Chapter 1

Factoring

1.1 The Greatest Common Factor

1. First, list all possible ways that we can express 42 as a product of twopositive integers:

42 = 1 · 42 42 = 2 · 21 42 = 3 · 1442 = 6 · 7

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 twopositive integers:

44 = 1 · 44 44 = 2 · 22 44 = 4 · 11Therefore, 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 twopositive integers:

51 = 1 · 51 51 = 3 · 17Therefore, the list of divisors of 51 is:

{1, 3, 17, 51}

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CHAPTER 1 FACTORING

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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1.1. THE GREATEST COMMON FACTOR

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 usingexponents.

600 = 23 · 31 · 521080 = 23 · 33 · 51

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CHAPTER 1 FACTORING

Write each prime factor that appears above to the highest power that appearsin common.

GCD = 23 · 31 · 51 Raise each factor to highestpower that appears in common.

Expand and simplify.

= 8 · 3 · 5 Expand: 23 = 8, 31 = 3,and 51 = 5

= 120 Multiply.

Therefore, GCD(600, 1080) = 120.

21. Prime factor each number and place the result in compact form usingexponents.

1800 = 23 · 32 · 522250 = 21 · 32 · 53

Write each prime factor that appears above to the highest power that appearsin common.

GCD = 21 · 32 · 52 Raise each factor to highestpower that appears in common.

Expand and simplify.

= 2 · 9 · 25 Expand: 21 = 2, 32 = 9,and 52 = 25

= 450 Multiply.

Therefore, GCD(1800, 2250) = 450.

23. Prime factor each number and place the result in compact form usingexponents.

600 = 23 · 31 · 52450 = 21 · 32 · 52

Write each prime factor that appears above to the highest power that appearsin common.

GCD = 21 · 31 · 52 Raise each factor to highestpower that appears in common.

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1.1. THE GREATEST COMMON FACTOR

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 hap-pens when we write each of the given monomials as a product of the greatestcommon factor and a second monomial:

16b4 = 8b4 · 256b9 = 8b4 · 7b5

Note how the set of second monomial factors (2 and 7b5) contain no additionalcommon factors.

27. To find the GCF of 35z2 and 49z7, we note that:

1. The greatest common factor (divisor) of 35 and 49 is 7.

2. The monomials 35z2 and 49z7 have the variable z in common.

3. The highest power of z in common is z2.

Thus, the greatest common factor is GCF(35z2, 49z7) = 7z2. Note what hap-pens when we write each of the given monomials as a product of the greatestcommon factor and a second monomial:

35z2 = 7z2 · 549z7 = 7z2 · 7z5

Note how the set of second monomial factors (5 and 7z5) contain no additionalcommon factors.

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CHAPTER 1 FACTORING

29. To find the GCF of 56x3y4 and 16x2y5, we note that:

1. The greatest common factor (divisor) of 56 and 16 is 8.

2. The monomials 56x3y4 and 16x2y5 have the variables x and y in common.

3. The highest power of x in common is x2. The highest power of y incommon is y4.

Thus, the greatest common factor is GCF(56x3y4, 16x2y5) = 8x2y4. Notewhat happens when we write each of the given monomials as a product of thegreatest common factor and a second monomial:

56x3y4 = 8x2y4 · 7x16x2y5 = 8x2y4 · 2y

Note how the set of second monomial factors (7x and 2y) contain no additionalcommon factors.

31. To find the GCF of 24s4t5 and 16s3t6, we note that:

1. The greatest common factor (divisor) of 24 and 16 is 8.

2. The monomials 24s4t5 and 16s3t6 have the variables s and t in common.

3. The highest power of s in common is s3. The highest power of t incommon is t5.

Thus, the greatest common factor is GCF(24s4t5, 16s3t6) = 8s3t5. Note whathappens when we write each of the given monomials as a product of the greatestcommon factor and a second monomial:

24s4t5 = 8s3t5 · 3s16s3t6 = 8s3t5 · 2t

Note how the set of second monomial factors (3s and 2t) contain no additionalcommon factors.

