The Greatest Common Factor

Chapter
6
Factoring
6.1
The Greatest Common Factor
1. First, list all possible ways that we can express 42 as a product of two
positive integers:
42 = 1 · 42
42 = 6 · 7
42 = 2 · 21
42 = 3 · 14
Therefore, the list of divisors of 42 is:
{1, 2, 3, 6, 7, 14, 21, 42}
3. First, list all possible ways that we can express 44 as a product of two
positive integers:
44 = 1 · 44
44 = 2 · 22
44 = 4 · 11
Therefore, the list of divisors of 44 is:
{1, 2, 4, 11, 22, 44}
5. First, list all possible ways that we can express 51 as a product of two
positive integers:
51 = 1 · 51
51 = 3 · 17
Therefore, the list of divisors of 51 is:
{1, 3, 17, 51}
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7. First, list the positive divisors of 36:
1, 2, 3, 4, 6, 9, 12, 18, 36
Secondly, list the positive divisors of 42:
1, 2, 3, 6, 7, 14, 21, 42
Finally, list the positive divisors that are in common.
1, 2, 3, 6
9. First, list the positive divisors of 78:
1, 2, 3, 6, 13, 26, 39, 78
Secondly, list the positive divisors of 54:
1, 2, 3, 6, 9, 18, 27, 54
Finally, list the positive divisors that are in common.
1, 2, 3, 6
11. First, list the positive divisors of 8:
1, 2, 4, 8
Secondly, list the positive divisors of 76:
1, 2, 4, 19, 38, 76
Finally, list the positive divisors that are in common.
1, 2, 4
13. We’re asked to find the greatest common divisor of 76 and 8. Therefore,
we must try to find the largest number that divides evenly (zero remainder)
into both 76 and 8. For some folks, the number 4 just pops into their heads.
However, if the number doesn’t just “pop into your head,” then you can:
i) List the positive divisors of 76:
1, 2, 4, 19, 38, 76
ii) List the positive divisors of 8:
1, 2, 4, 8
iii) List the positive divisors that are in common.
1, 2, 4
The greatest common divisor is therefore 4.
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6.1. THE GREATEST COMMON FACTOR
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15. We’re asked to find the greatest common divisor of 32 and 36. Therefore,
we must try to find the largest number that divides evenly (zero remainder)
into both 32 and 36. For some folks, the number 4 just pops into their heads.
However, if the number doesn’t just “pop into your head,” then you can:
i) List the positive divisors of 32:
1, 2, 4, 8, 16, 32
ii) List the positive divisors of 36:
1, 2, 3, 4, 6, 9, 12, 18, 36
iii) List the positive divisors that are in common.
1, 2, 4
The greatest common divisor is therefore 4.
17. We’re asked to find the greatest common divisor of 24 and 28. Therefore,
we must try to find the largest number that divides evenly (zero remainder)
into both 24 and 28. For some folks, the number 4 just pops into their heads.
However, if the number doesn’t just “pop into your head,” then you can:
i) List the positive divisors of 24:
1, 2, 3, 4, 6, 8, 12, 24
ii) List the positive divisors of 28:
1, 2, 4, 7, 14, 28
iii) List the positive divisors that are in common.
1, 2, 4
The greatest common divisor is therefore 4.
19. Prime factor each number and place the result in compact form using
exponents.
600 = 23 · 31 · 52
1080 = 23 · 33 · 51
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Write each prime factor that appears above to the highest power that appears
in common.
GCD = 23 · 31 · 51
Raise each factor to highest
power that appears in common.
Expand and simplify.
Expand: 23 = 8, 31 = 3,
and 51 = 5
Multiply.
=8·3·5
= 120
Therefore, GCD(600, 1080) = 120.
21. Prime factor each number and place the result in compact form using
exponents.
1800 = 23 · 32 · 52
2250 = 21 · 32 · 53
Write each prime factor that appears above to the highest power that appears
in common.
GCD = 21 · 32 · 52
Raise each factor to highest
power that appears in common.
Expand and simplify.
Expand: 21 = 2, 32 = 9,
and 52 = 25
Multiply.
= 2 · 9 · 25
= 450
Therefore, GCD(1800, 2250) = 450.
23. Prime factor each number and place the result in compact form using
exponents.
600 = 23 · 31 · 52
450 = 21 · 32 · 52
Write each prime factor that appears above to the highest power that appears
in common.
GCD = 21 · 31 · 52
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Raise each factor to highest
power that appears in common.
