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3. Lesson 5-3 Dividing Polynomials Lesson 5-4 Factoring Polynomials
Lesson 5-1 Monomials
Lesson 5-2 Polynomials
Lesson 5-3 Dividing Polynomials
Lesson 5-4 Factoring Polynomials
Lesson 5-5 Roots of Real Numbers
Lesson 5-6 Radical Expressions
Lesson 5-7 Rational Exponents
Lesson 5-8 Radical Equations and Inequalities
Lesson 5-9 Complex Numbers
Contents

4. Example 1 Simplify Expressions with Multiplication
Example 2 Simplify Expressions with Division
Example 3 Simplify Expressions with Powers
Example 4 Simplify Expressions Using Several Properties
Example 5 Express Numbers in Scientific Notation
Example 6 Multiply Numbers in Scientific Notation
Example 7 Divide Numbers in Scientific Notation
Lesson 1 Contents

5. Definition of exponents
Commutative Property
Answer:
Definition of exponents
Example 1-1a

6. Answer:
Example 1-1b

7. Subtract exponents.
Remember that a simplified expression cannot contain negative exponents. 1
Answer:
Simplify.
Example 1-2a

8. Answer:
Example 1-2b

9. Power of a power
Answer:
Example 1-3a

10. Power of a power
Answer:
Example 1-3b

11. Power of a quotient
Power of a product
Answer:
Example 1-3c

12. Negative exponent
Power of a quotient
Answer:
Example 1-3d

13. Simplify each expression. a. b.d.
Answer:
Answer:
Answer:
Answer:
Example 1-3e

14. Method 1
Raise the numerator and the denominator to the fifth power before simplifying.
Example 1-4a

15. Answer:
Example 1-4a

16. Simplify the fraction before raising to the fifth power.
Method 2
Simplify the fraction before raising to the fifth power.
Answer:
Example 1-4b

17. Answer:
Example 1-4c

18. 4,560,000 Express 4,560,000 in scientific notation. Answer:
Write 1,000,000 as a power of 10.
Example 1-5a

19. Express 0.000092 in scientific notation.
Use a negative exponent.
Answer:
Example 1-5b

20. Express each number in scientific notation. a. 52,000
Answer:
Answer:
Example 1-5c

21. Express the result in scientific notation.
Associative and Commutative Properties
Answer:
Example 1-6a

22. Express the result in scientific notation.
Associative and Commutative Properties
Answer:
Example 1-6b

23. Evaluate. Express the result in scientific notation. a.b.
Answer:
Answer:
Example 1-6c

24.  number of red blood cells in sample
Biology There are about red blood cells in one milliliter of blood. A certain blood sample contains red blood cells. About how many milliliters of blood are in the sample?
Divide the number of red blood cells in the sample by the number of red blood cells in 1 milliliter of blood.
 number of red blood cells in sample
 number of red blood cells in 1 milliliter
Answer: There are about 1.66 milliliters of blood in the sample.
Example 1-7a

25. Biology A petri dish started with. germs in it
Biology A petri dish started with germs in it. A half hour later, there are How many times as great is the amount a half hour later?
Answer:
Example 1-7b

26. End of Lesson 1

27. Example 1 Degree of a Polynomial Example 2 Subtract and Simplify
Example 3 Multiply and Simplify
Example 4 Multiply Two Binomials
Example 5 Multiply Polynomials
Lesson 2 Contents

28. Answer: This expression is not a polynomial because is not a monomial.
Determine whether is a polynomial. If it is a polynomial, state the degree of the polynomial.
Answer: This expression is not a polynomial
because is not a monomial.
Example 2-1a

29. Determine whether is a polynomial.
If it is a polynomial, state the degree of the polynomial.
Answer: This expression is a polynomial because each term is a monomial. The degree of the first term is 5 and the degree of the second term is or 9. The degree of the polynomial is 9.
Example 2-1b

30. Determine whether each expression is a polynomial
Determine whether each expression is a polynomial. If it is a polynomial, state the degree of the polynomial. a. b.
Answer: yes, 5
Answer: no
Example 2-1c

31. Simplify Distribute the –1. Group like terms. Combine like terms.
Answer:
Example 2-2a

32. Simplify
Answer:
Example 2-2b

33. Distributive Property
Answer:
Multiply the monomials.
Example 2-3a

34. Answer:
Example 2-3b

35. Multiply monomials and add like terms.+
Outer terms
Inner terms
Last terms
First terms
Answer:
Multiply monomials and add like terms.
Example 2-4a

