Chapter 4 Polynomial and Rational Functions

Chapter Polynomial and Rational Functions - Polynomial Functions Pages 09 0 Check for Understanding. A zero is the value of the variable for which a polynomial function in one variable equals zero. A root is a solution of a polynomial equation in one variable. When a polynomial function is the related function to the polynomial equation, the zeros of the function are the same as the roots of the equation.. The ordered pair (, 0) represents the points on the -ais. Therefore, the -intercept of a graph of a function represents the point where f() 0. 3. Acomple number is any number in the form a bi, where a and b are real numbers and i is the imaginary unit. In a pure imaginary number, a 0 and b 0. Eamples: i, 3i; Noneamples: 5, i. y 5. 3; 6. 5; 8 7. no; f() 3 5 3 8 f(5) (5) 3 5(5) 3(5) 8 f(5) 5 5 5 8 f(5) 33 8. yes; f() 3 5 3 8 f(6) (6) 3 5(6) 3(6) 8 f(6) 6 80 8 8 f(6) 0 9. ( (5))( 7) 0 ( 5)( 7) 0 35 0; even; 0. ( 6)( i)( (i))( i) ( (i)) 0 ( 6)( i)( i)( i)( i) 0 ( 6)( i )( i ) 0 ( 6)( )( ) 0 ( 3 6 )( ) 0 5 6 5 3 30 0; odd;. ; 9 0 ( 7)( 7) 0 7 0 7 0 7 7 f() 60 0 (0, 9) 0 f() 9 (7, 0) 8. 3; a 3 a 8a 0 a(a a 8) 0 a(a )(a ) 0 a 0 a 0 a 0 a a 30 0 0 f(a) f(a) a a 8a 0 (, 0) (0, 0) (, 0) a 3. ; t 0 (t )(t ) 0 (t )(t )(t ) 0 t 0 t 0 t 0 t t t t i (, 0) f(t) f (t) t (, 0) t a. r 6 V() Bh r 36 V() (36 )() b. V() (36 )() V() (36 )() V() 7 3 c. V() 7 3 V() 7() () 3 V() 50.65 units 3 Pages 0 Eercises 5. ; 5 6. 7; 3 7. 3; 5 8. 5; 5 9. 6; 0. ;. Yes; the coefficients are comple numbers and the eponents of the variable are nonnegative integers. 0 Chapter

. No; a, which is a negative eponent. a 3. yes; f(a) a 3a a f(0) (0) 3(0) (0) f(0) 0. no; f(a) a 3a a f() () 3() () f() 3 f() 5. yes; f(a) a 3a a f() () 3() () f() 3 f() 0 6. yes; f(a) a 3a a f() () 3() () f() 56 08 8 f() 0 7. no; f(a) a 3a a f(3) (3) 3(3) (3) f(3) 8 7 36 f(3) 7 8. yes; f(a) a 3a a f(3) (3) 3(3) (3) f(3) 8 7 36 f(3) 0 9. f(b) b 3b b f() () 3() () f() 6 f() ; no 30. f() 3 f() () () 3 () () f() f() 0; yes 3a. 3; 3b. ; 3c. ; 3. ( ( ))( 3) 0 ( )( 3) 0 6 0; even; 33. ( ())( )( 5) 0 ( )( )( 5) 0 ( )( 5) 0 3 5 5 0; odd; 3 3. ( ())( (0.5))( ) 0 ( )( 0.5)( ) 0 (.5 )( ) 0 3.5 0 0 3.5 9 0 3 3 8 8 0; odd; 3 35. ( (3))( (i))( i) 0 ( 3)( i)( i) 0 ( 3)( i ) 0 ( 3)( ) 0 3 3 0; odd; 36. ( (5i))( (i))( i)( 5i) 0 ( 5i)( i)( i)( 5i) 0 ( 5i)( 5i)( i)( i) 0 ( 5)( ) 0 6 5 0; even; 0 37. ( ())( )( )( ()( 5) 0 ( )( )( )( )( 5) 0 ( )( 6)( 5) 0 ( 7 6)( 5) 0 5 5 7 3 85 6 80 0; odd; 5 38. ( ())( )( 3)( (3)) 0 ( )( )( 3)( 3) 0 ( )( 9) 0 0 9 0 39. ; 8 0 f() 8 (0, 8) (9, 0) f(a) 0 0 0 0a 0 0 60 80 (9, 0) f(a) a 8 (0, 8) f() 8 (8, 0) 0. ; a 8 0 (a 9)(a 9) 0 a 9 0 a 9 0 a 9 a 9. ; b 36 0 b 36 b 6i f(t) (, 0) (, 0) (0, 8) t f(t) t 3 t t 8 80 60 f(b) b 36 0 0 f(b) (0, 36) b. 3; t 3 t t 8 0 t (t ) (t ) 0 (t )(t ) 0 (t )(t )(t ) 0 t 0 t 0 t 0 t t t Chapter 0

3. 3; n 3 9n 0 n(n 9) 0 n(n 3)(n 3) 0 n 0 n 3 0 n 3 0 n 3 n 3 (3, 0) 0 0 0 0 00 (.5, 0) 50 50 00 f(n) (0, 0) (3, 0) n f(n) n 3 9n. 3; 6c 3 3c 5c 0 c(6c 3c 5) 0 c(c 3)(6c 5) 0 c 0 c 3 0 6c 5 0 c 3 c 5 6 f(c) (0, 0) (3, 0) c f(c) 6c 3 3c 5c c.5 7. ; m 7m 0 (m )(m ) 0 m 0 m 0 m m m 0.5i m i f(m) f(m) m 7m f(u) (0, ) 8. (u )(u ) 0 (u )(u )(u ) 0 u 0 u 0 u 0 u u u (, 0) (, 0) (0, ) m u f(u) (u )(u ) 9a. y 9b. y 5. ; a a 0 (a )(a ) 0 (a )(a )(a ) 0 a 0 a 0 a 0 a a a a i f(a) 9c. y 9d. y (, 0) f(a) a a (, 0) (0, ) a 6. ; 0 9 0 ( 9)( ) 0 ( 3)( 3)( )( ) 0 3 0 3 0 0 0 3 3 0 0 f() 0 (0, 9) (3, 0)(, 0) (, 0) (3, 0) 0 f() 0 9 9e. y 9f. not possible 03 Chapter

50. [5, 5] sc by [, 8] sc 50a. 50b. ;, 50c. There are real roots. However, there is a double root at and a double root at. 5a. V() 99,000 3 55,000 65,000 5b. r 0.5 r 0.5.5 V() 99,000 3 55,000 65,000 V(.5) 99,000(.5) 3 55,000(.5) 65,000(.5) V(.5) 50,566.65 7,737.5 7,750 V(.5) 98,05.5; about $98,05.3 5. and 3 are two of its zeros. 53a. d(t) at d(t) at d(30) (6.)(30) d(60) (6.)(60) d(30) 7380 ft d(60) 9,50 ft d(t) at d(0) (6.)(0) d(0) 8,080 ft 53b. It quadruples; (t) t. 5. Let the width of the sidewalk. The length of the pool would be 70 feet. The width of the pool would be 50 feet. A w 00 (70 )(50 ) 00 3500 0 0 0 00 0 60 75 0 ( 55)( 5) 55 0 5 0 55 5 Use 5 since 55 is an unreasonable solution. 5 ft 55. Let the number of pizzas. (60 6)(6 0.0) 000 6. 9 560 000 6. 9 0 0 30 5 0 ( 5)( 5) 0 5 0 5 0 5 5 6 0.0 6 0.0(5) $0 56. ( B)( C) 0 C B BC 0 (C B) BC 0 C B B from B C 0 BC C B C C C Sample answer: ; 57. y ( )( ) 58a. Let the width. The length (5 ) or 6. A() (6 ) 58b. A() (6 ) A() 6 [5, 30] sc5 by [, 00] sc0 3 6 6 3 3 3 yd by 3 yd 59. The graph of y 3 is the graph of y 3 shifted unit up. 60. (6, 9) 6. 5 5 5(3) (9)(5) 9 3 5 5 or 0; no 6. AB 3 9 3 5 7 6 (3) ()(5) (9) ()(7) 3(3) (5) 3(9) (7) () ()(6) 3() (6) 5 0 9 8 63. y 9 y y 9 y 9 6. Parallel; the lines have the same slope. y 9 Chapter 0

