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1 Chapter Polnomials and Polnomial Functions Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. Prerequisite Skills for the chapter Polnomials and Polnomial Functions. and. 4. a b c p 4 p () 4 5 ()()()() p Volume 5 4 pr 5 4 p p p < (using the p ke on a calculator) So, the volume of the sphere is 88p cm, or about cm. 8. Volume 5 s So, the volume of the cube is 64 in ( 4)( 5) ( )( 0). 8 5 ( )( 6) Lesson. Use Properties of Eponents Guided Practice for the lesson Use Properties of Eponents. (4 ) Power of a power propert (8)(8) 5 (8) Product of powers propert 5 (8) Power of a quotient propert p 04 9 p p 047 Quotient of powers propert p 0 5 Negative eponent propert p Product of powers propert = 6. (7 z 5 )( 4 z ) 5 7 (4) z 5 () Product of powers propert 5 7 z 4 5 7z4 Negative eponent propert 7. s t (s) 4 5 (t 4 ) Power of a quotient propert s6 t 8 Power of a power propert 5 s 6 t 8 Negative eponent propert 5 (4 ) ( 6 ) Power of a quotient propert 5 4 p p p 6 p Power of a product and power of a power properties Quotient of powers propert 5 4 Negative eponent propert Eercises for the lesson Use Properties of Eponents Skill Practice. a. Product of powers propert b. Negative eponent propert c. Power of a product propert. No, 5. is not less then 0; the number should be p 5 ( ) Product of powers propert (4 ) Power of a power propert 5 6 Negative eponent propert Algebra Worked-Out Solution Ke 7

2 5. (5)(5) 4 5 (5) ( 4) Product of powers propert 5 (5) ( 4 ) 5 8 Power of a power propert ( 5) Quotient of powers propert Negative eponent propert Power of a quotient propert Power of a quotient propert Negative eponent propert p ( ()) Product of powers propert () Quotient of powers propert Product of powers propert 5 Negative eponent propert. 6 p 6 0 p Product of powers propert Negative eponent propert Power of a power propert Power of a quotient propert 5 0 Negative eponent propert (4. 0 )( ) 5 (4..5)(0 0 6 ) (. 0 )( ) 5 (. 6.7)(0 0 7 ) ( )(8.9 0 ) 5 (6. 8.9)(0 5 0 ) ( )( ) 5 (7. 9.4)( ) (. 0 4 ) 5. (0 4 ) (4.0 0 ) (0 ) ( )( ) 5 ( )(08 04) w w 6 5 w 6 5 w 8 5 w Quotient of powers propert Negative eponent propert 5. ( ) 5 5 ( ) 5 ( ) 5 Power of a product propert Power of a power propert ( p q ) 5 p q Power of a product propert 5 p q Negative eponent propert 7. (w )(w 6 ) 5 w 6 () Product of powers propert 5 w 9 5 w9 Negative eponent propert Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 74 Algebra Worked-Out Solution Ke

3 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. Power of a product propert 5 5 p s() p t 4() Power of a power propert 5 5 p s6 p t 8. (5s t 4 ) 5 5 (s ) (t 4 ) s 6 5 5t Negative eponent propert 9. a b 5 5 () (a ) (b 5 ) Power of a product propert 5 7 a() b 5() Power of a power propert 5 7 a9 b 5 5 7a 9 b () 5 5 Negative eponent propert Quotient of powers propert Negative eponent propert. c d 9cd 5 9 c d () Quotient of powers propert 5 c d. 4r4 s 5 4r 4 s r4 4 s 5 (5) Quotient of powers propert 5 6 r0 s p p s0 Zero eponent propert 5 s0 6. a b 4 a 5 b 5 a 5 b 4 () Quotient of powers propert 5 a b 5 a b 4. 4z p 8z7 5 8 z z Negative eponent propert 5 7 z 7 Quotient of powers propert 5 4 z 4 5. p Product of powers propert 5 (4) Quotient of powers propert Negative eponent propert 6. B; 6 5 () 5 7. The error is that the eponents were divided and not subtracted The error is that the eponents were multiplied instead of added. 5 p The error is that the bases were multiplied. () p () 4 5 () Substitute for s in A 5 Ï 4 s. A 5 Ï 4 s 5 Ï 4 5 Ï 4 5 Ï 4 p 9 5 Ï 6 4. Substitute for r and for h in V 5 πr h. V 5 πr h 5 π() 5 π p 5 π 4. Substitute for l, 5 for w, and for h in V 5 l p w p h. V 5 l p w p h 5 () 5 () z z p? 5 z z 5? z 8 5? 5 z 5 5 z 5? ?? Algebra Worked-Out Solution Ke 75

4 45. (a 5 b 4 ) 5 a 4 b p? a 5 p b 4 p 5 a 4 b p? a 0 b 8 5 a 4 b p? a 0 b 8 a 4 b 5? a 0 4 p b 8 () 5 a 4 b 9 5 b9 a 4 5? 46. Sample answer: 6 5 ( 4 4 )( 8 ) 5 ( 5 )( 0 ) 5 ( 5 7 )( 7 9 ) 47. a m 5 a0 a m 5 a 0 m 5 a m 48. a m a n 5 a m p a n a n p a m 5 a n m 5 a m n Problem Solving 49. Pacific: V 5 ( )(4.0 0 ) 5 ( )(0 4 0 ) meters Atlantic: V 5 ( )(.9 0 ) 5 (7.68.9)(0 0 ) meters Indian: V 5 ( )(.96 0 ) 5 ( )(0 0 ) meters Arctic: V 5 (.4 0 )(. 0 ) 5 (.4.)(0 0 ) meters 50. (0.0000)(5,000,000) 5 (. 0 5 )( ) 5 (..5)( ) The continent has moved.75 0 miles. 5. Bead: d 5 6 mm r 5 mm V 5 4 πr 5 4 π() 5 6π Pearl : d 5 9 mm r 5 9 mm V 5 4 πr 5 4 π π 4 π Volume of pearl Volume of bead 5 6 π The volume of the pearl is about.75 times greater than the volume of the bead. 5. a. V 5 4 πr 5 4πr b. V 5 πr h c. h 5 (r) 5 6r d. Volume of clinder 5 πr (6r) 5 6πr Occupied volume 5 4πr 6πr 5 The tennis balls take up of the can s volume. 5. a. V 5 πr h 5 π(9.5) (.55) ø 44 mm m b. Answers will var. Volume of classroom c. pennies 5 Volume of penn The answer is an overestimate because of the shape of the pennies. There will be gaps between them. 54. a. Volume of inner core 5 4 πr Earth s total volume 5 4 π(5r) 5 4 π(5r ) 5 5 p 4 πr 5 5 p Volume of inner core ratio 5 5 : b. Volume of outer core 5 4 π 6 r r 4 πr πr 4 πr πr p 4 πr ratio 5 4 πr Lesson. Evaluate and Graph Polnomial Functions Investigating Algebra Activit for the lesson Evaluate and Graph Polnomial Functions. As approaches `, f() approaches `. As approaches `, f() approaches `.. As approaches `, f() approaches `. As approaches `, f() approaches `. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 76 Algebra Worked-Out Solution Ke

