18. 3x−18.

The greatest common factor is 3.

3x−18=3⋅x−3⋅6 Factor out 3. 3x−18=3(               ) Divide each term of the product by 3. 3x 3 =x   and    −18 3 =−6 (Try to perform this division mentally.) 3x−18=3(x−6) 3x−18=3⋅x−3⋅6 Factor out 3. 3x−18=3(               ) Divide each term of the product by 3. 3x 3 =x   and    −18 3 =−6 (Try to perform this division mentally.) 3x−18=3(x−6)

Example 2

Factor 9 x 3 +18 x 2 +27x. 9 x 3 +18 x 2 +27x.

Notice that 9x 9x is the greatest common factor.

9 x 3 +18 x 2 +27x=9x⋅ x 2 +9x⋅2x+9x⋅3. Factor out 9x. 9 x 3 +18 x 2 +27x=9x(               ) Mentally divide 9x into each term of the product. 9 x 3 +18 x 2 +27x=9x( x 2 +2x+3) 9 x 3 +18 x 2 +27x=9x⋅ x 2 +9x⋅2x+9x⋅3. Factor out 9x. 9 x 3 +18 x 2 +27x=9x(               ) Mentally divide 9x into each term of the product. 9 x 3 +18 x 2 +27x=9x( x 2 +2x+3)

Example 3

Factor 10 x 2 y 3 −20x y 4 −35 y 5 . 10 x 2 y 3 −20x y 4 −35 y 5 .

Notice that 5 y 3 5 y 3 is the greatest common factor. Factor out 5 y 3 . 5 y 3 .

10 x 2 y 3 −20x y 4 −35 y 5 =5 y 3 (               ) 10 x 2 y 3 −20x y 4 −35 y 5 =5 y 3 (               )

Mentally divide 5 y 3 5 y 3 into each term of the product and place the resulting quotients inside the (               ). (               ).

10 x 2 y 3 −20x y 4 −35 y 5 =5 y 3 (2 x 2 −4xy−7 y 2 ) 10 x 2 y 3 −20x y 4 −35 y 5 =5 y 3 (2 x 2 −4xy−7 y 2 )

Example 4

Factor −12 x 5 +8 x 3 −4 x 2 . −12 x 5 +8 x 3 −4 x 2 .

We see that the greatest common factor is −4 x 2 . −4 x 2 .

−12 x 5 +8 x 3 −4 x 2 =−4 x 2 (               ) −12 x 5 +8 x 3 −4 x 2 =−4 x 2 (               )

Mentally dividing −4 x 2 −4 x 2 into each term of the product, we get

−12 x 5 +8 x 3 −4 x 2 =−4 x 2 (3 x 3 −2x+1) −12 x 5 +8 x 3 −4 x 2 =−4 x 2 (3 x 3 −2x+1)

Practice Set A

Exercise 1

Factor 4x−48. 4x−48.

Solution

4(x−12) 4(x−12)

Exercise 2

Factor 6 y 3 +24 y 2 +36y. 6 y 3 +24 y 2 +36y.

Solution

6y( y 2 +4y+6) 6y( y 2 +4y+6)

Exercise 3

Factor  10 a 5 b 4 −14 a 4 b 5 −8 b 6 .  10 a 5 b 4 −14 a 4 b 5 −8 b 6 .

Solution

2 b 4 (5 a 5 −7 a 4 b−4 b 2 ) 2 b 4 (5 a 5 −7 a 4 b−4 b 2 )

Exercise 4

Factor  −14 m 4 +28 m 2 −7m.  −14 m 4 +28 m 2 −7m.

Solution

−7m(2 m 3 −4m+1) −7m(2 m 3 −4m+1)

Consider this problem: factor Ax+Ay. Ax+Ay. Surely, Ax+Ay=A(x+y). Ax+Ay=A(x+y). We know from the very beginning of our study of algebra that letters represent single quantities. We also know that a quantity occurring within a set of parentheses is to be considered as a single quantity. Suppose that the letter A A is representing the quantity (a+b). (a+b). Then we have

              Ax+              Ay=               A(x+y)               Ax+              Ay=               A(x+y)

(a+b)x+(a+b)y=(a+b)(x+y) (a+b)x+(a+b)y=(a+b)(x+y)

When we observe the expression

(a+b)x+(a+b)y (a+b)x+(a+b)y

we notice that (a+b) (a+b) is common to both terms. Since it is common, we factor it out.

