8-8 Differences of Squares Factor each polynomial. 2 1. x − 9 SOLUTION: 2. 4a2 − 25 SOLUTION: 2 3. 9m − 144 SOLUTION: 4. 2p 3 − 162p SOLUTION: 4 5. u − 81 SOLUTION: eSolutions Manual - Powered by Cognero Page 1 6. 2d4 − 32f 4 SOLUTION: 4 4 7. 20r − 45n SOLUTION: 8. 256n4 − c4 SOLUTION: 3 2 9. 2c + 3c − 2c − 3 SOLUTION: 10. f 3 − 4f 2 − 9f + 36 SOLUTION: 3 2 11. 3t + 2t − 48t − 32 SOLUTION: 12. w3 − 3w2 − 9w + 27 SOLUTION: EXTENDED RESPONSE After an accident, skid marks may result from sudden breaking. The formula s2 = d approximates a vehicle’s speed s in miles per hour given the length d in feet of the skid marks on dry concrete. 13. If skid marks on dry concrete are 54 feet long, how fast was the car traveling when the brakes were applied? SOLUTION: Since the speed cannot be negative, the car was traveling 36 mph when the brakes were applied. 14. If the skid marks on dry concrete are 150 feet long, how fast was the car traveling when the brakes were applied? SOLUTION: Since the speed cannot be negative, the car was traveling 60 mph when the brakes were applied. Factor each polynomial. 2 15. q − 121 SOLUTION: 16. r4 − k4 SOLUTION: 4 17. 6n − 6 SOLUTION: 18. w4 − 625 SOLUTION: 2 2 19. r − 9t SOLUTION: 20. 2c2 − 32d2 SOLUTION: 3 21. h − 100h SOLUTION: 22. h4 − 256 SOLUTION: 3 2 23. 2x − x − 162x + 81 SOLUTION: 24. x2 − 4y2 SOLUTION: 4 4 25. 7h − 7p SOLUTION: 26. 3c3 + 2c2 − 147c − 98 SOLUTION: 2 4 4 27. 6k h − 54k SOLUTION: 28. 5a3 − 20a SOLUTION: 3 2 29. f + 2f − 64f − 128 SOLUTION: 30. 3r3 − 192r SOLUTION: 3 31. 10q − 1210q SOLUTION: 32. 3xn4 − 27x3 SOLUTION: 3 5 3 33. p r − p r SOLUTION: 34. 8c3 − 8c SOLUTION: 3 2 35. r − 5r − 100r + 500 SOLUTION: 36. 3t3 − 7t2 − 3t + 7 SOLUTION: 2 37. a − 49 SOLUTION: 38. 4m3 + 9m2 − 36m − 81 SOLUTION: 4 39. 3m + 243 SOLUTION: 40. 3x3 + x2 − 75x − 25 SOLUTION: 3 2 41. 12a + 2a − 192a − 32 SOLUTION: 42. x4 + 6x3 − 36x2 − 216x SOLUTION: 3 2 43. 15m + 12m − 375m − 300 SOLUTION: 44. GEOMETRY The drawing shown is a square with a square cut out of it. a. Write an expression that represents the area of the shaded region. b. Find the dimensions of a rectangle with the same area as the shaded region in the drawing. Assume that the dimensions of the rectangle must be represented by binomials with integral coefficients. SOLUTION: a. Use the formula for the area of square A = s · s to write an expression for the area of the shaded region. b. Write the area of the shaded region in factor form to find two binomial dimensions for a rectangle with the same area. The dimensions of the rectangle would be (4n + 6) by (4n − 4). 45. DECORATIONS An arch decorated with balloons was used to decorate the gym for the spring dance. The shape of the arch can be modeled by the equation y = −0.5x2 + 4.5x, where x and y are measured in feet and the x-axis represents the floor. a. Write the expression that represents the height of the arch in factored form. b. How far apart are the two points where the arch touches the floor? c. Graph this equation on your calculator. What is the highest point of the arch? SOLUTION: a. b. Find the two places along the x-axis (the floor) where the height of the arch is 0. or Subtract the two values to find out how far apart they are: 9 – 0 = 9. The two points where the arch touches the floor are 9 ft. apart. c. Use a graphing calculator to find the height of the arch at its highest point. [–2, 10] scl: 2 by [–2, 10] scl: 2 The arch is 10.125 ft. high. 46. CCSS SENSE-MAKING Zelda is building a deck in her backyard. The plans for the deck show that it is to be 24 feet by 24 feet. Zelda wants to reduce one dimension by a number of feet and increase the other dimension by the same number of feet. If the area of the reduced deck is 512 square feet, what are the dimensions of the deck? SOLUTION: Let x be the number of feet added and subtracted to each dimension of the deck. Replace A with 512, l with 24 + x, and w with 24 - x. Since the amount added and subtracted to the dimensions cannot be negative, 8 feet is the amount that is added and subtracted to the dimensions. 