5 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved.. As approaches `, f() approaches `. As approaches `, f() approaches `. 4. As approaches `, f() approaches `. As approaches `, f() approaches `. 5. a. f () 5 n, where n is odd: As approaches `, f () approaches `. As approaches `, f () approaches `. b. f () 5 n, where n is odd: As approaches `, f () approaches `. As approaches `, f () approaches `. c. f () 5 n, where n is even : As approaches `, f () approaches `. As approaches `, f () approaches `. d. f () 5 n, where n is even : As approaches `, f () approaches `. As approaches `, f () approaches `. 6. As approaches `, f () approaches `. As approaches `, f () approaches `. f() is going to resemble its highest power. Because 6 is the highest power, the end behavior will be the same as f() 5 6. Guided Practice for the lesson Evaluate and Graph Polnomial Functions. The function is a polnomial function written as f () 5 in standard form. It has degree (linear) and a leading coefficient of.. The function is not a polnomial function because the term 5 has an eponent that is not a whole number.. The function is a polnomial function written as h () 5 6 π in standard form. It has a degree of (quadratic) and a leading coefficient of f () f () 5 () 4 () () g () g (4) 5 (4) 5(4) 6(4) f () f () The degree is odd and the leading coefficient is negative s E The wind speed needed to generate the wave is about 5 miles per hour. Energ per square foot (foot-pounds) E s Wind speed (miles per hour) Algebra Worked-Out Solution Ke 77

6 Eercises for the lesson Evaluate and Graph Polnomial Functions Skill Practice. f () is a degree 4 (quartic) polnomial function. The leading coefficient is 5 while the constant term is 6.. The end behavior of a polnomial is the behavior the function demonstrates as it approaches 6`.. The function is a polnomial function written as f () 5 8 in standard form. It has a degree (quadratic) and a leading coefficient of. 4. The function is a polnomial function written as f () in standard form. It has a degree 4 (quartic) and a leading coefficient of The function is a polnomial that is alread written in standard form. It has degree 4 (quartic) and a leading coefficient of π. 6. The function is not a polnomial function because the term 5 has an eponent that is not a whole number. 7. The function is a polnomial function alread written in standard form. It has degree (cubic) and a leading coefficient of The function is not a polnomial function because the term has an eponent that is not a whole number. 9. f () f () 5 5() () 0() f () f () 5 5() 4 () () 8() g () g () 5 () 5 4() h() h (5) 5 6(5) 5(5) h() h(4) 5 ( 4)4 4 (4) (4) g() g() 5 4() 5 6() () 0() f () f () f (6) f (4) f() f () f () f (4) 5 48 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 78 Algebra Worked-Out Solution Ke

7 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved.. The error is that a zero was not placed as the coefficient of the missing term f () 5 Þ 7 4. A; The ends approach opposite directions so the function must be odd. It imitates an odd function whose coefficient is positive. 5. The degree is even and the leading coefficient is positive. 6. The degree is odd and the leading coefficient is negative. 7. The degree is even and the leading coefficient is negative. 8. f () ` as ` f () ` as ` 9. f () ` as ` f () ` as ` 0. f () ` as ` f () ` as `. f () ` as ` f () ` as `. f () ` as ` f () ` as `. f () ` as ` f () ` as ` 4. f () ` as ` f () ` as ` 5. f () ` as ` f () ` as ` 6. f () ` as ` f () ` as ` 7. Sample answer: f () Algebra Worked-Out Solution Ke 79

8 B; f() 5 From the graph, f() ` as `; and f() ` as `. So, the degree is odd and the leading coefficient is negative. The constant term 5 f(0) When g() 5 f(): g() ` as ` and g() ` as `. 5. f() 5 a a a a 0 When a 5 and a 0 5 5: f() 5 a a 5 When f() 5 0: () a () a () a a 5 When f() 5 : () a () a () 5 5 4a a 5 8 Solve the sstem of equations: a a 5 (4) 4a 4a 5 4a a 5 8 4a a 5 8 a 5 0 a 5 0 Substitute a into one of the original equations. Solve for a. a 0 5 a 5 7 The cubic function is f() So, f(5) 5 (5) 7(5) 0(5) Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 80 Algebra Worked-Out Solution Ke

9 5. a. The graph of 5 is smmetric in the origin. b. The eponent in 5 0 is even, so ou can predict that the graph will be smmetric in the -ais. Graphing the function confirms this prediction. The graph of 5 is smmetric in the -ais. The eponent in 5 is odd, so ou can predict that the graph will be smmetric in the origin. Graphing the function confirms this prediction. The graph of 5 is smmetric in the origin. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. The graph of 5 4 is smmetric in the -ais. The graph of 5 5 is smmetric in the origin. The graph of 5 6 is smmetric in the -ais. c. You know that a graph has smmetr in the -ais if for ever point (a, b) on the graph, the point (a, b) is also on the graph. For a function f(), saing that the point (a, b) is on its graph is equivalent to saing that f(a) 5 b, and saing that the point (a, b) is also on its graph is equivalent to saing that f(a) 5 b. In other words, the graph of f() is smmetric in the -ais provided f(a) 5 f(a) for an a in the domain of the function. For the case of f() 5 n where n is a positive even integer, the graph of f() is smmetric in the -ais because f(a) 5 (a) n 5 () n p a n 5 p a n 5 a n 5 f(a). Similarl, the graph of g() 5 n 5 n 5 f() where n is a positive even integer must be smmetric in the -ais because g(a) 5 f(a) 5 f(a) 5 g(a) provided a Þ 0. You know that a graph has smmetr in the origin if for ever point (a, b) on the graph, the point (a, b) is also on the graph. For a function f(), saing that the point (a, b) is on its graph is equivalent to saing that f(a) 5 b, and saing that the point (a, b) is also on its graph is equivalent to saing that f(a) 5 b. In other words, the graph of f() is smmetric in the origin provided f(a) 5 f(a) for an a in the domain of the function. For the case of f() 5 n where n is a positive odd integer, the graph of f() is smmetric in the origin because f(a) 5 (a) n 5 () n p a n 5 p a n 5 a n 5 f(a). Similarl, the graph of g() 5 n 5 n 5 f() where n is a positive odd integer must be smmetric in the origin because g(a) 5 f(a) 5 f(a) 5 g(a) provided a Þ 0. Algebra Worked-Out Solution Ke 8

10 Problem Solving 54. w d 0.090d 0.48d w (5) 0.090(5) 0.48(5) carats 55. t S a. As t `, P() `. As t `, P() `. b. Quarterl periodicals P Skateboarding participants (millions) S t Years since 99 There were 8 million skateboarders 6 ears after 99 (998). 56. a. The function is degree (cubic). b. t c. M,600,96,800 5,84 8,0 t M,80,96 5,460 5,44 Indoor movie screens (thousands) M Years since t s Snowboarding participants (millions) S t Years since 99 The number of snowboarders was more than million in 00. t t Years since 980 c. p(t) 5 0.8t 4 6.4t 86.8t 9t 450 p(0) 5 0.8(0) 4 6.4(0) 86.8(0) 9(0) 450 5,780 68,480 78, ,700 periodicals The model ma not be appropriate to use to predict the number of periodicals in 00 because the model ma not be accurate outside the range of ears from 980 to a. S 5 0.t.49t 4.6t 6 S(5) 5 0.(5).49(5) 4.6(5) grams H 5 0.5t.7t 0.6t 4 H(5) 5 0.5(5).7(5) 0.6(5) grams Difference 5 S(5) H(5) grams b. W Weight (grams) S Das after hatching c. The chick is more likel to be a Sarus chick than a hooded chick. At das the Sarus chick s weight is about 0 grams, higher than the hooded chicks. Even though neither chick s average weight at das equals 0 grams, the Sarus chick is closer. H t Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 8 Algebra Worked-Out Solution Ke