36. Answer:
Example 2-4b

37. Distributive Property
Multiply monomials.
Answer:
Combine like terms.
Example 2-5a

38. Answer:
Example 2-5b

39. End of Lesson 2

40. Example 1 Divide a Polynomial by a Monomial
Example 2 Division Algorithm
Example 3 Quotient with Remainder
Example 4 Synthetic Division
Example 5 Divisor with First Coefficient Other than 1
Lesson 3 Contents

41. Sum of quotients
Divide.
Answer:
Example 3-1a

42. Answer:
Example 3-1b

43. Use long division to find
Answer:
Example 3-2a

44. Use long division to find
Answer: x + 2
Example 3-2b

45. Multiple-Choice Test Item Which expression is equal to AB C D
Example 3-3a

46. Read the Test Item
Since the second factor has an exponent of –1, this is a division problem.
Solve the Test Item
Example 3-3b

47. The quotient is –a + 3 and the remainder is –3.
Therefore,
Answer: D
Example 3-3c

48. Multiple-Choice Test Item Which expression is equal to AB C D
Answer: B
Example 3-3d

49. Use synthetic division to find
Step 1 Write the terms of the dividend so that the degrees of the terms are in descending order. Then write just the coefficients as shown.
x3 – 4x2 + 6x – 4
   
1 – 4 6 – 4
Step 2 Write the constant r of the divisor x – r to the left. In this case, r = 2. Bring the first coefficient, 1, down as shown.
1 – 4 6 – 4 1
Example 3-4a

50. Step 3. Multiply the first coefficient by. Write the product
Step 3 Multiply the first coefficient by . Write the product under the second coefficient. Then add the product and the second coefficient. 1
1 –4 6 – 4 2 –2
Step 4 Multiply the sum, –2, by Write the product under the next coefficient and add: 2
1 –4 6 – 4 –2 1 –4 2
Example 3-4b

51. Step 5 Multiply the sum, 2, by. Write the product
Step 5 Multiply the sum, 2, by Write the product under the next coefficient and add: The remainder is 0. 2
1 –4 6 – 4 –2 1 –4 4 2
The numbers along the bottom are the coefficients of the quotient. Start with the power of x that is one less than the degree of the dividend.
Answer: The quotient is
Example 3-4c

52. Use synthetic division to find
Answer: x + 7
Example 3-4d

53. Use synthetic division to find
Use division to rewrite the divisor so it has a first coefficient of 1.
Divide numerator and denominator by 2.
Simplify the numerator and denominator.
Example 3-5a

54. Since the numerator does not have a y3 term, use a coefficient of 0 for y3.–
The result is
Now simplify the fraction.
Example 3-5b

55. Rewrite as a division expression.
Multiply by the reciprocal.
Multiply.
Answer: The solution is
Example 3-5c

56. Use synthetic division to find
Answer:
Example 3-5d

57. End of Lesson 3

58. Example 3 Two or Three Terms Example 4 Quotient of Two Trinomials
Example 1 GCF
Example 2 Grouping
Example 3 Two or Three Terms
Example 4 Quotient of Two Trinomials
Lesson 4 Contents

59. Distributive Property
Factor
The GCF is 5ab.
Answer:
Distributive Property
Example 4-1a

60. Factor
Answer:
Example 4-1b

61. Factor the GCF of each binomial.
Group to find the GCF.
Factor the GCF of each binomial.
Answer:
Distributive Property
Example 4-2a

63. Factoring with boxes Factor Answer: Remove common factor from top row.
Fill in the circles by finding the missing factors.
Answer:
Example 4-2a

64. Factor
Answer:
Example 4-2b

65. Factor
To find the coefficient of the y terms, you must find two numbers whose product is 3(–5) or –15 and whose sum is –2. The two coefficients must be 3 and –5 since and .
Rewrite the expression using –5y and 3y in place of –2y and factor by grouping.
Substitute –5y + 3y for –2y.
Example 4-3a

66. Factor out the GCF of each group.
Associative Property
Factor out the GCF of each group.
Answer:
Distributive Property
Example 4-3b

67. p2 – 9 is the difference of two squares.
Factor
Factor out the GCF.
Answer:
p2 – 9 is the difference of two squares.
Example 4-3c

