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Chapter 5 Polynomials and Their Operations

 

5.1 Rules of Exponents

Learning Objectives

  1. Simplify expressions using the rules of exponents.
  2. Simplify expressions involving parentheses and exponents.
  3. Simplify expressions involving 0 as an exponent.

Product, Quotient, and Power Rule for Exponents

If a factor is repeated multiple times, then the product can be written in exponential formAn equivalent expression written using a rational exponent. x n. The positive integer exponent n indicates the number of times the base x is repeated as a factor.

For example,

Here the base is 5 and the exponent is 4. Exponents are sometimes indicated with the caret (^) symbol found on the keyboard: 5^4 = 5*5*5*5.

Next consider the product of 2 3 and 25,

Expanding the expression using the definition produces multiple factors of the base, which is quite cumbersome, particularly when n is large. For this reason, we will develop some useful rules to help us simplify expressions with exponents. In this example, notice that we could obtain the same result by adding the exponents.

In general, this describes the product rule for exponentsxmxn=xm+n; the product of two expressions with the same base can be simplified by adding the exponents.. If m and n are positive integers, then

In other words, when multiplying two expressions with the same base, add the exponents.

 

Example 1: Simplify: 1051018.

Solution:

Answer: 1023

 

In the previous example, notice that we did not multiply the base 10 times itself. When applying the product rule, add the exponents and leave the base unchanged.

 

Example 2: Simplify: x6x12x.

Solution: Recall that the variable x is assumed to have an exponent of 1: x=x1.

Answer: x19

 

The base could be any algebraic expression.

 

Example 3: Simplify: (x+y)9 (x+y)13.

Solution: Treat the expression (x+y) as the base.

Answer: (x+y)22

 

The commutative property of multiplication allows us to use the product rule for exponents to simplify factors of an algebraic expression.

 

Example 4: Simplify: 2x8y3x4y7.

Solution: Multiply the coefficients and add the exponents of variable factors with the same base.

Answer: 6x12y8

 

Next, we will develop a rule for division by first looking at the quotient of 27 and 23.

Here we can cancel factors after applying the definition of exponents. Notice that the same result can be obtained by subtracting the exponents.

This describes the quotient rule for exponentsxmxn=xmn; the quotient of two expressions with the same base can be simplified by subtracting the exponents.. If m and n are positive integers and x0, then

In other words, when you divide two expressions with the same base, subtract the exponents.

 

Example 5: Simplify: 12y154y7.

Solution: Divide the coefficients and subtract the exponents of the variable y.

Answer: 3y8

 

Example 6: Simplify: 20x10(x+5)610x9(x+5)2.

Solution:

Answer: 2x(x+5)4

 

Now raise 23 to the fourth power as follows:

After writing the base 23 as a factor four times, expand to obtain 12 factors of 2. We can obtain the same result by multiplying the exponents.

In general, this describes the power rule for exponents(xm)n=xmn; a power raised to a power can be simplified by multiplying the exponents.. Given positive integers m and n, then

In other words, when raising a power to a power, multiply the exponents.

 

Example 7: Simplify: (y6)7.

Solution:

Answer: y42

 

To summarize, we have developed three very useful rules of exponents that are used extensively in algebra. If given positive integers m and n, then

Product rule: xmxn=xm+n
Quotient rule: xmxn=xmn,x0
Power rule: (xm)n=xmn

 

Try this! Simplify: y5(y4)6.

Answer: y29

Video Solution

(click to see video)

Power Rules for Products and Quotients

Now we consider raising grouped products to a power. For example,

After expanding, we have four factors of the product xy. This is equivalent to raising each of the original factors to the fourth power. In general, this describes the power rule for a product(xy)n=xnyn; if a product is raised to a power, then apply that power to each factor in the product.. If n is a positive integer, then

 

Example 8: Simplify: (2ab)7.

Solution: We must apply the exponent 7 to all the factors, including the coefficient, 2.

If a coefficient is raised to a relatively small power, then present the real number equivalent, as we did in this example: 27=128.

Answer: 128a7b7

 

In many cases, the process of simplifying expressions involving exponents requires the use of several rules of exponents.

 

Example 9: Simplify: (3xy3)4.

Solution:

Answer: 81x4y12

 

Example 10: Simplify: (4x2y5z)3.

Solution:

Answer: 64x6y15z3

 

Example 11: Simplify: [5( x+y)3]3.

Solution:

Answer: 125(x+y)9

 

Next, consider a quotient raised to a power.

Here we obtain four factors of the quotient, which is equivalent to the numerator and the denominator both raised to the fourth power. In general, this describes the power rule for a quotient(xy)n=xnyn; if a quotient is raised to a power, then apply that power to the numerator and the denominator.. If n is a positive integer and y0, then

In other words, given a fraction raised to a power, we can apply that exponent to the numerator and the denominator. This rule requires that the denominator is nonzero. We will make this assumption for the remainder of the section.

 

Example 12: Simplify: (3ab)3.

Solution: First, apply the power rule for a quotient and then the power rule for a product.

Answer: 27a3b3

 

In practice, we often combine these two steps by applying the exponent to all factors in the numerator and the denominator.

 

Example 13: Simplify: (a b 22 c 3)5.

Solution: Apply the exponent 5 to all of the factors in the numerator and the denominator.

Answer:  a5b1032c15

 

Example 14: Simplify: (5 x 5 ( 2x1 ) 43 y 7)2.

Solution:

Answer: 25x10(2x1)89y14

 

It is a good practice to simplify within parentheses before using the power rules; this is consistent with the order of operations.

 

Example 15: Simplify: (2 x 3 y 4zx y 2)4.

Solution:

Answer: 16x8y8z4

 

To summarize, we have developed two new rules that are useful when grouping symbols are used in conjunction with exponents. If given a positive integer n, where y is a nonzero number, then

Power rule for a product: (xy)n=xnyn
Power rule for a quotient: (xy)n=xnyn

 

Try this! Simplify: (4 x 2 ( xy ) 33y z 5)3.

Answer: 64x6(xy)927y3z15

Video Solution

(click to see video)

Zero as an Exponent

Using the quotient rule for exponents, we can define what it means to have 0 as an exponent. Consider the following calculation:

Eight divided by 8 is clearly equal to 1, and when the quotient rule for exponents is applied, we see that a 0 exponent results. This leads us to the definition of zero as an exponentx0=1; any nonzero base raised to the 0 power is defined to be 1., where x0:

It is important to note that 00 is undefined. If the base is negative, then the result is still +1. In other words, any nonzero base raised to the 0 power is defined to be 1. In the following examples, assume all variables are nonzero.

 

Example 16: Simplify:

a. (5)0

b. 50

Solution:

a. Any nonzero quantity raised to the 0 power is equal to 1.

b. In the example 50, the base is 5, not −5.

Answers: a. 1; b. −1

 

Example 17: Simplify: (5x3y0z2)2.

Solution: It is good practice to simplify within the parentheses first.

Answer: 25x6z4

 

Example 18: Simplify: (8 a 10 b 55 c 12 d 14)0.

Solution:

Answer: 1

 

Try this! Simplify: 5x0 and (5x)0.

Answer: 5x0=5 and (5x)0=1

Video Solution

(click to see video)

Key Takeaways

  • The rules of exponents allow you to simplify expressions involving exponents.
  • When multiplying two quantities with the same base, add exponents: xmxn=xm+n.
  • When dividing two quantities with the same base, subtract exponents: xmxn=xmn.
  • When raising powers to powers, multiply exponents: (xm)n=xmn.
  • When a grouped quantity involving multiplication and division is raised to a power, apply that power to all of the factors in the numerator and the denominator: (xy)n=xnyn and (xy)n=xnyn.
  • Any nonzero quantity raised to the 0 power is defined to be equal to 1: x0=1.

Topic Exercises

Part A: Product, Quotient, and Power Rule for Exponents

Write each expression using exponential form.

1. (2x)(2x)(2x)(2x)(2x)

2. (3y)(3y)(3y)

3. 10aaaaaaa

4. 12xxyyyyyy

5. 6(x1)(x1)(x1)

6. (9ab)(9ab)(9ab)(a2b)(a2b)

Simplify.

7. 2725

8. 393

9. 24

10. (2)4

11. 33

12. (3)4

13. 1013105104

14. 10810710

15. 51252

16. 10710

17. 1012109

18. (73)5

19. (48)4

20. 106(105)4

Simplify.

21. (x)6

22. a5(a)2

23. x3x5x

24. y5y4y2

25. (a5)2(a3)4a

26. (x+1)4(y5)4y2

27. (x+1)5(x+1)8

28. (2ab)12(2ab)9

29. (3x1)5(3x1)2

30. (a5)37(a5)13

31. xy2x2y

32. 3x2y37xy5

33. 8a2b2ab

34. 3ab2c39a4b5c6

35. 2a2b4c (3abc)

36. 5a2(b3)3c3(2)2a3(b2)4

37. 2x2(x+y)53x5(x+y)4

38. 5xy6(2x1)6x5y(2x1)3

39. x2yxy3x5y5

40. 2x10y3x2y125xy3

41. 32x4y2z3xy4z4

42. (x2)3(x3)2(x4)3

43. a10( a 6)3a3

44. 10x9( x 3)52x5

45. a6b3a2b2

46. m10n7m3n4

47. 20x5y12z310x2y10z

48. 24a16b12c36a6b11c

49. 16 x4(x+2)34x(x+2)

50. 50y2(x+y)2010y(x+y)17

Part B: Power Rules for Products and Quotients

Simplify.