33. To find the GCF of 18y7, 45y6, and 27y5, we note that:

1. The greatest common factor (divisor) of 18, 45, and 27 is 9.

2. The monomials 18y7, 45y6, and 27y5 have the variable y in common.

3. The highest power of y in common is y5.

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1.1. THE GREATEST COMMON FACTOR

Thus, the greatest common factor is GCF(18y7, 45y6, 27y5) = 9y5. Note whathappens when we write each of the given monomials as a product of the greatestcommon factor and a second monomial:

18y7 = 9y5 · 2y245y6 = 9y5 · 5y27y5 = 9y5 · 3

Note how the set of second monomial factors (2y2, 5y, and 3) contain noadditional 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 whathappens when we write each of the given monomials as a product of the greatestcommon factor and a second monomial:

9a6 = 3a4 · 3a26a5 = 3a4 · 2a15a4 = 3a4 · 5

Note how the set of second monomial factors (3a2, 2a, and 5) contain noadditional common factors.

37. The greatest common factor (GCF) of 25a2, 10a and 20 is 5. Factor outthe 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 1 FACTORING

39. The greatest common factor (GCF) of 35s2, 25s and 45 is 5. Factor outthe 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 outthe 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. Factorout 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. Factorout 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.

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1.1. THE GREATEST COMMON FACTOR

47. The greatest common factor (GCF) of 14b7, 35b6 and 56b5 is 7b5. Factorout 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 54y5z3, 30y4z4 and 36y3z5 is 6y3z3.Factor out the GCF.

54y5z3 + 30y4z4 + 36y3z5 = 6y3z3 · 9y2 + 6y3z3 · 5yz + 6y3z3 · 6z2= 6y3z3(9y2 + 5yz + 6z2)

Check: Multiply. Distribute the 6y3z3.

6y3z3(9y2 + 5yz + 6z2) = 6y3z3 · 9y2 + 6y3z3 · 5yz + 6y3z3 · 6z2= 54y5z3 + 30y4z4 + 36y3z5

That’s the original polynomial, so we factored correctly.

51. The greatest common factor (GCF) of 45s4t3, 40s3t4 and 15s2t5 is 5s2t3.Factor out the GCF.

45s4t3 + 40s3t4 + 15s2t5 = 5s2t3 · 9s2 + 5s2t3 · 8st+ 5s2t3 · 3t2= 5s2t3(9s2 + 8st+ 3t2)

Check: Multiply. Distribute the 5s2t3.

5s2t3(9s2 + 8st+ 3t2) = 5s2t3 · 9s2 + 5s2t3 · 8st+ 5s2t3 · 3t2= 45s4t3 + 40s3t4 + 15s2t5

That’s the original polynomial, so we factored correctly.

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CHAPTER 1 FACTORING

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 topull 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 ofthe commutative property of multiplication, the order does not matter. Theanswers 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 topull 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 ofthe commutative property of multiplication, the order does not matter. Theanswers 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 and12 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 fromboth 48a and −42.

= 6(8a− 7)(2a+ 5)

Note that the order of factors differs from the first solution, but because ofthe commutative property of multiplication, the order does not matter. Theanswers are the same.

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1.1. THE GREATEST COMMON FACTOR

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 and12 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 fromboth 56a and −21.

= 7(8a− 3)(2a− 1)

Note that the order of factors differs from the first solution, but because ofthe commutative property of multiplication, the order does not matter. Theanswers are the same.

61. We “group” the first and second terms, noting that we can factor x out ofboth of these terms. Then we “group” the third and fourth terms, noting thatwe 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 ofboth of these terms. Then we “group” the third and fourth terms, noting thatwe 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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CHAPTER 1 FACTORING

65. We “group” the first and second terms, noting that we can factor x out ofboth of these terms. Then we “group” the third and fourth terms, noting thatwe 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 ofboth of these terms. Then we “group” the third and fourth terms, noting thatwe 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 ofboth of these terms. Then we “group” the third and fourth terms, noting thatwe 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 outof both of these terms. Then we “group” the third and fourth terms, notingthat 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)

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