Name: Class: Date: ID: A Alg 2 Mid Term Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Solve 4x 2 5x 2 0. A x 5 8 7 8 C x 5 8 7 8 B x 5 8 7 8 i D x 5 8 7 8 i 2. Solve x 2 x 1. A x 1 2 3 2 i C x 1 2 3 2 i B x 1 2 5 2 i D x 1 2 5 2 i 3. Solve 8x 2 135. A x 3 30 4 B x 3 2 15 2 C x 8.4275 D x 16.875 4. Find the number and type of solutions for x 2 5x 10. A The equation has two real solutions. B Cannot determine without graphing. C The equation has two nonreal complex solutions. D The equation has one real solution. 5. During the eruption of Mount St. Helens in 1980, debris was ejected at a speed of over 440 feet per second (300 miles per hour). The elevation in feet above sea level of a rock ejected at an angle of 75 above horizontal is given by the function yt 16t 2 425t 8200 where t is the time in seconds after the eruption. The rock s horizontal distance in feet from the point of ejection is given by xt 113t. If the rock landed at an elevation of 6000 feet, what is the horizontal distance from the point of ejection to where it landed rounded to the nearest hundred feet? A 7000 ft C 3500 ft B 5000 ft D 3100 ft 6. Solve 9x 2 + 49 = 0. A No real solution C ± 7 3 B ± 3 7 D ± 9 49 1
Name: ID: A 7. Which equation shows x 2 8x 9 0 after the method of completing the square has been applied? A x 2 2 5 B x 4 2 7 C ( x 8) 2 55 D x 2 8x 9 8. When the quadratic formula is applied to 2x 2 3x 4 0, what is the numerator of the simplified answer? A 3 41 B 3 41 C 3 38 D 3 23 9. Which of the following quadratic equations has no real solutions? A x 2 3x 5 0 B 4x 2 4x 1 0 C 2x 2 4x 3 0 D x 2 6x 2 0 10. What is the simplified form of i 8 9? A 3 C 3 B 3i D 3i 11. What is the simplified form of i 32 169? A 13i C 13 B 13i D 13 12. Simplify i 45. A 3 5 B 3i 5 13. Simplify i 19 396. A 19.9 C 2i 27 B 6 11 D 6 11 14. What is the simplified form of i 20 196? A 14i C 14 B 14i D 14 15. What is the simplified form of i 17 400? A 20i C 20 B 20i D 20 2
Name: ID: A 16. What is the simplified form of i 29 64? A 8 C 8 B 8i D 8i 17. What does the imaginary number i represent? A 1 B 1 C 1 D 1 18. Simplify i 5 9 i 5 9. A 5i 2 81 C 86 B i 5 81 D 76 19. Simplify 3 2i 2 3i. A 5i C 7 B 12 5i D 6 9i 20. Add. Write the result in the form a bi. (7 9i) ( 6 + 5i) A 12 15i C 2 i B 1 4i D 13 14i 21. Simplify (3 2i) 2. A 13 12i C 5 12i B 13 D 9 4i 22. Find the product (x 2i) 2. A x 2 4ix 4 B x 2 4ix 4 C x 2 4 D x 2 4 23. Which of these expressions is equal to 4 7i? A B C D 6 i 2 8i 6 i 2 8i 6 i 2 8i 6 i 2 8i 3
Name: ID: A 24. What value should be added in the blanks to complete the square? x 2 8x y 2 10y 25 15 A 4 B 8 C 16 D 64 25. Which is a graph of an even function with a positive leading coefficient? A C B D 4
Name: ID: A 26. Graph f(x) x 3 6x 2 9x. Identify the intercepts and give the domain and range. A C B The x-intercepts are 3 and 0. The y-intercept is 0. The domain and range are all real numbers. D The x-intercepts are 1.5 and 0. The y-intercept is 0. The domain is all real numbers. The range is approximately y 3.25. The x-intercepts are 3 and 0. The y-intercept is 0. The domain and range are all real numbers. The x- and y-intercepts are both zero. The domain and range are all real numbers. 5
Name: ID: A 27. If f(x) is an odd function with a negative leading coefficient, g(x) is an even function with a negative leading coefficient, and h(x) is the product of f(x) and g(x), which of the following could be the graph of h(x)? A C B D A C B D 28. Which is the graph of the polynomial function p x x 1 x 1 x 4? 6
Name: ID: A 29. The graph of the polynomial function px is shown. What are the zeros of px? (Assume that the zeros of px are integers and that the graph of px does not cross the x-axis at places other than those shown.) A x 3 and x 2 B x 3, x 0, and x 2 C x 0 D x 2, x 0, and x 3 30. Which of the following is a true statement about the graph of p(x) x 4 x 2 3x 2 6x? A B C D The graph crosses the x-axis four times and is never tangent to the x-axis. The graph crosses the x-axis three times and is never tangent to the x-axis. The graph crosses the x-axis two times and is tangent to the x-axis once. The graph crosses the x-axis three times and is tangent to the x-axis once. 31. Multiply (2 x)(2 x). A 4 x 2 C 4 4x x 2 B 4 2x D 4 4x x 2 32. Subtract. 5 4x 3 x 3 2x A x 3 x 2 6x 8 C x 3 x 2 6x 8 B x 3 3x 2 6x 8 D x 3 x 2 6x 8 33. Multiply. 5x 3(x 3 5x 2) A 5x 4 3x 3 25x 2 5x 6 C 5x 3 22x 2 5x 6 B 5x 3 28x 2 25x 6 D 5x 4 3x 3 25x 2 25x 6 7
Name: ID: A 34. Multiply. (6r 4s ) 2 A 36r 2 16s 2 C 36r 2 24r s 16s 2 B 12r 2 8s 2 D 36r 2 48r s 16s 2 35. Which polynomial represents the measure in degrees of angle PMQ? A a 2 5a 34 C 4a 2 2a 34 B 3a 4 2a 34 D 6a 5 34 36. Which expression represents the perimeter of the triangle below? A 3 4m C 5 4m B 3 6m D 5 6m 37. Find the product 5x 3(x 3 5x 2). A 5x 4 3x 3 25x 2 5x 6 C 5x 4 3x 3 25x 2 25x 6 B 5x 3 28x 2 25x 6 D 5x 3 22x 2 5x 6 38. Find the product of 3x 2 x 1 and 4x 5. A 3x 2 5x 4 C 12x 3 19x 2 x 5 B 12x 3 4x 2 4x D 12x 3 19x 2 9x 5 39. Which is NOT a factor of x 4 2x 3 7x 2 8x 12? A x 1 C x 2 B x 1 D x 2 40. Write an equivalent expression for a 2 2ab b 2. A (a b)(a b) C (a b) 2 B a 2 b 2 D a 2 b 2 8
Name: ID: A 41. Write an equivalent expression for x 2 2xy y 2. A (x y) 2 C (x y)(x y) B (x y) 2 D x 2 y 2 42. Write an equivalent expression for (x y)(x y). A x 2 2xy y 2 C (x y) 2 B x 2 2xy y 2 D x 2 y 2 43. Write an equivalent expression for (a b) a 2 ab b 2. A (a b)(a b) 2 C (a b) 3 B a 3 b 3 D a 3 b 3 44. Suppose ab 4 and (a b) 2 18. What is a 2 b 2? A 18 C 10 B 14 D 4 45. Suppose pq 3 and (p q) 2 20. What is p 2 q 2? A 20 C 14 B 26 D 3 46. Expand (3p 2q) 4. A 81p 4 216p 3 q 216p 2 q 2 96pq 3 16q 4 B 81p 4 16q 4 C 3p 4 2q 4 D 81p 4 54p 3 q 36p 2 q 2 24pq 3 16q 4 47. What is the coefficient of the x 4 -term in the expanded form of 2x 7 6? A 11,760 B 784 C 784 D 11,760 48. Divide. x 3 x 6 x 2 A x 2 3 C x 2 x 3 B x 2 2x 3 D x 1 x 2 49. Divide: x 2 8x 5 x A x 2 8 5 x C x 5 B x 8 5 x D x 8 9