6.1. THE GREATEST COMMON FACTOR
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Expand and simplify.
= 2 · 3 · 25
Expand: 21 = 2, 31 = 3,
and 52 = 25
= 150
Multiply.
Therefore, GCD(600, 450) = 150.
25. To find the GCF of 16b4 and 56b9 , we note that:
1. The greatest common factor (divisor) of 16 and 56 is 8.
2. The monomials 16b4 and 56b9 have the variable b in common.
3. The highest power of b in common is b4 .
Thus, the greatest common factor is GCF(16b4 , 56b9 ) = 8b4 . Note what happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
16b4 = 8b4 · 2
56b9 = 8b4 · 7b5
Note how the set of second monomial factors (2 and 7b5 ) contain no additional
common factors.
27. To find the GCF of 35z 2 and 49z 7 , we note that:
1. The greatest common factor (divisor) of 35 and 49 is 7.
2. The monomials 35z 2 and 49z 7 have the variable z in common.
3. The highest power of z in common is z 2 .
Thus, the greatest common factor is GCF(35z 2 , 49z 7) = 7z 2 . Note what happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
35z 2 = 7z 2 · 5
49z 7 = 7z 2 · 7z 5
Note how the set of second monomial factors (5 and 7z 5 ) contain no additional
common factors.
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29. To find the GCF of 56x3 y 4 and 16x2 y 5 , we note that:
1. The greatest common factor (divisor) of 56 and 16 is 8.
2. The monomials 56x3 y 4 and 16x2 y 5 have the variables x and y in common.
3. The highest power of x in common is x2 . The highest power of y in
common is y 4 .
Thus, the greatest common factor is GCF(56x3 y 4 , 16x2 y 5 ) = 8x2 y 4 . Note
what happens when we write each of the given monomials as a product of the
greatest common factor and a second monomial:
56x3 y 4 = 8x2 y 4 · 7x
16x2 y 5 = 8x2 y 4 · 2y
Note how the set of second monomial factors (7x and 2y) contain no additional
common factors.
31. To find the GCF of 24s4 t5 and 16s3 t6 , we note that:
1. The greatest common factor (divisor) of 24 and 16 is 8.
2. The monomials 24s4 t5 and 16s3 t6 have the variables s and t in common.
3. The highest power of s in common is s3 . The highest power of t in
common is t5 .
Thus, the greatest common factor is GCF(24s4 t5 , 16s3 t6 ) = 8s3 t5 . Note what
happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
24s4 t5 = 8s3 t5 · 3s
16s3 t6 = 8s3 t5 · 2t
Note how the set of second monomial factors (3s and 2t) contain no additional
common factors.
33. To find the GCF of 18y 7 , 45y 6 , and 27y 5 , we note that:
1. The greatest common factor (divisor) of 18, 45, and 27 is 9.
2. The monomials 18y 7 , 45y 6 , and 27y 5 have the variable y in common.
3. The highest power of y in common is y 5 .
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Thus, the greatest common factor is GCF(18y 7 , 45y 6, 27y 5 ) = 9y 5 . Note what
happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
18y 7 = 9y 5 · 2y 2
45y 6 = 9y 5 · 5y
27y 5 = 9y 5 · 3
Note how the set of second monomial factors (2y 2 , 5y, and 3) contain no
additional common factors.
35. To find the GCF of 9a6 , 6a5 , and 15a4 , we note that:
1. The greatest common factor (divisor) of 9, 6, and 15 is 3.
2. The monomials 9a6 , 6a5 , and 15a4 have the variable a in common.
3. The highest power of a in common is a4 .
Thus, the greatest common factor is GCF(9a6 , 6a5 , 15a4 ) = 3a4 . Note what
happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
9a6 = 3a4 · 3a2
6a5 = 3a4 · 2a
15a4 = 3a4 · 5
Note how the set of second monomial factors (3a2 , 2a, and 5) contain no
additional common factors.
37. The greatest common factor (GCF) of 25a2 , 10a and 20 is 5. Factor out
the GCF.
25a2 + 10a + 20 = 5 · 5a2 + 5 · 2a + 5 · 4
= 5(5a2 + 2a + 4)
Check: Multiply. Distribute the 5.
5(5a2 + 2a + 4) = 5 · 5a2 + 5 · 2a + 5 · 4
= 25a2 + 10a + 20
That’s the original polynomial, so we factored correctly.
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CHAPTER 6. FACTORING
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39. The greatest common factor (GCF) of 35s2 , 25s and 45 is 5. Factor out
the GCF.