65. [f g]() f(g()) f 6 6 6 36 6 3 [g f ]() g( f()) g( ) ( ) 6 6 66. The pictograph shows two more small car symbols in the row for 999 than it does for 000. These two small cars represent the 70 additional cars that were sold in 999 compared to 000. Since the two small cars represent 70 real cars, each small car symbol must represent 7 0 or 35 real cars. The correct choice is A. - Quadratic Equations Pages 8 9 Check for Understanding. Add to each side of the equation to get t 6t. Determine the value needed to make t 6t a perfect square trinomial. Add this value (9) to each side. Take the square root of each side of the equation and solve the two resulting equations. t 3 3. Quadratic Formula; Since the leading coefficient does not equal and the discriminant equals 85 which is not a perfect square, the Quadratic Formula would be the best way to get an eact answer. Completing the square can also be used, but errors in arithmetic are more likely. A graph will give only approimate solutions. 3 (3) p 5)(7) ( 3 9 or (5) 0 3a. equals 0 3b. negative number 3c. positive number. Graphing f() f() 5 (5, 0) (, 0) 6 8 Factoring 5 0 ( 5)( ) 0 5 0 0 5 Completing the Square 5 0 5 5 ( ) 9 3 3 Quadratic Formula ()(5 ) () 36 3 5 6 3 3 3 5 See students work. 5. 8 0 0 8 0 8 6 0 6 ( ) 36 6 6 6 0 6. a a 0 a a 0 a a a a 6 6 a 8 9 6 7 a a 7 a 7 a 3 a 7 7. b ac ()(36) 0; real m 0 () m or 6 8. b ac (6) ()(3) 6; imaginary (6) 6 t () 6 i 3 i 9. p 6p 5 0 (p 5)(p ) 0 p 5 0 p 0 p 5 p 0. r r 0 0 () () )(0) ( r r r () i6 r i6 05 Chapter

. P I 0.0I 600 I 0.0I 0.0I I 600 0 I 600I 80,000 0 (I 00)(I 00) 0 I 00 0 I 00 0 I 00 amps I 00 amps Pages 9 Eercises. z z 0 z z z z (z ) 5 z 5 z 5 z 5 z 6 z 3. p 3p 88 0 p 3p 88 p 3p 9 88 9 p 3 36 p 3 9 p 3 9 p p 8. 0 0 0 0 5 5 ( 5) 5 5 5 7 3 5. d 3 d 8 0 d 3 d 8 d 3 9 d 6 8 9 6 d 3 8 6 d 3 8 8 d 3 8 8 d 3 8 8 d 6. 3g g g g 3 g g 3 (g ) 8 3 g 6 3 g 6 3 d 7. t 3t 7 0 t 3t 7 t 3t 9 7 9 t 3 3 7 t 3 8. 37 37 t 3 9. b ac (6) ()(5) 36 imaginary; the discriminant is negative. 0. b ac 7 (6)(3) or ; real 7 m m 7 (6) m 3, 3. b ac (5) ()(9) or ; imaginary 5 s () 5 i s. b ac (8) (36)(9) or 0; real 8 0 d (36) d 8 7 or 7 6 3. b ac () ()(9) or 0; imaginary 0 () i35 8 i35. 3p p 8 3p p 8 0 b ac (3)(8) or ; real p p p (3) 7 6 7 3 5. k 5k 9 k 5k 9 0 b ac 5 ()(9) or 97; real 5 97 k () 5 97 k 6. 7 i5 7. 5 i 8. s b b ac s s s a 5 (5) )(9) (3 (3) 5 83 6 5 i83 6 Chapter 06

9. 3 8 0 ( 7)( ) 0 7 0 0 7 30. w 9w 5 0 (w )(w 5) 0 w 0 w 5 0 w w 5 w 3. r r 5 r r 5 0 (r 5)(r ) 0 r 5 0 r 0 r 5 r r 5 3. p b b ac a ()(8) () 8 p p p 33. i7 p i7 6 (6) )( ( ) () 6 3 6 6 3a. P 0.0A 0.05A 07 P 0.0(5) 0.05(5) 07 P 6.5.5 07 P.5 mm Hg 3b. P 0.0A 0.05A 07 5 0.0A 0.05A 07 0 0.0A 0.05A 8 A b b ac A A a 0.05 0.05.0)( (08) (0.0) 0.05 0.7 5 0.0 0.05 0.7 5 0.0 A or A A 0 A 5 0 years old 3c. P 50 5 00 P 75 0.0A 0.05A 07 50 5 5 50 75 00 A 0.05 0.7 5 0.0 As a woman gets older, the normal systolic pressure increases. 35. b ac 0 8 ()(c) 0 6 c 0 c 6 c 6 36a. A bh A (6) A 9 ( )(6 ) (9) ( )(6 ) 96 36b. ( )(6 ) 96 9 56 96 56 96 0 0 f() 0 0 0 0 0 0 36c. roots:, () or 8 6 6 () or 8 ft by ft () or 6 6 () or 8 37a. d(t) v 0 t gt d(t) d(t) 5t (3)t d(t) 5t 6t f() 37b. 0 and about 0.3 37c. The -intercepts indicate when the woman is at the same height as the beginning of the jump. 37d. d(t) 5t 6t 50 5t 6t 37e. 50 5t 6t 6t 5t 50 0 t b b ac t t a 5 (5) 6)(5 (0) (6) 5 35 3 5 35 3 t t.93 s about.93 s t 5 35 3 t.6 t d(t) 5t 6t 07 Chapter

38. a b c 0 b c a a 0 a b a c b a b a c a b a b a c b a a b a a c b a b a b ac a b y 0 0 0 a b ac a b bac a 39. ; 8a 3a 0 (3a )(6a ) 0 3a 0 6a 0 3a 6a a 3 a 6 0.. f() ( 9) y ( 9) (y 9) y 9 y 9 f () 9. 3 y 375 (5 y) (35) 3 y 375 3(5) y 375 y 60 (5, 60) 3. m 6. 8 m 0 9 595.. f(a) f(a) 8a 3a 3 y 375 0 y 690 7 35 5 60 69.8 3. 69 m 60 $63. 3y 8 3y 8 y 3 ; 8 3 5. 0 ( 5)( ) The correct choice is A. a y y -3 The Remainder and Factor Theorems Page 6 Check for Understanding. The Remainder Theorem states that if a polynomial P() is divided by r, the remainder is P(r). If a division problem has a remainder of 0, then the divisor is a factor of the dividend. This leads to the Factor Theorem which states that the binomial r is a factor if and only if P(r) 0.. ( 3 7 8) ( 5); ; 3. The degree of a polynomial is one more than the degree of its depressed polynomial.. Isabel; if f(3) 0, then ( (3)) or ( 3) is a factor. 5. 6. 5 7 5 5 0 5 6 3 0, R6 3 7. f() 5 f(3) (3) (3) 5 9 6 5 or 0; yes 8. f() f(3) (3) (3) 8 9 or 9; no 9. f() 3 5 5 f() () 3 5() 5 5 5 or 0 5 5 5 5 0 is a factor 5 ( 5)( ) ( 5), ( ), ( ) 0. f() 3 6 6 f() () 3 6() () 6 6 6 or 0 6 6 5 6 5 6 0 is a factor 5 6 ( )( 3) ( ), ( ), ( 3). 0 7 k 6 6 k k 6 6 k a. b. c. d. f() 7 9 9 7 ( 8 6 ) ( 0 7 5 ),, 8 6, or 0 7 5 3. h r V r h V r (r ) 5 r (r ) 5 r 3 r 0 r 3 r 5 0 (r 3 r 5) 0 5 5 5 5 5 0 r 0 h r r in. h or 5 in. Chapter 08

Pages 6 8 Eercises. 7 0 9 5. 7 9 3 0 3 6 9, R 6. 0 0 6 3 6 3 3 3 6, R3 7. 0 8 0 6 8 6 8 0 3 8 8. 3 5 3 5 0 3 5 0 3 3 5 0, R 9. 0 3, R 3 0. f(). f() 5 3 f() () f() () 5 3 or ; no 3 3 or 0; yes. f() 6 8 f() () 6() 8 8 or 0; yes 3. f() 3 6 f() () 3 6 8 6 or ; no. f() 3 3 f() () 3 () () 3 3 or 3; no 5. f() 3 3 f() () 3 3() 3 or 0; yes 6a-d. 0 3 0 7. (6) 36 36 36 or 0 8. 7 7 6 7 6 7 0 6 7 ( )( 7) ( )( )( 7) 9. 6 3 0 3 ( )( ) ( ), ( ), ( ) 30. 9 9 0 9 0 9 0 9 ( 7)( 7) ( ), ( 7), ( 7) 3 9 7 8 3 8 7 6 9 r 3 8 6 5 8 8 0 d 3. 5 8 8 0 ( )( ) ( ), ( ), ( ) 3. 8 0 8 0 0 ( )( ) ( ), ( ), ( ) 33. 5 5 0 5 ( )( ) ( ), ( ), ( ) 3. 3 3 0 ( )( ) ( ), ( ), ( ) 35. 0 9 0 0 6 0 0 8 6 5 0 8 0 5 5 3 0 8 5 0 8 8 6 8 8 3 0 3 3 3 30 5 6 5 6 56 times 36. 0 ( )( ) time;, 37. f() 3 k f() () 3 () k 0 k k 38. f() 3 k f() () 3 k() () 0 8 k 0 k 8 k 39. f() 3 8 k f() () 3 8() k() 0 8 7 k 0 k 68 3 k 0. f() 3 k f() () 3 () k() 0 k 0 k k 09 Chapter