51. (2x)5

52. (3y)4

53. (xy)3

54. (5xy)3

55. (4abc)2

56. (72x)2

57. (53y)3

58. (3abc)3

59. (2xy3z)4

60. (5y(2x1)x)3

61. (3x2)3

62. (2x3)2

63. (xy5)7

64. (x2y10)2

65. (3x2y)3

66. (2x2y3z4)5

67. (7ab4c2)2

68. [x5y4( x+y)4]5

69. [2y( x+1)5]3

70. (a b 3)3

71. (5 a 23b)4

72. (2 x 33 y 2)2

73. ( x 2 y 3)3

74. (a b 23 c 3 d 2)4

75. (2 x 7y ( x1 ) 3 z 5)6

76. (2x4)3(x5)2

77. (x3y)2(xy4)3

78. (2a2b3)2(2a5b)4

79. (a2b)3(3ab4)4

80. (2x3( x+y)4)5(2x4( x+y)2)3

81. (3 x 5 y 4x y 2)3

82. (3 x 5 y 4x y 2)2

83. (25 x 10 y 155 x 5 y 10)3

84. (10 x 3 y 55x y 2)2

85. (24a b 36bc)5

86. (2 x 3y16 x 2y)2

87. (30a b 33abc)3

88. (3 s 3 t 22 s 2t)3

89. (6x y 5 ( x+y ) 63 y 2z ( x+y ) 2)5

90. (64 a 5 b 12 c 2 ( 2ab1 ) 1432 a 2 b 10 c 2 ( 2ab1 ) 7)4

91. The probability of tossing a fair coin and obtaining n heads in a row is given by the formula P=(12)n. Determine the probability, as a percent, of tossing 5 heads in a row.

92. The probability of rolling a single fair six-sided die and obtaining n of the same faces up in a row is given by the formula P=(16)n. Determine the probability, as a percent, of obtaining the same face up two times in a row.

93. If each side of a square measures 2x3 units, then determine the area in terms of the variable x.

94. If each edge of a cube measures 5x2 units, then determine the volume in terms of the variable x.

Part C: Zero Exponents

Simplify. (Assume variables are nonzero.)

95. 70

96. (7)0

97. 100

98. 30(7)0

99. 86753090

100. 523023

101. 30(2)2(3)0

102. 5x0y2

103. (3)2x2y0z5

104. 32(x3)2y2(z3)0

105. 2x3y0z3x0y3z5

106. 3ab2c03a2(b3c2)0

107. (8xy2)0

108. (2 x 2 y 3)0

109. 9x0y43y3

Part D: Discussion Board Topics

110. René Descartes (1637) established the usage of exponential form: a2, a3, and so on. Before this, how were exponents denoted?

111. Discuss the accomplishments accredited to Al-Karismi.

112. Why is 00 undefined?

113. Explain to a beginning student why 343296.

Answers

1: (2x)5

3: 10a7

5: 6(x1)3

7: 212

9: −16

11: −27

13: 1022

15: 510

17: 103

19: 432

21: x6

23: x9

25: a23

27: (x+1)13

29: (3x1)3

31: x3y3

33: 16a3b2

35: 6a3b5c2

37: 6x7(x+y)9

39: x8y9

41: 27x5y6z5

43: a25

45: a4b

47: 2x3y2z2

49: 4x3(x+2)2

51: 32x5

53: x3y3

55: 16a2b2c2

57: 12527y3

59: 16x4y481z4

61: 27x6

63: x7y35

65: 27x6y3

67: 49a2b8c4

69: 8y3(x+1)15

71: 625a881b4

73: x6y9

75: 64x42y6(x1)18z30

77: x9y14

79: 81a10b19

81: 27x12y6

83: 125x15y15

85: 1024a5b10c5

87: 1000b6c3

89: 32x5y15(x+y)20z5

91: 318%

93: A=4x6

95: 1

97: −1

99: 1

101: −4

103: 9x2z5

105: 6x3y3z6

107: 1

109: 3y

5.2 Introduction to Polynomials

Learning Objectives

  1. Identify a polynomial and determine its degree.
  2. Evaluate a polynomial for given values of the variables.
  3. Evaluate a polynomial using function notation.

Definitions

A polynomialAn algebraic expression consisting of terms with real number coefficients and variables with whole number exponents. is a special algebraic expression with terms that consist of real number coefficients and variable factors with whole number exponents.

Polynomials do not have variables in the denominator of any term.

The degree of a termThe exponent of the variable; if there is more than one variable in the term, the degree of the term is the sum their exponents. in a polynomial is defined to be the exponent of the variable, or if there is more than one variable in the term, the degree is the sum of their exponents. Recall that x0=1; any constant term can be written as a product of x0 and itself. Hence the degree of a constant term is 0.

Term Degree

3x2

2

6x2y

2+1=3

7a2b3

2+3=5

8

0 , since 8=8x0

2x

1, since x=x1

The degree of a polynomialThe largest degree of all of its terms. is the largest degree of all of its terms.

Polynomial Degree

4x53x3+2x1

5

6x2y5xy3+7

4 , because 5xy3 has degree 4.

12x+54

1, because x=x1

We classify polynomials by the number of terms and the degree as follows:

Expression Classification Degree

5x7

MonomialPolynomial with one term. (one term)

7

8x61

BinomialPolynomial with two terms. (two terms)

6

3x2+x1

TrinomialPolynomial with three terms. (three terms)

2

5x32x2+3x6

PolynomialAn algebraic expression consisting of terms with real number coefficients and variables with whole number exponents. (many terms)

3

In this text, we will call polynomials with four or more terms simply polynomials.

 

Example 1: Classify and state the degree: 7x24x51.

Solution: Here there are three terms. The highest variable exponent is 5. Therefore, this is a trinomial of degree 5.

Answer: Trinomial; degree 5

 

Example 2: Classify and state the degree: 12a5bc3.

Solution: Since the expression consists of only multiplication, it is one term, a monomial. The variable part can be written as a5b1c3; hence its degree is 5+1+3=9.

Answer: Monomial; degree 9

 

Example 3: Classify and state the degree: 4x2y6xy4+5x3y3+4.

Solution: The term 4x2y has degree 3; 6xy4 has degree 5; 5x3y3 has degree 6; and the constant term 4 has degree 0. Therefore, the polynomial has 4 terms with degree 6.

Answer: Polynomial; degree 6

 

Of particular interest are polynomials with one variableA polynomial where each term has the form anxn, where an is any real number and n is any whole number., where each term is of the form anxn. Here an is any real number and n is any whole number. Such polynomials have the standard form

Typically, we arrange terms of polynomials in descending order based on the degree of each term. The leading coefficientThe coefficient of the term with the largest degree. is the coefficient of the variable with the highest power, in this case, an.

 

Example 4: Write in standard form: 3x4x2+5x3+72x4.

Solution: Since terms are separated by addition, write the following:

In this form, we can see that the subtraction in the original corresponds to negative coefficients. Because addition is commutative, we can write the terms in descending order based on the degree of each term as follows:

Answer: 2x4+5x34x2+3x+7

 

We can further classify polynomials with one variable by their degree as follows:

Polynomial Name

5

Constant (degree 0)

2x+1

Linear (degree 1)

3x2+5x3

Quadratic (degree 2)

x3+x2+x+1

Cubic (degree 3)

7x4+3x37x+8

Fourth-degree polynomial

In this text, we call any polynomial of degree n4 an nth-degree polynomial. In other words, if the degree is 4, we call the polynomial a fourth-degree polynomial. If the degree is 5, we call it a fifth-degree polynomial, and so on.

Evaluating Polynomials

Given the values for the variables in a polynomial, we can substitute and simplify using the order of operations.

 

Example 5: Evaluate: 3x1, where x=32.

Solution: First, replace the variable with parentheses and then substitute the given value.

Answer: −11/2

 

Example 6: Evaluate: 3x2+2x1, where x=1.

Solution:

Answer: 0

 

Example 7: Evaluate: 2a2b+ab27, where a=3 and b=2.

Solution:

Answer: 41

 

Example 8: The volume of a sphere in cubic units is given by the formula V=43πr3, where r is the radius. Calculate the volume of a sphere with radius r=32 meters.

Solution:

Answer: 92π cubic meters

 

Try this! Evaluate: x3x2+4x2, where x=3.

Answer: −50

Video Solution

(click to see video)

Polynomial Functions

Polynomial functions with one variable are functions that can be written in the form

where an is any real number and n is any whole number. Some examples of the different classes of polynomial functions are listed below:

Polynomial function Name

f(x)=5

Constant functionA polynomial function with degree 0. (degree 0)

f(x)=2x+1

Linear functionA polynomial function with degree 1. (degree 1)

f(x)=5x2+4x3

Quadratic functionA polynomial function with degree 2. (degree 2)

f(x)=x31

Cubic functionA polynomial function with degree 3. (degree 3)

f(x)=4x5+3x47

Polynomial function

Since there are no restrictions on the values for x, the domain of any polynomial function consists of all real numbers.

 

Example 9: Calculate: f(5), given f(x)=2x2+5x+10.

Solution: Recall that the function notation f(5) indicates we should evaluate the function when x=5. Replace every instance of the variable x with the value 5.

Answer: f(5)=15

 

Example 10: Calculate: f(1), given f(x)=x3+2x24x+1.

Solution: Replace the variable x with −1.

Answer: f(1)=8

 

Try this! Given g(x)=x32x2x4, calculate g(1).

Answer: g(1)=6

Video Solution

(click to see video)

Key Takeaways

  • Polynomials are special algebraic expressions where the terms are the products of real numbers and variables with whole number exponents.
  • The degree of a polynomial with one variable is the largest exponent of the variable found in any term.
  • The terms of a polynomial are typically arranged in descending order based on the degree of each term.
  • When evaluating a polynomial, it is a good practice to replace all variables with parentheses and then substitute the appropriate values.
  • All polynomials are functions.

Topic Exercises

Part A: Definitions

Classify the given polynomial as linear, quadratic, or cubic.

1. 2x+1

2. x2+7x+2

3. 23x2+x

4. 4x

5. x2x3+x+1

6. 510x3

Classify the given polynomial as a monomial, binomial, or trinomial and state the degree.

7. x31

8. x2y2

9. xx5+1

10. x2+3x1

11. 5ab4

12. 13x12

13. 5x3+2x+1

14. 8x29

15. 4x55x3+6x

16. 8x4x5+2x3

17. 9x+7

18. x5+x4+x3+x2x+1

19. 6x1+5x48

20. 4x3x2+3

21. 7

22. x2

23. 4x2y3x3y3+xy3

24. a3b26ab

25. a3b3

26. x2yy2x

27. xy3

28. a5bc2+3a95a4b3c

29. 3x10y2zxy12z+9x13+30

30. 7x0

Write the following polynomials in standard form.

31. 16x+7x2

32. x9x28

33. 7x3+x7x2+x5x5

34. a3a9+6a5a+3a4

Part B: Evaluating Polynomials

35. Fill in the following chart:

36. Fill in the following chart:

Evaluate.