35s2 + 25s + 45 = 5 · 7s2 + 5 · 5s + 5 · 9
= 5(7s2 + 5s + 9)
Check: Multiply. Distribute the 5.
5(7s2 + 5s + 9) = 5 · 7s2 + 5 · 5s + 5 · 9
= 35s2 + 25s + 45
That’s the original polynomial, so we factored correctly.
41. The greatest common factor (GCF) of 16c3 , 32c2 and 36c is 4c. Factor out
the GCF.
16c3 + 32c2 + 36c = 4c · 4c2 + 4c · 8c + 4c · 9
= 4c(4c2 + 8c + 9)
Check: Multiply. Distribute the 4c.
4c(4c2 + 8c + 9) = 4c · 4c2 + 4c · 8c + 4c · 9
= 16c3 + 32c2 + 36c
That’s the original polynomial, so we factored correctly.
43. The greatest common factor (GCF) of 42s3 , 24s2 and 18s is 6s. Factor
out the GCF.
42s3 + 24s2 + 18s = 6s · 7s2 + 6s · 4s + 6s · 3
= 6s(7s2 + 4s + 3)
Check: Multiply. Distribute the 6s.
6s(7s2 + 4s + 3) = 6s · 7s2 + 6s · 4s + 6s · 3
= 42s3 + 24s2 + 18s
That’s the original polynomial, so we factored correctly.
45. The greatest common factor (GCF) of 35s7 , 49s6 and 63s5 is 7s5 . Factor
out the GCF.
35s7 + 49s6 + 63s5 = 7s5 · 5s2 + 7s5 · 7s + 7s5 · 9
= 7s5 (5s2 + 7s + 9)
Check: Multiply. Distribute the 7s5 .
7s5 (5s2 + 7s + 9) = 7s5 · 5s2 + 7s5 · 7s + 7s5 · 9
= 35s7 + 49s6 + 63s5
That’s the original polynomial, so we factored correctly.
Second Edition: 2012-2013
6.1. THE GREATEST COMMON FACTOR
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47. The greatest common factor (GCF) of 14b7 , 35b6 and 56b5 is 7b5 . Factor
out the GCF.
14b7 + 35b6 + 56b5 = 7b5 · 2b2 + 7b5 · 5b + 7b5 · 8
= 7b5 (2b2 + 5b + 8)
Check: Multiply. Distribute the 7b5 .
7b5 (2b2 + 5b + 8) = 7b5 · 2b2 + 7b5 · 5b + 7b5 · 8
= 14b7 + 35b6 + 56b5
That’s the original polynomial, so we factored correctly.
49. The greatest common factor (GCF) of 54y 5 z 3 , 30y 4 z 4 and 36y 3z 5 is 6y 3 z 3 .
Factor out the GCF.
54y 5 z 3 + 30y 4 z 4 + 36y 3 z 5 = 6y 3 z 3 · 9y 2 + 6y 3 z 3 · 5yz + 6y 3 z 3 · 6z 2
= 6y 3 z 3 (9y 2 + 5yz + 6z 2 )
Check: Multiply. Distribute the 6y 3 z 3 .
6y 3 z 3 (9y 2 + 5yz + 6z 2 ) = 6y 3 z 3 · 9y 2 + 6y 3 z 3 · 5yz + 6y 3 z 3 · 6z 2
= 54y 5 z 3 + 30y 4 z 4 + 36y 3 z 5
That’s the original polynomial, so we factored correctly.
51. The greatest common factor (GCF) of 45s4 t3 , 40s3 t4 and 15s2 t5 is 5s2 t3 .
Factor out the GCF.
45s4 t3 + 40s3 t4 + 15s2 t5 = 5s2 t3 · 9s2 + 5s2 t3 · 8st + 5s2 t3 · 3t2
= 5s2 t3 (9s2 + 8st + 3t2 )
Check: Multiply. Distribute the 5s2 t3 .
5s2 t3 (9s2 + 8st + 3t2 ) = 5s2 t3 · 9s2 + 5s2 t3 · 8st + 5s2 t3 · 3t2
= 45s4 t3 + 40s3 t4 + 15s2 t5
That’s the original polynomial, so we factored correctly.
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53. In this case, the greatest common factor (GCF) is 2w − 3.
7w(2w − 3) − 8(2w − 3) = 7w · (2w − 3) − 8 · (2w − 3)
= (7w − 8)(2w − 3)
Because of the commutative property of multiplication, it is equally valid to
pull the GCF out in front.