. d(t) v 0 t at 5 t (0.)t 0 0.t t 5 5 0. 5 5 0. 5 0 t 5 0 t 5 s. 7 a b 5 5 a 5 5 a 5 a b 5 5 a 5 a b 0 0 0 a 0 5 5 a 5 a b 5 a b 0 5 a b 0 5 a b 0 5 a 0 0 0 b 0 a 5 0 b 0 a 5 b 0 3a. V() (3 )( )(5 ) V() ( 7 )(5 ) V() 3 7 60 3b. V() V() 3 7 60 3c. V w h V 3 5 or 60 3 5 V 3 5 (60) 36 V() 3 7 60 36 3 7 60 3d. 36 3 7 60 0 3 7 b. 5 0 5 0 50 V() V() 3 38 80 8 c. V() 3 38 80 3 38 80 d. 3 38 80 0 3 38 80 0 3 9 90 9 90 3 7 56 0 in. 5. P(3 i) 0 and P(3 i) 0 implies that these are both roots of a b c. Since this polynomial is of degree it has only these two roots. 3 i 3 i ( 3) 6 6 9 6 6 5 0 a, b 6, c 5 6. r 5r 8 0 r 5r 8 r 5r 5 8 5 r 5 5 7 r 5 57 r 5 57 7a. f() 3 f() () () 3 () () f() 6 3 8 or ; no 7b. f(0) (0) (0) 3 (0) (0) f(0) 0 0 0 0 or 0; yes 7c. f() () () 3 () () f() 6 3 8 or 36; no 7d. f() () () 3 () () f() 56 56 6 6 or 0; yes 8. f() 5 3 [3, 0] sc by [00, 00] sc5 about 0.60 ft a. (0 ) or 0 w 8 h V() (0 )(8 )() V() (80 38 )() V() 3 38 80 (0, 3); point of inflection [, ] sc by [50, 0] sc0 9. wider than parent graph and moved unit left Chapter 0

50. Let number of 00 foot units of Pipe A and y number of 00 foot units of Pipe B. 6y 8 y 8 y 6 0 y 0 P(, y) 3 0y P(0, 0) 3(0) 0(0) or 0 P(0, 8) 3(0) 0(8) or 30 P(3, 6) 3(3) 0(6) or 3 P(7, ) 3(7) 0() or 38 P(8, 0) 3(8) 0(0) or 7 3 00 foot units of A, or 300 ft of A 6 00 foot units of B, or 600 ft of B 5. y 3z 6 y 3z 6 7y 3z 7y 3z 0 3 9y 3 z y 3z 6 7y 3z 0 6 9y 6 ( 7y 3z) (0) 3(3 9y z) 3(3) y 6z 0 9 7y 6z 39 5 3y 39 5(6 9y) 5(6) 6(5 3y) 6(39) 30 5y 30 30 78y 3 33y 0 8 3 9y z 3 y 6 6 9y 6 y 3z 6 8 6 ( 9 ) 6 8 3z 6 6 5 6 3z 9 6 9 6 9, 6 8, 9 8 5. M 7 N 9 z 9, 6, 6 3 or (.5, ) or (,.5).5 slope of MN.5 ( ) or (3) slope of RI or 7 () y (0, 8) 0 (0, 0) y 8 y 6 (3, 6) 6y 8 y 0 (8, 0) (7, ) Since the slopes are the same, MN L RI. 53. a b a b a b ac bc a c b c a c b c I. true II. true III. false The correct choice is D. 8 - The Rational Root Theorem Page 3 Graphing Calculator Eploration. 3;,,. ;, 3. () positive; f() () () 3 3() () f() 3 3 ; 3 or () positive; f() () 3 3() f() 3 3 ; or 0. In the first function, there are negative zeros, but according to Descartes Rule of Signs, there should be 3 or negative zeros. This is because the is a double zero. In the second function, there is one negative zero, but according to Descartes Rule of Signs, there should be or 0 zeros. This is because is a double root. 5. ne number represents two zeros of the function. Page 33 Check for Understanding. possible values of p:,, 3, 6 possible values of q: possible rational roots:,, 3, 6. If the leading coefficient is, then q must equal. Therefore, p q becomes p or p, and p is defined as a factor of a n. 3. Sample answer: f() 3 3; f() () 3 () () 3 f() 3 3; 0 3 or possible positive zeros and no possible negative zeros. Sample answer: You can factor the polynomial, graph the function, complete the square, or use the Quadratic Formula if it is a second-degree function, or use the Factor Theorem and the Rational Root Theorem. I would factor the polynomial if it can be factored easily. If not and it is a second-degree function, I would use the Quadratic Formula. therwise, I would graph the function on a graphing utility and use the Rational Root Theorem to find the eact zeros. 5. p q, rational root r 3 0 5 6 3 6 3 Chapter

6. p:, 3 q:, P q :, 3,, 3 r 3 8 3 5 3 0 9 6 0 6.5 3 9 9 60 3 3 0 rational roots: 3,, 7. or 0; f() 8() 3 6() 3() 6 f() 8 3 6 3 6; p q :,, 3, 6,, 3,, 3, 8, 3 8 r 8 6 3 6 8 5 8 0 3 0 8 0 3 0 ( )( 3) 0 0 3 0 3 3 or,, 8. ; f() () 3 7() 7() 5 f() 3 7 7 5; or 0 p q :, 3, 5, 5 r 7 7 5 8 5 0 6 6 3 5 0 5 3 0 5, 3, 9. r 5 V 3 r h 5 3 (5 )(5 ) 356 (5 )(5 ) 356 3375 5 5 3 3 5 5 8 0 Possible rational roots:, 9, 8 f() 3 5 5 8 0 f() 8 f() 30 f(9) 0 f(9) 59 f(8) 6,7 f(8),70 represents 9 cm. Pages 3 35 p 0. q :,, 3, 6 Eercises r 5 6 3 8 3 0 3 0 ( 3)( ) 0 3, rational roots: 3,,. p q :,, 3, 6, 9, 8 r 8 3 9 0 9 0 does not factor rational root:. p q :, r 5 9 7 5 0 3 5 r 5 0 0 ( )( ) 0, rational roots:, 3. p q :,,, 5, 0, 0 r 5 0 8 6 8 3 0 0 3 0 0 ( 5)( ) 0 5, rational roots:,, 5. p:, 3 q:, p q :, 3,, 3 r 0 6 3 0 0 6 0 3 6 0 3 3 3 3 rational root: Chapter

5. p: q:,, 3, 6 p q :,, 3, 6 r 6 35 7 6 38 8 0 6 3 38 8 r 6 38 8 3 6 36 6 0 6 36 6 0 6 0 does not factor rational roots: 3, 6. ; 3 or ; f() () () 3 7() () 5 f() 3 7 5; negative positive 7. f() 3 7 6 0 or negative r 0 7 6 6 6 0 6 0 ( 3)( ) 0 3, rational zeros:,, 3 8. positive f() 3 8 negative f() 3 8 0 ( 8) ( )( ) 0,, rational zeros:, 0, 9. positive f() 3 3 0 or 0 negative r 3 0 3 6 8 6 8 0 ( )( ) 0, rational zeros:,, 3 0. or 0 positive f() 0 3 7 7 negative r 0 7 7 0 0 0 0 5 0 (5 )( ) 0 5, rational zeros:, 5,. or 0 positive f() 3 9 8 or 0 negative r 9 8 3 6 8 0 3 3 6 8 r 3 6 8 8 0 8 0 ( )( ) 0, rational zeros:,,,. or 0 positive f() 5 or 0 negative r 0 5 0 0 3 r 0 0 0 ( )( ) 0, rational zeros:,,, 3a. f() ( )( )( ) 0 ( )( )( ) 0 0 ( ) 0 0 3b. f() ( )( )( ) f() ( )( ) f() 3 3 8 3c. positive f() 3 3 8 3 or negative 3d. There are negative zeros, but according to Descartes Rule of Signs, there should be 3 or. This is because is actually a zero twice. a. Let the length. w h V() w h V() ( )( ) V() ( )( ) V() 3 9 3 Chapter

b. V() 3 9 08 3 9 c. 08 3 9 0 3 9 08 r 9 08 5 8 0 w h w or 8 h () or 3 in. 8 in. 3 in. 5a. Sample answer: 3 3 0 5b. Sample answer: 3 0 5c. Sample answer: 3 0 6a. Let the length. h 9 V() 3 Bh V() 3 ( )( 9) V() 3 3 3 6b. V() 3 3 3 6300 3 3 3 6c. 6300 3 3 3 0 3 3 3 6300 0 3 9 8,900 r 9 0 8,900 30 630 0 30 h 9 h 30 9 or base: 30 in. by 30 in., height: in. 7. d 0.0000008 (00 ) 0.8 0.0000008 (00 ) 0 0.0006 0.0000008 3 0.8 0 8 3 600 8,000,000 0 3 00,000,000 r 00 0,000,000 00 00 0,000 0 00 ft 8. The graphs are reflections of each other over the -ais. The zeros are the same. 9. 7 56 7 56 8 0 8 30. b ac 6 ()(5) 36; imaginary 3. ( )( ())( )( ()) 0 ( )( )( )( ) 0 ( )( ) 0 5 0 3. y.3 8.3 y.3(008) 8.3 y $0.0 33. 3 3 ( 3) (3 ) 6 3 6 0 ( 3)( ) 0 3 0 0 3 The correct choice is A. Page 35 Mid-Chapter Quiz. ( )( ())( i)( (i)) 0 ( )( )( i)( i) 0 ( )( ) 0 3 0. 3; 3 30 0 ( 30) 0 ( 6)( 5) 0 0 6 0 5 0 6 5 3. 5 50 5 5 50 5 5 6 5 5 5 5 5 5 5 5 0 5. b ac (39) (6)(5) ; real roots b b 39 39 (6) 5 b 39 or b 39 b 5 b 3 5. 3 8 8 0 6. 0 6 8 0 ; no 7. 5 6 6 6 0 6 ( 3)( ) ( 3)( )( ) 8. p q :, 3 r 6 0 3 3 3 0 3 0 does not factor rational root: 3 Chapter