Name: ID: A 50. Divide: (5x 6x 3 8) (x 2). A 6x 2 12x 29 64 (x 2) C 6x 2 12x 29 50 (x 2) B 6x 2 12x 29 D 6x 2 5 8 (x 2) 51. Divide: (3x 2 11x 9) x 2 A 3x 10 4 x 2 C 6x 22 9 x 2 B 3x 5 1 x 2 D 3x 17 43 x 2 52. Given fx 3x 2 16x 12 and gx 3x 2, find f(x) g(x). A x 6 C 3x 2 13x 10 B x 6 D 3x 2 16 3 x 6 53. When you divide to simplify the expression 6x 3 5x 2 2x 7, what is the fractional part of the quotient? 2x 3 A 5 B 5 2x 3 7 C 2x 3 D 3x 2 2x 4 54. Completely factor 3x 4 15x 3 18x 2. A x 2 3x 2 2 1x 9 C 3x2 x 1 x 6 B 3 x 2 1 x2 6 D cannot be factored 55. What is the complete factorization of 10x 3 35x 2 20x? A 2x 1 x 4 C 5x2x 1 x 4 B 5x 2x 2 7x 4 D x2x 1 5x 20 56. Factor 27x 2 z 36xz 12z completely. A z3x 12 C 3z3x 2 B 12z(2x 2 3x 1) D 3z3x 2 3x 2 57. Factor 8x 3 125. A 2x 5 4x 2 10x 25 C 2x 5 4x 2 10x 25 B 2x 5 4x 2 10x 25 D 2x 5 4x 2 10x 25 10
Name: ID: A 58. When x 3 125 is written as a product of a binomial and a trinomial, what is the trinomial factor? A x 2 5x 25 B x 2 5x 25 C x 2 10x 25 D x 2 10x 25 59. Which of the following is equal to x 6 64? A 64x 6 B (x 3 8)(x 3 8) C (x 3 8) 2 D (x 3 8) 2 60. Write the simplest polynomial function with the zeros 2 i, 5, and 2. A Px x 6 4x 5 4x 4 36x 3 25x 2 80x 100 0 B Px x 5 2x 4 8x 3 20x 2 65x 50 0 C Px x 5 2x 4 8x 3 20x 2 15x 50 0 D Px x 5 2x 4 10x 3 16x 2 25x 30 0 61. Which polynomial function has zeros 1, 1 i, and 1 i? A P(x) x 3 x 2 2x 1 C P(x) x 3 3x 2 4x 2 B P(x) x 3 4x 2 3x 1 D P(x) x 3 2x 2 3x 3 62. What polynomial function has zeros 1, 1 i, and 1 i? A P(x) x 3 x 2 2x 1 C P(x) x 3 3x 2 4x 2 B P(x) x 3 4x 2 3x 1 D P(x) x 3 2x 2 3x 3 63. Solve the polynomial equation 2x 5 14x 4 12x 3 0 by factoring. A The roots are 0, 1, and 6. C The roots are 2 and 12. B The roots are 0, 1, and 6. D The roots are 1 and 6. 64. Which of the following lists all the roots of x 4 4? A 2 C 2, i 2 B 2i D 2 2, 2 2 i 65. If 3 and 3 3 are two of the roots of a third degree polynomial with integer coefficients, which of the following is the other root? A 3 C 3 3 B 3 3 D There is no other real root. 11
Name: ID: A 66. Which is a list of all the roots of x 5 7x 4 18x 3? A 0, 2 C 9, 0, 2 B 2, 0, 9 D 9, 2, 0, 2, 9 67. Which is a third degree polynomial with 3 and 2 as its only zeros? A f(x) x 2 x 6 C f(x) x 3 4x 2 3x 18 B f(x) x 3 2x 2 5x 6 D f(x) x 3 x 2 8x 12 68. Which is a list of all the roots of x 3 x 2 4 4x? A 1, 2, 2 C 1, 2i, 2i B 1, 2, 2 D 1, 2i, 2i 69. Which lists all the roots of x 4 x 2 6? A 2, 3 C 3, 2i B 2, 3i D 2i, 3i 70. What is the solution set of x 5 7x 4 18x 3? A {0, 2} C {9, 0, 2} B {2, 0, 9} D {9, 2, 0, 2, 9} 71. What is the degree of the simplest polynomial with integer coefficients that has A 3 C 5 B 4 D 6 5 and 5i as zeros? 72. Which is a list of all the roots of x 3 x 2 4 4x? A 1, 2, 2 C 1, 2i, 2i B 1, 2, 2 D 1, 2i, 2i 73. What is the degree of the simplest polynomial with integer coefficients that has A 3 C 5 B 4 D 6 74. What is the degree of the simplest polynomial with integer coefficients that has A 3 C 5 B 4 D 6 2, 2, and 2 2i as zeros? 5 and 5i as zeros? 75. What polynomial function has zeros 1, 1 i, and 1 i? A P(x) x 3 x 2 2x 1 C P(x) x 3 3x 2 4x 2 B P(x) x 3 4x 2 3x 1 D P(x) x 3 2x 2 3x 3 12
Name: ID: A Multiple Response Identify one or more choices that best complete the statement or answer the question. 1. Which of the following equations, when rewritten in the form x p 2 q, have a value of q that is a perfect square? A x 2 2x 5 13 B x 2 8x 9 5 C 2x 2 12x 29 81 D 5x 2 20x 14 6 E 3x 2 36x 88 4 2. Identify the quadratic equations below that have non-real solutions. A x 2 3x 25 7 B x 2 7x 1 13 C x 2 2x 5 D 2x 2 x 13 0 E 2x 2 4x 9 11 3. Which of the following sums, differences, and products can be simplified to 6 3i? A B C D E F 9 5i 3 2i 4 2i 2 5i 9 5i 3 2i 4 2i 2 5i 3i1 2i 3i1 2i 13
Name: ID: A 4. Choose all the statements that are true about the graph. A The x-intercept is 9, B The y-intercept is 2. C fx is increasing when x 1. D fx is decreasing when x 1. E fx has a local maximum at 1, 2. F fx has a local minimum at 1, 2. G fx is negative when x 9. H fx is positive when x 2. 14
Name: ID: A 5. Which of the following statements are true about the polynomial function px? (The zeros of px are integers, and the graph of px does not cross the x-axis at places other than those shown.) A B C D E F The degree of px is even. The degree of px is 4. The leading coefficient of px is negative. The degree of px is at least 6. The graph of px has a y-intercept of 150. px has four distinct zeros. 6. Which of the following statements present(s) valid reasoning? A x 6 81 can be rewritten as x 2 3 3 3 and factored as a sum of two cubes. B 49c 2 154c 121 can be rewritten as 7c 2 27c 11 11 2 and factored as a perfect square trinomial. C 36p 4 96p 64 can be rewritten as 6p 2 2 2 6p 2 8 8 2 and factored as a perfect square trinomial. D x 4 16 can be rewritten as x 2 2 4 2 and factored as a difference of squares. E x 18 8 can be rewritten as x 6 3 2 3 and factored as a difference of cubes. F x 9 64 cannot be factored as the sum of two cubes because x 9 is a perfect cube and 64 is a perfect square. 15
Name: ID: A 7. Use the graphs of fx and gx to determine all of the solutions of the equation fx gx. Approximate to the nearest integer. A x 3 B x 2 C x 0 D x 3 E x 5 F x 10 G x 12 H x 25 Short Answer 1. Find the value of c that makes the expression a perfect square trinomial. x 2 15x c 2. Rewrite 4x 2 16x 21 12 in the form x p 2 q by completing the square. Show your work. 3. How many real solutions does 3x 2 18x 77 2 have? Justify your answer by rewriting the equation in the form x p 2 q. 4. Why is it imprecise to say that the equation 25x 2 27 0 has no solution? Find all solutions of the equation. 1 5. Simplify 5 4i 10 i. 16
