Chapter 7 powerpoint

1. Splash Screen

3. Lesson 1 Menu Five-Minute Check (over Chapter 6) Main Ideas and Vocabulary Example 1: Identify Monomials Key Concept: Product of Powers Example 2: Product of Powers Key Concept: Power of a Power Example 3: Power of a Power Key Concept: Power of a Product Example 4: Power of a Product Concept Summary: Simplifying Expressions Example 5: Simplify Expressions

5. Lesson 1 Ex1 Identify Monomials Determine whether each expression is a monomial. Explain your reasoning.

7. Key Concept 7-1a

8. Lesson 1 Ex2 Product of Powers A. Simplify ( r 4 )(–12 r 7 ). ( r 4 )(–12 r 7 ) = (1)(–12)( r 4 )( r 7 ) Group the coefficients and the variables. = –12( r 4+7 ) Product of Powers Answer: = –12 r 11 Simplify.

9. Lesson 1 Ex2 Product of Powers B. Simplify (6 cd 5 )(5 c 5 d 2 ). Answer: = 30 c 6 d 7 Simplify. (6 cd 5 )(5 c 5 d 2 ) = (6)(5)( c ● c 5 )( d 5 d 2 ) Group the coefficients and the variables. = 30( c 1+5 )( d 5+2 ) Product of Powers

12. Key Concept 7-1b

13. Lesson 1 Ex3 Power of a Power Simplify [(2 3 ) 3 ] 2 . Answer: = 2 18 or 262,144 Simplify. [(2 3 ) 3 ] 2 = (2 3●3 ) 2 Power of a Power = (2 9 ) 2 Simplify. = 2 9●2 Power of a Power

15. Key Concept 7-1c

16. Lesson 1 Ex4 GEOMETRY Find the volume of a cube with side length 5 xyz . Answer: = 125 x 3 y 3 z 3 Simplify. Power of a Product Volume = s 3 Formula for volume of a cube = (5 xyz ) 3 Replace s with 5 xyz. = 5 3 x 3 y 3 z 3 Power of a Product

18. Concept Summary 7-1d

19. Lesson 1 Ex5 Simplify [(8 g 3 h 4 ) 2 ] 2 (2 gh 5 ) 4 . [(8 g 3 h 4 ) 2 ] 2 (2 gh 5 ) 4 = (8 g 3 h 4 ) 4 (2 gh 5 ) 4 Power of a Power = (8 4 )( g 3 ) 4 ( h 4 ) 4 (2) 4 g 4 (h 5 ) 4 Power of a Product = 4096 g 12 h 16 (16) g 4 h 20 Power of a Power = 4096(16) g 12 ● g 4 ● h 16 ● h 20 Commutative Property Answer: = 65,536 g 16 h 36 Power of Powers Simplify Expressions

21. End of Lesson 1

22. Lesson 2 Menu Five-Minute Check (over Lesson 7-1) Main Ideas and Vocabulary Key Concept: Quotient of Powers Example 1: Quotient of Powers Key Concept: Power of a Quotient Example 2: Power of a Quotient Key Concept: Zero Exponent Example 3: Zero Exponent Key Concept: Negative Exponent Example 4: Negative Exponents Example 5: Standardized Test Practice: Apply Properties of Exponents

24. Key Concept 7-2a

25. Lesson 2 Ex1 Quotient of Powers Group powers that have the same base. Answer: = xy 9 Quotient of Powers

27. Key Concept 7-2b BrainPOP: Multiplying and Dividing Monomials

28. Lesson 2 Ex2 Power of a Quotient Power of a Quotient Power of a Quotient Answer: Power of a Power

30. Key Concept 7-2c

31. Lesson 2 Ex3 Zero Exponent Answer: 1 A.

32. Lesson 2 Ex3 Zero Exponent B. a 0 = 1 Answer: n Quotient of Powers Simplify.

35. Key Concept 7-2d

36. Lesson 2 Ex4 Negative Exponent Properties Negative Exponents A. Simplify . Assume that no denominator is equal to zero. Answer:

37. Lesson 2 Ex4 Group powers with the same base. Negative Exponents Quotient of Powers and Negative Exponent Properties B. Simplify . Assume that p , q and r are not equal to zero.