9. positive F() 3 3 3 or negative 3 5 8 0 0 ( )( ) 0, rational zeros:,, 0. Let r radius. h r 6 V 3 r h 7 3 r (r 6) 0 3 r 3 r 7 0 r 3 6r 8 r 3 h r 6 h 3 6 or 9 r 3 cm, h 9 cm -5 r 3 5 8 0 r 5 8 0 r 6 0 8 3 9 7 0 Locating Zeros of a Polynomial Function Pages 39 0 Check for Understanding. If the function is negative for one value and positive for another value, the function must cross the -ais in at least one point between the two values. f() f(b) f(a) a (a, f(a)) (b, f(b)) b. Use synthetic division to find the values of the polynomial function for consecutive integers. When the values of the function change from positive to negative or from negative to positive, there is a zero between the integers. 3. Use synthetic division to find the values of the polynomial function for consecutive integers. An integer that produces no sign change in the quotient and the remainder is an upper bound. To find a lower bound of a function, find an upper bound for the function of. The lower bound is the negative of the upper bound for the function of.. Nikki; the sign changes between and. 5. r 6 0 5 3 0 3 5 6 3 5 0 5 3 and 5, and 0 6. r 3 5 8 0 3 0 3 0 and, at, 3 and 7. r 0 3 0 0 3 5 0 0 3 3 6 5 approimate zero:.3 8. r 3 0 0 zeros:, 9. Sample answer: r 0 0 8 7 5 0 upper bound: f() 8 r 0 0 8 0 0 0 8 lower bound: 0 0. Sample answer: r 0 0 3 5 0 7 upper bound: f() 3 r 0 0 3 5 0 7 lower bound: a. Let amount of increase. V() (5 )(30 )(5 ) V() (750 55 )(5 ) V() 3 60 05 3750 5 Chapter

b. V w h.5v.5(3750) V 5(30)(5).5V 565 V 3750 V() 3 60 05 3750 565 3 60 05 3750 c. 565 3 60 05 3750 0 3 60 05 875 r 60 05 875 6 086 789 6 9 3.7 5 5.7 30 30.7 6.7 3.7 5 5.7 6.7 about 6.7 cm by 3.7 cm by 6.7 cm Pages 0 Eercises. r 0 0 3 0 0 0 6 3 3 9 5 and 3. r 5 7 8 0 5 3 3 0 and, and 3. r 0 8 5 8 3 3 0 0 0 0 0 0 0 at, at 5. r 0 8 0 0 3 3 3 9 8 6 7 7 3 0 0 8 0 0 7 7 3 8 6 3 3 3 9 3 and, and, and, and 3 6. r 0 3 3 0 0 3 3 and, 0 and, and 7. r 0 3 3 3 6 9 60 83 9 5 3 6 9 0 0 3 3 3 0 3 9 6 35 no real zeros 8. r 6 5 3 6 6 8 5 6 6 7 yes; f(6), f(5) 7 9 5. Use the TABLE feature of a graphing calculator. 9. 0.7, 0.7 0..6, 0...5. 0., 3. 3.,..3, 0.9, 7. 5.. 6. Sample answers: r 3 5 3 6 5 upper bound: f() 3 3 5 r 3 5 0 3 5 lower bound: 0 7. Sample answers: r 0 upper bound: f() r lower bound: 8. Sample answers: r 6 6 3 5 3 3 0 6 6 5 3 3 7 5 58 6 8 85 5 3 9 58 6 0 8 95 upper bound: 6 f() 6 3 6 3 r 6 6 3 7 9 3 0 8 8 30 7 lower bound: Chapter 6

9. Sample answers: r 5 3 0 6 3 7 7 upper bound: f() 3 5 3 0 r 5 3 0 7 3 3 9 3 9 7 7 8 5 0 3 5 6 3 38 lower bound: 6 30. Sample answers: r 3 3 5 6 5 5 3 0 3 39 upper bound: f() 3 3 3 5 r 3 3 5 6 5 8 3 lower bound: 3. Sample answers: r 5 3 0 0 5 6 3 3 3 8 upper bound: f() 5 5 3 3 0 5 r 5 3 0 0 5 7 7 7 8 3 9 38 76 37 3 9 7 08 7 8 9 753 5 0 3 35 75 860 6 3 57 7 57 399 808 lower bound: 7 3a. 3b., 5 3c. 3 or ; f() 3 3 3 5 negative real zero 3d. r 3 3 5 5 8 3 0 39 3 0 3 5 5 6 0 3 3 5 6 5 8 3 3 6 6 5 30 and, 3 and 3e. Sample answers: upper bound: (See table in 3d.) f() 3 3 3 5 r 3 3 5 6 5 8 3 lower bound: 3f.., 3. (Use TABLE feature of a graphing calculator.) 33a. 890: P(0) 0.78(0) 33(0) 3 7500(0) 7,500(0),0,000,0,000 90: P(0) 0.78(0) 33(0) 3 7500(0) 7,500(0),0,000,39,00 930: P(0) 0.78(0) 33(0) 3 7500(0) 7,500(0),0,000,855,00 950: P(60) 0.78(60) 33(60) 3 7500(60) 7,500(60),0,000,909,00 970: P(80) 0.78(80) 33(80) 3 7500(80) 7,500(80),0,000,387,00 The model is fairly close, although it is less accurate at for 950 and 970. 33b. 980 890 90 P(90) 0.78(90) 33(90) 3 7500(90) 7,500(90),0,000 P(90) 53,800 33c. The population becomes 0. 33d. No; there are still many people living in Manhattan. 3. Sample answer: f() ( )( )( ) f() ( )( ) f() 3 ;, f() f() 3 35a. 37. 60 3 60 60 35b. f() 60 3 60 60 37. 35c. f() f() 60 3 60 60 37. about 35d. 0. (Use TABLE feature of a graphing calculator.) 36. Sample answer: f() 0 0 7 Chapter

37a. f () 00,000 80,000 60,000 0,000,000 f() 0.5 5 3.5 000 8 6 0. 7 9 7(6) 3(9) or 5 3 6 3. (, y) 3 8, (.5, ). y 0 y ; ; 5. A 37b. f(0) 0.5(0) 5 3.5(0) 000 f(0) 000 deer 37c. 90 905 5 f(5) 0.5(5) 5 3.5(5) 000 f(5) 67,8.5 about 67,8 deer 37d. in 930 38a. 8.58 6 8 3 8 38b. 8.58 6 8 3 8 0 6 8 3 8 8.58 about. (Use TABLE feature of a graphing calculator.) 38c. rate. rate 0. rate about 0% 39. or 0; f() 3 5 8 5 negative zero r 5 8 5 3 5 0 5 0 ( )( 5) 0 0.5 5 rational zeros: 3, 0.5, 5 0. d(t) v 0 t gt 750 t (9.8)t.9t t 750 0 t b b ac a ().9)( (750) t t (.9) 3,3 6 9.8 3,3 6 9.8 t t 9.3 about 9.3 s. y y t 3,3 6 9.8 t 8.5 C B y cannot be true. The correct choice is B. -6 y Rational Equations and Partial Fractions Page 7 Check for Understanding. Multiply by the LCD, 6(b ). Then, solve the resulting equation.. If a possible solution causes a denominator to equal 0, it is not a solution of the equation. 3. Decomposing a fraction means to find two fractions whose sum or difference equals the original fraction.. ( ) ( ) ( ) ( ) 0 ( )( ) 0 0 0 If you solve the equation, you will get. However, if, the denominators will equal 0. 5. b 5 b b 5 b b ()b b 5 b b b 5 0 (b 5)(b ) 0 b 5 0 b 0 b 5 b D 9 3 6. b 5 b 3 9 b 5 (b 5)(b 3) 3 b 3 9(b 3) 3(b 5) 9b 7 3b 5 6b 0 b 7 (b 5)(b 3) Chapter 8