Name: ID: A 6. a. Find the product 3 4i 3 4i. b. Find the product 3 4i 3 4i. c. Using your results from parts a and b as a guide, write a general form for the products a bi a bi and a bi a bi, where a and b are real numbers. 7. Draw a graph of an odd function with exactly two real zeros and a positive leading coefficient. 8. Px as x and Px as x. Qx as x and Qx as x. If Rx Px Qx, describe the end behavior of R(x). 9. Identify the zeros of p x x 3 x 2 16x 16, and describe the function s end behavior. Then graph the function using the zeros and the end behavior and plotting any additional points as needed. 17
Name: ID: A 10. The function pt 1 63 t 2 t 1 t 7 t 9 models the annual profit, in tens of thousands of dollars, for a small company from 2000 to 2010, where t is the number of years since 2000. a. State the function s domain, and then graph the function. b. Identify and interpret the zeros of pt. 11. Multiply (b 4)(b 2 3b 2). 12. What are all the values of a that make x a a factor of x 3 5x 2 2x 24? 13. What are all the values of a that make x a a factor of x 4 5x 3 21x 2 23x 8? 14. What are all the real values of a that make x a a factor of x 4 3x 3 6x 2 28x 24? 15. Use synthetic division to factor fx 1 4 x 4 9 4 x 2 x 3. 16. Give three examples of sets of values for a and b so that ax 2 bx 4 is a perfect square trinomial. What is the general relationship between a and b? Explain. 17. Expand p 3r 5. 18. Expand 2x 2 3 6. 19. Use the binomial theorem to write the binomial expansion of x 2 y 3. 18
Name: ID: A 20. Without expanding the power, what is the x 5 y 2 -term in the expanded form of x 2y 7? Show your work. 21. Divide. (15x 2 10x 5) 5x 22. Write 2x 2 5x 1 as the sum of a quotient and a remainder where the degree of the remainder s numerator x 2 is less than that of its denominator. 23. Factor 16x 3 54 completely. 24. What are all the roots of 2x 3 6x 2 36x? 25. What is the degree of the simplest polynomial with integer coefficients that has 3, 3, 3 3, and 3 3i as some of it zeros? 26. Let p(x) x 3 2x 2 4x 8. a. Identify the zeros of the function. List all zeros as many times as they occur. b. Sketch a graph of the function. 27. x 3 x 2 x 1 0 is a polynomial equation. Part A: Explain how you know, without factoring, the number of roots and the minimum number of real roots. Part B: Factor the polynomial to support your answer to Part A. Explain which factor(s), if any, indicate(s) that there are complex roots. 19
Name: ID: A 28. Sheila says that x 2 18x 81 has one real root, 9. She uses the corollary of the fundamental theorem of algebra to conclude that the polynomial must have one non-real root. Is Sheila correct? Explain. Then state all complex roots of the polynomial. 29. Factor x 4 9 into four linear factors. Problem 1. The graph shows a function that models the value V, in millions of dollars, of a stock portfolio as a function of time t, in months, over an 18-month period. a. For what values of t is the function increasing? For what values of t is the function decreasing? Approximate the endpoints of the intervals to the nearest 0.5 month. b. Interpret the intervals found in part a in terms of the situation. c. Identify the coordinates of any local maximums and local minimums. Approximate the t-values to the nearest 0.5 month and the V-values to the nearest 0.25 million dollars. d. Explain the significance of any local maximums and minimums found in part c. e. What does the fact that the function is always positive indicate about the appropriateness of this model? 20
Name: ID: A 2. a. Find the zeros of px xx 5x 3x 2 2. b. Describe the end behavior of px. Justify your answer algebraically. c. Could the following graph represent px? Explain. (The zeros of the graphed function are integers.) 3. The quartic function Tt 0.04t 4 1.16t 3 9.93t 2 22.19t 50.32 models the average monthly high temperature T, in degrees Fahrenheit, of a city t months after January 1. a. Use a graphing utility to graph the function and then sketch the graph on the coordinate plane below. Describe the function s end behavior based on the graph. b. Does the end behavior make sense in this context? Explain. c. What is a reasonable domain for this function? Explain. Essay 1. Is x 3 a factor of 2x 3 4x 2 x 4? How do you know? 21
Name: ID: A 2. Suppose you know that 3 is a zero of the function gx 4x 3 x 2 27x 18. Part A: What must be a factor of the polynomial in gx? Part B: Divide the polynomial by this factor. What type of polynomial results? Part C: List the possible rational zeros of the resulting polynomial function from Part B. Are these also possible zeros of gx? Explain. Part D: Find all real zeros of gx. 3. Explain how to use patterns to write a b 6 in expanded form. Other 1. Determine whether each given equation has equal values of p and q when written in the form x p 2 q. a. x 2 6x 4 0 Yes No b. x 2 8x 1 11 Yes No c. 2x 2 12x 37 13 Yes No d. 3x 2 42x 127 1 Yes No e. 2x 2 8x 5 9 Yes No 22
Alg 2 Mid Term Review Answer Section MULTIPLE CHOICE 1. ANS: B PTS: 1 DIF: DOK 2 NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b 2. ANS: A PTS: 1 DIF: DOK 2 NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b 3. ANS: B PTS: 1 DIF: DOK 1 NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b 4. ANS: C PTS: 1 DIF: DOK 1 OBJ: Analyzing Quadratic Equations by Using the Discriminant NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b LOC: MTH.C.10.06.04.009 MTH.C.10.06.04.010 TOP: The Quadratic Formula KEY: quadratic formula 5. ANS: C PTS: 1 DIF: DOK 2 OBJ: Application NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b LOC: MTH.C.10.06.04.01.005 TOP: The Quadratic Formula KEY: quadratic formula 6. ANS: A PTS: 1 DIF: DOK 1 OBJ: Using Square Roots to Solve Quadratic Equations NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b LOC: MTH.C.10.06.04.01.002 TOP: Solving Quadratic Equations by Using Square Roots KEY: quadratic 1