38. Lesson 2 Ex4 Negative Exponent Properties Negative Exponents Simplify. Answer: Multiply fractions.

39. Lesson 2 Ex4 Group powers with the same base. Negative Exponents Simplify. C. Simplify . Assume that no denominator is equal to zero. Answer: = Negative Exponent Property

43. Lesson 2 Ex5 Read the Test Item A ratio is a comparison of two quantities. It can be written in fraction form. Apply Properties of Exponents Refer to the figure in Example 5 on page 370 of your textbook. Write the ratio of the circumference of the circle to the area of the square in simplest form. A B C D

45. Lesson 2 Ex5 Answer: C Apply Properties of Exponents Simplify.

47. End of Lesson 2

48. Lesson 3 Menu Five-Minute Check (over Lesson 7-2) Main Ideas and Vocabulary Example 1: Identify Polynomials Example 2: Write a Polynomial Example 3: Degree of a Polynomial Example 4: Arrange Polynomials in Ascending Order Example 5: Arrange Polynomials in Descending Order

50. Lesson 3 Ex1 Identify Polynomials State whether each expression is a polynomial. If it is a polynomial, identify it as a monomial, binomial, or trinomial .

55. Lesson 3 Ex2 Write a Polynomial GEOMETRY Write a polynomial to represent the area of the shaded region. Words The area of the shaded region is the area of the rectangle minus the area of the triangle. Variables area of the shaded region = A height of rectangle = 2 h area of rectangle = b (2 h )

56. Lesson 3 Ex2 Write a Polynomial Area of shaded region = rectangle area – triangle area. Answer: Answer Equation

58. Lesson 3 Ex3 Degree of a Polynomial Find the degree of each polynomial. Interactive Lab: Exploring Polynomials

62. Lesson 3 Ex4 A. Arrange the terms of 16 + 14 x 3 + 2 x – x 2 so that the powers of x are in ascending order. 16 + 14 x 3 + 2 x – x 2 = 16 x 0 + 14 x 3 + 2 x 1 – x 2 x 0 = 1 Answer: = 16 + 2 x – x 2 + 14 x 3 0 < 1 < 2 < 3 Arrange Polynomials in Ascending Order

63. Lesson 3 Ex4 B. Arrange the terms of 7 y 2 + 4 x 3 + 2 xy 3 – x 2 y 2 so that the powers of x are in ascending order. 7 y 2 + 4 x 3 + 2 xy 3 – x 2 y 2 = 7 y 2 + 4 x 3 + 2 x 1 y 3 – x 2 y 2 x = x 1 Answer: = 7 y 2 + 2 xy 3 – x 2 y 2 + 4 x 3 0 < 1 < 2 < 3 Arrange Polynomials in Ascending Order

66. Lesson 3 Ex5 A. Arrange 8 + 7 x 2 – 12 x y 3 – 4 x 3 y so that the powers of x are in descending order. 8 + 7 x 2 – 12 x y 3 – 4 x 3 y = 8 x 0 + 7 x 2 – 12 x 1 y 3 – 4 x 3 y x 0 = 1 and x = x 1 Answer: = – 4 x 3 y + 7 x 2 – 12 xy 3 + 8 3 > 2 > 1 > 0 Arrange Polynomials in Descending Order

67. Lesson 3 Ex5 B. Arrange a 4 + ax 2 – 2 a 3 x y 3 – 9 x 4 y so that the powers of x are in descending order. a 4 + ax 2 – 2 a 3 x y 3 – 9 x 4 y = a 4 x 0 + a 1 x 2 – 2 a 3 x 1 y 3 – 9 x 4 y 1 x 0 = 1 and x = x 1 Answer: = – 9 x 4 y + ax 2 – 2 a 3 x y 3 + a 4 4 > 2 > 1 > 0 Arrange Polynomials in Descending Order

70. End of Lesson 3

71. Lesson 4 Menu Five-Minute Check (over Lesson 7-3) Main Ideas Example 1: Add Polynomials Example 2: Subtract Polynomials Example 3: Real-World Example

73. Lesson 4 Ex1 Add Polynomials Find (7 y 2 + 2 y – 3) + (2 – 4 y + 5 y 2 ). Method 1 Horizontal (7 y 2 + 2 y – 3) + (2 – 4 y + 5 y 2 ) = (7 y 2 + 5 y 2 ) + [2 y + (–4 y) + [(– 3) + 2] Group like terms. = 12 y 2 – 2y – 1 Add like terms.

