6.5
Common Factors and Factoring
by Grouping
6.5
OBJECTIVES
1. Remove the greatest common factor (GCF)
2. Factor by grouping
When the integers 3 and 5 are multiplied, the product is 15. We call 3 and 5 factors of 15.
Writing 3 5 15 indicates multiplication, but when we write 15 3 5, we say we
have factored 15. In general, factoring is the reverse of multiplication. We can extend this
idea to algebra.
From the last section on multiplying polynomials and special products, we know that
(2x 3)(x 2) 2x 2 x 6
NOTE Recall that a prime
number is any integer greater
than 1 that has only itself and 1
as factors. Writing
15 3 5
as a product of prime factors is
called the completely factored
form for 15.
But what if we begin with 2x2 x 6? How do we write the polynomial as a product of
other polynomials?
There are a number of methods we can use to factor polynomials: factoring out the
greatest common factor (GCF), factoring by grouping, factoring the difference of two
squares, and factoring the sum or difference of two cubes.
The first step in factoring is always to factor out the greatest common factor (GCF), if
any.
Definitions: Greatest Common Factor
NOTE In fact, we will see that
factoring out the GCF is the first
method to try in any of the
factoring problems we will
discuss.
The greatest common factor (GCF) of a polynomial is the monomial with the
highest degree and the largest numerical coefficient that is a factor of each term
of the polynomial.
Once the GCF is found, we apply the distributive property to write the original polynomial as a product of the GCF and the polynomial formed by dividing each term by that
GCF. Example 1 illustrates this approach.
Example 1
Factoring Out a Monomial
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(a) Factor 4x3 12x.
NOTE Here 4 is the GCF of the
numerical coefficients, and the
highest common power of x is
1, so the GCF of 4x3 12x is 4x.
Note that the numerical coefficient of the GCF is 4 and the variable factor is x (the highest power common to each term). So
4x3 12x 4x x2 4x 3
4x(x2 3)
NOTE Here 6 is the GCF of the
numerical coefficients, the
highest common power of a is
2, and the highest common
power of b is 2.
(b) Factor 6a3b2 12a2b3 24a4b4.
Here the GCF is 6a2b2, and we can write
6a3b2 12a2b3 24a4b4
6a2b2 a 6a2b2 2b 6a2b2 4a2b2
6a2b2 (a 2b 4a2b2)
415
416
CHAPTER 6
POLYNOMIALS AND POLYNOMIAL FUNCTIONS
(c) Factor 8m4n2 16m2n2 24mn3 32mn4.
Here the GCF is 8mn2, and we have
8m4n2 16m2n2 24mn3 32mn4
8mn2 m3 8mn2 2m 8mn2 3n 8mn2 4n2
8mn2(m3 2m 3n 4n2)
Notice that in Example 1(b) it is also true that
6a3b2 12a2b3 24a4b4 3ab(2a2b 4ab2 8a3b3)
However, this is not in completely factored form because we agree that this means factoring out the GCF (that monomial with the largest possible coefficient and degree). In this
case, we must remove 6a2b2.
CHECK YOURSELF 1
Write each of the following in completely factored form.
(a) 7x3y 21x2y2 28xy3
(b) 15m4n4 5mn3 20mn2 25m2n2
A related factoring method is called factoring by grouping. We introduce this method in
Example 2.
Example 2
Finding a Common Factor
(a) Factor 3x(x y) 2(x y).
We see that the binomial x y is a common factor and can be removed.
NOTE Because of the
3x(x y) 2(x y)
commutative property, the
factors can be written in either
order.
(x y) 3x (x y) 2
(x y)(3x 2)
(b) Factor 3x2(x y) 6x(x y) 9(x y).
We note that here the GCF is 3(x y). Factoring as before, we have
CHECK YOURSELF 2
Completely factor each of the polynomials.
(a) 7a(a 2b) 3(a 2b)
(b) 4x2(x y) 8x(x y) 16(x y)
If the terms of a polynomial have no common factor (other than 1), factoring by grouping is the preferred method, as illustrated in Example 3.
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3(x y)(x2 2x 3)
COMMON FACTORS AND FACTORING BY GROUPING
SECTION 6.5
417
Example 3
Factoring by Grouping Terms
Suppose we want to factor the polynomial
NOTE Notice that our example
has four terms. That is the clue
for trying the factoring by
grouping method.
ax ay bx by
As you can see, the polynomial has no common factors. However, look at what happens if
we separate the polynomial into two groups of two terms.
ax ay bx by
ax ay bx by
(1)
(2)
Now each group has a common factor, and we can write the polynomial as
a(x y) b(x y)
In this form, we can see that x y is the GCF. Factoring out x y, we get
a(x y) b(x y) (x y)(a b)
CHECK YOURSELF 3
Use the factoring by grouping method.
x2 2xy 3x 6y
Be particularly careful of your treatment of algebraic signs when applying the factoring
by grouping method. Consider Example 4.