7. ANS: B Complete the square. x 2 8x 9 0 x 2 8x 9 x 2 8x 16 9 16 x 4 2 7 A B C D Feedback You added b 2 2 to both sides of the equation. When completing the square, add b to 2 both sides of the equation. That s correct! 2 b You added b 2 to both sides of the equation. When completing the square, add to 2 both sides of the equation. To complete the square of a trinomial ax 2 bx c 0, where a 1, write the trinomial 2 2 in the form x 2 bx b 2 c b 2. PTS: 1 DIF: DOK 1 NAT: A-REI.B.4a MP.7 STA: MCC9-12.A.REI.4a KEY: completing the square 8. ANS: B 2x 2 3x 4 0 x 3 32 42 4 22 3 3 41 4 9 32 4 A B C D Feedback Do not forget that the b-term in the numerator of the quadratic formula is negative. That s correct! It seems you multiplied b by 2 instead of squaring it. It seems you forgot the negative sign in front of the c-value, 4, when multiplying underneath the square root. PTS: 1 DIF: DOK 1 NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b KEY: quadratic equations quadratic formula 2
9. ANS: C If the discriminant of the quadratic formula is negative, the solutions of the quadratic equation are non-real numbers. For the equation 2x 2 4x 3 0, the discriminant is as follows. b 2 4ac 4 2 4 2 3 16 24 8 So, the solutions of the equation are non-real numbers. A B C D Feedback The equation x 2 3x 5 0 has real solutions because the discriminant is positive. The equation 4x 2 4x 1 0 has real solutions because the discriminant is zero. That s correct! The equation x 2 6x 2 0 has real solutions because the discriminant is positive. PTS: 1 DIF: DOK 1 NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b KEY: solving quadratic equations using the discriminant of the quadratic formula 10. ANS: C PTS: 1 DIF: DOK 1 NAT: N-CN.A.1 STA: MCC9-12.N.CN.1 11. ANS: A PTS: 1 DIF: DOK 1 NAT: N-CN.A.1 STA: MCC9-12.N.CN.1 12. ANS: A PTS: 1 DIF: DOK 1 NAT: N-CN.A.1 STA: MCC9-12.N.CN.1 13. ANS: D PTS: 1 DIF: DOK 1 NAT: N-CN.A.1 STA: MCC9-12.N.CN.1 14. ANS: A PTS: 1 DIF: DOK 1 NAT: N-CN.A.1 STA: MCC9-12.N.CN.1 15. ANS: D PTS: 1 DIF: DOK 1 NAT: N-CN.A.1 STA: MCC9-12.N.CN.1 16. ANS: C PTS: 1 DIF: DOK 1 NAT: N-CN.A.1 STA: MCC9-12.N.CN.1 17. ANS: C The imaginary number i is defined to be 1. Feedback A Recall that i 2 1. When you square 1, do you get 1? B Recall that i 2 1. When you square 1, do you get 1? C That s correct! D 1 represents i, not i. PTS: 1 DIF: DOK 1 NAT: N-CN.A.1 STA: MCC9-12.N.CN.1 KEY: the imaginary number i 18. ANS: C PTS: 1 DIF: DOK 1 NAT: N-CN.A.2 STA: MCC9-12.N.CN.2 3
19. ANS: B PTS: 1 DIF: DOK 1 NAT: N-CN.A.2 STA: MCC9-12.N.CN.2 20. ANS: B PTS: 1 DIF: DOK 1 OBJ: Adding and Subtracting Complex Numbers NAT: N-CN.A.2 STA: MCC9-12.N.CN.2 LOC: MTH.C.10.03.01.002 TOP: Operations with Complex Numbers 21. ANS: C PTS: 1 DIF: DOK 1 NAT: N-CN.A.2 STA: MCC9-12.N.CN.2 LOC: MTH.C.10.03.01.003 TOP: Operations with Complex Numbers 22. ANS: B PTS: 1 DIF: DOK 2 NAT: N-CN.C.8 STA: MCC9-12.N.CN.8 23. ANS: B 6 i 2 8i 6 2 i 8i 4 7i 4 7i The other expressions produce different results: (6 i) (2 8i) 4 9i (6 i) (2 8i) 8 9i (6 i) (2 8i) 8 9i A B C D Feedback To subtract two complex numbers, subtract the real parts and subtract the imaginary parts. That s correct! To add two complex numbers, add the real parts and add the imaginary parts. To add two complex numbers, add the real parts and add the imaginary parts. PTS: 1 DIF: DOK 1 NAT: N-CN.A.2 STA: MCC9-12.N.CN.2 KEY: adding complex numbers subtracting complex numbers 24. ANS: C 8 The value to be added is 2 Feedback 2 16. Then x 2 8x 16 is a perfect square because it equals x 4 2. A Find the value that makes x 2 8x a perfect square trinomial. B Find the value that makes x 2 8x a perfect square trinomial. C That s correct! D Find the value that makes x 2 8x a perfect square trinomial. PTS: 1 DIF: DOK 1 NAT: G-GPE.A.1 STA: MCC9-12.G.GPE.1 KEY: equation of a circle completing the square 25. ANS: B PTS: 1 DIF: DOK 1 NAT: F-IF.B.4 STA: MCC9-12.F.IF.4 4
26. ANS: A PTS: 1 DIF: DOK 2 OBJ: Graphing Cubic Functions NAT: F-IF.B.4 F-IF.C.7c STA: MCC9-12.F.IF.4 LOC: MTH.C.10.07.07.008 MTH.C.10.07.07.009 MTH.C.10.07.07.03.002 MTH.C.10.07.07.03.011 TOP: Cubic Functions and Equations KEY: cubic domain and range intercepts 27. ANS: A PTS: 1 DIF: DOK 3 NAT: F-BF.B.3 F-IF.B.4 STA: MCC9-12.F.BF.3 28. ANS: B The zeros of the function are x 1, x 1, and x 4, and these are the x-intercepts of the function s graph. The result of expanding x 1x 1x 4 is x 3 4x 2 x 4. The leading term, x 3, has a positive coefficient and an odd exponent, so px approaches as x approaches, and px approaches as x approaches. Of the four given graphs, only the one shown below has these characteristics. A B C D Feedback Check the zeros of px. That s correct! Check the end behavior of px. Check the zeros and end behavior of px. PTS: 1 DIF: DOK 1 NAT: F-IF.C.7c* STA: MCC9-12.F.IF.7c KEY: graphs of polynomial functions zeros end behavior 29. ANS: B The zeros of px are the x-intercepts of the function s graph. The x-intercepts are x 3, x 0, and x 2. A B C D Feedback Identify the graph s x-intercepts. That s correct! Identify the graph s x-intercepts. Identify the graph s x-intercepts. PTS: 1 DIF: DOK 1 NAT: F-IF.C.7c* STA: MCC9-12.F.IF.7c KEY: graphs of polynomial functions zeros 5
30. ANS: C The number of unique real zeros that a polynomial function has is equal to the number of times the graph of the function intersects the x-axis. p(x) x 4 x 2 3x 2 6x 3xx 4 x 2 x 2 This function has 3 unique real zeros, 0, 4, and 2, so the graph intersects the x-axis three times. Since two of those zeros, 0 and 4, each occur once, the graph crosses the x-axis at x 0 and x 4. Since one of those zeros, 2, occurs twice, the graph is tangent to the x-axis at x 2. A B C D Feedback Identify all of the zeros of the function and how many times each zero occurs. Identify all of the zeros of the function and how many times each zero occurs. That s correct! Identify all of the zeros of the function and how many times each zero occurs. PTS: 1 DIF: DOK 1 NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 KEY: zeros of polynomial functions graphs of polynomial functions 31. ANS: A PTS: 