7. t 3 6 t t t t t 3 t t (t)(t ) 6 t t (t )(t ) 3(t) 6 t 6 3t 6 t 3t 0 t(t 3) 0 t 0 t 3 0 t 3 But t 0, so t 3. 8. 3 p 3p p (p ) (p ) 3 p A B p p p (t)(t ) 3p A(p ) B(p ) Let p. 3() A( ) B( ) B B Let p. 3() A( ) B( ) A A 3 p p p p 9. 5 ; 6 eclude: 0 5 6 5 5 3 6 Test : 5 ( ) ( ) 6 true Test : 5 6 6 6 false Test : 5 6 5 true Solution: 0, 3 5 0. a 7 6 ; eclude: 6(a ) 30 7(a ) 6a 6 30 7a 7 3 a 5 Test a : 7 6 3 7 6 true 5 Test a : 7 6 6 7 6 false 5 Test a 36: 36 7 6 7 7 6 Solution: a, a 3 a. 3 60 0 3 b. 3 60 0 3 57. 57. 8 9 true 3 60 0 57.(3 ) 00 7. 57. 0.50 ; 0.50 h Pages 7 50 Eercises. t t 8 0 t t 8t (0)t t 8t 0 t 8t 0 (t 6)(t ) 0 t 6 0 t 0 t 6 t 3 m 3. m m m m m m m 3 m 3m m 0 m 3m 0 m(m 3) m 0 m 3 0 m 3 But m 0, so m 3.. y 3 y y y y 3 y (y)(y ) y y (y)(y ) y 3(y ) y 5y 6 y y 5y 6 0 ( y 3)(y ) 0 y 3 0 y 0 y 3 y But y, so y 3. 5. n 0 n n5 n 0 n n 5 n 5 n n5 (n )(n ) n (n )(n ) 0 (n 5)(n ) (n 5)(n ) n 3n 5 n 3n 5 6n 0 n 5 3 3 6. b b b b b (b )(b ) 3 b b b 3(b ) b 3b 6 b (b )(b ) 7a 5 3a 7. 3a 3 a a 7a 5 3a 3(a ) (a ) (a ) 7a 5 3(a ) (a ) ()(a )(a ) 3a (a ) ()(a )(a ) (a )7a 3(a )5 6(a )3a 8a 8a 5a 5 8a 8a 0a 5a 5 0 a 5a 3 0 (a )(a 3) 0 a 0 a 3 0 a a 3 9 Chapter

8. a a a a a a ( a)(a ) a a a a a( a) a a a a 0 0 all reals ecept q q 9. q 3 q 3 q q q 3 q 3 q ( a)(a ) (q 3)(q 3) (q 3)(q 3) q(q 3) q(q 3) (q 3)(q 3) q 6q q 6q q 9 0 q q 9 9) ()( 88 8 8 3 3 0. 3 m 6m 9 3m 3 3m m 9 3 3 m 6m (m) 3m 3m m (m) (6m 9) 3(3m 3) m 36 9m 9 5m 3 7 3. m 3 5 ( )( )( ) 7 3 ( )( )( ) ( )( ) 7( )( ) 3( )( ) ( ) 7( ) 3( 3 ) 8 9 3 5 3 5 3 a. (n )(n ) b., 6 n c. n n n 6 (n )(n ) n n (n )(n ) (n )(n ) (n )(n 6) (n ) n n n n n n n 8 0 n ()( 8) 5 6 6 3. ( ) 6 A B 6 A( ) B() Let. 6 A( ) B() B B Let 0. 0 6 A(0 ) B(0) 6 A 3 A 6 3. 5 m 5m m (m )( m ) 5 m A B m m m 5m A(m ) B(m ) Let m. 5() A( ) B( ) 6 B.5 B Let m. 5() A( ) B( ) A 3.5 A 5 m 3.5.5 m m m y y y (3y )( y ) y A B 3y y 3y y 5. 3y 6. 9 y A(y ) B(3y ) Let y. () A( ) B(3() ) B B Let y 3. 3 A 3 3 A 3 A B3 3 y 3y y 3y y 9 9 9 9 ( 3)( 3) 9 9 A B 9 ( 3) ( 3) Let 3. 9 9(3) A(3 3) B(3 3) 8 6B 3 B Let 3. 9 9(3) A(3 3) B(3 3) 36 6A 6 A 9 9 6 3 9 3 3 6 3, 3 3 7a. a(a 6) Chapter 0

7b. a a a a 6 a (a)(a 6) a a a 6 (a )(a 6) (a )(a) a 8a a a a 3 a (a)(a 6) 7c. 0, 6 7d. Test a : 6 3 5 false 7 Test a : 6 3 5 true Test a : 6 0 false Test a 7: 7 7 7 7 6 5 3 true 7 Solution: 0 a 3, 6 a 8. w 3 9 ; eclude: 0 w 3w 9 w 9 9 Test w : 3 9 true Test w : 3 9 5 9 false Test w 0: ( 3 9 0) 0 3 0 9 true 0 Solution: w 0, w 9 ( 3) ( ) 9. ( 5) ( 6) 0; eclude 5, 6 ( 3)( ) 0 3 0 0 3 Test 0: ( ( 0 3) ( 0 ) 0 0 5) ( 0 6) 0 80 (3.5 3)(3.5 ) (3.5 5)(3.5 6) 0. 5 9. 375 (.5 3)(.5 ) (.5 5)(.5 6) 0. 75. 5 (5.5 3)(5.5 ) (5.5 5)(5.5 6) 3. 75 0. 5 6.5 3)(6.5 ) (6.5 5)(6.5 6) true Test 3.5: 0 0 Test.5: 0 0 Test 5.5: 0 0 Test 6.5:( 0 8. 75 0. 375 0 Solution: 3, 5 false true false false 6 5 6 ( 5)( ) 30. 0 0; eclude 5, 6 0 6 (5) 6 (5 5)(5 ) 9 0 0 () 6 ( 5)( ) Test 5: 0 Test : 0 Test 0: 0 7 0 6 0 (0) 5 6 0 5.5 6 (.5 5)(.5 ).5.75 6 6 0 5 0 0 7 true 0 false true Test.5: 0 Test 6: 6 0 false true Solution:,, 5 3. a 5 8 a ; eclude: 0 a 5 8 a 5 a 7 a 5 Test a : ( ) 8( Test a : ( ) Test a : ( ) 5 8( ) ) 7 8 false 7 8 true 5 8( ) 7 6 Solution: 0 a 7 false Chapter

8 3. b b 5 8 b b 5 ; eclude:, 5(b ) 5(b ) 8(b )(b ) 5b 30 6b b 8 0 6b b 0 (6b )(b ) 6b 0 b 0 b 6 b 8 Test b : ( ) 5 Test b 0.8: (0. 8) Test b 0.6: (0. 6) Test b 0: (0) Test b 3: (3) 3 8 5 false 8 (0.8 ) 5 0 8 3 5 8 (0.6 ) 5 5 8 5 8 0 5 8 5 8 3 5 8 8 5 true false Solution: b, 6 b 7 33. y 7; eclude 7 7(y ) y 0 y 7 Test y : 7 7 7 false 7 Test y 0.5: 0.5 7 7 true 7 Test y : 7 7 7 false Solution: y 0 3. Let the number. 0 5 0 5 5 5 5 0 0 (5 )( 0) 0 5 0 0 0 5 0 true false 35. 5 5 0.30 0.30; eclude 5 0.30( 5) 0.30.5 0.7 3.5 5 Test 6: 6 6 5 0.30 0.36 0.30 true Test 0: 0 0 5 0.30 0. 0.30 false Test 6: 6 6 5 0.30 8 0.30 true Solution: 5 or 5 36a. 8 d 3 36b. 8 d i 8 (3d i ) d i 3 i 3 d i 3 d i 3d i 3 d i 0 3 cm 37. Sample answer: (3di ) 3 38. Let capacity of larger truck. 5 3 5( 3) 5 5 3 5 5 tons 39a. 0 r 39b. 0 r 0 r r 0 r (0r) r 0 (0r) r 0 0 r r 30 r (30) or 60; 60 ohms, 30 ohms 0. Let the number of quiz questions to be answered. 0 0.70 0 0.70(0 ) 0.70 0.3 3 0 questions. Let the speed of the wind. 06 738 0 0 0 0 06(00 ) 738(00 ),00 06 7,600 738 6,800 800 36 ; 36 mph Chapter