74. Lesson 4 Ex1 Add Polynomials Method 2 Vertical Answer: 12 y 2 – 2 y – 1 Notice that terms are in descending order with like terms aligned. 12 y 2 – 2 y – 1 7 y 2 + 2 y – 3 (+) 5 y 2 – 4 y + 2

76. Lesson 4 Ex2 Subtract Polynomials Find (6 y 2 + 8 y 4 – 5 y ) – (9 y 4 – 7 y + 2 y 2 ). Method 1 Horizontal Subtract 9 y 4 – 7 y +2 y 2 by adding its additive inverse. (6 y 2 + 8 y 4 – 5 y ) – (9 y 4 – 7 y + 2 y 2 ) = (6 y 2 + 8 y 4 – 5 y ) + (–9 y 4 + 7 y – 2 y 2 ) = [8 y 4 + (–9y 4 )] + [6 y 2 + (–2 y 2 )] + (–5 y + 7 y ) = – y 4 + 4 y 2 + 2 y

77. Lesson 4 Ex2 Subtract Polynomials Method 2 Vertical Align like terms in columns and subtract by adding the additive inverse. Answer: 4 y 2 – y 4 + 2 y or – y 4 + 4 y 2 + 2 y 6 y 2 + 8 y 4 – 5 y (–) 2 y 2 + 9 y 4 – 7 y Add the opposite. 6 y 2 + 8 y 4 – 5 y (+) –2 y 2 – 9 y 4 + 7 y 4 y 2 – y 4 + 2 y

79. Lesson 4 Ex3 A. VIDEO GAMES The total amount of toy sales T (in billions of dollars) consists of two groups: sales of video games V and sales of traditional toys R . In recent years, the sales of traditional toys and total sales could be modeled by the following equations, where n is the number of years since 1996. R = 0.5 n 3 + 1.9 n 2 + 3 n + 19 T = 0.45 n 3 + 1.85 n 2 + 4.4 n + 22.6

80. Lesson 4 Ex3 Find an equation that models the sales of video games V . Subtract the polynomial for R from the polynomial for T . Answer: V = –0.05 n 3 + 0.05 n 2 + 1.4 n + 3.6 Total 0.5 n 3 + 1.9 n 2 – 3 n – 19 – Trad. (–)0.45 n 3 + 1.85 n 2 – 4.4 n – 22.6 Video –0.05 n 3 + 0.05 n 2 + 1.4 n + 3.6 Add the opposite. 0.5 n 3 + 1.9 n 2 – 3 n – 19 (+) –0.45 n 3 – 1.85 n 2 + 4.4 n + 22.6 – 0.05 n 3 + 0.05 n 2 + 1.4 n + 3.6

81. Lesson 4 Ex3 B. What did this equation predict for the amount of video game sales is the year 1998? Answer: The year 1998 is 1998 – 1996 or 2 years after the year 1996. If this trend continues, the number of video game sales in 1998 would have been –0.05(2) 3 + 0.05(2) 2 + 1.4(2) + 3.6 or 6.2 billion dollars.

84. End of Lesson 4

85. Lesson 5 Menu Five-Minute Check (over Lesson 7-4) Main Ideas Example 1: Multiply a Polynomial by a Monomial Example 2: Simplify Expressions Example 3: Real-World Example Example 4: Polynomials on Both Sides

87. Lesson 5 Ex1 Multiply a Polynomial by a Monomial Method 1 Horizontal Find 6 y (4 y 2 – 9 y – 7). 6 y (4 y 2 – 9 y – 7) = 6 y (4 y 2 ) – 6 y (9 y) – 6 y( 7) Distributive Property = 24 y 3 – 54 y 2 – 42 y Multiply.

88. Lesson 5 Ex1 Multiply a Polynomial by a Monomial Method 2 Vertical Answer: 24 y 3 – 54 y 2 – 42 y 24 y 3 – 54 y 2 – 42 y Multiply. 4 y 2 – 9 y – 7 (x) 6 y Distributive Property

90. Lesson 5 Ex2 Simplify Expressions Simplify 3(2 t 2 – 4 t – 15) + 6 t (5 t + 2). 3(2 t 2 – 4 t – 15) + 6 t (5 t + 2) = 3(2 t 2 ) – 3(4 t) – 3(15) + 6 t (5 t) + 6 t (2) Distributive Property = 6 t 2 – 12 t – 45 + 30 t 2 + 12 t Product of Powers = (6 t 2 + 30 t 2 ) + [(– 12 t) + 12 t ] – 45 Commutative and Associative Properties Answer: = 36 t 2 – 45 Combine like terms.