1 DIF: DOK 1 NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 32. ANS: B PTS: 1 DIF: DOK 1 NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 33. ANS: D PTS: 1 DIF: DOK 1 OBJ: Multiplying Polynomials NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 LOC: MTH.C.10.05.08.03.02.002 TOP: Multiplying Polynomials 34. ANS: D PTS: 1 DIF: DOK 1 OBJ: Finding Products in the Form (a + b)^2 NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 LOC: MTH.C.10.05.08.03.01.003 MTH.C.10.05.08.03.01.005 TOP: Special Products of Binomials 35. ANS: C PTS: 1 DIF: DOK 2 NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 36. ANS: A PTS: 1 DIF: DOK 2 NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 37. ANS: C PTS: 1 DIF: DOK 1 OBJ: Multiplying Polynomials NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 LOC: MTH.C.10.05.08.03.02.002 TOP: Multiplying Polynomials 38. ANS: C PTS: 1 DIF: DOK 1 NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 39. ANS: A PTS: 1 DIF: DOK 2 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 40. ANS: C PTS: 1 DIF: DOK 1 NAT: A-APR.C.4 STA: MCC9-12.A.APR.4 KEY: polynomial identities 41. ANS: A PTS: 1 DIF: DOK 1 NAT: A-APR.C.4 STA: MCC9-12.A.APR.4 KEY: polynomial identities 42. ANS: D PTS: 1 DIF: DOK 1 NAT: A-APR.C.4 STA: MCC9-12.A.APR.4 KEY: polynomial identities 6
43. ANS: B PTS: 1 DIF: DOK 1 NAT: A-APR.C.4 STA: MCC9-12.A.APR.4 KEY: polynomial identities 44. ANS: C PTS: 1 DIF: DOK 2 NAT: A-APR.C.4 STA: MCC9-12.A.APR.4 KEY: polynomial identities 45. ANS: B PTS: 1 DIF: DOK 2 NAT: A-APR.C.4 STA: MCC9-12.A.APR.4 KEY: polynomial identities 46. ANS: A PTS: 1 DIF: DOK 1 NAT: A-APR.C.5 STA: MCC9-12.A.APR.5 TOP: Use Combinations and the Binomial Theorem KEY: binomial theorem expand 47. ANS: D Use the binomial theorem. a b n n C 0 a n b 0 n C 1 a n 1 b 1 n C 2 a n 2 b 2 n C n 1 a 1 b n 1 n C n a 0 b n For 2x 7 6, 2x corresponds to a, 7 corresponds to b, and 6 corresponds to n in the binomial theorem. Since the expression that corresponds to a contains the variable x, the x 4 -term in the expanded form corresponds to the term in which a is raised to the fourth power. Since n 6, this occurs in the third term, C a n 2 n 2 b 2. Substitute 2x for a, 7 for b, and 6 for n in this term and simplify. C a n 2 n 2 b 2 6 C 2 2x 6 2 7 2 6 C 2 2x 4 7 2 The value of 6 C 2 is the third number in the sixth row of Pascal s triangle, 15. Substitute 15 for 6 C 2 and simplify. C 2x 4 6 2 7 2 152x 4 7 2 15 16x 4 49 11,760x 4 The coefficient of the x 4 -term in the expanded form of 2x 7 6 is 11,760. A B C D Feedback The coefficient of the x 4 -term is not negative. Be sure to include a factor of 6 C 2 in the coefficient. Also, the coefficient of the x 4 -term is not negative. Be sure to include a factor of 6 C 2 in the coefficient. That s correct! PTS: 1 DIF: DOK 1 NAT: A-APR.C.5(+) STA: MCC9-12.A.APR.5 KEY: binomial theorem 48. ANS: B PTS: 1 DIF: DOK 1 NAT: A-APR.D.6 STA: MCC9-12.A.APR.6 7
49. ANS: B PTS: 1 DIF: DOK 1 NAT: A-APR.D.6 STA: MCC9-12.A.APR.6 TOP: Divide Polynomials KEY: monomial polynomial long division remainder 50. ANS: C PTS: 1 DIF: DOK 1 OBJ: Using Long Division to Divide Polynomials NAT: A-APR.D.6 STA: MCC9-12.A.APR.6 LOC: MTH.C.10.05.08.03.03.002 TOP: Dividing Polynomials 51. ANS: B PTS: 1 DIF: DOK 1 OBJ: Using Synthetic Division to Divide by a Linear Binomial NAT: A-APR.D.6 STA: MCC9-12.A.APR.6 LOC: MTH.C.10.05.08.03.03.003 TOP: Dividing Polynomials 52. ANS: B PTS: 1 DIF: DOK 1 NAT: A-APR.D.6 STA: MCC9-12.A.APR.6 53. ANS: B 3x 2 2x 4 2x 3 6x 3 5x 2 2x 7 6x 3 9x 2 4x 2 2x 4x 2 6x 8x 7 8x 12 5 6x 3 5x 2 2x 7 2x 3 3x 2 2x 4 5 5, so the fractional part of the quotient is 2x 3 2x 3. A B C D Feedback This is the remainder. Use the remainder and the divisor to form the fractional part of the quotient. That s correct! The fractional part of the quotient is not the constant term in 6x 3 5x 2 2x 7 over the divisor, 2x 3. This is the polynomial part of the quotient, not the fractional part of the quotient. PTS: 1 DIF: DOK 1 NAT: A-APR.D.6 STA: MCC9-12.A.APR.6 KEY: rational expressions dividing polynomials 54. ANS: C PTS: 1 DIF: DOK 1 NAT: A-SSE.A.2 STA: MCC9-12.A.SSE.2 55. ANS: C PTS: 1 DIF: DOK 1 NAT: A-SSE.A.2 STA: MCC9-12.A.SSE.2 8
56. ANS: C PTS: 1 DIF: DOK 1 OBJ: Factoring by GCF and Recognizing Patterns NAT: A-SSE.A.2 STA: MCC9-12.A.SSE.2 LOC: MTH.C.10.05.08.03.04.002 MTH.C.10.05.08.03.04.012 TOP: Choosing a Factoring Method 57. ANS: C PTS: 1 DIF: DOK 1 NAT: A-SSE.A.2 STA: MCC9-12.A.SSE.2 58. ANS: B Notice that x 3 125 is the difference of two cubes, x 3 and 5 3. The general form for factoring the difference of two cubes is a 3 b 3 a b a 2 ab b 2. Rewriting x 3 125 this way results in x 3 5 3 x 5 x 2 5x 25. The trinomial factor is x 2 5x 25. Feedback A This is the trinomial factor from rewriting x 3 125. B That s correct! C The coefficient of x is incorrect. D The coefficient of x is incorrect. PTS: 1 DIF: DOK 1 NAT: A-SSE.A.2 MP.7 STA: MCC9-12.A.SSE.2 KEY: rewriting expressions difference of cubes factoring 59. ANS: B Notice that x 6 x 3 2 and 64 8 2, so x 6 64 x 3 2 8 2. The given expression is a difference of two squares. The factors of a difference of two squares are the sum of the roots and the difference of the roots, a 2 b 2 (a b)(a b). x 6 64 x 3 2 8 2 x 3 8 x 3 8 A B C D Feedback Notice that the given expression is a difference of two squares. That s correct! Notice that the given expression is a difference of two squares. Notice that the given expression is a difference of two squares. PTS: 1 DIF: DOK 1 NAT: A-SSE.A.2 MP.7 STA: MCC9-12.A.SSE.2 KEY: rewriting expressions properties of exponents 60. ANS: C PTS: 1 DIF: DOK 2 OBJ: Writing a Polynomial Function with Complex Zeros NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 LOC: MTH.C.10.06.05.008 TOP: Fundamental Theorem of Algebra 61. ANS: C PTS: 1 DIF: DOK 2 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 LOC: MTH.C.10.06.05.007 TOP: Fundamental Theorem of Algebra 9