. a b c a b (a)(b)(c) c (a)(b)(c) bc ac ab bc ab ac bc a(b c) bc b a c 3a. y z 3 0 5 0 5 6 0 9 0 0 9 0 3b. 3 (360) 6 (360) 360 6 360 0 36. Let number of gallons of gasoline. 0 m g 5,0 00m g 583 3 y 5,000 y 5.7; about 5.7 mpg 5. T d s 0 3 s 6 6 5 s 5 6 6 5 s 5 (3)(s 5)(s 5) 0 3 (3)(s 5)(s 5) s 3(s 5)(s 5) 6(3)(s 5) 6(3)(s 5) 3s 800 78s 390 78s 390 3s 56s 800 0 8s 39s 00 0 (8s 5)(s 8) 0 8s 5 0 s 8 0 5 s 8 s 8 8 mph 6. 3 5 5y y 3 5 y 5 y 5y 7. 0 5,000 750 gallons 750 $.0 $900 $.0 $900 $00 583 3 gallons Let y number of miles per gallon. y 5,000 m gm 583 3 g 0 r 3 5 3 0 5 0 3 0 3 5 3 0 5 5 5 8. 5 0 30 0 5 5 5 5 5 5 no 9. ; 8 5 0 (6 5)( 3) 0 6 5 0 3 0 5 6 3 50. 3 0 3 3 3 5. y 3 3? (6) 3 6 3 5 false no 5. y b a 5a. (b) a 5b. b (a) b a yes b a yes 5c. (a) (b) 5d. (a) (b) a b no a b no 53a. Let short answer questions and y essay questions. y 0 0 y y 60 0 y 0 6 y 0 8 y 60 (0, 5) (8, ) 0 8 6 0 (0, 0) y 0 (0, 0) S(, y) 5 5y S(0, 0) 5(0) 5(0) or 0 S(0, 5) 5(0) 5(5) or 75 S(8, ) 5(8) 5() or 0 S(0, 0) 5(0) 5(0) or 00 8 short answer and essay for a score of 0 points 3 and, and, and 3 Chapter

53b. Let short answer questions and y essay questions. y y 0 0 y 0 6 0 y 0 y 0 (0, 0) y 0 8 (, 8) 0 8 (0, 0) y 0 6 0 (0, 0) S(, y) 5 5y S(0, 0) 5(0) 5(0) or 0 S(0, 0) 5(0) 5(0) or 50 S(, 8) 5() 5(8) or 80 S(0, 0) 5(0) 5(0) or 00 short answer and 8 essay for a score of 80 points 5. 3 5 3 5 (3) (5) (5) (3) (3) (5) (5) 0 0 0 0 55. y y m( ) y ( (3)) y 6 y 7 0 56a. m 300 0 5000 0 60 56b. $000; $50 m 50 y 3000 50( 0) y 50 000 C() 50 000 56c. C() Cost $000 $3000 $000 $000 0 0 Televisions Produced 57. A of JKL (9)(7) or 3.5 C() 50 000 A of small triangle (5)(3) or 7.5 A of shaded region 3.5 7.5 or The answer is. 6 8-7 Radical Equations and Inequalities Pages 5 55 Check for Understanding. To solve the equation, you need to get rid of the radical by squaring both sides of the equation. If the radical is not isolated first, a radical will remain in the equation.. The process of raising to a power sometimes creates a new equation with more solutions than the original equation. These etra or etraneous solutions do not solve the original equation. 3. When solving an equation with one radical, you isolate the radical on one side and then square each side. When there is more than one radical epression in an equation, you isolate one of the radicals and then square each side. Then you isolate the other radical and square each side. In both cases, once you have eliminated all radical signs, you solve for the variable.. t t t 3 t 3 3 3 5. 3 3 Check: 3 3 3 9 3 733 3 79 3 79 3 733 9 3 3 3 6. 5 Check: 5 3 5 3 9 5 9 3 5 3 no real solution 7. 6 0 6 0 3.5 Check: 6 0 7 0 6 7 Check: t 7 0 7 7 8. a a 3 7 a 7 a 3 a 9 a 3 a 3 a 3 76 96(a 3) 9 a 3 a Check: a a 3 7 3 7 6 9 7 3 7 Chapter

9. 5 8 5 0 5 6 5 5 60 0.8 Test : 5() 8 8 meaningless Test 0: 5(0) 8 8 true Test 3: 5(3) 8 69 8 false Solution: 0.8 0. 3 a 5 0 a 5 0 a 5 7 a 5 a 5 9 a.5 a 5 a 3.5 Test a 0: 3 (0) 5 0 3 5 0 meaningless Test a : 3 () 5 0 3 0 true Test a : 3 () 5 0 3 5 0 false Solution:.5 a 3.5 a. v v 6h 0 90 0 6h 90 00 6h b. 90 00 6h Check: 90 00 6h 800 00 6h 90 00 5) 6( 8000 6h 90 800 5 h; 5 ft 90 90 Pages 55 57 Eercises. 8 5 Check: 8 5 8 5 7 8 5 7 5 5 5 5 3. 3 y 7 Check: 3 y 7 y 7 6 3 7 7 y 7 3 6. 8n 5 8n 5 3 Check: 8n 5 87 5 8n 5 9 5 8n 3 n 7 5. 6 6 8 6 0 8 0 0 Check: 6 0 6 0 6 6. 3m 5 3m 5 3m 5 3m 6 m 6 3 m 3 3 Check: 3m 5 3 3 3 5 Check: 3m 5 3 3 3 5 7. 9u 7u 0 9u 7u 0 u 6 u 8 Check: 9u 7u 0 9(8) 7(8) 0 76 76 no real solution 8. 3 6u 5 3 3 6u 5 5 6u 5 5 6u 0 u 0 Check: 3 6u 5 3 3 6(0 ) 5 3 3 5 3 5 3 3 3 9. m 3m m 5 0 m 3m m 5 m 3m m 0m 5 3 3m m Check: m 3m m 5 0 () ) 3( () 5 0 9 5 0 3 3 0 0 0 0. k 9 k 3 k 9 3 k k 9 3 3k k 6 3k 36 (3k) 36 k 3 k Check: k 9 k 3 3 9 3 3 3 3 3 3 3 3 3 5 Chapter

. a a a a a a 0 (a ) 00 a 5 a Check: a a 5 6 5. 3 7 3 3 3 7 3 9 6 7 7 6 7 36( 7) 68 56 0 ( )( 6) 0 0 6 0 6 Check: 3 7 3 3() () 7 3 6 3 3 3 3 Check: 3 7 3 3(6) (6) 7 3 96 3 3 3 3 3. 3 7b 0 3 7b 8(7b ) 6 7b 8 7b 9 b 9 7 Check: 3 7b 0 3 7 9 7 0 3 8 0 0 0 0. 3t 0 Check: 3t 0 3t 3 6 3 0 3t 6 6 0 t 6 3 0 0 0 5. 7 9 9 9 96( ) 76 9 7 Check: 7 9 7 7 7 9 3 7 no real solution 6. 6 5 5 6 5 0 6 6 30 0 6 3 6 9 6 3 3 Check: 6 5 3 3 6 5 9 5 3 5 5 5 7. 3 0 3 0 3 0 0 ( 5)( ) 0 5 0 0 5 Check: 3 0 3(5) 0 5 5 6 5 5 3 Check: 3 0 3() 0 9 3 Solution: 8a. 3t t 6 3t 6 t 3t 36 t t 0 t 5t 50 0 (t 5)(t 0) t 5 0 t 0 0 t 5 t 0 Check: 3t t 6 3(5) 5 6 5 6 5 6 6 6 Check: 3t t 6 3(0) 0 6 6 0 6 0 6 6 0 8b. 5 Chapter 6

9. 7 5 Test 0: (0) 7 5 7 5 7 5 3 meaningless 6 Test : () 7 5 7 0 8 7 5 7 5 7 false Test 7: (7) 7 5 7 5 true Solution: 6 30. b 6 Test b 5: 5 6 b 36 6 b 3 meaningless b 0 Test b 0: 0 6 b 6 6 true Test b 33: 33 6 37 6 false Solution: b 3 3. a 5 Test a 0: 0 5 a 5 6 5 a meaningless a 5 0 Test a 6: 6 5 a 5 true Test a : 5 7 false Solution: 5 a 3. 5 6 Test 0: (0) 5 6 5 36 5 6 meaningless 0.5 Test 5: (5) 5 6 5 0 5 6 5 true.5 Test : () 5 6 39 6 false Solution:.5 0.5 33. 5y 9 Test y 0: 5(0) 9 5y 9 6 9 5y 5 meaningless y 5 Test y : 5(0) 9 5y 9 0 5y 9 true y.8 Test y 6: 5(6) 9 false Solution:.8 y 5 3. m 3m m 3m m m m 0 m 3m 0 3m m 3 Test m 3: 3 3(3) 5 meaningless Test m.6:.6 3(. 6) 0. 0.8 meaningless Test m.:. 3(. ) 0.8 0. false Test m 0: 0 3(0) true Solution: m 35. c 5 7 Test c 0: (0) 5 7 c 5 9 5 7 c 5 meaningless c 7 Test c 5: (5) 5 7 c 5 0 5 7 c 5 false c.5 Test c 8: (8) 5 7 5 7 true Solution: c 7 36a. t s g 3 (7.) g 3. g 36b. 3. g 9. g 9g. g.6 m/s 37. 5 3 3 5 3 ( ) 3 5 3 6 9 ( 5) 3 6 9 3 5 75 5 6 9 3 6 8 3 0 Use a graphing calculator to find the zero. [, 0] sc by [0, 0] sc about 7.88 38a. s 30fd s 30(0.6 )(5) s 50 s. mph 38b. s 30fd 35 30(0.6 )d 5 8d 68.06 d; about 68 ft 7 Chapter