92. Lesson 5 Ex3 A. ENTERTAINMENT Admission to the Super Fun Amusement Park is $10. Once in the park, super rides are an additional $3 each and regular rides are an additional $2. Sarita goes to the park and rides 15 rides, of which s of those 15 are super rides. Find an expression for how much money Sarita spent at the park. Words Variable If s = the number of super rides, then 15 – s is the number of regular rides. Let M be the amount of money Sarita spent at the park. Amount of money = admission + super rides  $3 per ride regular rides $2 per ride. + 

93. Lesson 5 Ex3 Equation M = 10 + s ● 3 + (15 – s ) ● 2 = 10 +3 s + 15(2) – s (2) Distributive Property = 10 + 3 s + 30 – 2 s Simplify. = 40 + s Simplify. Answer: An expression for the amount of money Sarita spent in the park is 40 + s, where s is the number of super rides she rode.

94. Lesson 5 Ex3 B. Evaluate the expression to find the cost if Sarita rode 9 super rides. 40 + s = 40 + 9 s = 9 = 49 Add. Answer: Sarita spent $49.

97. Lesson 5 Ex4 Solve b (12 + b ) – 7 = 2 b + b (–4 + b ). b (12 + b ) – 7 = 2 b + b (–4 + b ) Original equation 12 b + b 2 – 7 = 2 b – 4 b + b 2 Distributive Property 12 b + b 2 – 7 = –2 b + b 2 Combine like terms. 12 b – 7 = –2 b Subtract b 2 from each side. Polynomials on Both Sides

98. Lesson 5 Ex4 12 b = –2 b + 7 Add 7 to each side. Divide each side by 14. Answer: Polynomials on Both Sides 14 b = 7 Add 2 b to each side.

99. Lesson 5 Ex4 Check b (12 + b ) – 7 = 2 b + b (–4 + b ) Original equation Simplify. Polynomials on Both Sides Multiply.  Subtract.

101. End of Lesson 5

102. Lesson 6 Menu Five-Minute Check (over Lesson 7-5) Main Ideas and Vocabulary Example 1: The Distributive Property Key Concept: FOIL Method Example 2: FOIL Method Example 3: FOIL Method Example 4: The Distributive Property

104. Lesson 6 Ex1 The Distributive Property Find ( y + 8)( y – 4). Method 1 Vertical Multiply by –4. – 4 y – 32 –4( y + 8) = –4 y – 32 Multiply by y. Combine like terms. y 2 + 4 y – 32 y + 8 ( ×) y – 4 y 2 + 8 y y ( y + 8) = y 2 + 8 y y + 8 ( ×) y – 4

105. Lesson 6 Ex1 The Distributive Property Method 2 Horizontal ( y + 8 )( y – 4) = y ( y – 4) + 8 ( y –4) Distributive Property = y ( y ) – y (4) + 8 ( y ) – 8 (4) Distributive Property = y 2 – 4 y + 8 y – 32 Multiply. = y 2 + 4 y – 32 Combine like terms. Answer: y 2 + 4 y – 32

107. Key Concept 7-6a

108. Lesson 6 Ex2 FOIL Method A. Find ( z – 6)( z – 12). ( z – 6)( z – 12) = z ( z ) Answer: z 2 – 18 z + 72 ( z – 6)( z – 12) = z ( z ) + z (–12) ( z – 6)( z – 12) = z ( z ) + z (–12) + (–6) z + (–6)(–12) ( z – 6)( z – 12) = z ( z ) + z (–12) + (–6) z = z 2 – 12 z – 6 z + 72 Multiply. = z 2 – 18 z + 72 Combine like terms. F ( z – 6)( z – 12) O I L F O I L

109. Lesson 6 Ex2 FOIL Method B. Find (5 x – 4)(2 x + 8). (5 x – 4)(2 x + 8) Answer: = 10 x 2 + 32 x – 32 Combine like terms. = 10 x 2 + 40 x – 8 x – 32 Multiply. = (5 x )(2 x ) + (5 x )(8) + (–4)(2 x ) + (–4)(8) F O I L

112. Lesson 6 Ex3 FOIL Method GEOMETRY The area A of a triangle is one half the height h times the base b . Write an expression for the area of the triangle. Explore Identify the height and the base. h = x – 7 b = 6 x + 7 Plan Now write and apply the formula. Area equals one half height times base. A = h ● b

113. Lesson 6 Ex3 FOIL Method Original formula Substitution FOIL method Multiply.

114. Lesson 6 Ex3 FOIL Method Combine like terms. Answer: The area of the triangle is 3 x 2 – 19 x – 14 square units. Distributive Property