62. ANS: C PTS: 1 DIF: DOK 2 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 LOC: MTH.C.10.06.05.007 TOP: Fundamental Theorem of Algebra 63. ANS: A PTS: 1 DIF: DOK 1 OBJ: Using Factoring to Solve Polynomial Equations NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 LOC: MTH.C.10.06.05.01.003 TOP: Finding Real Roots of Polynomial Equations 64. ANS: C PTS: 1 DIF: DOK 2 NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 65. ANS: B PTS: 1 DIF: DOK 2 NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 66. ANS: C PTS: 1 DIF: DOK 2 NAT: A-SSE.B.3 A-APR.B.3 STA: MCC9-12.A.SSE.3 67. ANS: C PTS: 1 DIF: DOK 2 NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 68. ANS: D PTS: 1 DIF: DOK 1 NAT: N-CN.C.8 A-SSE.B.3 A-APR.B.3 STA: MCC9-12.N.CN.8 69. ANS: C PTS: 1 DIF: DOK 2 NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 70. ANS: C PTS: 1 DIF: DOK 2 NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 71. ANS: B PTS: 1 DIF: DOK 2 NAT: N-CN.C.9 A-SSE.B.3 A-APR.B.3 STA: MCC9-12.N.CN.9 72. ANS: D PTS: 1 DIF: DOK 1 NAT: N-CN.C.9 A-SSE.B.3 A-APR.B.3 STA: MCC9-12.N.CN.9 73. ANS: C PTS: 1 DIF: DOK 2 NAT: N-CN.C.9 A-SSE.B.3 A-APR.B.3 STA: MCC9-12.N.CN.9 74. ANS: B PTS: 1 DIF: DOK 2 NAT: N-CN.C.9 A-SSE.B.3 A-APR.B.3 STA: MCC9-12.N.CN.9 75. ANS: C PTS: 1 DIF: DOK 2 NAT: N-CN.C.9 A-APR.B.2 STA: MCC9-12.N.CN.9 LOC: MTH.C.10.06.05.007 TOP: Fundamental Theorem of Algebra 10
MULTIPLE RESPONSE 1. ANS: A, C, D A: x 2 2x 5 13 x 2 2x 8 x 2 2x 1 9 ( x 1) 2 9 B: x 2 8x 9 5 x 2 8x 4 x 2 8x 16 12 ( x 4) 2 12 C: 2x 2 12x 29 81 2x 2 12x 110 x 2 6x 55 x 2 6x 9 64 ( x 3) 2 64 D: 5x 2 20x 14 6 5x 2 20x 20 x 2 4x 4 x 2 4x 4 0 ( x 2) 2 0 E: 3x 2 36x 88 4 3x 2 36x 84 x 2 12x 28 x 2 12x 36 8 ( x 6) 2 8 Since 9, 64, and 0 are all perfect squares, A, C, and D meet the criteria. Since 12 and 8 are not perfect squares, B and E do not meet the criteria. Correct Incorrect Feedback That s correct! Complete the square to rewrite each equation in the desired form. 11
PTS: 2 DIF: DOK 2 NAT: A-REI.B.4a MP.7 STA: MCC9-12.A.REI.4a KEY: completing the square 12
2. ANS: C, D After putting the equation in standard form, use the discriminant to determine whether each equation has real solutions or non-real solutions. A: x 2 3x 25 7 x 2 3x 18 0 b 2 4ac 3 2 41 18 9 72 81 Since the discriminant is not negative, the equation x 2 3x 25 7 has real solutions. B: x 2 7x 1 13 x 2 7x 12 0 b 2 4ac 7 2 41 12 49 48 97 Since the discriminant is not negative, the equation x 2 7x 1 13 has real solutions. C: x 2 2x 5 x 2 2x 5 0 b 2 4ac 2 2 41 5 4 20 16 Since the discriminant is negative, the equation x 2 2x 5 has non-real solutions. D: 2x 2 x 13 0 b 2 4ac 1 2 42 13 1 104 103 Since the discriminant is negative, the equation 2x 2 x 13 0 has non-real solutions. E: 2x 2 4x 9 11 2x 2 4x 2 0 b 2 4ac 4 2 42 2 16 16 0 13
Since the discriminant is not negative, the equation 2x 2 4x 9 11 has real solutions. Correct Incorrect Feedback That s correct! Use the discriminant to determine if each equation has real solutions or non-real solutions. PTS: 2 DIF: DOK 1 NAT: A-REI.B.4b STA: MCC9-12.A.REI.4b KEY: solving quadratic equations using the discriminant of the quadratic formula 3. ANS: B, C, F A: 9 5i 3 2i 12 7i B: 4 2i 2 5i 6 3i C: 9 5i 3 2i 6 3i D: 4 2i 2 5i 2 7i E: 3i1 2i 6 3i F: 3i1 2i 6 3i So, 4 2i 2 5i, 9 5i 3 2i, and 3i1 2i can be simplified to 6 3i. Correct Incorrect Feedback That s correct! Use the rules for adding, subtracting, and multiplying complex numbers to simplify each expression. PTS: 2 DIF: DOK 1 NAT: N-CN.A.2 STA: MCC9-12.N.CN.2 KEY: adding complex numbers subtracting complex numbers multiplying complex numbers 14
4. ANS: A, C, G A: As labeled on the graph, the x-intercept is 9. B: The y-intercept is 3. C: fx is always increasing, so fx is increasing when x 1. D: fx is always increasing, so fx is not decreasing when x 1. E: fx has no local maxima. F: fx has no local minima. G: fx is below the x-axis when x 9, so fx is negative when x 9. H: f0 3, so fx is not positive when x 2. Correct Incorrect Feedback That s correct! Carefully examine the graph to determine its key features. PTS: 2 DIF: DOK 1 NAT: F-IF.B.4* STA: MCC9-12.F.IF.4 KEY: cube root function graph increasing decreasing intercepts positive function negative function 5. ANS: A, D, F A, C: In the graph, notice that px approaches as x approaches and as x approaches. Since the end behavior is the same as x approaches and as x approaches, the degree of px is even. Since px approaches at both ends, the leading coefficient is positive. B, D, F: In the graph, notice that px crosses the x-axis at x 2 and x 3 and is tangent to the x-axis at x 1 and x 5. So, px has zeros of even multiplicity at x 1 and x 5 and zeros of odd multiplicity at x 2 and x 3. Since px has two zeros of even multiplicity and two zeros of odd multiplicity, its degree is at least 6. E: In the graph, px passes through the point (0, 150). So, the graph of px has a y-intercept of 150. Correct Incorrect Feedback That s correct! Examine the key properties of the graph of px. PTS: 2 DIF: DOK 2 NAT: F-IF.C.7c* STA: MCC9-12.F.IF.7c KEY: graphs of polynomial functions intercepts zeros end behavior 15