38c. No; it is not a linear function. 39a. T g 39b. t g T 9.8 T 8 T.0 s T. s 39c. Let the new length of the pendulum. g g g g g g g g It must be multiplied by. 3 a a r r 0. T Tb b 5 687 67,0 0,000 3 r b 50, 65 7, 969 3.03 0 3 rb 3.9 50,65r b 3.3 0 9 r b 3.83 0 r,33,33.8; about,33,3 mi. 9 a b 9 a b 9 0, so a b 0 no real solution when a b 0. T t c t c p 08 t ( 00) t (00) 50 08 t 00 t 0 0 500 t 00t 0,000 00 5 t 6 t 83t 73,056 t 00t 50,000 3,056 3t 99.88 t; about 99.88 psi a 3. a a a a 3 3 a ; eclude: (6)(a ) a 3 (a 3 (6)(a ) ) 6(a ) a()(a ) 3(3) 6a a a 9 0 a a 3 0 (a 3)(a ) a 3 0 a 0 a 3 a 3 t 83t 73,056 t 00t 50,000. p q : 6, 3,, 3 6 6 0 r 6 6 5 6 0 5 6 0 ( 3)( ) 0 3 0 0 3 3,,, 5a. point discontinuity 5b. jump discontinuity 5c. infinite discontinuity 6a. p w v p 0 56 w 0 p 0 6b. - and y-aes 6c. It increases. 6d. It is halved. 7. 6 r 5 5 5 6 6 6 0 0 0 5 3 0 056 p w (0) ()() 6(5) (0) 0() (5) (3)()() 6() (3) 0() () 8 0 0w 0 8. (a b c) (6) a b c a 3b c 3 a 3b c 3 a 7b (a b c) (6) a b c a 8b c a 8b c b b a 7b a b c 6 a 7() 7 c 6 a 7 c (7,, ) 9. y 3.5 75. y 3.5(00) 75. y 0 students 50. 7y 3 0 y 7 3 7 perpendicular slope: 7 y 5 7 ( ) y 7 3 5. A r A r () 5 The correct choice is C. Chapter 8

-8 Modeling Real-World Data with Polynomial Functions Pages 6-6 Check for Understanding a. Sample answer: y b. Sample answer: c. Sample answer:. You need to recognize the general shape so that you can tell the graphing calculator which type of polynomial function to use as a model. 3. Sample answer: If companies use less packaging materials, consumers keep items longer, and old buildings are restored instead of demolished, the amount of waste will decrease more rapidly. If consumers buy more products, companies package items in larger containers, and many old buildings are destroyed, the amount of waste will increase instead of decrease.. quartic 5. Sample answer: f().98.95 3 5.9 0..89 6. Sample answer: f() 3.007 0.00 7.896 7a. Sample answer: f() 0.8 58.0 7b. Sample answer: 00 950 60 f() 0.8 58.0 0.8(60) 58.0 86.8% 7c. Sample answer: f() 0.8 58.0 89 0.8 58.0 6 950 6 0 y y. f().5 5 3. f() 8 3 9. Sample answer: f().03 5.6 3 6.08 0.3 0.9 5. Sample answer: f() 0.09 3.70.63 65. 6. Sample answer: f().05 0.09 3 6.69.03 697.7 7. Sample answer: f() 0.0 3 8.79 3.35 7.3 8a. Sample answer: f().99.7.76 8b. Sample answer: f() 0.96 3 0.56 0.36.05 8c. Sample answer: Cubic; the value of r for the cubic function is closer to. 9a. Sample answer: f() 0.6.73 9b. Sample answer: 00 900 0 f() 0.6.73 f(0) 0.6(0).73 f(0) 36.59 37 9c. Sample answer: 05 900 5 f() 0.6.73 f(5) 0.6(5).73 f(5) 38.8 38 0. Sample answer: 3 5 6 7 8 9 f() 3 6 3 3 9 09 37 a. Sample answer: f() 0.008 0.38 3 0.6 0.097 8.96 b. Sample answer: 99 99 f() 0.008 0.38 3 0.6 0.097 8.96 f() 0.008() 0.38() 3 0.6() 0.097() 8.96 f() 0.663 about %. A sith-degree polynomial; there are 5 changes in direction. 3a. Sample answer: f() 0.09 0.00 8.696 Pages 6-6 Eercises 8. cubic 9. quadratic 0. linear. quadratic 9 Chapter

3b. Sample answer: f() 0.09 0.00 8.696 00 0.09 0.00 8.696 0 0.09 0.00 5.30 7. r 0 3 3 0 0 0 0, 8a. Let number of weeks. P (0 0)(0.8 0.03) P 57.6. 0.3 [5, 5] sc by [50, 0] sc5 root: (.7, 0) 985 007 3c. Sample answer: 998 985 3 f() 0.09 0.00 8.696 f(3) 0.09(3) 0.00(3) 8.696 f(3) 67.0 No; according to the model, there should have been an attendance of only about 67 million. Since the actual attendance was much higher than the projected number, it is likely that the race to break the homerun record increased the attendance. a. Sample answer: f() 0.033 3.7.368 5.563 b. Sample answer: 996 990 6 f() 0.033 3.7.368 5.563 f(6) 0.033(6) 3.7(6).368(6) 5.563 f(6) 3.83 about 3.8 million c. Sample answer: f() 0.033 3.7.368 5.563 00 0.033 3.7.368 5.563 0 0.033 3.7.368 9.37 [5, 5] sc5 by [300, 50] sc50 root: (.8, 0) 990 00 about 00 5. 5 b 0 Check: 5 b 0 5 b 5 3 0 5 b 5 5 0 3 b 5 5 0 0 0 6 p 6. p 3 p 3 6 p p 3 p 3 (p 3)(p 3) (p 3)(p 3) 6(p 3) p(p 3) (p 3)(p 3) 6p 8 p 3p p 9 9p 9 p [0, 0] sc by [0, 60] sc5 maimum: (, 58.8) weeks 8b. $58.80 per tree 9. 0.0 0.90 0.90 0.0(0.90) 0.90 0.09 0.99 The correct choice is B. -8B Fitting a Polynomial Function to a Set of Points Page 66. y 7 3 7 5. y 7 3 7 5; yes 3. Sample answer: y 5 6 5 0 3 8. Infinitely many; suppose that you are given a set of n points in a coordinate plane, no two of which are on the same vertical line. You can pick an infinite number of other points with different - coordinates. You could find polynomial functions that went through the original n points and any number of the other points. 5. There is no problem with using L 0 with list L for the eample. However, if you are using a different list which happens to have 0 as one of its elements, using L 0 will result in an error message, since 0 0 is undefined. Chapter Study Guide and Assessment Page 67 Understanding and Using the Vocabulary. Quadratic Formula. Integral Root Theorem 3. zero. Factor Theorem 5. polynomial function 6. lower bound Chapter 30

7. Etraneous 8. comple roots 9. comple numbers 0. quadratic equation Pages 68-70 Skills and Concepts. no; f(a) a 3 3a 3a f(0) (0) 3 3(0) 3(0) f(0). yes; f(a) a 3 3a 3a f() () 3 3() 3() f() 0 3. no; f(a) a 3 3a 3a f() () 3 3() 3() f() 8. f(t) t t 3t f(3) (3) (3) 3(3) f(3) 73 no 5. 3; 3 3 0 ( 3) 0 ( 3)( ) 0 0 3 0 0 3 6. b ac (7) ()() 8; real 7 8 7 9 9 7 7 7. b ac (0) (3)(5) 0; real m m m f() f() 3 3 () 0 0 (3) 0 0 6 5 0 3 8. b ac () ()(6) 3; imaginary 3 () i3 9. b ac 3 ()(8) 73; real y y 3 73 () 3 73 9 0. b ac ()() 0; real a 0 () a a. b ac () (5)(0) 99; imaginary 99 r r (5) i99 0. f() 3 0 8 f() () 3 () 0() 8 8 0 8 or 0; yes 3. f() 3 5 7 f(5) (5) 3 5(5) 7(5) 50 5 35 or 6; no. f() 3 7 f 3 7 8 7 or ; no 5. f() 0 9 f(3) (3) 0(3) 9 8 90 9 or 0; yes 6. p q :, r 0 0 ( )( ) 0 0 0 rational roots:,, 7. p q : rational root: 8. p:,, q:, r 0 0 0 0 r 0 6 p q :,, ; 0 0 0 does not factor rational root: 3 Chapter