Example 4
Factoring by Grouping Terms
Factor 2x3 3x2 6x 9.
We group the polynomial as follows.
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2x3 3x2 6x 9
(1)
(2)
Remove the common factor of 3 from the second two terms.
NOTE Notice that
x2(2x 3) 3(2x 3)
9 (3)(3).
(2x 3)(x2 3)
CHECK YOURSELF 4
Factor by grouping.
3y3 2y2 6y 4
It may also be necessary to change the order of the terms as they are grouped. Look at
Example 5.
POLYNOMIALS AND POLYNOMIAL FUNCTIONS
Example 5
Factoring by Grouping Terms
Factor x2 6yz 2xy 3xz.
Grouping the terms as before, we have
x2 6yz 2xy 3xz
(1)
(2)
Do you see that we have accomplished nothing because there are no common factors in the
first group?
We can, however, rearrange the terms to write the original polynomial as
x2 2xy 3xz 6yz
(1)
(2)
x(x 2y) 3z(x 2y)
(x 2y)(x 3z)
We can now remove the common factor of x 2y in
group (1) and group (2).
Note: It is often true that the grouping can be done in more than one way. The factored
form will be the same.
CHECK YOURSELF 5
We can write the polynomial of Example 5 as
x2 3xz 2xy 6yz
Factor, and verify that the factored form is the same in either case.
CHECK YOURSELF ANSWERS
1. (a) 7xy(x2 3xy 4y2); (b) 5mn2(3m3n2 n 4 5m)
2. (a) (a 2b)(7a 3); (b) 4(x y)(x2 2x 4)
3. (x 2y)(x 3)
4. (3y 2)(y2 2)
5. (x 3z)(x 2y)
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CHAPTER 6
418
Name
6.5
Exercises
Section
Date
Find the greatest common factor for each of the following sets of terms.
1. 10, 12
ANSWERS
2. 15, 35
1.
3. 16, 32, 88
2.
4. 55, 33, 132
3.
5. x2, x5
4.
6. y7, y9
5.
6.
7. a3, a6, a9
8. b4, b6, b8
7.
8.
4
5
9
9. 5x , 10x
3
10. 8y , 24y
9.
10.
11. 8a4, 6a6, 10a10
12. 9b3, 6b5, 12b4
11.
12.
13. 9x2y, 12xy2, 15x2y2
14. 12a3b2, 18a2b3, 6a4b4
13.
14.
15. 15ab3, 10a2bc, 25b2c3
15.
16. 9x2, 3xy3, 6y3
16.
17.
17. 15a2bc2, 9ab2c2, 6a2b2c2
18. 18x3y2z3, 27x4y2z3, 81xy2z
18.
19.
19. (x y) , (x y)
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2
3
20. 12(a b) , 4(a b)
4
3
20.
21.
In exercises 21 to 38, completely factor each polynomial.
22.
21. 6x 9y
23.
22. 7a 21b
24.
23. 4x2 12x
24. 5a2 25a
419
ANSWERS
25.
25. 18m2n 27mn2
26. 24c2d 3 30c3d 2
27. 5x3 15x2 25x
28. 28r3 21r2 7r
29. 12m3n 6mn 18mn2
30. 18w2z 27wz 36wz2
31. 4a3b2 8a2b 12ab2 4ab
32. 9r2s3 27r3s2 6r2s2 3rs
33. x(y z) 3(y z)
34. 2a(c d) b(c d)
35. 3(m n) 5(m n)2
36. 4(r 2s) 3(r 2s)2
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37. 5x2(x y) 10x(x y) 15(x y)
37.
38.
38. 7a2(a 2b) 21a(a 2b) 14(a 2b)
39.
40.
Determine if the factoring in each of the following is correct.
41.
42.
39. x2 x 6 (x 3)(x 2)
40. x2 x 12 (x 4)(x 3)
41. x2 x 12 (x 6)(x 2)
42. x2 2x 8 (x 8)(x 1)
43. 2x2 5x 3 (2x 1)(x 3)
44. 6x2 13x 6 (3x 2)(2x 3)
43.
44.
45.
46.
In exercises 45 to 50, factor each polynomial by grouping.
48.