37. 2x3, where x=3

38. x23x+5, where x=2

39. 12x+13, where x=13

40. x2+5x1, where x=12

41. 2x2+3x5, where x=0

42. 8x527x3+81x17, where x=0

43. y32y+1, where y=2

44. y4+2y232, where y=2

45. a3+2a2+a3, where a=3

46. x3x2, where x=5

47. 34x212x+36, where x=23

48. 58x214x+12, where x=4

49. x2y+xy2, where x=2 and y=3

50. 2a5bab4+a2b2, where a=1 and b=2

51. a2b2, where a=5 and b=6

52. a2b2, where a=34 and b=14

53. a3b3, where a=2 and b=3

54. a3+b3, where a=5 and b=5

For each problem, evaluate b24ac, given the following values.

55. a=1, b=2, and c=1

56. a=2, b=2, and c=12

57. a=3, b=5, c=0

58. a=1, b=0, and c=4

59. a=14, b=4, and c=2

60. a=1, b=5, and c=6

The volume of a sphere in cubic units is given by the formula V=43πr3, where r is the radius. For each problem, calculate the volume of a sphere given the following radii.

61. r = 3 centimeters

62. r = 1 centimeter

63. r = 1/2 feet

64. r = 3/2 feet

65. r = 0.15 in

66. r = 1.3 inches

The height in feet of a projectile launched vertically from the ground with an initial velocity v0 in feet per second is given by the formula h=16t2+v0t, where t represents time in seconds. For each problem, calculate the height of the projectile given the following initial velocity and times.

67. v0=64 feet/second, at times t = 0, 1, 2, 3, 4 seconds

68. v0=80 feet/second, at times t = 0, 1, 2, 2.5, 3, 4, 5 seconds

The stopping distance of a car, taking into account an average reaction time, can be estimated with the formula d=0.05v2+1.5, where d is in feet and v is the speed in miles per hour. For each problem, calculate the stopping distance of a car traveling at the given speeds.

69. 20 miles per hour

70. 40 miles per hour

71. 80 miles per hour

72. 100 miles per hour

Part C: Polynomial Functions

Given the linear function f(x)=23x+6, evaluate each of the following.

73. f(6)

74. f(3)

75. f(0)

76. f(3)

77. Find x when f(x)=10.

78. Find x when f(x)=4.

Given the quadratic function f(x)=2x23x+5, evaluate each of the following.

79. f(2)

80. f(1)

81. f(0)

82. f(2)

Given the cubic function g(x)=x3x2+x1, evaluate each of the following.

83. g(2)

84. g(1)

85. g(0)

86. g(1)

The height in feet of a projectile launched vertically from the ground with an initial velocity of 128 feet per second is given by the function h(t)=16t2+128t, where t is in seconds. Calculate and interpret the following.

87. h(0)

88. h(12)

89. h(1)

90. h(3)

91. h(4)

92. h(5)

93. h(7)

94. h(8)

Part D: Discussion Board Topics

95. Find and share some graphs of polynomial functions.

96. Explain how to convert feet per second into miles per hour.

97. Find and share the names of fourth-degree, fifth-degree, and higher polynomials.

Answers

1: Linear

3: Quadratic

5: Cubic

7: Binomial; degree 3

9: Trinomial; degree 5

11: Monomial; degree 5

13: Trinomial; degree 3

15: Trinomial; degree 5

17: Binomial; degree 1

19: Not a polynomial

21: Monomial; degree 0

23: Trinomial; degree 6

25: Monomial; degree 6

27: Binomial; degree 2

29: Polynomial; degree 14

31: 7x26x+1

33: x75x5x3x2+x+7

35:

37: 3

39: 1/2

41: −5

43: −3

45: −15

47: 7/6

49: 6

51: −11

53: −35

55: 0

57: 25

59: 14

61: 36π cubic centimeters

63: π/6 cubic feet

65: 0.014 cubic inches

67:

Time Height
t = 0 seconds h = 0 feet
t = 1 second h = 48 feet
t = 2 seconds h = 64 feet
t = 3 seconds h = 48 feet
t = 4 seconds h = 0 feet

69: 21.5 feet

71: 321.5 feet

73: 2

75: 6

77: x=6

79: 19

81: 5

83: −15

85: −1

87: The projectile is launched from the ground.

89: The projectile is 112 feet above the ground 1 second after launch.

91: The projectile is 256 feet above the ground 4 seconds after launch.

93: The projectile is 112 feet above the ground 7 seconds after launch.

5.3 Adding and Subtracting Polynomials

Learning Objectives

  1. Add polynomials.
  2. Subtract polynomials.
  3. Add and subtract polynomial functions.

Adding Polynomials

Recall that we combine like terms, or terms with the same variable part, as a means to simplify expressions. To do this, add the coefficients of the terms to obtain a single term with the same variable part. For example,

Notice that the variable part, x2, does not change. This, in addition to the commutative and associative properties of addition, allows us to add polynomialsThe process of combining all like terms of two or more polynomials..

 

Example 1: Add: 3x+(4x5).

Solution: The property +(a+b)=a+b, which was derived using the distributive property, allows us to remove the parentheses so that we can add like terms.

Answer: 7x5

 

Example 2: Add: (3x2+3x+5)+(2x2x2).

Solution: Remove the parentheses and then combine like terms.

Answer: 5x2+2x+3

 

Example 3: Add: (5x2y2xy2+7xy)+(4x2y+7xy23xy).

Solution: Remember that the variable parts have to be exactly the same before we can add the coefficients.

Answer: x2y+5xy2+4xy

 

It is common practice to present the terms of the simplified polynomial expression in descending order based on their degree. In other words, we typically present polynomials in standard form, with terms in order from highest to lowest degree.

 

Example 4: Add: (a4a3+a58)+(9a5+a47a+5+a3).

Solution:

Answer: 8a5+a43a36a3

 

Try this! Add: (65x3+x2x)+(x2+x+6x31).

Answer: x3+2x2+5

Video Solution

(click to see video)

Subtracting Polynomials

When subtracting polynomialsThe process of subtracting all the terms of one polynomial from another and combining like terms., we see that the parentheses become very important. Recall that the distributive property allowed us to derive the following:

In other words, when subtracting an algebraic expression, we remove the parentheses by subtracting each term.

 

Example 5: Subtract: 10x(3x+5).

Solution: Subtract each term within the parentheses and then combine like terms.

Answer: 7x5

 

Subtracting a quantity is equivalent to multiplying it by −1.

 

Example 6: Subtract: (3x2+3x+5)(2x2x2).

Solution: Distribute the −1, remove the parentheses, and then combine like terms.

Answer: x2+4x+7

 

Multiplying the terms of a polynomial by −1 changes all the signs.

 

Example 7: Subtract: (5x32x2+7)(4x3+7x23x+2).

Solution: Distribute the −1, remove the parentheses, and then combine like terms.

Answer: 9x39x2+3x+5

 

Example 8: Subtract 6x23x1 from 2x2+5x2.

Solution: Since subtraction is not commutative, we must take care to set up the difference correctly. First, write the quantity (2x2+5x2); from this, subtract the quantity (6x23x1).

Answer: 4x2+8x1

 

Example 9: Simplify: (2x23x+5)(x23x+1)+(5x24x8).

Solution: Apply the distributive property, remove the parentheses, and then combine like terms.

Answer: 6x24x4

 

Try this! Subtract: (8x2y5xy2+6)(x2y+2xy21).

Answer: 7x2y7xy2+7

Video Solution

(click to see video)

Adding and Subtracting Polynomial Functions

We use function notation to indicate addition and subtraction of functions as follows:

Addition of functions: (f+g)(x)=f(x)+g(x)
Subtraction of functions: (fg)(x)=f(x)g(x)

When using function notation, be careful to group the entire function and add or subtract accordingly.

 

Example 10: Calculate: (f+g)(x), given f(x)=x23x+5 and g(x)=3x2+2x+1.

Solution: The notation (f+g)(x) indicates that you should add the functions f(x)+g(x) and collect like terms.

Answer: (f+g)(x)=2x2x+6

 

Example 11: Calculate: (fg)(x), given f(x)=2x3 and g(x)=2x2+2x+5.

Solution: The notation (fg)(x) indicates that you should subtract the functions f(x)g(x):

Answer: (fg)(x)=2x28

 

We may be asked to evaluate the sum or difference of two functions. We have the option to first find the sum or difference and use the resulting function to evaluate for the given variable, or to first evaluate each function and then find the sum or difference.

 

Example 12: Calculate: (fg)(5), given f(x)=x2+x7 and g(x)=4x+10.

Solution: First, find (fg)(x)=f(x)g(x).

Therefore,

Next, substitute 5 for the variable x.

Answer: (fg)(5)=7

Alternate Solution: Since (fg)(5)=f(5)g(5), we can find f(5) and g(5) and then subtract the results.

Therefore, we have

Answer: (fg)(5)=7

Key Takeaways

  • When adding polynomials, remove the associated parentheses and then combine like terms.
  • When subtracting polynomials, distribute the −1 and subtract all the terms before removing the parentheses and combining like terms.
  • The notation (f+g)(x) indicates that you add the functions.
  • The notation (fg)(x) indicates that you subtract the functions.

Topic Exercises

Part A: Addition of Polynomials

Add.

1. (2x+1)+(x+7)

2. (6x+5)+(3x1)

3. (23x+12)+(13x2)

4. (13x34)+(56x+18)

5. (2x+1)+(x3)+(5x2)

6. (2x8)+(3x2+7x5)

7. (x23x+7)+(3x28x5)

8. (5x21+x)+(x+7x29)

9. (12x213x+16)+(32x2+23x1)

10. (35x2+14x6)+(2x238x+52)

11. (x2+5)+(3x22x+1)+(x2+x3)

12. (a3a2+a8)+(a3+a2+6a2)

13. (a38)+(3a3+5a22)

14. (4a5+5a3a)+(3a42a2+7)

15. (2x2+5x12)+(7x5)

16. (3x+5)+(x2x+1)+(x3+2x23x+6)

17. (6x57x3+x215)+(x4+2x36x+12)

18. (1+7x5x3+4x4)+(3x3+5x2+x)

19. (x2y27xy+7)+(4x2y23xy8)

20. (x2+xyy2)+(7x25xy+2y2)

21. (2x2+3xy7y2)+(5x23xy+8y2)

22. (a2b2100)+(2a2b23ab+20)

23. (ab23a2b+ab3)+(2a2b+ab27ab1)

24. (10a2b7ab+8ab2)+(6a2bab+5ab2)

25. Find the sum of 2x+8 and 7x1.

26. Find the sum of 13x15 and 16x+110.

27. Find the sum of x210x+8 and 5x22x6.

28. Find the sum of a25a+10 and 9a2+7a11.

29. Find the sum of x2y2xy+6 and x2y2+xy7.

30. Find the sum of x29xy+7y2 and 3x23xy+7y2.

Part B: Subtraction of Polynomials

Subtract.