(a+b)(               ) (a+b)(               )

As usual, we determine what to place inside the parentheses by dividing each term of the product by (a+b). (a+b).

(a+b)x (a+b) =x and (a+b)y (a+b) =y (a+b)x (a+b) =x and (a+b)y (a+b) =y

Thus, we get

(a+b)x+(a+b)y=(a+b)(x+y) (a+b)x+(a+b)y=(a+b)(x+y)

This is a forerunner of the factoring that will be done in Section 5.4. 5.4.

Sample Set B

Example 5

Factor (x−7)a+(x−7)b. (x−7)a+(x−7)b.

Notice that (x−7) (x−7) is the greatest common factor. Factor out (x−7). (x−7).

(x−7)a+(x−7)b = (x−7)( ) Then, (x−7)a (x−7) = a and  (x−7)b (x−7) =b. (x−7)a+(x−7)b = (x−7)(a+b) (x−7)a+(x−7)b = (x−7)( ) Then, (x−7)a (x−7) = a and  (x−7)b (x−7) =b. (x−7)a+(x−7)b = (x−7)(a+b)

Example 6

Factor 3 x 2 (x+1)−5x(x+1) 3 x 2 (x+1)−5x(x+1) .

Notice that x x and (x+1) (x+1) are common to both terms. Factor them out. We’ll perform this factorization by letting A=x(x+1). A=x(x+1). Then we have

3xA−5A = A(3x−5) But A = x(x+1), so 3 x 2 (x+1)−5x(x+1) = x(x+1)(3x−5) 3xA−5A = A(3x−5) But A = x(x+1), so 3 x 2 (x+1)−5x(x+1) = x(x+1)(3x−5)

Practice Set B

Exercise 5

Factor (y+4)a+(y+4)b. (y+4)a+(y+4)b.

Solution

(y+4)(a+b) (y+4)(a+b)

Exercise 6

Factor 8 m 3 (n−4)−6 m 2 (n−4). 8 m 3 (n−4)−6 m 2 (n−4).

Solution

2 m 2 (n−4)(4m−3) 2 m 2 (n−4)(4m−3)

Exercises

For the following problems, factor the polynomials.

Exercise 7

9a+18 9a+18

Solution

9( a+2 ) 9( a+2 )

Exercise 8

6a+24 6a+24

Exercise 9

8b+12 8b+12

Solution

4( 2b+3 ) 4( 2b+3 )

Exercise 10

16x+12 16x+12

Exercise 11

4x−6 4x−6

Solution

2( 2x−3 ) 2( 2x−3 )

Exercise 12

8x−14 8x−14

Exercise 13

21y−28 21y−28

Solution

7( 3y−4 ) 7( 3y−4 )

Exercise 14

16f−36 16f−36

Exercise 15

12 x 2 +18x 12 x 2 +18x

Solution

6x( 2x+3 ) 6x( 2x+3 )

Exercise 16

10 y 2 +15y 10 y 2 +15y

Exercise 17

8 y 2 +18 8 y 2 +18

Solution

2( 4 y 2 +9 ) 2( 4 y 2 +9 )

Exercise 18

7 x 2 −21 7 x 2 −21

Exercise 19

3 y 2 −6 3 y 2 −6

Solution

3( y 2 −2 ) 3( y 2 −2 )

Exercise 20

2 x 2 −2 2 x 2 −2

Exercise 21

6 y 2 −6y 6 y 2 −6y

Solution

6y( y−1 ) 6y( y−1 )

Exercise 22

a x 2 −a a x 2 −a

Exercise 23

b y 2 +b b y 2 +b

Solution

b( y 2 +1 ) b( y 2 +1 )

Exercise 24

7b y 2 +14b 7b y 2 +14b

Exercise 25

5 a 2 x 2 +10x 5 a 2 x 2 +10x

Solution

5x( a 2 x+2 ) 5x( a 2 x+2 )

Exercise 26

24a x 2 +28a 24a x 2 +28a

Exercise 27

10 x 2 +5x−15 10 x 2 +5x−15

Solution

5( 2 x 2 +x−3 ) 5( 2 x 2 +x−3 )