11 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 60. a b. Width (kg and lb) (0.94) (0.94) ø ( ) ( ) Length (cm and in.) The function from part (a) is a vertical shrink of the original function. Graphing Calculator Activit for the lesson Evaluate and Graph Polnomial Functions. 0 0, , , , , , , , Sample answer: The window for g() should be the same -interval, but the -interval will be shifted b c units. 0. Sample answer: 0 0, Lesson. Add, Subtract, and Multipl Polnomials Guided Practice for the lesson Add, Subtract, and Multipl Polnomials. (t 6t ) (5t t 8) 5 t 5t 6t t 8 5 6t 7t 6. (8d 9d ) (d d 4) 5 8d 9d d d 4 5 9d d d 8d 4 5 8d d 8d. ( )( 5) 5 ( ) ( )() ( )(5) (a 5)(a )(a 6) 5 (a a 0)(a 6) 5 (a a 0)a (a a 0)6 5 a a 0a 6a 8a 60 5 a a 8a ( 4) 5 () () (4) ()(4) (4) T 5 C p D t 7.6t t t 87.6t 8, t t t t 9.98t 0,057t 8,94 T t 9.98t 0,057t 8,94 Eercises for the lesson Add, Subtract, and Multipl Polnomials Skill Practice. When ou add or subtract polnomials, ou add or subtract the coefficients of like terms.. Subtracting two polnomials is simpl adding the opposite of the subtracted polnomial to the first polnomial.. ( 5) (7 ) ( 5) (4 8 9) (4 9 5) (4 5 ) (z 5z 7) (5z z 6) 5 z 5z 5z z z 6z 7. (s s) (4s s 7s 0) 5 s 4s s s 7s 0 5 7s s 8s 0 8. (a 8) (a 4a a 4) 5 a 8 a 4a a 4 5 a a 4a a a a a 9. (5c 7c ) (c 6c 8) 5 c 5c 7c 6c 8 5 c 5c c 9 0. (4t t 4t) ( 7t 5t 8) 5 4t t 4t 7t 5t 8 5 4t t 7t 4t 5t 8 5 4t 4t 9t 8. (5b 6b b 4 ) (9b 4b 4 7) 5 5b 6b b 4 9b 4b b 4 4b 4 6b 9b 5b 7 5 b 4 5b 5b 7 Algebra Worked-Out Solution Ke 8

12 . ( ) ( ) ( 4 ) ( 4 ) (8v 4 v v 4) (v v 8v) 5. B; 5 8v 4 v v 4 v v 8v 5 8v 4 v v v v 8v 4 5 8v 4 v 0v 7v 4 (8 4 4 ) ( 4 8 0) ( 5 7) 5 () 5() 7() (6 ) 5 6(5 ) (5 ) ( 7)( 6) 5 ( 7)( ) ( 7)(6) (z )(z ) 5 (z )(z) (z )() 5 z z 9z 5 z 8z 0. (w 4)(w 6w ) 5 (w 4)(w ) (w 4)(6w) (w 4)() 5 w 4w 6w 4w w 44 5 w 0w w 44. (a )(a 0a ) 5 (a )(a ) (a )(0a) (a )() 5 a a 0a 0a 4a 6 5 a a 6a 6. (5c 4)(c c ) 5 (5c 4)(c ) (5c 4)(c) (5c 4)() 5 0c 4 8c 5c 4c 5c 5 0c 4 5c c 4c. ( 4 )( 8 ) 5 ( 4 )( ) ( 4 )(8) ( 4 )() (d 4d )(d 7d 6) 5 (d 4d )(d ) (d 4d )(7d) (d 4d )(6) 5 d 4 d 9d 7d 8d d 6d 4d 8 5 d 4 9d 5d d 8 5. ( 6 )(4 5) 5 ( 6 )(4 ) ( 6 )() ( 6 )(5) The error is that the negative sign was not carried through the subtracted polnomial ( 7 ). ( 4) ( 7 ) ( 7) is not the difference of the cubes of () and (7). It is the quantit ( 7) multiplied together three times. ( 7) 5 () () (7) ()(7) (7) 5 8 (4 ) 6(49) ( 4)( 6)( 5) 5 ( 4)( 5) 5 ( 4)() ( 4)(5) ( )( 7)( ) 5 ( 6 7)( ) 5 ( 6 7)() ( 6 7)() (z 4)(z )(z 8) 5 (z 6z 8)(z 8) 5 (z 6z 8)(z) (z 6z 8)(8) 5 z 6z 8z 8z 48z 64 5 z z 40z 64. (a 6)(a 5)(a ) 5 (a 7a 0)(a ) 5 (a 7a 0)(a) (a 7a 0)() 5 a 7a 0a a 7a 0 5 a 5a 7a 0. (p )( p )( p ) 5 (p 0p )( p ) 5 (p 0p )( p) (p 0p )() 5 p 0p p p 0p 5 p p p. (b )(b )(b ) 5 (b 5b )(b ) 5 (b 5b )(b) (b 5b )() 5 b 5b b b 5b 5 b 7b 7b Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 84 Algebra Worked-Out Solution Ke

13 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 4. (s )(s )(4s ) 5 (6s s )(4s ) 5 (6s s )(4s) (6s s )() 5 4s 4s 8s 8s s 6 5 4s s 5s 6 5. (w 6)(4w )(w 5) 5 (4w 5w 6)(w 5) 5 (4w 5w 6)(w) (4w 5w 6)(5) 5 w 75w 8w 0w 5w 0 5 w 95w 4w 0 6. (4 )( 7)(5 4) 5 (8 6 7)(5 4) 5 (8 6 7)(5) (8 6 7)(4) (q 8)(9q )(q ) 5 (7q 78q 6)(q ) 5 (7q 78q 6)(q) (7q 78q 6)() 5 7q 78q 6q 54q 56q 5 7q q 7q 8. ( 5)( 5) 5 () (5) (w 9) 5 (w) (w p 9) (9) 5 w 8w ( 4) 5 ( ) ( ) (4) ( )(4) (4) (c 5) 5 (c) (c p 5) (5) 5 4c 0c 5 4. (t 4) 5 (t) (t) (4) (t)(4) (4) 5 7t 08t 44t (5p )(5p ) 5 (5p) () 5 5p (7 ) 5 (7) (7) ( ) (7)( ) ( ) (a 9b)(a 9b) 5 (a) (9b) 5 4a 8b 46. (z 7) 5 (z) (z) (7) (z)(7) (7) 5 7z 89z 44z D; ( ) 5 () ( p ) () V 5 lwh l 5 ; w 5 ; h 5 5 ( )()( ) 5 ( )( ) 5 ( )() ( )() V 5 πr h r 5 4 ; h 5 5 π( 4) ( ) 5 π( 8 6)( ) 5 π(( 8 6)() ( 8 6)()) 5 π( ) 5 π( 8 48) 5 π π 8π 48π 50. V 5 s s ( 5) 5 () () (5) ()(5) (5) V 5 Bh B 5 ( ) ; h 5 ( 4) 5 ( ) ( 4) 5 (4 9)( 4) 5 ((4 9)() (4 9)(4)) 5 ( ) 5 ( 0 6) (a b)(a b) 0 a b (a b)(a) (a b)(b) 0 a b a ab ab b 0 a b a b 5 a b 5. (a b) 0 a ab b (a b)(a b) 0 a ab b (a b)(a) (a b)(b) 0 a ab b a ab ab b 0 a ab b a ab b 5 a ab b 54. (a b) 0 a a b ab b (a b)(a b)(a b) 0 a a b ab b (a ab b )(a b) 0 a a b ab b (a ab b )(a) (a ab b )(b) 0 a a b ab b a a b ab a b ab b 0 a a b ab b a a b ab b 5 a a b ab b 55. (a b) 0 a a b ab b (a b)(a b)(a b) 0 a a b ab b (a ab b )(a b) 0 a a b ab b (a ab b )(a) (a ab b )(b) 0 a a b ab b a a b ab a b ab b 0 a a b ab b a a b ab b 5 a a b ab b Algebra Worked-Out Solution Ke 85