31. (5x3)(2x1)

32. (4x+1)(7x+10)

33. (14x34)(34x+18)

34. (35x+37)(25x32)

35. (x2+7x5)(4x25x+1)

36. (6x2+3x12)(6x2+3x12)

37. (3x3+4x8)(x2+4x+10)

38. (12x2+13x34)(32x216x+12)

39. (59x2+15x13)(13x2+310x+59)

40. (a34a2+3a7)(7a32a26a+9)

41. (3a3+5a22)(a3a+8)

42. (5x5+4x3+x26)(4x43x3x+3)

43. (35xx3+5x4)(5x3+2x27x)

44. (x56x3+9x)(4x4+2x25)

45. (2x2y24xy+9)(3x2y23xy5)

46. (x2+xyy2)(x2+xyy2)

47. (2x2+3xy7y2)(5x23xy+8y2)

48. (ab23a2b+ab3)(2a2b+ab27ab1)

49. (10a2b7ab+8ab2)(6a2bab+5ab2)

50. (10a2b2+5ab6)(5a2b2+5ab6)

51. Subtract 3x+1 from 5x9.

52. Subtract x25x+10 from x2+5x5.

53. Find the difference of 3x7 and 8x+6.

54. Find the difference of 2x2+3x5 and x29.

55. The cost in dollars of producing customized coffee mugs with a company logo is given by the formula C=150+0.10x, where x is the number of cups produced. The revenue from selling the cups in the company store is given by R=10x0.05x2, where x is the number of units sold.

a. Find a formula for the profit. (profit = revenue − cost)

b. Find the profit from producing and selling 100 mugs in the company store.

56. The cost in dollars of producing sweat shirts is given by the formula C=10q+1200, where C is the cost and q represents the quantity produced. The revenue generated by selling the sweat shirts for $37 each is given by R=37q, where q represents the quantity sold. Determine the profit generated if 125 sweat shirts are produced and sold.

57. The outer radius of a washer is 3 times the radius of the hole.

a. Derive a formula for the area of the face of the washer.

b. What is the area of the washer if the hole has a diameter of 10 millimeters?

58. Derive a formula for the surface area of the following rectangular solid.

Part C: Addition and Subtraction of Polynomial

Simplify.

59. (2x+3)(5x8)+(x7)

60. (3x5)(7x11)(5x+2)

61. (3x2)(4x1)+(x+7)

62. (5x3)(2x+1)(x1)

63. (5x23x+2)(x2+x4)+(7x22x6)

64. (2x3+x28)(3x2+x6)(2x1)

65. (2x7)(x2+3x7)+(6x1)

66. (6x210x+13)+(4x29)(9x2)

67. (a2b2)(2a2+3ab4b2)+(5ab1)

68. (a23ab+b2)(a2+b2)(3ab5)

69. (12x234x+14)(32x34)+(54x12)

70. (95x213x+2)(310x245)(x+52)

Part D: Addition and Subtraction of Polynomial Functions

Find (f+g)(x) and (fg)(x), given the following functions.

71. f(x)=4x1 and g(x)=3x+1

72. f(x)=x+5 and g(x)=2x3

73. f(x)=3x25x+7 and g(x)=2x2+5x1

74. f(x)=x3+2x26x+2 and g(x)=2x3+2x25x1

75. f(x)=12x+13 and g(x)=15x232x+16

76. f(x)=x25x+13 and g(x)=23x2x12

Given f(x)=2x3 and g(x)=x2+3x1, find the following.

77. (f+g)(x)

78. (g+f)(x)

79. (fg)(x)

80. (gf)(x)

81. (g+g)(x)

82. (f+g)(3)

83. (f+g)(2)

84. (f+g)(0)

85. (fg)(0)

86. (fg)(2)

87. (gf)(2)

88. (gf)(12)

Given f(x)=5x23x+2 and g(x)=2x2+6x4, find the following.

89. (f+g)(x)

90. (g+f)(x)

91. (fg)(x)

92. (gf)(x)

93. (f+g)(2)

94. (fg)(2)

95. (f+g)(0)

96. (fg)(0)

Answers

1: x+8

3: x32

5: 8x4

7: 4x211x+2

9: x2+13x56

11: 5x2x+3

13: 2a3+5a210

15: 2x2+12x17

17: 6x5+x45x3+x26x3

19: 5x2y210xy1

21: 3x2+y2

23: 5a2b+2ab26ab4

25: 9x+7

27: 6x212x+2

29: 2x2y21

31: 3x2

33: 12x78

35: 3x2+12x6

37: 3x3+x218

39: 29x2110x89

41: 2a3+5a2+a10

43: 5x4+4x3+x2+2x+1

45: x2y2xy+14

47: 7x2+6xy15y2

49: 4a2b+3ab26ab

51: 2x10

53: 5x13

55: a. P=0.05x2+9.9x150; b. $340

57: a. A=8πr2; b. 628.32 square millimeters

59: 2x+4

61: 6

63: 11x26x

65: x2+5x1

67: a2+2ab+3b21

69: 12x2x+12

71: (f+g)(x)=x and (fg)(x)=7x2

73: (f+g)(x)=x2+6 and (fg)(x)=5x210x+8

75: (f+g)(x)=15x2x+12 and (fg)(x)=15x2+2x+16

77: (f+g)(x)=x2+5x4

79: (fg)(x)=x2x2

81: (g+g)(x)=2x2+6x2

83: (f+g)(2)=10

85: (fg)(0)=2

87: (gf)(2)=4

89: (f+g)(x)=7x2+3x2

91: (fg)(x)=3x29x+6

93: (f+g)(2)=20

95: (f+g)(0)=2

5.4 Multiplying Polynomials

Learning Objectives

  1. Multiply a polynomial by a monomial.
  2. Multiply a polynomial by a binomial.
  3. Multiply a polynomial by any size polynomial.
  4. Recognize and calculate special products.
  5. Multiply polynomial functions.

Multiplying by a Monomial

Recall the product rule for exponents: if m and n are positive integers, then

In other words, when multiplying two expressions with the same base, add the exponents. This rule applies when multiplying a monomial by a monomial. To find the product of monomials, multiply the coefficients and add the exponents of variable factors with the same base. For example,

To multiply a polynomial by a monomial, apply the distributive property and then simplify each term.

 

Example 1: Multiply: 5x(4x2).

Solution: In this case, multiply the monomial, 5x, by the binomial, 4x2. Apply the distributive property and then simplify.

Answer: 20x2+10x

 

Example 2: Multiply: 2x2(3x25x+1).

Solution: Apply the distributive property and then simplify.

Answer: 6x410x3+2x2

 

Example 3: Multiply: 3ab2(a2b3+2a3b6ab4).

Solution:

Answer: 3a3b56a4b3+18a2b3+12ab2

 

To summarize, multiplying a polynomial by a monomial involves the distributive property and the product rule for exponents. Multiply all of the terms of the polynomial by the monomial. For each term, multiply the coefficients and add exponents of variables where the bases are the same.

 

Try this! Multiply: 5x2y(2xy23xy+6x2y1).

Answer: 10x3y3+15x3y230x4y2+5x2y

Video Solution

(click to see video)

Multiplying by a Binomial

In the same way that we used the distributive property to find the product of a monomial and a binomial, we will use it to to find the product of two binomials.

Here we apply the distributive property multiple times to produce the final result. This same result is obtained in one step if we apply the distributive property to a and b separately as follows:

This is often called the FOILWhen multiplying binomials we apply the distributive property multiple times in such a way as to multiply the first terms, outer terms, inner terms, and last terms. method. We add the products of the first terms of each binomial ac, the outer terms ad, the inner terms bc, and finally the last terms bd. This mnemonic device only works for products of binomials; hence it is best to just remember that the distributive property applies.

 

Example 4: Multiply: (2x+3)(5x2).

Solution: Distribute 2x and then distribute 3.

Simplify by combining like terms.

Answer: 10x2+11x6

 

Example 5: Multiply: (12x14)(12x+14).

Solution: Distribute 12x and then distribute 14.

Answer: 14x2116

 

Example 6: Multiply: (3y21)(2y+1).

Solution:

Answer: 6y3+3y22y1

 

After applying the distributive property, combine any like terms.

 

Example 7: Multiply: (x25)(3x22x+2).

Solution: After multiplying each term of the trinomial by x2 and 5, simplify.

Answer: 3x42x313x2+10x10

 

Example 8: Multiply: (2x1)3.

Solution: Perform one product at a time.

Answer: 8x312x2+6x1

 

At this point, it is worth pointing out a common mistake:

The confusion comes from the product to a power rule of exponents, where we apply the power to all factors. Since there are two terms within the parentheses, that rule does not apply. Care should be taken to understand what is different in the following two examples:

 

Try this! Multiply: (2x3)(7x25x+4).

Answer: 14x331x2+23x12

Video Solution

(click to see video)

Product of Polynomials

When multiplying polynomials, we apply the distributive property many times. Multiply all of the terms of each polynomial and then combine like terms.

 

Example 9: Multiply: (2x2+x3)(x22x+5).

Solution: Multiply each term of the first trinomial by each term of the second trinomial and then combine like terms.

Aligning like terms in columns, as we have here, aids in the simplification process.

Answer: 2x43x3+5x2+11x15

 

Notice that when multiplying a trinomial by a trinomial, we obtain nine terms before simplifying. In fact, when multiplying an n-term polynomial by an m-term polynomial, we will obtain n × m terms.

In the previous example, we were asked to multiply and found that

Because it is easy to make a small calculation error, it is a good practice to trace through the steps mentally to verify that the operations were performed correctly. Alternatively, we can check by evaluatingWe can be fairly certain that we have multiplied the polynomials correctly if we check that a few values evaluate to the same results in the original expression and in the answer. any value for x in both expressions to verify that the results are the same. Here we choose x = 2:

Because the results could coincidentally be the same, a check by evaluating does not necessarily prove that we have multiplied correctly. However, after verifying a few values, we can be fairly confident that the product is correct.