45. ab ac b2 bc
46. ax 2a bx 2b
47. 6r2 12rs r 2s
48. 2mn 4m2 3n 6m
49. ab2 2b2 3a 6
50. r 2s2 3s2 2r2 6
49.
50.
420
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47.
ANSWERS
In exercises 51 to 54, factor each polynomial by grouping. Hint: Consider a rearrangement
of terms.
51.
52.
51. x 10y 5xy 2x
52. a 12b 3ab 4a
53.
53. m2 6n3 2mn2 3mn
54. r2 3rs2 12s3 4rs
54.
2
2
55.
In exercises 55 to 58, factor. Hint: Consider three groups of two terms.
56.
55. x3 x2 3x x2y xy 3y
56. m3 m2 4m 2m2n 2mn 8n
57. a a b 3a 3ab 3a 3b
58. r 2r s r 2rs 3r 6s
57.
3
2
2
3
2
2
59. The GCF of 2x 6 is 2. The GCF of 5x 10 is 5. Find the greatest common factor
58.
59.
of the product (2x 6)(5x 10).
60.
60. The GCF of 3z 12 is 3. The GCF of 4z 8 is 4. Find the GCF of the product
61.
(3z 12)(4z 8).
61. The GCF of 2x3 4x is 2x. The GCF of 3x 6 is 3. Find the GCF of the product
62.
(2x3 4x)(3x 6).
63.
62. State, in a sentence, the rule that the previous three exercises illustrated.
64.
63. For the monomials x4y2, x8y6, and x9y4, explain how you can determine the GCF by
inspecting exponents.
64. It is not possible to use the grouping method to factor 2x3 6x2 8x 4. Is it
65.
66.
correct to conclude that the polynomial is prime? Justify your answer.
67.
65. Area of a rectangle. The area of a rectangle with width t is given by 33t t2. Factor
the expression and determine the length of the rectangle in terms of t.
66. Area of a rectangle. The area of a rectangle of length x is given by 3x2 5x. Find
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the width of the rectangle.
67. For centuries, mathematicians have found factoring numbers into prime factors a
fascinating subject. A prime number is a number that cannot be written as a product
of any whole numbers but 1 and itself. The list of positive primes begins with 2
because 1 is not considered a prime number and then goes on: 3, 5, 7, 11, . . . . What
are the first 10 primes? What are the primes less than 100? If you list the numbers
from 1 to 100 and then cross out all numbers that are factors of 2, 3, 5, and 7, what is
left? Are all the numbers not crossed out prime? Write a paragraph to explain why
this might be so. You might want to investigate the sieve of Eratosthenes, a system
from 230 B.C. for finding prime numbers.
421
68. If we could make a list of all the prime numbers, what number would be at the end of
the list? Because there are an infinite number of prime numbers, there is no “largest
prime number.” But is there some formula that will give us all the primes? Here are
some formulas proposed over the centuries:
n2 n 17
2n2 29
n2 n 11
In all these expressions, n 1, 2, 3, 4, . . . , that is, a positive integer beginning with 1.
Investigate these expressions with a partner. Do the expressions give prime numbers
when they are evaluated for these values of n? Do the expressions give every
prime in the range of resulting numbers? Can you put in any positive number
for n?
69. How are primes used in coding messages and for security? Work together to decode
the messages. The messages are coded using this code: After the numbers are
factored into prime factors, the power of 2 gives the number of the letter in the
alphabet. This code would be easy for a code breaker to figure out, but you might
make up a code that would be more difficult to break.
(a) 1310720, 229376, 1572864, 1760, 460, 2097152, 336
(b) 786432, 143, 4608, 278528, 1344, 98304, 1835008, 352, 4718592, 5242880
(c) Code a message using this rule. Exchange your message with a partner to
decode it.
Answers
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1. 2
3. 8
5. x2
7. a3
9. 5x4
11. 2a4
13. 3xy
2
2
15. 5b
17. 3abc
19. (x y)
21. 3(2x 3y)
23. 4x(x 3)
25. 9mn(2m 3n)
27. 5x(x2 3x 5)
29. 6mn(2m2 1 3n)
2
31. 4ab(a b 2a 3b 1)
33. (y z)(x 3)
35. (m n)(3 5m 5n)
37. 5(x y)(x2 2x 3)
39. Correct
41. Incorrect
43. Correct
45. (b c)(a b)
47. (r 2s)(6r 1)
49. (a 2)(b2 3)
51. (x 2)(x 5y)
53. (m 3n)(m 2n2)
55. (x y)(x2 x 3)
2
57. (a b)(a 3a 3)
59. 10
61. 6x
63.
65. 33 t
67.
69.