7w(2w − 3) − 8(2w − 3) = (2w − 3) · 7w − (2w − 3) · 8
= (2w − 3)(7w − 8)
Note that the order of factors differs from the first solution, but because of
the commutative property of multiplication, the order does not matter. The
answers are the same.
55. In this case, the greatest common factor (GCF) is 5r − 1.
9r(5r − 1) + 8(5r − 1) = 9r · (5r − 1) + 8 · (5r − 1)
= (9r + 8)(5r − 1)
Because of the commutative property of multiplication, it is equally valid to
pull the GCF out in front.
9r(5r − 1) + 8(5r − 1) = (5r − 1) · 9r + (5r − 1) · 8
= (5r − 1)(9r + 8)
Note that the order of factors differs from the first solution, but because of
the commutative property of multiplication, the order does not matter. The
answers are the same.
57. In this case, the greatest common factor (GCF) is 6(2a + 5).
48a(2a + 5) − 42(2a + 5) = 6(2a + 5) · 8a − 6(2a + 5) · 7
= 6(2a + 5)(8a − 7)
Alternate solution: It is possible that you might fail to notice that 15 and
12 are divisible by 3, factoring out only a common factor 2a + 5.
48a(2a + 5) − 42(2a + 5) = 48a · (2a + 5) − 42 · (2a + 5)
= (48a − 42)(2a + 5)
However, you now need to notice that you can continue, factoring out a 6 from
both 48a and −42.
= 6(8a − 7)(2a + 5)
Note that the order of factors differs from the first solution, but because of
the commutative property of multiplication, the order does not matter. The
answers are the same.
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59. In this case, the greatest common factor (GCF) is 7(2a − 1).
56a(2a − 1) − 21(2a − 1) = 7(2a − 1) · 8a − 7(2a − 1) · 3
= 7(2a − 1)(8a − 3)
Alternate solution: It is possible that you might fail to notice that 15 and
12 are divisible by 3, factoring out only a common factor 2a − 1.
56a(2a − 1) − 21(2a − 1) = 56a · (2a − 1) − 21 · (2a − 1)
= (56a − 21)(2a − 1)
However, you now need to notice that you can continue, factoring out a 7 from
both 56a and −21.
= 7(8a − 3)(2a − 1)
Note that the order of factors differs from the first solution, but because of
the commutative property of multiplication, the order does not matter. The
answers are the same.
61. We “group” the first and second terms, noting that we can factor x out of
both of these terms. Then we “group” the third and fourth terms, noting that
we can factor −9 out of both of these terms.
x2 + 2x − 9x − 18 = x (x + 2) − 9 (x + 2)
Note that we can now factor x + 2 out of both of these terms.
= (x − 9)(x + 2)
63. We “group” the first and second terms, noting that we can factor x out of
both of these terms. Then we “group” the third and fourth terms, noting that
we can factor 6 out of both of these terms.
x2 + 3x + 6x + 18 = x (x + 3) + 6 (x + 3)
Note that we can now factor x + 3 out of both of these terms.
= (x + 6)(x + 3)
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65. We “group” the first and second terms, noting that we can factor x out of
both of these terms. Then we “group” the third and fourth terms, noting that
we can factor −3 out of both of these terms.
x2 + 6x + 3x + 18 = x (x − 6) − 3 (x − 6)
Note that we can now factor x − 6 out of both of these terms.
= (x − 3)(x − 6)
67. We “group” the first and second terms, noting that we can factor x out of
both of these terms. Then we “group” the third and fourth terms, noting that
we can factor 3 out of both of these terms.
x2 − 9x + 3x − 27 = x (x − 9) + 3 (x − 9)
Note that we can now factor x − 9 out of both of these terms.
= (x + 3)(x − 9)
69. We “group” the first and second terms, noting that we can factor x out of
both of these terms. Then we “group” the third and fourth terms, noting that
we can factor −7 out of both of these terms.
8x2 + 3x − 56x − 21 = x (8x + 3) − 7 (8x + 3)
Note that we can now factor 8x + 3 out of both of these terms.
= (x − 7)(8x + 3)
71. We “group” the first and second terms, noting that we can factor 9x out
of both of these terms. Then we “group” the third and fourth terms, noting
that we can factor −5 out of both of these terms.
9x2 + 36x − 5x − 20 = 9x (x + 4) − 5 (x + 4)
Note that we can now factor x + 4 out of both of these terms.
= (9x − 5)(x + 4)
Second Edition: 2012-2013