 

Try this! Multiply: (x22x3)2.

Answer: x44x32x2+12x+9

Video Solution

(click to see video)

Special Products

In this section, the goal is to recognize certain special products that occur often in our study of algebra. We will develop three formulas that will be very useful as we move along. The three should be memorized. We begin by considering the following two calculations:

This leads us to two formulas that describe perfect square trinomialsThe trinomials obtained by squaring the binomials (a+b)2=a2+2ab+b2 and (ab)2=a22ab+b2.:

We can use these formulas to quickly square a binomial.

 

Example 10: Multiply: (3x+5)2.

Solution: Here a=3x and b=5. Apply the formula:

Answer: 9x2+30x+25

 

This process should become routine enough to be performed mentally.

 

Example 11: Multiply: (x4)2.

Solution: Here a=x and b=4. Apply the appropriate formula as follows:

Answer: x28x+16

 

Our third special product follows:

This product is called difference of squaresa2b2=(a+b)(ab), where a and b represent algebraic expressions.:

The binomials (a+b) and (ab) are called conjugate binomialsThe binomials (a+b) and (ab).. Therefore, when conjugate binomials are multiplied, the middle term eliminates, and the product is itself a binomial.

 

Example 12: Multiply: (7x+4)(7x4).

Solution:

Answer: 49x216

 

Try this! Multiply: (5x+2)2.

Answer: 25x220x+4

Video Solution

(click to see video)

Multiplying Polynomial Functions

We use function notation to indicate multiplication as follows:

Multiplication of functions: (fg)(x)=f(x)g(x)

 

Example 13: Calculate: (fg)(x), given f(x)=5x2 and g(x)=x2+2x3.

Solution: Multiply all terms of the trinomial by the monomial function f(x).

Answer: (fg)(x)=5x4+10x315x2

 

Example 14: Calculate: (fg)(1), given f(x)=x+3 and g(x)=4x23x+6.

Solution: First, determine (fg)(x).

We have

Next, substitute −1 for the variable x.

Answer: (fg)(1)=52

 

Because (fg)(1)=f(1)g(1), we could alternatively calculate f(1) and g(1) separately and then multiply the results (try this as an exercise). However, if we were asked to evaluate multiple values for the function (fg)(x), it would be best to first determine the general form, as we have in the previous example.

Key Takeaways

  • To multiply a polynomial by a monomial, apply the distributive property and then simplify each of the resulting terms.
  • To multiply polynomials, multiply each term in the first polynomial with each term in the second polynomial. Then combine like terms.
  • The product of an n-term polynomial and an m-term polynomial results in an m × n term polynomial before like terms are combined.
  • Check results by evaluating values in the original expression and in your answer to verify that the results are the same.
  • Use the formulas for special products to quickly multiply binomials that occur often in algebra.

Topic Exercises

Part A: Product of a Monomial and a Polynomial

Multiply.

1. 5x(3x2y)

2. (2x3y2)(3xy4)

3. 12(4x3)

4. 34(23x6)

5. 3x(5x2)

6. 4x(2x1)

7. x2(3x+2)

8. 6x2(5x+3)

9. 2ab(4a2b)

10. 5a2b(a2b2)

11. 6x2y3(3x3y+xy2)

12. 3ab3(5ab3+6a2b)

13. 12x2y(4xy10)

14. 3x4y2(3x8y3)

15. 2x2(5x3)(3x4)

16. 4ab(a2b3c)(a4b2c4)

17. 2(5x23x+4)

18. 45(25x250xy+5y2)

19. 3x(5x22x+3)

20. x(x2+x1)

21. x2(3x25x7)

22. x3(4x27x+9)

23. 14x4(8x32x2+12x5)

24. 13x3(32x523x3+92x1)

25. a2b(a23ab+b2)

26. 6a2bc3(2a3b+c2)

27. 23xy2(9x3y27xy+3xy3)

28. 3x2y2(12x210xy6y2)

29. Find the product of 3x and 2x23x+5.

30. Find the product of 8y and y22y+12.

31. Find the product of 4x and x43x3+2x27x+8.

32. Find the product of 3xy2 and 2x2y+4xyxy2.

Part B: Product of a Binomial and a Polynomial

Multiply.

33. (3x2)(x+4)

34. (x+2)(x3)

35. (x1)(x+1)

36. (3x1)(3x+1)

37. (2x5)(x+3)

38. (5x2)(3x+4)

39. (3x+1)(x1)

40. (x+5)(x+1)

41. (y23)(y+23)

42. (12x+13)(32x23)

43. (34x+15)(14x+25)

44. (15x+310)(35x52)

45. (y22)(y+2)

46. (y31)(y2+2)

47. (a2b2)(a2+b2)

48. (a23b)2

49. (x5)(2x2+3x+4)

50. (3x1)(x24x+7)

51. (2x3)(4x2+6x+9)

52. (5x+1)(25x25x+1)

53. (x12)(3x2+4x1)

54. (13x14)(3x2+9x3)

55. (x+3)3

56. (x2)3

57. (3x1)3

58. (2x+y)3

59. (5x2)(2x34x2+3x2)

60. (x22)(x32x2+x+1)

Part C: Product of Polynomials

Multiply.

61. (x2x+1)(x2+2x+1)

62. (3x22x1)(2x2+3x4)

63. (2x23x+5)(x2+5x1)

64. (a+b+c)(abc)

65. (a+2bc)2

66. (x+y+z)2

67. (x3)4

68. (x+y)4

69. Find the volume of a rectangular solid with sides measuring x , x+2, and x+4 units.

70. Find the volume of a cube where each side measures x5 units.

Part D: Special Products

Multiply.

71. (x+2)2

72. (x3)2

73. (2x+5)2

74. (3x7)2

75. (x+2)2

76. (9x+1)2

77. (a+6)2

78. (2a3b)2

79. (23x+34)2

80. (12x35)2

81. (x2+2)2

82. (x2+y2)2

83. (x+4)(x4)

84. (2x+1)(2x1)

85. (5x+3)(5x3)

86. (15x13)(15x+13)

87. (32x+25)(32x25)

88. (2x3y)(2x+3y)

89. (4xy)(4x+y)

90. (a3b3)(a3+b3)

91. A box is made by cutting out the corners and folding up the edges of a square piece of cardboard. A template for a cardboard box with a height of 2 inches is given. Find a formula for the volume, if the initial piece of cardboard is a square with sides measuring x inches.

92. A template for a cardboard box with a height of x inches is given. Find a formula for the volume, if the initial piece of cardboard is a square with sides measuring 12 inches.

Part E: Multiplying Polynomial Functions

For each problem, calculate (fg)(x), given the functions.

93. f(x)=8x and g(x)=3x5

94. f(x)=x2 and g(x)=5x+1

95. f(x)=x7 and g(x)=6x1

96. f(x)=5x+3 and g(x)=x2+2x3

97. f(x)=x2+6x3 and g(x)=2x23x+5

98. f(x)=3x2x+1 and g(x)=x2+2x1

Given f(x)=2x3 and g(x)=3x1, find the following.

99. (fg)(x)

100. (gf)(x)

101. (fg)(0)

102. (fg)(1)

103. (fg)(1)

104. (fg)(12)

Given f(x)=5x1 and g(x)=2x24x+5, find the following.

105. (fg)(x)

106. (gf)(x)

107. (fg)(0)

108. (fg)(1)

109. (fg)(1)

110. (fg)(12)

111. (ff)(x)

112. (gg)(x)

Part F: Discussion Board Topics

113. Explain why (x+y)2x2+y2.

114. Explain how to quickly multiply a binomial with its conjugate. Give an example.

115. What are the advantages and disadvantages of using the mnemonic device FOIL?

Answers

1: 15x3y

3: 2x32

5: 15x26x

7: 3x3+2x2

9: 8a2b4ab2

11: 18x5y4+6x3y5

13: 2x3y2+5x2y

15: 30x9

17: 10x2+6x8

19: 15x36x2+9x

21: 3x45x37x2

23: 2x712x6+18x554x4

25: a4b3a3b2+a2b3

27: 6x4y318x2y3+2x2y5

29: 6x39x2+15x

31: 4x5+12x48x3+28x232x

33: 3x2+10x8

35: x21

37: 2x2+x15

39: 3x2+4x1

41: y249

43: 316x2+720x+225

45: y3+2y22y4

47: a4b4

49: 2x37x211x20

51: 8x327

53: 3x3+52x23x+12

55: x3+9x2+27x+27

57: 27x327x2+9x1

59: 10x424x3+23x216x+4

61: x4+x3+x+1

63: 2x4+7x312x2+28x5

65: a2+4ab2ac+4b24bc+c2

67: x412x3+54x2108x+81

69: x3+6x2+8x

71: x2+4x+4

73: 4x2+20x+25

75: x24x+4

77: a2+12a+36

79: 49x2+x+916

81: x4+4x2+4

83: x216

85: 25x29

87: 94x2425

89: 16x2y2

91: V=2x216x+32 cubic inches

93: (fg)(x)=24x240x

95: (fg)(x)=6x243x+7

97: (fg)(x)=2x4+9x319x2+39x15

99: (fg)(x)=6x211x+3

101: (fg)(0)=3

103: (fg)(1)=2

105: (fg)(x)=10x322x2+29x5

107: (fg)(0)=5

109: (fg)(1)=12

111: (ff)(x)=25x210x+1

5.5 Dividing Polynomials

Learning Objectives

  1. Divide by a monomial.
  2. Divide by a polynomial using the division algorithm.
  3. Divide polynomial functions.

Dividing by a Monomial

Recall the quotient rule for exponents: if x is nonzero and m and n are positive integers, then

In other words, when dividing two expressions with the same base, subtract the exponents. This rule applies when dividing a monomial by a monomial. In this section, we will assume that all variables in the denominator are nonzero.

 

Example 1: Divide: 28y37y.

Solution: Divide the coefficients and subtract the exponents of the variable y.

Answer: 4y2

 

Example 2: Divide: 24x7y58x3y2.

Solution: Divide the coefficients and apply the quotient rule by subtracting the exponents of the like bases.

Answer: 3x4y3

 

When dividing a polynomial by a monomial, we may treat the monomial as a common denominator and break up the fraction using the following property:

Applying this property results in terms that can be treated as quotients of monomials.