14 56. a. p() q() 5 ( 4 7 4) ( 5) degree b. p() q() 5 ( 4 7 4) ( 5) degree c. p() p q() 5 ( 4 7 4)( 5) degree 6 d. p() q() : degree m 5 ( 4 7 4)( ) ( 4 7 4)(5) p() q() : degree m p() p q() : degree (m n) 57. a. 5 5 ( )( 4 ) Check: ( 4 )() ( 4 ) ( )( 5 4 ) Check: ( 5 4 )() ( 5 4 ) b. n 5 ( )( n n n... n n ) 5 ( )( n n n... ) Check: ( n n n... )() ( n n n... ) 5 n n n... n n n... 5 n n n... n (n ) 5 n n n n f() 5 ( a)( b)( c)( d) 5 ( a b ab)( c)( d) 5 (( a b ab)() ( a b ab)(c))( d) 5 ( a b ab c ac bc abc)( d) 5 ( a b ab c ac bc abc)() ( a b ab c ac bc abc)(d) 5 4 a b ab c ac bc abc d ad bd abd cd acd bcd abcd 5 4 (a b c d ) (ab ac bc ad bd cd ) (abc abd acd bcd ) abcd coefficient of 5 (a b c d ) constant term 5 abcd Problem Solving 59. T 5 M F 5 (0.09t 4.8t 0t 5000) (0.9t t 50t 600) 5 0.8t 6.8t 460t R 5 P p D 5 (6.8t 6.7t 65)(4.t 4.44) 5 (6.8t 6.7t 65)(4.t) (6.8t 6.7t 65)(4.44) t 5.587t 089.5t 0.808t 7.948t t.06t 85.0t 76.6 R() ().06() 85.0() The total revenue in 00 from DVD sales was about 70 million dollars, or $,70,000, F s P sF s(0.06s 0.789) s 0.00s The model is P s 0.00s. P(0) (0) 0.00(0) So, ou need 0.05 horsepower to keep the biccle moving 0 miles per hour. 6. cm r r 0.5 cm M(r) 5 volume of cookie volume of marshmallow 5 π(r ) h 4 πr 5 π(r ) π 5 π p 4 r (r r ) π p 4 r Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 86 Algebra Worked-Out Solution Ke

15 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 5 π 4 r r r 5 πr π r πr π D(r) 5 4 π(r ) π(r ) 5 π 4 (r ) π (r ) 5 π p 4 (r r r ) π (r r ) 5 π 4 r 4r 4r 4 r r 5 πr 5 πr πr 7 6 π C(r) 5 D(r) M(r) 5 π r 5 π r π r 7 6 π π r π r π r π 5 π r 5 π r π r 7 6 π π r π r π r π 5 π r π r π 6. N 5 total men s plaers total women s plaers 5 (L m p S m ) (L w p S w ) L m p S m 5 (0.7t 0.8t.0t.5) p (5.57t 8) 5 (0.7t 0.8t.0t.5)(5.57t) (0.7t 0.8t.0t.5)(8) t t 4.96t 0.85t 57 L w p S w 5 (0.066t 0.47t 0.75t.) p (.t 85) 5 (0.066t 0.47t 0.75t.) p (.t) (0.066t 0.47t 0.75t.)(85) t t 7t 7.95t 45.5 N t t 4.96t 0.85t t t 7t 7.95t t t 5.96t 7.75t A model is N 5.5t 4 5.5t 6t 8t The model for the total number of people is written b adding the number of men plaers to the number of women plaers. The number of men and women plaers is written b multipling the number of teams b the average team size for each group. 64. s 5 t 5 t 5 s 5 C t 0.08t 0.465t 48.8 C (s 5) 0.08(s 5) 0.465(s 5) (s s (5) s(5) 5) 0.08(s 0s 5) 0.465s s s s s 0.8s s s 0.04s 0.667s 5.7 Quiz for the lessons Use Properties of Eponents, Evaluate and Graph Polnomial Functions and Add, Subtract, and Multipl Polnomials. 5 p 5 5 () ( 4 ) 5 4 p ( ) p p ( 4 )( 8 ) 5 4 () () (a b 5 ) 5 (a ) (b 5 ) 5 a 6 p b 5 5 b5 a ( 4) c d c 5 d 5 c 5 p d () 5 c d 5 c d Algebra Worked-Out Solution Ke 87

16 ( 6) ( 4 8) ( 4 0) ( 9 5) ( 5)( 5 7) 5 ( 5 7)() ( 5 7)(5) ( )( 6)( ) 5 ( 8)( ) 5 ( 8)() ( 8) (7,8,000,000,000) 4 (94,000,000) ø $4,800 Lesson.4 Factor and Solve Polnomial Equations Guided Practice for the lesson Factor and Solve Polnomial Equations ( 7 0) 5 ( 5)( ) ( 5) 5 ( 5)( 5). 6b 5 686b 5 b (8b 4) 5 b (b 7)(4b 4b 49) 4. w 7 5 (w )(w w 9) ( 7) 9( 7) 5 ( 7)( 9) 5 ( 7)( )( ) 6. 6g (4g ) (5) 5 (4g 5)(4g 5) 5 (g 5)(g 5)(4g 5) 7. 4t 6 0t 4 4t 5 4t (t 4 5t 6) 5 4t (t )(t ) ( 4 0 9) 5 0 4( 9)( ) 5 0 4( )( )( )( ) , 5, 5, 5, or ( 4 7 ) 5 0 ( )( 4) 5 0 ( )( )( ) , 5 Ï, 5 Ï, 5, or ( 9 5) or or Ï 9 4 p p Ï 4 p 4 Volume (ft ) 5 Interior length (ft) p Interior width (ft) p Interior height (ft) 40 5 (6 ) p ( ) p ( ) ( ) ( ) 0 5 (8 )( ) The onl real solutions is 5. The basin is ft long, 6 ft wide, and ft high. Eercises for the lesson Factor and Solve Polnomial Equations Skill Practice. The epression is in quadratic form because it can be written as u 5u where u 5.. The factorization of a polnomial is factored completel if it is written as a product of unfactorable polnomials with integer coefficients ( ) 4. 0b 54b 5 6b (5b 9) 5. c 9c 8c 5 c(c 9c 8) 5 c(c )(c 6) 6. z 6z 7z 5 z (z 6z 7) 5 z(z )(z 6) ( 6) 5 ( 4)( 4) 8. 54m 5 8m 4 9m 5 9m (6m m ) 9. A; ( 7 ) 5 ( 4 6) 5 ( 4)( 4) 5 ( )( )( 4) ( )( 4) ( 4)( 4 6) Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 88 Algebra Worked-Out Solution Ke