68. This is the sum of two cubes.
Factor
This is the sum of two cubes.
Sum of two cubes formula with and
Answer:
Simplify.
Example 4-3d

69. Difference of two squares
Factor
This polynomial could be considered the difference of two squares or the difference of two cubes. The difference of two squares should always be done before the difference of two cubes.
Difference of two squares
Answer:
Sum and difference of two cubes
Example 4-3e

70. Factor each polynomial. a.b. c. d.
Answer:
Answer:
Answer:
Answer:
Example 4-3f

71. Factor the numerator and the denominator.
Simplify
Factor the numerator and the denominator.
Divide. Assume a  –5, –2.
Answer: Therefore,
Example 4-4a

72. Simplify
Answer:
Example 4-4b

73. End of Lesson 4

74. Example 2 Simplify Using Absolute Value
Example 1 Find Roots
Example 2 Simplify Using Absolute Value
Example 3 Approximate a Square Root
Lesson 5 Contents

75. Answer: The square roots of 16x6 are  4x3.
Simplify
Answer: The square roots of 16x6 are  4x3.
Example 5-1a

76. Answer: The opposite of the principal square root of
Simplify
Answer: The opposite of the principal square root of
Example 5-1b

77. Answer: The principal fifth root is 3a2b3.
Simplify
Answer: The principal fifth root is 3a2b3.
Example 5-1c

78. Answer: n is even and b is negative. Thus, is not a real number.
Simplify
Answer: n is even and b is negative.
Thus, is not a real number.
Example 5-1d

79. Answer: not a real number
Simplify. a. b. c. d.
Answer:  3x4
Answer:
Answer: 2xy2
Answer: not a real number
Example 5-1e

80. Simplify
Note that t is a sixth root of t6. The index is even, so the principal root is nonnegative. Since t could be negative, you must take the absolute value of t to identify the principal root.
Answer:
Example 5-2a

81. Since the index is odd, you do not need absolute value.
Simplify
Since the index is odd, you do not need absolute value.
Answer:
Example 5-2b

82. Simplify. a. b.
Answer:
Answer:
Example 5-2c

83. given by the formula where L is the length
Physics The time T in seconds that it takes a pendulum to make a complete swing back and forth is
given by the formula where L is the length
of the pendulum in feet and g is the acceleration due to gravity, 32 feet per second squared. Find the value of T for a 1.5-foot-long pendulum.
Explore You are given the values of L and g and must find the value of T. Since the units on g are feet per second squared, the units on the time T should be seconds.
Plan Substitute the values for L and g into the formula. Use a calculator to evaluate.
Example 5-3a

84. Original formula Solve Use a calculator.
Answer: It takes the pendulum about 1.36 seconds to make a complete swing.
Example 5-3b

85. Examine The closest square to and
 is approximately 3, so the answer should
be close to
The answer is reasonable.
Example 5-3c

86. given by the formula where L is the length
Physics The time T in seconds that it takes a pendulum to make a complete swing back and forth is
given by the formula where L is the length
of the pendulum in feet and g is the acceleration due to gravity, 32 feet per second squared. Find the value of T for a 2-foot-long pendulum.
Answer: about 1.57 seconds
Example 5-3d

87. End of Lesson 5

88. Example 1 Square Root of a Product Example 2 Simplify Quotients
Example 3 Multiply Radicals
Example 4 Add and Subtract Radicals
Example 5 Multiply Radicals
Example 6 Use a Conjugate to Rationalize a Denominator
Lesson 6 Contents

89. Factor into squares where possible.
Simplify
Factor into squares where possible.
Product Property of Radicals
Answer:
Simplify.
Example 6-1a

90. Simplify
Answer:
Example 6-1b

91. Simplify Quotient Property Factor into squares. Product Property
Example 6-2a

92. Rationalize the denominator.
Answer:
Example 6-2b

93. Rationalize the denominator.
Simplify
Quotient Property
Rationalize the denominator.
Product Property
Example 6-2c

94. Multiply.
Answer:
Example 6-2d

95. Simplify each expression.b.
Answer:
Answer:
Example 6-2e

96. Product Property of Radicals
Simplify
Product Property of Radicals
Factor into cubes.
Product Property of Radicals
Answer:
Multiply.
Example 6-3a