 

Example 3: Divide: 5x4+25x315x25x2.

Solution: Break up the fraction by dividing each term in the numerator by the monomial in the denominator and then simplify each term.

Answer: x2+5x3

 

Check your division by multiplying the answer, the quotientThe result after dividing., by the monomial in the denominator, the divisorThe denominator of a quotient., to see if you obtain the original numerator, the dividendThe numerator of a quotient..

 

Example 4: Divide: 9a4b7a3b2+3a2b3a2b.

Solution:

Answer: 3a2+73ab1. The check is optional and is left to the reader.

 

Try this! Divide: (16x58x4+5x3+2x2)÷(2x2).

Answer: 8x34x2+52x+1

Video Solution

(click to see video)

Dividing by a Polynomial

The same technique outlined for dividing by a monomial does not work for polynomials with two or more terms in the denominator. In this section, we will outline a process called polynomial long divisionThe process of dividing two polynomials using the division algorithm., which is based on the division algorithm for real numbers. For the sake of clarity, we will assume that all expressions in the denominator are nonzero.

 

Example 5: Divide: x3+3x28x4x2.

Solution: Here x2 is the divisor and x3+3x28x4 is the dividend.

Step 1: To determine the first term of the quotient, divide the leading term of the dividend by the leading term of the divisor.

Step 2: Multiply the first term of the quotient by the divisor, remembering to distribute, and line up like terms with the dividend.

Step 3: Subtract the resulting quantity from the dividend. Take care to subtract both terms.

Step 4: Bring down the remaining terms and repeat the process from step 1.

Notice that the leading term is eliminated and that the result has a degree that is one less than the dividend. The complete process is illustrated below:

Polynomial long division ends when the degree of the remainderThe expression that is left after the division algorithm ends. is less than the degree of the divisor. Here the remainder is 0. Therefore, the binomial divides the polynomial evenly and the answer is the quotient shown above the division line.

To check the answer, multiply the divisor by the quotient to see if you obtain the dividend:

Answer: x2+5x+2

 

Next, we demonstrate the case where there is a nonzero remainder.

Just as with real numbers, the final answer adds the fraction where the remainder is the numerator and the divisor is the denominator to the quotient. In general, when dividing we have

If we multiply both sides by the divisor we obtain

 

Example 6: Divide: 6x25x+32x1.

Solution: Since the denominator is a binomial, begin by setting up polynomial long division.

To start, determine what monomial times 2x1 results in a leading term 6x2. This is the quotient of the given leading terms: (6x2)÷(2x)=3x. Multiply 3x times the divisor 2x1 and line up the result with like terms of the dividend.

Subtract the result from the dividend and bring down the constant term +3.

Subtracting eliminates the leading term and 5x(3x)=5x+3x=2x. The quotient of 2x and 2x is −1. Multiply 2x1 by −1 and line up the result.

Subtract again and notice that we are left with a remainder.

The constant term 2 has degree 0, and thus the division ends. We may write

Answer: 3x1+22x1. To check that this result is correct, we multiply as follows:

 

Occasionally, some of the powers of the variables appear to be missing within a polynomial. This can lead to errors when lining up like terms. Therefore, when first learning how to divide polynomials using long division, fill in the missing terms with zero coefficients, called placeholdersTerms with zero coefficients used to fill in all missing exponents within a polynomial..

 

Example 7: Divide: 27x3+643x+4.

Solution: Notice that the binomial in the numerator does not have terms with degree 2 or 1. The division is simplified if we rewrite the expression with placeholders:

Set up polynomial long division:

We begin with 27x3÷3x=9x2 and work the rest of the division algorithm.

Answer: 9x212x+16

 

Example 8: Divide: 3x42x3+6x2+23x7x22x+5.

Solution:

Begin the process by dividing the leading terms to determine the leading term of the quotient 3x4÷x2=3x2. Take care to distribute and line up the like terms. Continue the process until the remainder has a degree less than 2.

The remainder is x2. Write the answer with the remainder:

Answer: 3x2+4x1+x2x22x+5

 

Polynomial long division takes time and practice to master. Work lots of problems and remember that you may check your answers by multiplying the quotient by the divisor (and adding the remainder if present) to obtain the dividend.

 

Try this! Divide: 20x432x3+7x2+8x10 5x3.

Answer: 4x34x2x+175x3

Video Solution

(click to see video)

Dividing Polynomial Functions

We may use function notation to indicate division as follows:

Division of functions: (f/g)(x)=f(x)g(x)

The quotient of two polynomial functions does not necessarily have a domain of all real numbers. The values for x that make the function in the denominator 0 are restricted from the domain. This will be discussed in more detail at a later time. For now, assume all functions in the denominator are nonzero.

 

Example 9: Calculate: (f/g)(x) given f(x)=6x536x4+12x36x2 and g(x)=6x2.

Solution: The notation indicates that we should divide:

Answer: (f/g)(x)=x3+6x22x+1

 

Example 10: Calculate: (f/g)(1), given f(x)=3x3+7x211x1 and g(x)=3x1.

Solution: First, determine (f/g)(x).

Therefore,

Substitute −1 for the variable x.

Answer: (f/g)(1)=4

Key Takeaways

  • When dividing by a monomial, divide all terms in the numerator by the monomial and then simplify each term. To simplify each term, divide the coefficients and apply the quotient rule for exponents.
  • When dividing a polynomial by another polynomial, apply the division algorithm.
  • To check the answer after dividing, multiply the divisor by the quotient and add the remainder (if necessary) to obtain the dividend.
  • It is a good practice to include placeholders when performing polynomial long division.

Topic Exercises

Part A: Dividing by a Monomial

Divide.

1. 81y59y2

2. 36y99y3

3. 52x2y4xy

4. 24xy52xy4

5. 25x2y5z35xyz

6. 77x4y9z22x3y3z

7. 125a3b2c10abc

8. 36a2b3c56a2b2c3

9. 9x2+27x33

10. 10x35x2+40x155

11. 20x310x2+30x2x

12. 10x4+8x26x24x

13. 6x59x3+3x3x

14. 36a126a9+12a512a5

15. 12x5+18x36x26x2

16. 49a8+7a521a37a3

17. 9x76x4+12x3x23x2

18. 8x9+16x724x4+8x38x3

19. 16a732a6+20a5a44a4

20. 5a6+2a5+6a312a23a2

21. 4x2y3+16x7y88x2y54x2y3

22. 100a10b30c550a20b5c40+20a5b20c1010a5b5c5

23. Find the quotient of 36x9y7 and 2x8y5.

24. Find the quotient of 144x3y10z2 and 12x3y5z.

25. Find the quotient of 3a418a3+27a2 and 3a2.

26. Find the quotient of 64a2bc316a5bc7 and 4a2bc3.

Part B: Dividing by a Polynomial

Divide.

27. (2x25x3)÷(x3)

28. (3x2+5x2)÷(x+2)

29. (6x2+11x+3)÷(3x+1)

30. (8x214x+3)÷(2x3)

31. x3x22x12x3

32. 2x3+11x2+4x5x+5

33. 2x3x24x+32x+3

34. 15x314x2+23x65x2

35. 14x49x3+22x2+4x17x1

36. 8x5+16x48x35x221x+102x+5

37. x2+8x+17x+5

38. 2x25x+5x2

39. 6x213x+92x+1

40. 12x2+x+13x+2

41. x3+9x2+19x+1x+4

42. 2x313x2+17x11x5

43. 9x312x2+16x153x2

44. 3x48x3+5x25x+9x2

45. (6x513x4+4x33x2+13x2)÷(3x+1)

46. (8x522x4+19x320x2+23x3)÷(2x3)

47. 5x5+12x4+12x37x219x+3x2+2x+3

48. 6x517x4+5x3+16x27x32x23x1

49. x5+7 x4x37 x249 x+9x2+7x1

50. 5x66x44x2+x+25x21

51. x327x3

52. 8x3+1252x+5

53. (15x59x420x3+12x2+15x9)÷(5x3)

54. (2x65x54x4+10x3+6x217x+5)÷(2x5)

55. x52x3+3x1x1

56. x43x2+5x13x+2

57. a24a+2

58. a5+1a5+1

59. a61a1

60. x51x1

61. x5+x4+6x3+12x24x2+x1

62. 50x630x55x4+15 x35x+15x23x+2

63. 5x515x3+25x255x

64. 36x6+12x46x26x2

65. 150x5y2z1510x3y6z5+4x3y2z410x3y2z5

66. 27m6+9m481m2+19m2

67. Divide 3x62x5+27x418x36x2+7x10 by 3x2.

68. Divide 8x6+4x514x45x3+x22x3 by 2x+1.

Part C: Dividing Polynomial Functions

Calculate (f/g)(x), given the functions.

69. f(x)=40x8 and g(x)=10x5

70. f(x)=54x5 and g(x)=9x3

71. f(x)=12x2+24x15 and g(x)=2x+5

72. f(x)=8x2+30x7 and g(x)=2x7

73. f(x)=18x236x+5 and g(x)=3x5

74. f(x)=7x2+29x6 and g(x)=7x1

75. f(x)=10x39x2+27x10 and g(x)=5x2

76. f(x)=15x3+28x211x+56 and g(x)=3x+8

77. f(x)=2x4+5x311x219x+20 and g(x)=x2+x5

78. f(x)=4x412x320x2+26x3 and g(x)=2x2+2x3

Given f(x)=6x3+4x211x+3 and g(x)=3x1, find the following.

79. (f/g)(x)

80. (f/g)(1)

81. (f/g)(0)

82. (f/g)(1)

Given f(x)=5x313x2+7x+3 and g(x)=x2, find the following.