17 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved.. 7m 5 (m) 5 (m )(9m m ). 5n 6 5 (5n) 6 5 (5n 6)(5n 0n 6) 4. 7a (a) 0 5 (a 0)(9a 0a 00) 5. 8c 4 5 (c) 7 5 (c 7)(4c 4c 49) 6. 9w 5 (64w ) 5 ((4w) ) 5 (4w )(6w 4w ) 7. 5z 0 5 5(z 64) 5 5(z 4 ) 5 5(z 4)(z 4z 6) 8. 5 ( ) ( ) 5 ( )( ) ( 7) 4( 7) 5 ( 7)( 4) 0. n 5n 9n 45 5 n (n 5) 9(n 5) 5 (n 5)(n 9) 5 (n 5)(n )(n ). m m 9m 5 m 9m m 5 m(m ) (m ) 5 (m )(m ). 5s 00s s 4 5 5s (s 4) (s 4) 5 (s 4)(5s ) 5 (s 4)(5s )(5s ). 4c 8c 9c 8 5 4c (c ) 9(c ) 5 (c )(4c 9) 5 (c )(c )(c ) ( ) 5 5 ( 5)( 5) 5. a 4 7a 6 5 (a )(a 6) 6. s 4 s 4 5 (s 8)(s ) 7. z 5 z 5 z (6z 4 ) 5 z (4z )(4z ) 5 z (4z )(z )(z ) 8. 6m 6 m 4 m 5 m (6m 4 m ) 5 m (6m )(6m ) 5 m (6m ) ( ) 5 (5 6)( 6) 0. The error is that when the binomial was factored, it was not factored properl ( )(4 6 9) Þ. The error is that 5 0 is not included in the solutions, and 6 was not factored completel before finding the zeros. ( 6) 5 0 ( 4)( 4) , 5 4, or ( 5) or s 5 50s 8s 50s 5 0 s(9s 5) 5 0 s(s 5)(s 5) 5 0 s 5 0, s 5 5, or s g g g 5 0 g (g ) (g ) 5 0 (g )(g ) 5 0 (g )(g )(g ) 5 0 g 5, g 5, or g 5 5. m 6m 4m m (m 6) 4(m 6) 5 0 (m 6)(m 4) 5 0 (m 6)(m )(m ) 5 0 m 5 6, m 5, or m w 4 40w (w 4 0w ) 5 0 4(w )(w ) 5 0 4(w )(w )(w ) 5 0 w 5 or w z z 4z 5 84z 5 0 4z (z ) 5 0 z 5 0, z 5 Ï, or z 5 Ï 8. 5b 5b b 5 6 5b 5b b b (b ) (b ) 5 0 (b )(5b ) 5 0 b ( 4) 9( 4) 5 0 ( 4)( 4 9) 5 0 ( )( )( )( ) 5 0 5, 5, 5 Ï, or 5 Ï Algebra Worked-Out Solution Ke 89

18 40. 48p 5 5 7p 4. C; 48p 5 7p 5 0 p (6p 9) 5 0 (p )(4p )(4p ) 5 0 p 5 0, p 5 4, or p ( 7 9) 5 0 (( 9) ( 9)) 5 0 ( 9)( ) 5 0 ( )( )( ) , 5, 5, or (8 ) 5 (4 )( 7) 4. n 4 4n 60 5 (n 6)(n 0) 44. 4b 4 500b 5 4b(b 5) 5 4b(b 5)(b 5b 5) 45. 6a 5a 84a 5 5 6a 84a 5a 5 5 a(a 7) 5(a 7) 5 (a 7)(a 5) 46. 8c 4 57c 0c 5 c (8c 57c 0) 5 c (6c )(c 0) 47. d 4 d 45 5 (d 5)(d 9) 5 (d 5)(d )(d ) (8 7) 5 4 ( )(4 6 9) ( ) 5 (4 )( 5) 50. z 5 z 4 6z 48 5 z 4 (z ) 6(z ) 5 (z )(z 4 6) 5 (z )(z 4)(z 4) 5 (z )(z 4)(z )(z ) 5. A 5 l p w l 5, w ( )( 4) ( 0)( ) 5 0 or 5, but a side length cannot be negative so V 5 l p w p h l 5, w 5, h ()( )( 4) 40 5 ( )( 4) ( 5) 8( 5) 0 5 ( 5)( 8) The onl real solution is V 5 πr h r 5 5, h 5 5π 5 π( 5) () 75 5 (4 0 5)() ( 5) 5( 5) 0 5 ( 5)(4 5) The onl real solution is ( ) 5 ( )( 4 9) 55. 7ac bc 7ad bd 5 c (7a b)d (7a b) 5 (7a b)(c d ) 5 (7a b)(c d )(c d) 56. n n 5 ( n )( n ) 5 ( n ) 57. a 5 b a b 4 a 4 b ab a b 5 a b (a b ) ab(a b ) (a b ) 5 (a b )(a b ab ) 5 (a b )(ab )(ab ) 5 (a b )(ab ) Problem Solving 58. V 5 l p w p h l 5 5, w 5, h ( 5)( )() (44 4 5) ( ) 5( ) 0 5 ( )(6 5) The onl real solution is 5. The block is meters high. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 90 Algebra Worked-Out Solution Ke

19 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 59. V 5 l p w p h l 5, w 5, h ( )( )(6 ) 5 ( 8 4)(6 ) ( ) 0( ) 0 5 (9 0)( ) 5 cm 5 9 cm cm The outer dimensions of the mold should be cm 9 cm 8 cm. 60. a. V 5 l p w p h Volume of platform 5 (8)(6)() 5 48 Volume of platform 5 (6)(4)() 5 4 Volume of platform 5 (4)()() 5 8 b c ft 5 Platform : 0 ft 5 ft.5 ft Platform : 5 ft 0 ft.5 ft Platform : 0 ft 5 ft.5 ft 6. l 5 h 5 5 w 5 5 V 5 l p w p h 50 5 ()( 5)( 5) 50 5 ( 0 5) ( 0) 5( 0) 0 5 ( 5)( 0) The dimensions of the prism should be 0 in. 5 in. 5 in Volume of steel 5 volume of outside volume of inside ()( )( 8) ( 6 6) ( 6) 6( 6) 0 5 ( 6)( 6) 0 5 ( 6)( 4)( 4) 5 4 feet The outer dimensions of the tank should be 4 ft ft ft. 6. V 5 l p w p h l 5, w 5, h ( )( )( 4) 64. a. 7 5 ( )(9 6 8) From the graph, the onl real real solution is.7. This result would ield a negative dimension of the platform. So, the volume cannot be 7 cubic feet. (.7, 0) Algebra Worked-Out Solution Ke 9