9. p:, 3 q:, p q :, 3,, 3 r 3 6 3 5 5 7 3 9 5 53 3 3 3 0 57 3 7 5 3 3 6 3 3 0 5 3 0 6 0 rational root: 3 30. p q :,, r 0 7 6 5 7 3 3 5 0 3 3 5 0 r 3 5 0 3 0 3 3 0 r 0 3 3 3 53 3 5 rational roots:, 3. p:,,, 8 q:, 3 p q :,,, 8, 3, 3, 3, 8 3 r 3 7 8 3 0 8 0 3 0 8 0 (3 )( ) 0 3 0 0 3 rational roots:, 3, 3. p:, q:,, p q :,,, r 8 0 8 0 9 33. p q :, 5 r 0 0 5 5 5 0 3 5 5 0 r 5 5 0 5 0 5 0 does not factor rational roots:, 3. positive f() 3 3 56 or 0 negative r 3 56 7 6 8 0 6 8 0 ( )( ) 0 0 0 rational zeros:,, 7 35. or 0 positive f() 3 9 negative r 9 8 0 8 0 6 9 0 ( 3)( 3) 0 3 0 3 0 3 3 rational zeros:, 3 36. or 0 positive f() 3 36 or 0 negative r 0 3 0 36 9 8 0 3 9 8 0 r 9 8 3 5 6 0 5 6 0 ( 3)( ) 0 3 0 0 3 rational zeros: 3,,, 3 8 0 0 does not factor rational root: Chapter 3

37. r 3 0 0 3 6 3 3 3 3 0 3 0 0 and 0 38. r 0 3 3 0 0 and, 3 and 39. r 3 3 0 3 3 5 5 3 0 3 and 0, 3 and 0. r 0 3 6 0 0 0 0 and 0. r 3 7 3 3 3 8 0 3 5 6 3 9 7 7 and, 0 and, and. r 9 5 6 9 3 58 6 0 9 5 6 9 6 8 9 7 0 0 and 3. Use the TABLE feature of a graphing calculator..9,.8,. 6. n n 5 0 6 n n 5(n) 0(n) n 5n 6 0 (n 6)(n ) 0 n 6 0 n 0 n 6 n 5. ( ) 3 3 ( ) 3 3 6. 5 6 m m 5 6 6(m )(m ) (m m 3m 3 m ) 3(m ) 6(m )(m ) 5(m )(m ) (m)(3)(m ) (m ) 5m 5 6m 6m m 0 m 8m 3 8 (8) )(3) ( () 8 5 m m 3 7. 3 y 5 y ; eclude: 0 3 y y 5 y y 3 y 5 y 3 5 Test y : 7 5 true 3 5 Test y 0.5: 0.5 0.5 8 0 false Test y : 3 5 5 true Solution: y, y 0 ; eclude, ( )( ) ( )( ) 8. ( ) ( )( ) ( ) 0 3 0 ( 3) 0 3 0 3 Test : 3 true Test 0.5: 0.5 0.5 false Test 0.5: 0.5 0.5 3 3 true Test : 3 0 false Test : 5 3 true Solution:, 0, 3 9. 5 0 Check: 5 0 5 5 3 0 5 5 5 0 3 5 5 0 0 0 33 Chapter

50. 3 a 8 5 Check: 3 a 8 5 3 a 3 3 (6. 5) 8 5 a 7 3 7 8 5 a 6 3 8 5 a 6.5 5 5 5. 3 8 35 9 6 8 8 35 6 8 8 8 3 8 9 Check: 3 8 35 3 8 5 3 3 9 36 3 3 6 6 6 5. 5 7 5 0 5 9 5 5 Test 0: 0 5 7 5 7 meaningless Test 0: 0 5 7 5 7 true Test 60: 60 5 7 55 7 false Solution: 5 5 53. a 7 6 a 7 0 a 7 a 7 a 7 a 3.5 a 3 a.5 Test a 5: (5) 7 6 3 6 meaningless Test a : () 7 6 3 6 false Test a 0: (0) 7 6 7 6 true Solution: a.5 5. cubic 55. f() 3 Page 7 Applications and Problem Solving 56. Let width of window. Let 6 height of window. A w 35 ( 6) 35 6 0 6 35 0 ( )( 5) 0 5 0 5 Since distance cannot be negative, 5 and 6. the window should be 5 in. by in. 57. Let width. Let 6 length. ( )( 6) ( 6)() 88 8 7 6 88 7 88 8 6 8 ft by ft 58a. 58b. g() 0.006 0.0 3 0.053.79 (0.006 3 0.0 0.053.79) ( 3 3.3 8.83 98.3) r 3.333 8.833 98.333.333 3.503 3.836 5 8.333 8.835 7.508 3.5 0.67.758 0 rational zeros: 0, about 3.5 59. T g.6 9 0.5 9.8.8 0.06 9.8 0.6 ; about 0.6 m Page 7 pen-ended Assessment. Sample answer: 3 00 00 g() g() 0.006 0.0 3 0.053.79 0 0 3 ( 3)( ) ( 3)( ) ( ) ( 3) 6 6 0 ( 3)( ) 0 3 0 0 3 a. Sample answer: b. Sample answer: ( ) 8 6 9 8 0 ( 6)( 3) 0 6 0 3 0 6 3 Check: 6 6 3 3 The solution is 6. Since, 3 is an etraneous root. 3a. Sample answer: 3 0.5 f() 0.5 0 0.5 f() 0 0.65 0 8 3b. Sample answer: f() 3 3c. Sample answer:, 0, Chapter 3

Chapter SAT & ACT Preparation Page 73 SAT and ACT Practice. There are two ways to solve this problem. You can use the distance formula or you can sketch a graph. d ( ) (y y ) ( ) ( ()) (3 3 9 6 5 or 5 y (, ) 3 (, 3) When you sketch the points and draw a right triangle as shown above, you can see that this is a 3--5 right triangle. Using the Pythagorean Theorem, you can calculate that the length of the hypotenuse is 5. 5 3 The correct choice is C.. Points on the graph of f() are of the form (, f()). To move the entire graph of f() up units, must be added to each of the second coordinates. Points on the translated graph are of the form (, f() ). The function which represents the translation of the graph up units is f(). The correct choice is E. 3. You need to find both the - and y-coordinates of point C. Use the properties of a parallelogram. First find the y-coordinate. Since opposite sides of a parallelogram are parallel and side AD is on the -ais, point C must have the same y-coordinate as point B. So the y-coordinate is b. This means you can eliminate answer choices A and B Now find the -coordinate. Since opposite sides of a parallelogram have equal length and side AD has length d, side BC must also have length d. Point B is a units from the y-ais, so point C must be a d units from the y-ais. The -coordinate of point C is a d. So point C has coordinates (a d, b). The correct choice is E..You may want to draw a diagram. s s 6 Use the formula for the perimeter of a rectangle, where represents the length and w represents the width. w P Replace w with s. Replace with s 6. (s 6) s 60 The correct choice is E. 5. First find the slope of the given line. Write the equation in the form y m b. 3 6y 6y 3 y The slope is. So the slope of the line perpendicular to this line is the negative reciprocal of this slope. The slope of the perpendicular line is. The correct choice is A. 6. Be sure to notice the small piece of given information: is an integer. You need to find the number, written in scientific notation, that could be 3. This means that the cube root of the number is an integer. Take the cube root of each of the answer choices and see which one is an integer. You can use your calculator or do the calculations by hand. Notice that.7 is one-tenth of 7, which is 3 3..7 0 3 7 0 3 7 0 3 0 or 30,000. 30,000 is an integer. When you try the same calculation with each of the other answer choices, the resulting power of 0 has a fractional eponent. So the number cannot be an integer. The correct choice is C. 7. This is a system of equations, but you do not need to solve for or y. You need to find the value of 6 6y. Notice that the first equation contains 5y and the second contains y. If you subtract the second from the first, you have 6y. Similarly, subtraction of the values gives a result of 6. Use the same strategy that you would for solving a system. Subtract the second equation from the first. 0 5y 5y 6 6y The correct choice is C. 35 Chapter

8. You can solve this problem using the midpoint formula or by sketching a graph. The midpoint formula: y () 3, 8,, y 3 y ( ) (, 3), (3, 5), 5 The correct choice is B. 9. The epression a a is a perfect square trinomial and can be factored as ( a). The square of a real quantity is never negative. The correct choice is A. ( a a a ( a ) a So the quantity in Column A equals the quantity in Column B plus the sum of the squares of and a. Since neither nor a equal 0, their squares must be greater than 0. So the quantity in Column A is always greater than the quantity in Column B. The correct choice is A. 0. Since the problem does not include a figure, draw one. Label the four points. 3 8 E F G H ne method of solving this problem is to plug-in numbers for the segment lengths. Since EG 5 3 EF, let EF 3. Then EG 5. This means that FG must equal, since EF FG EG. HF 5FG 5() 0 HG HF FG 0 8 EF H G 3 8 The answer is.375 or 3/8. Chapter 36