83. (f/g)(x)

84. (f/g)(3)

85. (f/g)(0)

86. (f/g)(7)

Part D: Discussion Board Topics

87. How do you use the distributive property when dividing a polynomial by a monomial?

88. Compare long division of real numbers with polynomial long division. Provide an example of each.

Answers

1: 9y3

3: 13x

5: 5xy4z2

7: 252a2b

9: 3x2+9x1

11: 10x25x+15

13: 2x4+3x21

15: 2x33x+1

17: 3x52x2+4x13

19: 4a38a2+5a14

21: 4x5y5+2y2+1

23: 18xy2

25: a26a+9

27: 2x+1

29: 2x+3

31: x2+2x+4

33: x22x+1

35: 2x3x2+3x+1

37: x+3+2x+5

39: 3x+5+42x+1

41: x2+5x1+5x+4

43: 3x22x+473x2

45: 2x45x3+3x22x+573x+1

47: 5x3+2x27x+1

49: x37+2x2+7x1

51: x2+3x+9

53: 3x44x2+3

55: x4+x3x2x+2+1x1

57: a2

59: a5+a4+a3+a2+a+1

61: x3+7x+5+2x+1x2+x1

63: x43x2+5x1x

65: 15x2z10y4+25z

67: x5+9x32x+183x2

69: (f/g)(x)=4x3

71: (f/g)(x)=6x3

73: (f/g)(x)=6x253x5

75: (f/g)(x)=2x2x+5

77: (f/g)(x)=2x2+3x4

79: (f/g)(x)=2x2+2x3

81: (f/g)(0)=3

83: (f/g)(x)=5x23x+1+5x2

85: (f/g)(0)=32

5.6 Negative Exponents

Learning Objectives

  1. Simplify expressions with negative integer exponents.
  2. Work with scientific notation.

Negative Exponents

In this section, we define what it means to have negative integer exponents. We begin with the following equivalent fractions:

Notice that 4, 8, and 32 are all powers of 2. Hence we can write 4=22, 8=23, and 32=25.

If the exponent of the term in the denominator is larger than the exponent of the term in the numerator, then the application of the quotient rule for exponents results in a negative exponent. In this case, we have the following:

We conclude that 23=123. This is true in general and leads to the definition of negative exponentsxn=1xn, given any integer n, where x is nonzero.. Given any integer n and x0, then

Here x0 because 10 is undefined. For clarity, in this section, assume all variables are nonzero.

Simplifying expressions with negative exponents requires that we rewrite the expression with positive exponents.

 

Example 1: Simplify: 102.

Solution:

Answer: 1100

 

Example 2: Simplify: (3)1.

Solution:

Answer: 13

 

Example 3: Simplify: 1y3.

Solution:

Answer: y3

 

At this point we highlight two very important examples,

If the grouped quantity is raised to a negative exponent, then apply the definition and write the entire grouped quantity in the denominator. If there is no grouping, then apply the definition only to the base preceding the exponent.

 

Example 4: Simplify: (2ab)3.

Solution: First, apply the definition of −3 as an exponent and then apply the power of a product rule.

Answer: 18a3b3

 

Example 5: Simplify: (3xy3)2.

Solution:

Answer: 19x2y6

 

Example 6: Simplify: x3y4.

Solution:

Answer: y4x3

 

The previous example suggests a property of quotients with negative exponentsxnym=ymxn, given any integers m and n, where x0 and y0.. If given any integers m and n, where x0 and y0, then

In other words, negative exponents in the numerator can be written as positive exponents in the denominator, and negative exponents in the denominator can be written as positive exponents in the numerator.

 

Example 7: Simplify: 2x5y3z2.

Solution: Take care with the coefficient −2; recognize that this is the base and that the exponent is actually +1: 2=(2)1. Hence the rules of negative exponents do not apply to this coefficient; leave it in the numerator.

Answer: 2y3z2x5

 

Example 8: Simplify: (3 x 4)3y2.

Solution: Apply the power of a product rule before applying negative exponents.

Answer: x12y227

 

Example 9: Simplify: (3 x 2)4(2 y 1 z 3)2.

Solution:

Answer: 4z681x8y2

 

Example 10: Simplify: (5 x 2y)3x5y3.

Solution: First, apply the power of a product rule and then the quotient rule.

Answer: 125x11y6

 

To summarize, we have the following rules for negative integer exponents with nonzero bases:

Negative exponents: xn=1xn
Quotients with negative exponents: xnym=ymxn

 

Try this! Simplify: (5x y 3)25x4y4.

Answer: y10125x6

Video Solution

(click to see video)

Scientific Notation

Real numbers expressed in scientific notationReal numbers expressed in the form a×10n, where n is an integer and 1a<10. have the form

where n is an integer and 1a<10. This form is particularly useful when the numbers are very large or very small. For example,

It is cumbersome to write all the zeros in both of these cases. Scientific notation is an alternative, compact representation of these numbers. The factor 10n indicates the power of 10 to multiply the coefficient by to convert back to decimal form:

This is equivalent to moving the decimal in the coefficient fifteen places to the right. A negative exponent indicates that the number is very small:

This is equivalent to moving the decimal in the coefficient eleven places to the left.

Converting a decimal number to scientific notation involves moving the decimal as well. Consider all of the equivalent forms of 0.00563 with factors of 10 that follow:

While all of these are equal, 5.63×103 is the only form considered to be expressed in scientific notation. This is because the coefficient 5.63 is between 1 and 10 as required by the definition. Notice that we can convert 5.63×103 back to decimal form, as a check, by moving the decimal to the left three places.

 

Example 11: Write 1,075,000,000,000 using scientific notation.

Solution: Here we count twelve decimal places to the left of the decimal point to obtain the number 1.075.

Answer: 1.075×1012

 

Example 12: Write 0.000003045 using scientific notation.

Solution: Here we count six decimal places to the right to obtain 3.045.

Answer: 3.045×106

 

Often we will need to perform operations when using numbers in scientific notation. All the rules of exponents developed so far also apply to numbers in scientific notation.

 

Example 13: Multiply: (4.36×105)(5.3×1012).

Solution: Use the fact that multiplication is commutative and apply the product rule for exponents.

Answer: 2.3108×108

 

Example 14: Divide: (3.24×108)÷(9.0×103).

Solution:

Answer: 3.6×1010

 

Example 15: The speed of light is approximately 6.7×108 miles per hour. Express this speed in miles per second.

Solution: A unit analysis indicates that we must divide the number by 3,600.

Answer: The speed of light is approximately 1.9×105 miles per second.

 

Example 16: By what factor is the radius of the sun larger than the radius of earth?

Solution: We want to find the number that when multiplied times the radius of earth equals the radius of the sun.

Therefore,

Answer: The radius of the sun is approximately 110 times that of earth.

 

Try this! Divide: (6.75×108)÷(9×1017).

Answer: 7.5×108

Video Solution

(click to see video)

Key Takeaways

  • Expressions with negative exponents in the numerator can be rewritten as expressions with positive exponents in the denominator.
  • Expressions with negative exponents in the denominator can be rewritten as expressions with positive exponents in the numerator.
  • Take care to distinguish negative coefficients from negative exponents.
  • Scientific notation is particularly useful when working with numbers that are very large or very small.

Topic Exercises

Part A: Negative Exponents

Simplify. (Assume variables are nonzero.)

1. 51

2. 52

3. (7)1

4. 71

5. 123

6. 532

7. (35)2

8. (12)5

9. (23)4

10. (13)3

11. x4

12. y1

13. 3x5

14. (3x)5

15. 1y3

16. 52x1

17. x1y2

18. 1(xy)4

19. x2y3z5

20. xy3

21. (ab)1

22. 1(ab)1

23. 5x3y2z4

24. 32x3y5z

25. 3x4y22x1y3

26. 10a2b32a8b10

27. (2a3)2

28. (3x2)1

29. (5a2b3c)2

30. (7r3s5t)3

31. (2r2s0t3)1

32. (2xy3z2)3

33. (5a2b3c0)4

34. (x2y3z4)7

35. (12 x 3)5

36. (2x y 2)2

37. (x2 y 1)4

38. (3 a 2b c 5)5

39. (20 x 3 y 25y z 1)1

40. (4 r 5 s 3 t 42 r 3s t 0)3

41. (2x y 3 z 1 y 2 z 3)3

42. (3 a 2bca b 0 c 4)2

43. (xyz x 4 y 2 z 3)4

44. (125 x 3 y 4 z 55 x 2 y 4 ( x+y ) 3)0

45. (xn)2

46. ( x n y n)2

The value in dollars of a new MP3 player can be estimated by using the formula V=100(t+1)1, where t is the number of years after purchase.

47. How much was the MP3 player worth new?

48. How much will the MP3 player be worth in 1 year?

49. How much will the MP3 player be worth in 4 years?

50. How much will the MP3 player be worth in 9 years?

51. How much will the MP3 player be worth in 99 years?

52. According to the formula, will the MP3 ever be worthless? Explain.

Part B: Scientific Notation

Convert to a decimal number.

53. 9.3×109

54. 1.004×104

55. 6.08×1010

56. 3.042×107

57. 4.01×107

58. 1.0×1010

59. 9.9×103

60. 7.0011×105

Rewrite using scientific notation.

61. 500,000,000

62. 407,300,000,000,000

63. 9,740,000

64. 100,230

65. 0.0000123

66. 0.000012

67. 0.000000010034

68. 0.99071

Perform the indicated operations.

69. (3×105)(9×104)

70. (8×1022)(2×1012)

71. (2.1×1019)(3.0×108)

72. (4.32×107)(1.50×1018)

73. 9.12×1093.2×1010

74. 1.15×1092.3×1011

75. 1.004×1082.008×1014

76. 3.276×10255.2×1015

77. 59,000,000,000,000 × 0.000032

78. 0.0000000000432 × 0.0000000000673

79. 1,030,000,000,000,000,000 ÷ 2,000,000

80. 6,045,000,000,000,000 ÷ 0.00000005

81. The population density of earth refers to the number of people per square mile of land area. If the total land area on earth is 5.751×107 square miles and the population in 2007 was estimated to be 6.67×109 people, then calculate the population density of earth at that time.

82. In 2008 the population of New York City was estimated to be 8.364 million people. The total land area is 305 square miles. Calculate the population density of New York City.

83. The mass of earth is 5.97×1024 kilograms and the mass of the moon is 7.35×1022 kilograms. By what factor is the mass of earth greater than the mass of the moon?

84. The mass of the sun is 1.99×1030 kilograms and the mass of earth is 5.97×1024 kilograms. By what factor is the mass of the sun greater than the mass of earth? Express your answer in scientific notation.

85. The radius of the sun is 4.322×105 miles and the average distance from earth to the moon is 2.392×105 miles. By what factor is the radius of the sun larger than the average distance from earth to the moon?

86. One light year, 9.461×1015 meters, is the distance that light travels in a vacuum in one year. If the distance to the nearest star to our sun, Proxima Centauri, is estimated to be 3.991×1016 meters, then calculate the number of years it would take light to travel that distance.