20 b c d. a 4 b 5 c Original equation a 4 4 b a 4 b 5 a c b 4 a b 4 a b 5 a c b 4 a Multipl each side b b. 4 Rewrite equation. Then, create a table with values for a. The three solids form a cube with a smaller cube taken out of the corner. If the solids were connected, V 5 a b because the side of the large cube 5 a and the side of the cube taken out 5 b. b. I: V 5 a p a p (a b) 5 a (a b) II: V 5 a p (b)(a b) 5 ab(a b) III: V 5 b p b(a b) 5 b (a b) c. V 5 I II II 5 a (a b) ab(a b) b (a b) 5 (a b)(a ab b ) 5 a b Problem Solving Workshop for the lesson Factor and Solve Polnomial Equations Intersections : (.56, ) (.56, ) (0,) 5.56,.56, or Intersections: (.4, 45) (.5, 45) (.96, 45) (, 45) 5.4,.5,.96, or Intersections: (.45, ) (4, ) 5.45 or Intersections: (.64, ) (, ) 5.64 or Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 9 Algebra Worked-Out Solution Ke

21 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved ( )( )( ) height 4 5 width 6 5 length V 5 l p w p h 5 ( 6)( 4)() The volume is 00 cubic feet when 5 6. Length 5 5 ft Width ft Height ft Intersection: 5 6, 5 00 Length 5 5 ft Width ft Height ft Intersection at: 5, height 5 5 ft width ft length ft Intersection: 5 5, In 975, pineapple consumption was about 9.97 pounds per person. Lesson.5 Appl the Remainder and Factor Theorems Guided Practice for the lesson Appl the Remainder and Factor Theorems. 8 q ww q ww f () ( 4)( ) 5 ( 4)( )( ) Algebra Worked-Out Solution Ke 9

22 f () ( 4)( 0) 5 ( 4)( 5)( ) f () ( )( 9) 5 ( )( )( ) The other zeros are and f () ( )( 6 7) 5 ( )( 7)( ) The other zeros are 7 and ( )(5 5 5) 5 0 ø 0.9 is the other positive solution. The compan could still make the same profit producing about 900,000 shoes. Eercises for the lesson Appl the Remainder and Factor Theorems Skill Practice. If a polnomial f () is divided b k, then the remainder is r 5 f (k).. The red numbers represent the coefficients of the quotient and the blue number represents the remainder q ww q ww q ww q ww q ww Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 94 Algebra Worked-Out Solution Ke

23 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved q ww q www q www The division was done correctl, however, there is an error in how the quotient is written The error is that a zero was not used as a place holder for the missing term Algebra Worked-Out Solution Ke 95

24 f () ( 6)( 4 5) 5 ( 6)( 5)( ) f () ( 4)( ) 5 ( 4)( )( ) f() ( 8)( 6 8) 5 ( 8)( 4)( ) f () ( 0)( 8 5) 5 ( 0)( )( 5) f () ( 9)( 7 ) 5 ( 9)( 4)( ) f () ( )( 0) 5 ( )( 5)( 6) f () ( )( ) 5 ( )( 7)( ) f() ( 5)( 4) 5 ( 5)( )( 4) f () ( )( 5 6) 5 ( )( 6)( ) The other zeros are 6 and f () ( 0)(4 5 4) 5 ( 0)(4 )( 4) The other zeros are and f () ( 7)(0 6) 5 ( 7)(5 )( ) The other zeros are 5 and. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 96 Algebra Worked-Out Solution Ke

25 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved f () ( 4)( 6) 5 ( 4)( )( 8) The other zeros are and f () ( 9)( 8 ) Ï 64 4() Ï 56 () Ï 4 The other zeros are about 0. and D; f () ( )(5 4) 5 6 Ï 4(5)(4) 5 6 Ï The other zeros are about 0.46 and f () ( 6)(4 9 9) 5 ( 6)(4 )( ) The zeros are 6,, and ( 4)( ) l q ww The missing dimension is ( )( 6) w q ww The missing dimension is a A; f () ( )( 8) 5 ( )( 6)( ) The other zeros are 6 and. b. The factors are ( ), ( 6), and ( ). c. The solutions are 5, 5 6, 5. 5 k k 0 0 For 5 to be a factor, the remainder must be zero. In the last column 0 must be added to 0 to get zero. k 0 must equal 6 in order to have a product of 0 to add to the 0. k so, k Algebra Worked-Out Solution Ke 97

26 a. b f() 5 (0 8) c. f() 5 ()(5 4) 5 ( )( )(5 7) Problem Solving 4. P P 5 ( 4)( ) 5 ( 4)( )( ) 5 is the onl other positive solution. So, the compan could still make the same profit b producing onl million T-shirts. 4. P P 5 ( )(4 6) 5 ( )(4 )(4 ) 5 is the onl other positive solution. The compan could produce million MP plaers and still make the same profit. 4. Let f() 5 average attendance per team. f() 5 A T q , A function is f() , Algebra Worked-Out Solution Ke 409, a. Total revenue 5 (number of radios sold)(price per radio) 5 ()(40 4 ) An epression is b. P 5 (40 4 ) A function is P c. When P 5 4: You know that.5 is a solution, so.5 is a factor of So, (.5)(4 6 6) 5 0. Use the quadratic formula to find ø.9, the other positive solution. So, about.9 million radios also make a profit of $4,000,000. d. No. The solution ø.88 does not make sense because ou cannot produce a negative number of radios. 45. Let P 5 the precent of visits to the national park that were overnight stas. P 5 S V q wwwww , ,5.87 So, P , The function for the percent of visits that were overnight stas is the overnight stas divided b the total visits. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved.

27 46. P f() Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved P 5 (6 48)( ) 5 ( 6)( 8)( ) 5 ( )()( 4)( ) 5 6( )( 4)( ) The zeros are at and 4. Production levels cannot be negative so 5 is the onl solution. Mied Review of Problem Solving for the lessons Use Properties of Eponents, Evaluate and Graph Polnomial Functions, Add, Subtract, and Multipl Polnomials, Factor and Solve Polnomial Equations and Appl the Remainder and Factor Theorems. a. 64,000,000, ards b a. T() 5 l p w p h 5 (4)()() football fields b. C() 5 (4 )( )( 4) 5 (4 0 4)( 4) c. I() 5 T() C() d. I() 5 8 ( ) I(8) 5 6(8) 48(8) The volume of the insulation is 96 cubic inches.. Cube: V 5 A V A 5 6 Sphere: V 5 4 π 5 π 6 A 5 4π 5 π A The cells echange at the same rate. V 5 π π a. C is a degree 4 (quartic). b. c. Cost (dollars) t, ear 0 4 C, cost t, ear C, cost C t Years since 995 No, the model will not accuratel predict cell phone bills beond 00. The model was not intended to be used past 00. The graph shows that the model is decreasing after 00. In fact, it takes on negative values after a. R 5 (00 0) A function for the total revenue is R() b. P 5 (00 0 ) A function for the profit is P() c. When P 5 60: You know that is a solution, so is a factor of So, ( )(0 0 0) 5 0. Use factoring to find that 5 and 5 are the other solutions. d. No. The solution 5 does not make sense because ou cannot produce a negative number of cameras. Algebra Worked-Out Solution Ke 99