116. Lesson 6 Ex4 A. Find (3 a + 4)( a 2 – 12 a + 1). (3 a + 4)( a 2 – 12 a + 1) = 3 a ( a 2 – 12 a + 1) + 4( a 2 – 12 a + 1) Distributive Property = 3 a 3 – 36 a 2 + 3 a + 4 a 2 – 48 a + 4 Distributive Property Answer: = 3 a 3 – 32 a 2 – 45 a + 4 Combine like terms. The Distributive Property

117. Lesson 6 Ex4 B. Find (2 b 2 + 7 b + 9)( b 2 + 3 b – 1) . (2 b 2 + 7 b + 9)( b 2 + 3 b – 1) = (2 b 2 )( b 2 + 3 b – 1)+ 7 b ( b 2 + 3 b – 1) + 9( b 2 + 3 b – 1) Distributive Property = (2 b 4 + 6 b 3 – 2 b 2 + 7 b 3 + 21 b 2 – 7 b + 9 b 2 + 27 b – 9) Distributive Property Answer: = 2 b 4 + 13 b 3 + 28 b 2 + 20 b – 9 Combine like terms. The Distributive Property

120. End of Lesson 6

121. Lesson 7 Menu Five-Minute Check (over Lesson 7-6) Main Ideas and Vocabulary Key Concept: Square of a Sum Example 1: Square of a Sum Key Concept: Square of a Difference Example 2: Square of a Difference Example 3: Real-World Example Key Concept: Product of a Sum and a Difference Example 4: Product of a Sum and a Difference

123. Key Concept 7-7a

124. Lesson 7 Ex1 Square of a Sum Find (7 z + 2) 2 . ( a + b ) 2 = a 2 + 2 a b + b 2 ( 7 z + 2 ) 2 = ( 7 z ) 2 + 2( 7 z )( 2 ) + ( 2 ) 2 a = 7 z and b = 2 Answer: = 49 z 2 + 28 z 2 + 4 Simplify.

126. Key Concept 7-7b

127. Lesson 7 Ex2 Square of a Difference Find (3 c – 4) 2 . ( a – b ) 2 = a 2 – 2 a b + b 2 ( 3 c – 4 ) 2 = ( 3 c ) 2 – 2( 3 c )( 4 ) + ( 4 ) 2 a = 3 c and b = 4 Answer: = 9 c 2 – 24 c + 16 Simplify.

129. Lesson 7 Ex3 GEOMETRY Write an expression that represents the area of a square that has a side length of 2 x + 12 units. The formula for the area of a square is A = s 2 . Answer: The area of the square is 4 x 2 + 48 x + 144 square units. A = s 2 Area of a square A = ( 2 x + 12 ) 2 s = (2 x + 12) A = ( 2 x ) 2 + 2( 2 x )( 12 ) + ( 12 ) 2 a = 2 x and b = 12 A = 4 x 2 + 48 x + 144 Simplify.

131. Key Concept 7-7c Animation: Product of a Sum and a Difference

132. Lesson 7 Ex4 Find (9 d + 4)(9 d – 4). ( a + b )( a – b ) = a 2 – b 2 ( 9 d + 4 )( 9 d – 4 ) = ( 9 d ) 2 – ( 4 ) 2 a = 9 d and b = 4 Answer: = 81 d 2 – 16 Simplify. Product of a Sum and a Difference

134. End of Lesson 7

135. CR Menu Five-Minute Checks Image Bank Math Tools Animation Menu Exploring Polynomials Multiplying and Dividing Monomials Polynomials

136. 5Min Menu Lesson 7-1 (over Chapter 6) Lesson 7-2 (over Lesson 7-1) Lesson 7-3 (over Lesson 7-2) Lesson 7-4 (over Lesson 7-3) Lesson 7-5 (over Lesson 7-4) Lesson 7-6 (over Lesson 7-5) Lesson 7-7 (over Lesson 7-6)

137. IB 1 To use the images that are on the following three slides in your own presentation: 1. Exit this presentation. 2. Open a chapter presentation using a full installation of Microsoft ® PowerPoint ® in editing mode and scroll to the Image Bank slides. 3. Select an image, copy it, and paste it into your presentation.

138. IB 2

139. IB 3

140. IB 4

141. Animation Menu 7-4 Use Algebra Tiles 7-6 Multiplying Polynomials 7-7 Product of a Sum and a Difference

142. Animation 1

143. Animation 2

144. Animation 3

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