6. ANS: B, E A: x 6 81 x 2 3 3 1 3 because 3 1 3 3 3 27 81. Thus, the reasoning is invalid. B: 49c 2 154c 121 7c 2 27c 11 11 2 because 7c 2 49c 2, 27c 11 154c, and 11 2 121. Thus, the reasoning is valid. C: 36p 4 96p 64 6p 2 2 2 6p 2 8 8 2 because 2 6p 2 8 96p 2 96p. Thus, the reasoning is invalid. D: x 2 2 4 2 x 4 16, which is not the original expression, so the reasoning is invalid. E: x 18 8 x 6 3 2 3 because x 6 3 x 18 and 2 3 8. Thus, the reasoning is valid. F: x 9 is a perfect cube and 64 is a perfect square. However, 64 is also a perfect cube because 4 3 64. Thus, x 9 64 x 3 3 4 3 and can be factored as a sum of cubes. The reasoning is invalid. Correct Incorrect Feedback That s correct! Carefully examine each statement. PTS: 2 DIF: DOK 2 NAT: A-SSE.A.2 MP.7 STA: MCC9-12.A.SSE.2 KEY: rewriting expressions properties of exponents difference of squares difference of cubes sum of cubes 7. ANS: B, C, E The x-coordinates of the intersection points of the graphs of fx and gx are the solutions of the equation fx gx. The x-coordinates of the intersection points are approximately x 2, x 0, and x 5. Correct Incorrect Feedback That s correct! Determine the x-coordinates of all the points where the graphs of fx and gx intersect. PTS: 2 DIF: DOK 1 NAT: A-REI.D.11* STA: MCC9-12.A.REI.11 KEY: solving equations graphically polynomial functions exponential functions SHORT ANSWER 1. ANS: 225 4 56.25 PTS: 1 DIF: DOK 1 NAT: A-REI.B.4a STA: MCC9-12.A.REI.4a TOP: Solve Quadratic Equations by Completing the Square KEY: perfect square trinomial term 16
2. ANS: 4x 2 16x 21 12 4x 2 16x 33 x 2 4x 33 4 x 2 4x 4 33 4 4 x 2 2 33 4 16 4 x 2 2 49 4 Rubric 1 point for correct form; 2 points for accurate work PTS: 3 DIF: DOK 1 NAT: A-REI.B.4a MP.7 STA: MCC9-12.A.REI.4a KEY: completing the square 3. ANS: 3x 2 18x 77 2 3x 2 18x 75 x 2 6x 25 x 2 6x 9 25 9 x 3 2 16 The original equation has no real solutions, as it can be rewritten to say the square of a binomial is equal to a negative number. Rubric 2 points for correctly completing the square; 1 point for correct conclusion PTS: 3 DIF: DOK 2 NAT: A-REI.B.4a MP.7 STA: MCC9-12.A.REI.4a KEY: completing the square real solutions 17
4. ANS: The statement is imprecise because while the equation has no solution in the set of real numbers, it does have two solutions in the set of complex numbers. 25x 2 27 0 25x 2 27 x 2 27 25 x 27 25 x 3i 3 5 Rubric 1 point for explanation; 1 point for each solution PTS: 3 DIF: DOK 3 NAT: N-CN.A.1 N-CN.C.7 MP.6 STA: MCC9-12.N.CN.1 KEY: imaginary numbers solving equations 5. ANS: 3 1 2 5 2 5 i PTS: 1 DIF: DOK 1 NAT: N-CN.A.2 STA: MCC9-12.N.CN.2 6. ANS: a. 3 4i 3 4i 7 24i b. 3 4i 3 4i 25 c. a bi a bi (a 2 b 2 ) 2abi a bi a bi a 2 b 2 Rubric a. 1 point b. 1 point c. 1 point for each identity PTS: 4 DIF: DOK 2 NAT: N-CN.A.2 STA: MCC9-12.N.CN.2 KEY: multiplying complex numbers 18
7. ANS: Sample answer: PTS: 1 DIF: DOK 1 NAT: F-IF.B.4 STA: MCC9-12.F.IF.4 8. ANS: Rx as x and Rx as x PTS: 1 DIF: DOK 3 NAT: F-IF.B.4 STA: MCC9-12.F.IF.4 19
9. ANS: px x 3 x 2 16x 16 (x 3 x 2 16x 16) [x 2 (x 1) 16(x 1)] (x 2 16)(x 1) x 4 x 4 x 1 The zeros of px are 4, 1, and 4. As x approaches, px approaches. As x approaches, px approaches. Besides plotting the points where the graph crosses the x-axis (as determined by the zeros), it may be helpful to plot these additional points: (0, 16) and (2, 36). Rubric 1 point for the correct zeros; 1 point for the correct end behavior; 2 points for the correct graph PTS: 4 DIF: DOK 2 NAT: F-IF.C.7c* A-APR.B.3 STA: MCC9-12.F.IF.7c KEY: graphs of polynomial functions zeros end behavior factoring 20
10. ANS: a. The interval 0 t 10 corresponds to the period 2000 2010, so the domain is 0 t 10. b. The zeros of pt are t 2, t 1, t 7, and t 9. The first zero, t 2, is outside the domain 0 t 10. The other three zeros indicate that the company broke even during the years 2001, 2007, and 2009. Rubric a. 2 points b. 1 point for the four zeros; 2 points for reasonable interpretation PTS: 5 DIF: DOK 2 NAT: F-IF.C.7c* F-IF.B.5* MP.4 STA: MCC9-12.F.IF.7c KEY: graphs of polynomial functions domain modeling 11. ANS: b 3 b 2 14b 8 PTS: 1 DIF: DOK 1 NAT: A-APR.A.1 STA: MCC9-12.A.APR.1 12. ANS: 2, 3, or 4 PTS: 1 DIF: DOK 2 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 13. ANS: 1 or 8 PTS: 1 DIF: DOK 2 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 14. ANS: 2, or 3 PTS: 1 DIF: DOK 2 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 15. ANS: fx 1 4 x 1 x 2 2 x 3 PTS: 1 DIF: DOK 2 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 21
16. ANS: Sample answers: a 1,b 4; 1x 2 4x 4 (x 2) 2 ; a 4,b 8; 4x 2 8x 4 (2x 2) 2 ; a 9,b 12; 9x 2 12x 4 (3x 2) 2. In general, for any value a 0 b 4 a Explanation: Suppose the trinomial ax 2 bx 4 is a perfect square of the form (px q) 2 p 2 x 2 2pqx q 2. Then a p 2, b 2pq, and 4 q 2. So, q 2 and b 4p or b 4p. The table below shows some integer values of p and related values of a and b. p a b 1 1 4 1 1 4 2 4 8 2 4 8 3 9 12 3 9 12 PTS: 1 DIF: DOK 3 NAT: A-APR.C.4 STA: MCC9-12.A.APR.4 TOP: Factor Special Products KEY: trinomial perfect square trinomial 17. ANS: p 5 15p 4 r 90p 3 r 2 270p 2 r 3 405pr 4 243r 5 PTS: 1 DIF: DOK 1 NAT: A-APR.C.5 STA: MCC9-12.A.APR.5 18. ANS: 64x 12 576x 10 2160x 8 4320x 6 4860x 4 2916x 2 729 PTS: 1 DIF: DOK 1 NAT: A-APR.C.5 STA: MCC9-12.A.APR.5 KEY: binomial theorem 19. ANS: x 6 3x 4 y 3x 2 y 2 y 3 PTS: 1 DIF: DOK 1 NAT: A-APR.C.5 STA: MCC9-12.A.APR.5 22
20. ANS: Comparing the expressions x 2y 7 and a b n shows that x corresponds to a, 2y corresponds to b, and 7 corresponds to n in the binomial theorem. The x 5 y 2 -term of the simplified expanded form is the third term produced by the binomial theorem. Use Pascal s triangle to determine that 7 C 2 21. C a 5 7 2 b 2 21x 5 2y 2 84x 5 y 2 The x 5 y 2 -term in the expanded form of x 2y 7 is 84x 5 y 2. Rubric 1 point for identifying the correct term of the binomial theorem to use; 1 point for substituting correctly for a, b, and n in the binomial theorem; 1 point for simplifying correctly PTS: 3 DIF: DOK 1 NAT: A-APR.C.5(+) STA: MCC9-12.A.APR.5 KEY: binomial theorem 21. ANS: 3x 2 1 x PTS: 1 DIF: DOK 1 NAT: A-APR.D.6 STA: MCC9-12.A.APR.6 LOC: NCTM.PSSM.00.MTH.9-12.ALG.2.c KEY: polynomial division 22. ANS: 2x 1 1 x 2 PTS: 1 DIF: DOK 2 NAT: A-APR.D.6 STA: MCC9-12.A.APR.6 TOP: Rewrite Rational Expressions KEY: rational function 23. ANS: 22x 3 4x 2 12x 9 PTS: 1 DIF: DOK 1 NAT: A-SSE.A.2 STA: MCC9-12.A.SSE.2 24. ANS: 6, 0, 3 PTS: 1 DIF: DOK 1 NAT: A-SSE.B.3 A-APR.B.3 STA: MCC9-12.A.SSE.3 25. ANS: 7 PTS: 1 DIF: DOK 2 NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 23