28 7. V 5 h(s ) 48 5 ( )( 6) 48 5 (9 6 6) ( 4) ( 4) 0 5 ( )( 4) 5 4 The height is 4 feet. Lesson.6 Find Rational Zeros Guided Practice for the lesson Find Rational Zeros. Factors of the constant term: 6, 6, 65, 65 Factors of the leading coefficient: 6 Possible rational zeros: 6, 6, 6 5, 6 5 Simplified list: 6, 6, 65, 65. Factors of the constant term: 6, 6, 6, 66 Factors of the leading coefficient: 6, 6 Possible rational zeros: 6, 6, 6, 6 6, 6, 6, 6, 6 6 Simplified list: 6, 6, 6, 66, 6, 6. Possible rational zeros: 6, 6, 6, 6 6, 6 9, 6 8 Test 5 : Because is a zero of f, f() can be writen as: f() 5 ( )( 8) 5 ( )( 6)( ) The zeros are, and Possible rational zeros: 6, 6, 67, 64 Test 5 : is not a zero. Test 5 : Because is a zero of f, f() can be written as: f() 5 ( )( 9 4) 5 ( )( 7)( ) The zeros of f are, and Possible rational zeros: 6, 6, 6, 6, 6, 6 4, 6 4, 6 6, 6 8, 6 8, 6, 6 6, 6 6, 6 4, 6 48 Reasonable values: 5 4, 5 6, 5 Check: 5 4 : is a zero. 4 f() 5 4 (48 4) 5 4 (4)( 8 ) 5 (4 )(6 )( ) The real zeros of f are 4, 6, and. 6. Possible rational zeros: 6, 6, 6, 66, 67, 64, 6, 64, 6, 6, 6 7, 6 Reasonable zeros: 5 7, 5, 5, 5 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 00 Algebra Worked-Out Solution Ke

29 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved f() 5 ( )( 0 ) 5 ( ) p g() Possible rational zeros of g(): 6, 6, 67, 6, 6, 6, 6 7, 6 Reasonable zeros: 5, f() 5 ( ) p g () 5 ( ) ( 4 4) Ï (4) 4()(4) Ï 8 () Ï The real zeros of f are:,, Ï, and Ï. 7. V 5 ( )( ) 6 5 ( ) Possible rational zeros: 6, 6, 6, 66, 69, The other solutions, which satisf 6, are 5 6 i Ï 5 and can be discarded because the are imaginar. The onl real solution is 5. The base is feet. The height is or feet. Eercises for the lesson Find Rational Zeros Skill Practice. If a polnomial function has integer coefficients, then ever rational zero of the function has the form p q where p is a factor of the constant term and q is a factor of the leading coefficient.. To shorten the list, draw the graph to see approimatel where the zeros are.. f() 5 8 Factors of constant term: 6, 6, 64, 67, 64, 68 Factors of leading coefficient: 6 Possible rational zeros: 6, 6, 6 4, 6 7, 6 4, 6 8 Simplified list: 6, 6, 64, 67, 64, g () Factors of constant term: 6, 6, 65, 60 Factors of leading coefficient: 6 Possible rational zeros: 6, 6, 6 5, 6 0 Simplified list: 6, 6, 65, f() Factors of constant term: 6, 6, 69 Factors of leading coefficient: 6, 6 Possible rational zeros: 6, 6, 6 9, 6, 6, 6 9 Simplified list: 6, 6, 69, 6, 6, h() 5 8 Factors of constant term: 6, 6, 6, 66, 69, 68 Factors of leading coefficient: 6, 6 Possible rational zeros: 6, 6, 6, 6 6, 6 9, 6 8, 6, 6, 6, 6 6, 6 9, 6 8 Simplified list: 6, 6, 6, 66, 69, 68, 6, 6, g () Factors of the constant term: 6, 6, 67, 64 Factors of the leading coefficient: 6, 6, 64 Possible rational zeros: 6, 6, 6 7, 6 4, 6, 6, 6 7, 6 4, 6 4, 6 4, 6 7 4, Simplified list: 6, 6, 6 7, 6 4, 6, 6 7, 6 4, Algebra Worked-Out Solution Ke 0

30 8. f() Factors of the constant term: 6, 6, 6, 66, 67, 64, 6, 64 Factors of the leading coefficient: 6, 6 Possible rational zeros: 6, 6, 6, 6 6, 6 7, 6 4, 6, 6 4, 6, 6, 6, 6 6, 6 7, 6 4, 6 4, 6 Simplified list: 6, 6, 6, 66, 67, 64, 6, 64, 6, 6, 6 7 4, 6 9. h() Factors of the constant term: 6, 6, 65, 65 Factors of the leading coefficient: 6, 6, 64, 68 Possible rational zeros: 6, 6, 6 5, 6 5, 6, 6, 6 5 5, 6, 6 4, 6 4, 6 5 5, 6 4 4, 6 8, 6 8, 6 5 8, Simplified list: 6, 6, 65, 65, 6, 6, 6 5, 6 5, 6 4, 6 4, 6 5 4, 6 5 7, 6 8, 6 8, 6 5 8, h() 5 6 Factors of the constant term: 6, 6, 6, 64, 66, 6 Factors of the leading coefficient: 6, 6, 6, 66 Possible rational zeros: 6, 6, 6, 6 4, 6 6, 6, 6, 6, 6, 6 4, 6 6, 6, 6, 6, 6, 6 4, 6 6, 6, 6 6, 6 6, 6 6, 6 4 6, 6 6, Simplified list: 6, 6, 6, 64, 66, 6, 6, 6, 6, 6, 6 4, 6 6. Possible rational zeros: 6, 6, 6, 64, 66, 68, 6, 64 Test 5 : f() 5 ( )( 4) 5 ( )( )( 8) Real zeros are,, and 8.. Possible rational zeros: 6, 6, 64, 67, 68, 64, 68, 656 Test 5 : Test 5 : 5 56 Test 5 : f() 5 ( )( 8) 5 ( )( 7)( 4) Real zeros are, 7, and 4.. Possible rational zeros: 6, 6, 6, 65, 66, 60, 65, 60 Test 5 : Test 5 : g() 5 ( )( 0) 5 ( )( 6)( 5) Real zeros are, 6, and Possible rational zeros: 6, 6, 6, 64, 66, 68, 69, 6, 68, 64, 66, 67 Test 5 : Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 0 Algebra Worked-Out Solution Ke

31 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. Test 5 : Test 5 : Test 5 : Test 5 : h() 5 ( )( 4) 5 ( )( 8)( ) The real zeros are, 8, and. 5. Possible rational zeros: 6, 6, 6, 64, 66, 68, 6, 64 Test 5 : Test 5 : h() 5 ( )( 6 0 4) 5 ( ) p g () Possible rational zeros of g (): 6, 6, 6, 64, 66, 68, 6, 64 Test: 5 : Test: 5 : h() 5 ( )( )( 4 ) Ï (4) 4()() () 5 6 i Ï, which are not real. The onl real solutions are and. 6. Possible rational zeros: 6, 6, 6, 64, 66, 68, 6, 64 Test 5 : Test 5 : Test 5 : Test 5 : Test 5 : Test 5 : f() 5 ( )( 5 6 8) 5 ( ) p g() Possible rational zeros of g (): 6, 6, 64, 68 Algebra Worked-Out Solution Ke 0