97. Simplify
Answer: 24a
Example 6-3b

98. Simplify Answer: Factor using squares. Product Property Multiply.
Combine like radicals.
Answer:
Example 6-4a

99. Simplify
Answer:
Example 6-4b

100. Simplify F O I L
Product Property
Answer:
Example 6-5a

101. Simplify
FOIL
Multiply.
Answer:
Subtract.
Example 6-5b

102. Simplify each expression. a.b.
Answer:
Answer: 41
Example 6-5c

103. Simplify Multiply by since is the conjugate of FOIL
Difference of squares
Example 6-6a

104. Multiply.
Answer:
Combine like terms.
Example 6-6b

105. Simplify
Answer:
Example 6-6c

106. End of Lesson 6

107. Example 2 Exponential Form
Example 1 Radical Form
Example 2 Exponential Form
Example 3 Evaluate Expressions with Rational Exponents
Example 4 Rational Exponent with Numerator Other Than 1
Example 5 Simplify Expressions with Rational Exponents
Example 6 Simplify Radical Expressions
Lesson 7 Contents

108. Write in radical form.
Answer:
Definition of
Example 7-1a

109. Write in radical form.
Answer:
Definition of
Example 7-1b

110. Write each expression in radical form. a.b.
Answer:
Answer:
Example 7-1c

111. Write using rational exponents.
Definition of
Answer:
Example 7-2a

112. Write using rational exponents.
Definition of
Answer:
Example 7-2b

113. Write each radical using rational exponents. a.b.
Answer:
Answer:
Example 7-2c

114. Evaluate
Method 1
Answer:
Simplify.
Example 7-3a

115. Method 2
Power of a Power
Multiply exponents.
Answer:
Example 7-3b

116. Evaluate . Method 1 Factor. Power of a Power Expand the square.
Find the fifth root.
Answer: The root is 4.
Example 7-3c

117. Method 2 Power of a Power Multiply exponents. Answer: The root is 4.
Example 7-3d

118. Evaluate each expression. a.b.
Answer:
Answer: 8
Example 7-3e

119. Answer: The formula predicts that he can lift at most 372 kg.
Weight Lifting The formula can be used to estimate the maximum total mass that a weight lifter of mass B kilograms can lift in two lifts, the snatch and the clean and jerk, combined.
According to the formula, what is the maximum that U.S. Weightlifter Oscar Chaplin III can lift if he weighs 77 kilograms?
Original formula
Use a calculator.
Answer: The formula predicts that he can lift at most 372 kg.
Example 7-4a

120. Weight Lifting The formula
Weight Lifting The formula can be used to estimate the maximum total mass that a weight lifter of mass B kilograms can lift in two lifts, the snatch and the clean and jerk, combined.
Oscar Chaplin’s total in the 2000 Olympics was 355 kg. Compare this to the value predicted by the formula.
Answer: The formula prediction is somewhat higher than his actual total.
Example 7-4b

121. Weight Lifting Use the formula where M is the maximum total mass that a weight lifter of mass B kilograms can lift.
a. According to the formula, what is the maximum that a weight lifter can lift if he weighs 80 kilograms?
b. If he actually lifted 379 kg, compare this to the value predicted by the formula.
Answer: 380 kg
Answer: The formula prediction is slightly higher than his actual total.
Example 7-4c

122. Simplify .
Multiply powers.
Answer:
Add exponents.
Example 7-5a

123. Simplify .
Multiply by
Example 7-5b

124. Answer:
Example 7-5c

125. Simplify each expression. a.b.
Answer:
Answer:
Example 7-5d

126. Simplify .
Rational exponents
Power of a Power
Example 7-6a

127. Quotient of Powers
Answer:
Simplify.
Example 7-6b

128. Simplify . Rational exponents Power of a Power Multiply. Answer:
Example 7-6c

129. Simplify .
is the conjugate of
Answer:
Multiply.
Example 7-6d

130. Simplify each expression. a.b. c.
Answer: 1
Answer:
Answer:
Example 7-6e

131. End of Lesson 7

132. Example 1 Solve a Radical Equation Example 2 Extraneous Solution
Example 3 Cube Root Equation
Example 4 Radical Inequality
Lesson 8 Contents

133. Add 1 to each side to isolate the radical.
Solve
Original equation
Add 1 to each side to isolate the radical.
Square each side to eliminate the radical.
Find the squares.
Add 2 to each side.
Example 8-1a

134. Answer: The solution checks. The solution is 38.
Original equation
Replace y with 38.
Simplify.
Answer: The solution checks. The solution is 38.
Example 8-1b