24 – 8 = 16 24 + 8 = 32 Therefore, the dimensions of the deck are 16 feet by 32 feet. 2 47. SALES The sales of a particular CD can be modeled by the equation S = −25m + 125m, where S is the number of CDs sold in thousands, and m is the number of months that it is on the market. a. In what month should the music store expect the CD to stop selling? b. In what month will CD sales peak? c. How many copies will the CD sell at its peak? SOLUTION: a. Find the values of m for which S equals 0. m = 0 or m = 5 The music store should expect the CD to stop selling in month 5. b. Use a graphing calculator to find the maximum. [0, 10] scl: 1 by [0, 250] scl: 25 The maximum occurs at the point x = 2.5. So the CD sales should peak in month 2.5. c. The maximum on the graph is S = 156.25, where S is measured in the thousands. The peak number of copies sold is 156.25 × 1000 = 156,250 copies. Solve each equation by factoring. Confirm your answers using a graphing calculator. 48. 36w2 = 121 SOLUTION: or The roots are and or about 1.833 and –1.833. Confirm the roots using a graphing calculator. Let Y1 = 36w2 and Y2 = 121. Use the intersect option from the CALC menu to find the points of intersection. Thus, the solutions are and . 49. 100 = 25x2 SOLUTION: or The roots are –2 and 2. Confirm the roots using a graphing calculator. Let Y1 = 100 and Y2 = 25x2. Use the intersect option from the CALC menu to find the points of intersection. Thus, the solutions are –2 and 2. 50. 64x2 − 1 = 0 SOLUTION: The roots are and or –0.125 and 0.125. Confirm the roots using a graphing calculator. Let Y1 = 64x2 – 1 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection. Thus, the solutions are and . 2 51. 4y − = 0 SOLUTION: The roots are and or -0.375 and 0.375. Confirm the roots using a graphing calculator. Let Y1 = 4y 2 – and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection. Thus, the solutions are and . 52. b2 = 16 SOLUTION: The roots are –8 and 8. Confirm the roots using a graphing calculator. Let Y1 = and Y2 = 16. Use the intersect option from the CALC menu to find the points of intersection. Thus, the solutions are –8 and 8. 2 53. 81 − x = 0 SOLUTION: or The roots are –45 and 45. Confirm the roots using a graphing calculator. Let Y1 = and Y2 =0. Use the intersect option from the CALC menu to find the points of intersection. Thus, the solutions are –45 and 45. 54. 9d2 − 81 = 0 SOLUTION: The roots are –3 and 3. Confirm the roots using a graphing calculator. Let Y1 = 9d2 – 81 and Y2 =0. Use the intersect option from the CALC menu to find the points of intersection. Thus, the solutions are –3 and 3. 2 55. 4a = SOLUTION: The roots are and or –0.1875 and 0.1875. Confirm the roots using a graphing calculator. Let Y1 = 4a2 and Y2 = . Use the intersect option from the CALC menu to find the points of intersection. Thus, the solutions are and . 56. MULTIPLE REPRESENTATIONS In this problem, you will investigate perfect square trinomials. a. TABULAR Copy and complete the table below by factoring each polynomial. Then write the first and last terms of the given polynomials as perfect squares. b. ANALYTICAL Write the middle term of each polynomial using the square roots of the perfect squares of the first and last terms. c. ALGEBRAIC Write the pattern for a perfect square trinomial. d. VERBAL What conditions must be met for a trinomial to be classified as a perfect square trinomial? SOLUTION: a. In this trinomial, a = 9, b = –24 and c = 16, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 9(16) or 144 with a sum of –24. Factors of 144 Sum –1, –144 –145 –2, –72 –74 –3, –48 –51 –4, –36 –40 –6, –24 –30 –8, –18 –26 –9, –16 –25 –12, –12 –24 The correct factors are –12 and –12. In this trinomial, a = 4, b = –20 and c = 25, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 4(25) or 100 with a sum of –20. Factors of 100 Sum –1, –100 –101 –2, –50 –52 –4, –25 –29 –10, –10 –20 The correct factors are –10 and –10. In this trinomial, a = 16, b = –20 and c = 9, so m + p is negative and mp is negative.