Exercise 28

12 x 2 −8x−16 12 x 2 −8x−16

Exercise 29

15 y 3 −24y+9 15 y 3 −24y+9

Solution

3( 5 y 3 −8y+3 ) 3( 5 y 3 −8y+3 )

Exercise 30

a x 2 +ax+a a x 2 +ax+a

Exercise 31

b y 3 +b y 2 +by+b b y 3 +b y 2 +by+b

Solution

b( y 3 + y 2 +y+1 ) b( y 3 + y 2 +y+1 )

Exercise 32

2 y 2 +6y+4xy 2 y 2 +6y+4xy

Exercise 33

9 x 2 +6xy+4x 9 x 2 +6xy+4x

Solution

x( 9x+6y+4 ) x( 9x+6y+4 )

Exercise 34

30 a 2 b 2 +40 a 2 b 2 +50 a 2 b 2 30 a 2 b 2 +40 a 2 b 2 +50 a 2 b 2

Exercise 35

13 x 2 y 5 c−26 x 2 y 5 c−39 x 2 y 5 13 x 2 y 5 c−26 x 2 y 5 c−39 x 2 y 5

Solution

13 x 2 y 5 ( −c−3 ) 13 x 2 y 5 ( −c−3 )

Exercise 36

−4 x 2 −12x−8 −4 x 2 −12x−8

Exercise 37

−6 y 3 −8 y 2 −14y+10 −6 y 3 −8 y 2 −14y+10

Solution

−2( 3 y 3 +4 y 2 +7y−5 ) −2( 3 y 3 +4 y 2 +7y−5 )

Exercise 38

Ab+Ac Ab+Ac

Exercise 39

Nx+Ny Nx+Ny

Solution

N( x+y ) N( x+y )

Exercise 40

Qx+Qy Qx+Qy

Exercise 41

Ax−Ay Ax−Ay

Solution

A( x−y ) A( x−y )

Exercise 42

(x+4)b+(x+4)c (x+4)b+(x+4)c

Exercise 43

(x−9)a+(x−9)b (x−9)a+(x−9)b

Solution

( x−9 )( a+b ) ( x−9 )( a+b )

Exercise 44

(2x+7)a+(2x+7)b (2x+7)a+(2x+7)b

Exercise 45

(9a−b)w−(9a−b)x (9a−b)w−(9a−b)x

Solution

( 9a−b )( w−x ) ( 9a−b )( w−x )

Exercise 46

(5−v)X+(5−v)Y (5−v)X+(5−v)Y

Exercise 47

3 x 5 y 4 −12 x 3 y 4 +27 x 5 y 3 −6 x 2 y 6 3 x 5 y 4 −12 x 3 y 4 +27 x 5 y 3 −6 x 2 y 6

Solution

3 x 2 y 3 ( x 3 y−4xy+9 x 3 −2 y 3 ) 3 x 2 y 3 ( x 3 y−4xy+9 x 3 −2 y 3 )

Exercise 48

8 a 3 b 15 +24 a 2 b 14 +48 a 3 b 6 −20 a 3 b 7 +80 a 4 b 6 −4 a 3 b 7 +4 a 2 b 8 a 3 b 15 +24 a 2 b 14 +48 a 3 b 6 −20 a 3 b 7 +80 a 4 b 6 −4 a 3 b 7 +4 a 2 b

Exercise 49

−8 x 3 y 2 −3 x 3 y 2 +16 x 4 y 3 +2 x 2 y −8 x 3 y 2 −3 x 3 y 2 +16 x 4 y 3 +2 x 2 y

Solution

− x 2 y( 11xy−16 x 2 y 2 −2 ) − x 2 y( 11xy−16 x 2 y 2 −2 )

Exercises for Review

Exercise 50

((Reference)) A quantity plus 21% 21% more of that quantity is 26.25. 26.25. What is the original quantity?

Exercise 51

((Reference)) Solve the equation 6(t−1)=4(5−s) if s=2. 6(t−1)=4(5−s) if s=2.

Solution

t=3 t=3

Exercise 52

((Reference)) Given that 4 a 3 4 a 3 is a factor of 8 a 3 −12 a 2 , 8 a 3 −12 a 2 , find the other factor.

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