135. Solve
Answer: 67
Example 8-1c

136. Solve Original equation Square each side. Find the squares.
Isolate the radical.
Divide each side by –4.
Example 8-2a

137. Evaluate the square roots.
Square each side.
Evaluate the squares.
Check
Original equation
Replace x with 16.
Simplify.
Evaluate the square roots.
Answer: The solution does not check, so there is no real solution.
Example 8-2b

138. Answer: no real solution
Solve .
Answer: no real solution
Example 8-2c

139. In order to remove the power, or cube root, you must
Solve
In order to remove the power, or cube root, you must
first isolate it and then raise each side of the equation to
the third power.
Original equation
Subtract 5 from each side.
Cube each side.
Evaluate the cubes.
Example 8-3a

140. Subtract 1 from each side.
Divide each side by 3.
Check
Original equation
Replace y with –42.
Simplify.
The cube root of –125 is –5.
Add.
Answer: The solution is –42.
Example 8-3b

141. Solve
Answer: 13
Example 8-3c

142. Solve
Since the radicand of a square root must be greater than or equal to zero, first solve to identify the values of x for which the left side of the inequality is defined.
Example 8-4a

143. Answer: The solution is
Now solve .
Original inequality
Isolate the radical.
Eliminate the radical.
Add 6 to each side.
Divide each side by 3.
Answer: The solution is
Example 8-4b

144. Only the values in the interval satisfy the inequality.
Check
Test some x values to confirm the solution. Let Use three test values: one less than 2, one between 2 and 5, and one greater than 5.
Since is not a real number, the inequality is not satisfied.
Since the inequality is satisfied.
Since the inequality is not satisfied.
Only the values in the interval satisfy the inequality.
Example 8-4c

145. Solve
Answer:
Example 8-4d

146. End of Lesson 8

147. Example 1 Square Roots of Negative Numbers
Example 2 Multiply Pure Imaginary Numbers
Example 3 Simplify a Power of i
Example 4 Equation with Imaginary Solutions
Example 5 Equate Complex Numbers
Example 6 Add and Subtract Complex Numbers
Example 7 Multiply Complex Numbers
Example 8 Divide Complex Numbers
Lesson 9 Contents

148. Simplify .
Answer:
Example 9-1a

149. Simplify .
Answer:
Example 9-1b

150. Simplify. a. b.
Answer:
Answer:
Example 9-1c

151. Simplify .
Answer: = 6
Example 9-2a

152. Simplify .
Answer:
Example 9-2b

153. Simplify. a. b.
Answer: –15
Answer:
Example 9-2c

154. Simplify
Multiplying powers
Power of a Power
Answer:
Example 9-3a

155. Simplify .
Answer: i
Example 9-3b

156. Subtract 20 from each side.
Solve
Original equation
Subtract 20 from each side.
Divide each side by 5.
Take the square root of each side.
Answer:
Example 9-4a

157. Solve
Answer:
Example 9-4b

158. Find the values of x and y that make the equation true.
Set the real parts equal to each other and the imaginary parts equal to each other.
Real parts
Divide each side by 2.
Imaginary parts
Answer:
Example 9-5a

159. Find the values of x and y that make the equation true.
Answer:
Example 9-5b

160. Commutative and Associative Properties
Simplify .
Commutative and Associative Properties
Answer:
Example 9-6a

161. Commutative and Associative Properties
Simplify .
Commutative and Associative Properties
Answer:
Example 9-6b

162. Simplify. a. b.
Answer:
Answer:
Example 9-6c

163. Answer: The voltage is volts.
Electricity In an AC circuit, the voltage E, current I, and impedance Z are related by the formula Find the voltage in a circuit with current j amps and impedance 3 – 6 j ohms.
Electricity formula
FOIL
Multiply.
Add.
Answer: The voltage is volts.
Example 9-7a

164. Electricity In an AC circuit, the voltage E, current I, and impedance Z are related by the formula E = I • Z. Find the voltage in a circuit with current 1 – 3 j amps and impedance j ohms.
Answer: 9 – 7 j
Example 9-7b

165. Simplify . and are conjugates. Multiply. Answer: Standard form
Example 9-8a

166. Simplify .
Multiply by
Multiply.
Answer:
Standard form
Example 9-8b

167. Simplify. a. b.
Answer:
Answer:
Example 9-8c

168. End of Lesson 9

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