26. ANS: a. p(x) (x 2)(x 2)(x 2); x 2, x 2, x 2 b. Rubric a. 0.5 point for each zero b. 0.5 point for showing the graph crossing the x-axis at x 2; 1 point for showing the graph tangent to the x-axis at x 2; 0.5 for showing downward end behavior at the left end; 0.5 point for showing upward end behavior at the right end PTS: 4 DIF: DOK 2 NAT: A-APR.B.3 STA: MCC9-12.A.APR.3 KEY: zeros of polynomial functions graphs of polynomial functions 27. ANS: Part A: There are 3 roots because x 3 x 2 x 1 is a degree 3 polynomial. There is a minimum of 1 real root because the degree is odd, and complex roots exist in pairs. Part B: x 3 x 2 x 1 x 2 (x 1) 1(x 1). (x 1)(x 2 1) Setting the factor x 1 0 shows that there is a real root at x 1. The equation x 2 1 0 has only complex number solutions, because both x 2 and 1 are positive for any real number x. There are 2 complex roots. PTS: 1 DIF: DOK 3 NAT: N-CN.C.9 STA: MCC9-12.N.CN.9 28. ANS: Sheila is incorrect. The quadratic polynomial x 2 18x 81 has the factored form x 9 2. This means the real root 9 is a repeated root. The polynomial has two complex roots, but both are 9. Rubric 1 point for stating that Sheila is incorrect; 1 point for explanation; 1 point for stating the second complex root PTS: 3 DIF: DOK 2 NAT: N-CN.C.9(+) MP.3 STA: MCC9-12.N.CN.9 KEY: fundamental theorem of algebra quadratic polynomials complex numbers 24
29. ANS: x 4 9 (x 2 ) 2 (3) 2 (x 2 3)(x 2 3) (x i 3)(x i 3)(x 3)(x 3) PTS: 1 DIF: DOK 2 NAT: N-CN.C.8 STA: MCC9-12.N.CN.8 PROBLEM 1. ANS: a. Vt is increasing for 0 t 3.5 and 13 t 18. Vt is decreasing for 3.5 t 13. b. The value of the portfolio increased during the first 3.5 months, and then the value decreased until the 13th month, at which point the value began to increase again. c. A local maximum occurs at approximately (3.5, 2.25), and a local minimum occurs at approximately (13, 0.75). d. The local maximum indicates that during this period, the value of the portfolio reached a peak of about $2,250,000 after about 3.5 months before losing value. The local minimum indicates that during this period, the value of the portfolio decreased to about $750,000 after about 13 months before beginning to regain value. e. The fact that the function is always positive over this period indicates that the portfolio always had some value. This makes sense because the portfolio can only lose all its value; the portfolio cannot have negative value. In this aspect, the model is appropriate. Rubric a. 1 point for correct intervals b. 1 point for appropriate explanation of significance c. 0.5 point for correct local maximum; 0.5 point for correct local minimum d. 1 point for appropriate explanation of significance e. 1 point for noting value is always nonnegative; 1 point for stating model is appropriate PTS: 6 DIF: DOK 3 NAT: F-IF.B.4* MP.4 MP.3 STA: MCC9-12.F.IF.4 KEY: interpreting graphs polynomial functions increasing decreasing relative maximums relative minimums 25
2. ANS: a. The zeros are x 5, x 3, x 0, and x 2. b. Expanding the factored form of px shows that the leading term is x 5. The degree of the polynomial is odd, and the leading coefficient is positive. So, px approaches as x approaches, and px approaches as x approaches. c. The graph cannot represent px because it does not have the same end behavior as px. Rubric a. 1 point b. 1 point for end behavior; 1 point for justification c. 1 point for answer; 1 point for explanation PTS: 5 DIF: DOK 2 NAT: F-IF.C.7c* STA: MCC9-12.F.IF.7c KEY: graphs of polynomial functions zeros end behavior 26
3. ANS: a. As t approaches, Tt approaches. As t approaches, Tt approaches. b. The end behavior of the function does not make sense in this context because it is not realistic to assume that the average monthly high temperature will increase without limit as t decreases from 0 or increases from 14. c. A reasonable domain is 0 t 12, since this covers a period of 1 year and allows for the full range of temperatures. Rubric a. 2 points for the graph; 2 points for the end behavior b. 2 points for answer and explanation c. 2 points for domain and explanation PTS: 8 DIF: DOK 3 NAT: F-IF.C.7e* F-IF.B.5* MP.2 MP.3 MP.4 STA: MCC9-12.F.IF.7e KEY: graphs of polynomial functions end behavior domain modeling NOT: Source (average high temperature for Boston, MA): http://www.weather.com/weather/wxclimatology/monthly/graph/usma0046 ESSAY 1. ANS: If a polynomial simplifies to 0 when a is substituted for x, then x a is a factor of the polynomial. 2(3) 3 4(3) 2 (3) 4 25 0 x 3 is not a factor of 2x 3 4x 2 x 4. PTS: 1 DIF: DOK 2 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 27
2. ANS: Part A: x 3 Part B: Dividing 4x 3 x 2 27x 18 by x 3 produces the quadratic expression 4x 2 11x 6. Part C: The possible zeros are 1 4,1 2,3 4,1,3 2,2,3,6. Sample answer: yes, 4x 3 x 2 27x 18 is a factor of gx, so all of the possible rational zeros of 4x 3 x 2 27x 18 would also be zeros for gx. Part D: 2, 3 4,3 PTS: 1 DIF: DOK 3 NAT: A-APR.B.2 STA: MCC9-12.A.APR.2 LOC: NCTM.PSSM.00.MTH.9-12.ALG.1.c TOP: Find Rational Zeros KEY: zeros polynomial functions 3. ANS: Sample answer: The pattern of coefficients comes from Pascal's triangle: 1 6 15 20 15 6 1. The pattern of the exponents of the variable a involves decreasing the value of its exponent by 1 for each term, beginning with 6 and ending with 0. The pattern of the exponents of the variable b involves increasing the value of its exponents by 1 for each term, beginning with 0 and ending with 6. The expanded form is a 6 6a 5 b 15a 4 b 2 20a 3 b 3 15a 2 b 4 6ab 5 b 6. PTS: 1 DIF: DOK 2 NAT: A-APR.C.5 STA: MCC9-12.A.APR.5 LOC: NCTM.PSSM.00.MTH.9-12.NOP.2.c NCTM.PSSM.00.MTH.9-12.GEO.4.e NCTM.PSSM.00.MTH.9-12.DAP.4.e NCTM.PSSM.00.MTH.9-12.PRS.1 NCTM.PSSM.00.MTH.9-12.PRS.4 NCTM.PSSM.00.MTH.9-12.CON.2 TOP: Use Combinations and the Binomial Theorem KEY: binomial multiply Pascal's triangle OTHER 1. ANS: a. No b. Yes c. Yes d. Yes e. No PTS: 2 DIF: DOK 2 NAT: A-REI.B.4a MP.7 STA: MCC9-12.A.REI.4a KEY: completing the square 28