87. It is estimated that there are about 1 million ants per person on the planet. If the world population was estimated to be 6.67 billion people in 2007, then estimate the world ant population at that time.

88. The sun moves around the center of the galaxy in a nearly circular orbit. The distance from the center of our galaxy to the sun is approximately 26,000 light years. What is the circumference of the orbit of the sun around the galaxy in meters?

89. Water weighs approximately 18 grams per mole. If one mole is about 6×1023 molecules, then approximate the weight of each molecule of water.

90. A gigabyte is 1×109 bytes and a megabyte is 1×106 bytes. If the average song in the MP3 format consumes about 4.5 megabytes of storage, then how many songs will fit on a 4-gigabyte memory card?

Answers

1: 15

3: 17

5: 8

7: 259

9: 8116

11: 1x4

13: 3x5

15: y3

17: y2x

19: x2z5y3

21: 1ab

23: 5y2x3z4

25: 6y5x5

27: a64

29: b625a4c2

31: t32r2

33: 625a8b12

35: 32x15

37: 16x4y4

39: x34yz

41: z128x3y3

43: x12z8y12

45: 1x2n

47: $100

49: $20

51: $1

53: 9,300,000,000

55: 60,800,000,000

57: 0.000000401

59: 0.0099

61: 5×108

63: 9.74×106

65: 1.23×105

67: 1.0034×108

69: 2.7×1010

71: 6.3×1011

73: 2.85×1019

75: 5×105

77: 1.888×109

79: 5.15×1011

81: About 116 people per square mile

83: 81.2

85: 1.807

87: 6.67×1015 ants

89: 3×1023 grams

5.7 Review Exercises and Sample Exam

Review Exercises

Rules of Exponents

Simplify.

1. 7376

2. 5956

3. y5y2y3

4. x3y2xy3

5. 5a3b2c6a2bc2

6. 55x2yz55xyz2

7. (3 a 2 b 42 c 3)2

8. (2 a 3b4 c 4)3

9. 5x3y0(z2)32x4(y3)2z

10. (25x6y5z)0

11. Each side of a square measures 5x2 units. Find the area of the square in terms of x.

12. Each side of a cube measures 2x3 units. Find the volume of the cube in terms of x.

Introduction to Polynomials

Classify the given polynomial as a monomial, binomial, or trinomial and state the degree.

13. 8a31

14. 5y2y+1

15. 12ab2

16. 10

Write the following polynomials in standard form.

17. 7x25x

18. 5x213x+2x3

Evaluate.

19. 2x2x+1, where x=3

20. 12x34, where x=13

21. b24ac, where a=12, b=3, and c=32

22. a2b2, where a=12 and b=13

23. a3b3, where a=2 and b=1

24. xy22x2y, where x=3 and y=1

25. Given f(x)=3x25x+2, find f(2).

26. Given g(x)=x3x2+x1, find g(1).

27. The surface area of a rectangular solid is given by the formula SA=2lw+2wh+2lh, where l, w, and h represent the length, width, and height, respectively. If the length of a rectangular solid measures 2 units, the width measures 3 units, and the height measures 5 units, then calculate the surface area.

28. The surface area of a sphere is given by the formula SA=4πr2, where r represents the radius of the sphere. If a sphere has a radius of 5 units, then calculate the surface area.

Adding and Subtracting Polynomials

Perform the operations.

29. (3x4)+(9x1)

30. (13x19)+(16x+12)

31. (7x2x+9)+(x25x+6)

32. (6x2y5xy23)+(2x2y+3xy2+1)

33. (4y+7)(6y2)+(10y1)

34. (5y23y+1)(8y2+6y11)

35. (7x2y23xy+6)(6x2y2+2xy1)

36. (a3b3)(a3+1)(b31)

37. (x5x3+x1)(x4x2+5)

38. (5x34x2+x3)(5x33)+(4x2x)

39. Subtract 2x1 from 9x+8.

40. Subtract 3x210x2 from 5x2+x5.

41. Given f(x)=3x2x+5 and g(x)=x29, find (f+g)(x).

42. Given f(x)=3x2x+5 and g(x)=x29, find (fg)(x).

43. Given f(x)=3x2x+5 and g(x)=x29, find (f+g)(2).

44. Given f(x)=3x2x+5 and g(x)=x29, find (fg)(2).

Multiplying Polynomials

Multiply.

45. 6x2(5x4)

46. 3ab2(7a2b)

47. 2y(5y12)

48. 3x(3x2x+2)

49. x2y(2x2y5xy2+2)

50. 4ab(a28ab+b2)

51. (x8)(x+5)

52. (2y5)(2y+5)

53. (3x1)2

54. (3x1)3

55. (2x1)(5x23x+1)

56. (x2+3)(x32x1)

57. (5y+7)2

58. (y21)2

59. Find the product of x21 and x2+1.

60. Find the product of 32x2y and 10x30y+2.

61. Given f(x)=7x2 and g(x)=x23x+1, find (fg)(x).

62. Given f(x)=x5 and g(x)=x29, find (fg)(x).

63. Given f(x)=7x2 and g(x)=x23x+1, find (fg)(1).

64. Given f(x)=x5 and g(x)=x29, find (fg)(1).

Dividing Polynomials

Divide.

65. 7y214y+287

66. 12x530x3+6x6x

67. 4a2b16ab24ab4ab

68. 6a624a4+5a23a2

69. (10x219x+6)÷(2x3)

70. (2x35x2+5x6)÷(x2)

71. 10x421x316x2+23x202x5

72. x53x428x3+61x212x+36x6

73. 10x355x2+72x42x7

74. 3x4+19x3+3x216x113x+1

75. 5x4+4x35x2+21x+215x+4

76. x44x4

77. 2x4+10x323x215x+302x23

78. 7x417x3+17x211x+2x22x+1

79. Given f(x)=x34x+1 and g(x)=x1, find (f/g)(x).

80. Given f(x)=x532 and g(x)=x2, find (f/g)(x).

81. Given f(x)=x34x+1 and g(x)=x1, find (f/g)(2).

82. Given f(x)=x532 and g(x)=x2, find (f/g)(0).

Negative Exponents

Simplify.

83. (10)2

84. 102

85. 5x3

86. (5x)3

87. 17y3

88. 3x4y2

89. 2a2b5c8

90. (5x2yz1)2

91. (2x3y0z2)3

92. (10 a 5 b 3 c 25a b 2 c 2)1

93. ( a 2 b 4 c 02 a 4 b 3c)3

The value in dollars of a new laptop computer can be estimated by using the formula V=1200(t+1)1, where t represents the number of years after the purchase.

94. Estimate the value of the laptop when it is 1½ years old.

95. What was the laptop worth new?

Rewrite using scientific notation.

96. 2,030,000,000

97. 0.00000004011

Perform the indicated operations.

98. (5.2×1012)(1.8×103)

99. (9.2×104)(6.3×1022)

100. 4×10168×107

101. 9×10304×1010

102. 5,000,000,000,000 × 0.0000023

103. 0.0003/120,000,000,000,000

Sample Exam

Simplify.

1. 5x3(2x2y)

2. (x2)4x3x

3. (2 x 2 y 3)2x2y

4. a. (5)0; b. 50

Evaluate.

5. 2x2x+5, where x=5

6. a2b2, where a=4 and b=3

Perform the operations.

7. (3x24x+5)+(7x2+9x2)

8. (8x25x+1)(10x2+2x1)

9. (35a12)(23a2+23a29)+(115a518)

10. 2x2(2x33x24x+5)

11. (2x3)(x+5)

12. (x1)3

13. 81x5y2z3x3yz

14. 10x915x5+5x25x2

15. x35x2+7x2x2

16. 6x4x313x22x12x1

Simplify.

17. 23

18. 5x2

19. (2x4y3z)2

20. (2 a 3 b 5 c 2a b 3 c 2)3

21. Subtract 5x2y4xy2+1 from 10x2y6xy2+2.

22. If each side of a cube measures 4x4 units, calculate the volume in terms of x.

23. The height of a projectile in feet is given by the formula h=16t2+96t+10, where t represents time in seconds. Calculate the height of the projectile at 1½ seconds.

24. The cost in dollars of producing custom t-shirts is given by the formula C=120+3.50x, where x represents the number of t-shirts produced. The revenue generated by selling the t-shirts for $6.50 each is given by the formula R=6.50x, where x represents the number of t-shirts sold.

a. Find a formula for the profit. (profit = revenuecost)

b. Use the formula to calculate the profit from producing and selling 150 t-shirts.

25. The total volume of water in earth’s oceans, seas, and bays is estimated to be 4.73×1019 cubic feet. By what factor is the volume of the moon, 7.76×1020 cubic feet, larger than the volume of earth’s oceans? Round to the nearest tenth.

Review Exercises Answers

1: 79

3: y10

5: 30a5b3c3

7: 9a4b84c6

9: 10x7y6z7

11: A=25x4

13: Binomial; degree 3

15: Monomial; degree 3

17: x25x+7

19: 22

21: 6

23: −7

25: f(2)=24

27: 62 square units

29: 12x5

31: 8x26x+15

33: 8y+8

35: x2y25xy+7

37: x5x4x3+x2+x6

39: 7x+9

41: (f+g)(x)=4x2x4

43: (f+g)(2)=14

45: 30x6

47: 10y224y

49: 2x4y25x3y3+2x2y

51: x23x40

53: 9x26x+1

55: 10x311x2+5x1

57: 25y2+70y+49

59: x41

61: (fg)(x)=7x323x2+13x2

63: (fg)(1)=45

65: y22y+4

67: a+4b+1

69: 5x2

71: 5x3+2x23x+4

73: 5x210x+1+32x7

75: x3x+5+15x+4

77: x2+5x10

79: (f/g)(x)=x2+x32x1

81: (f/g)(2)=1

83: 1100

85: 5x3

87: y37

89: 2a2c8b5

91: x98z6

93: 8a6b3c3

95: $1,200

97: 4.011×108

99: 5.796×1019

101: 2.25×1020

103: 2.5×1018

Sample Exam Answers

1: 10x5y

3: 4x2y5

5: 60

7: 4x2+5x+3

9: 23a259

11: 2x2+7x15

13: 27x2y

15: x23x+1

17: 18

19: y64x8z2

21: 5x2y2xy2+1

23: 118 feet

25: 16.4