32 Test 5 4: f() 5 ( )( 4)( ) Ï 4()() 5 6 i Ï 7 which are not real. The real solutions are and Possible rational zeros: 6, 6, 64, 68 Test 5 : f() 5 ( )( 6 8) 5 ( ) p g () Possible rational zeros of g (): 6, 6, 64, 68 Test 5 : f() 5 ( )( )( 8) 5 ( )( )( 4)( ) The real zeros are,, 4, and. 8. Possible rational zeros: 6, 65, 65 Test 5 : Test 5 : g () 5 ( )( 5 5) 5 ( ) p f() Possible zeros of f(): 6, 65, 65 Test 5 5: g() 5 ( )( 5)( 4 5) Ï (4) 4()(5) () 5 6 i, which are not real. The onl real zeros are and Possible rational zeros: 6, 6, 64, 68, 66, 6, 6 4 Reasonable rational zeros: 5 Test 5 : f() 5 ( )(4 4 6) 5 ( )(4)( 4) Ï () 4()(4) () 5 6 Ï 7 Real zeros:, Ï 7, Ï 7 0. Possible rational zeros: 6, 6, 65, 65, 6, 6, 6 5, 6 5, 6 4, 6 4, 6 5 5, Reasonable zeros: 5, 5, 5 5 Test 5 : f() 5 ( )(4 6 5) 5 ( )( )( 5) The real zeros are,, and 5.. Possible rational zeros: 6, 6, 6 5, 65, 6, 6, 6 5, 6 5, 6, 6 5, 6 6, Reasonable zeros: 5, 5 5, 5 Test 5 : Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 04 Algebra Worked-Out Solution Ke

33 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. f() 5 ( )(6 7 5) 5 ( )( )( 5) Real zeros are,, and 5.. Possible rational zeros: 6, 6, 6 4, 68, 66, 6, 6, 6 4, 6 8, 6 6 Reasonable zeros: 5, 5, 5 4 Test 5 : f() 5 ( 8 4) 5 ()( 6 8) 5 ( )( )( 4) Real zeros are,, and 4.. C; Possible zeros: 6, 6, 69, 6, 6, Possible zeros: 6, 6, 64, 68, 6 Reasonable zeros: 5, 5, and 5 Test 5 : f() 5 ( )( 4) 5 ( )( )( ) The real zeros are,, and. 5. Possible rational zeros: 6, 6, 69, 6, 6, 6 9 Reasonable zeros: 5, 5, 5 Test 5 : g() 5 ( )( 9 9) 5 ( )( )( ) The real zeros are,, and. 6. Possible rational zeros: 6, 6, 65, 6 5, 6, 6, 6 5 5, 6 Reasonable zeros: 5 5, 5, and 5 Test 5 5 : h() 5 5 ( 8 6) 5 5 ()( 4 ) 5 ( 5)( )( ) The real zeros are 5,, and. 7. Possible rational zeros: 6, 6, 6, 64, 66, 6, 6, 6, 6 4 Reasonable zeros: 5 4, 5, and 5 Test 5 4: f() 5 ( 4)( 8 ) 5 ( 4)( )( ) The real zeros are 4,, and. 8. Possible real zeros: 6, 6, 6, 64, 66, 6, 6, 6, 6 4 Reasonable zeros: 5 6, 5, 5 Test 5 6: f() 5 ( 6)( ) 5 ( 6)( )( ) The real zeros are 6,, and. Algebra Worked-Out Solution Ke 05

34 9. Possible real zeros: 6, 6, 67, 64, 6, 6 7 Reasonable zeros: 5 7, 5, and 5 Test 5 7 : g() 5 7 ( 4) 5 7 ()( ) 5 ( 7)( )( ) The real zeros are 7,, and. 0. Possible real solutions: 6, 6, 64, 6 Resonable zeros: 5 4 and 5 Test 5 4: g() 5 ( 4)( ) 5 ( 4) p f() Possible real solutions of f(): 6, 6 Test 5 : 0 g() 5 ( 4) ( ) 5 ( 4) ()( ) 5 ( 4)( )( ) Ï () 4()() () 5 6 i Ï, which are imaginar numbers The onl real solutions are 4 and.. Possible rational zeros: 6, 6, 64, 6 Reasonable zeros: 5 and 5 Test 5 : h() 5 ( )( 5 ) 5 ( ) p f() Possible rational zeros of f: 6, 6, 6 Test 5 : h() 5 ( )( )( ) Ï () 4()() () 5 6 i Ï 7, which are not real. 4 The real solutions are 5 and 5.. Possible rational zeros: 6, 6, 6, 64, 66, 6, 6, 6, 6 4 Reasonable zeros: 5, and 5 4 Test 5 : h() 5 ( )( 5 4) 5 ( ) p g() Possible rational zeros of g: 6, 6, 64, 6, 6, 6 4 Test 5 4: h() 5 ( )( 4)( ) Ï 4()() 5 6 i Ï, which are not real. () 6 The onl real solutions are and 4. Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved. 06 Algebra Worked-Out Solution Ke

35 Copright Houghton Mifflin Harcourt Publishing Compan. All rights reserved.. Possible rational zeros: 6, 6, 6, 65, 66, 60, 65, 60, 6, 6, 6 5 5, 6. Reasonable zeros: 5, 5, 5 5, and 5 Test 5 : f() 5 ( )( 6 5) 5 ( ) p g() Possible rational zeros of g: 6, 6, 65, 65, 6, 6, 6 5 5, 6 Test 5 : f() 5 ( )( )( 5) 5 ( )( )( 5)( ) The real zeros are,, 5, and. 4. Possible rational zeros: 6, 6, 6, 64, 66, 6 Reasonable zeros: 5, 5, 5, 5, and 5 Test 5 : f() 5 ( )( ) 5 ( ) p g() Possible rational zeros of g: 6, 6, 6, 66 Test 5 : f () 5 ( )( )( 6 6) 5 ( )( ) p h() Possible rational zeros of h: 6, 6, 6, 66 Test 5 : f() 5 ( )( )( )( 5 6) 5 ( )( )( )( )( ) The real zeros of f are,,,, and. 5. Possible rational zeros: 6, 6, 6, 6 Reasonable zeros: 5, 5, and 5 Test 5 : h() 5 ( )( 4 ) 5 ( ) p f() Possible rational zeros of f: 6, 6 Test 5 : h() 5 ( ) ( ) 5 ( ) ()( ) 5 ( )( )( )( ) Ï () 4()() () 5 6 i Ï, which are not real. The onl real solutions of h are,, and. 6. The error is that onl the positive factors were listed as possible rational zeros. Possible zeros: 6, 6, 67, The error is that the factors of the coefficient were divided b the factors of the constant term. The possible zeros are: 6, 65, 6, 6 5, 6, 6 5, 6 6, Sample answer: f() 5 4 Possible zeros: 6, 6, 6, 6, 6 4, 6 4 Algebra Worked-Out Solution Ke 07