SOLVING EQUATIONS BY FACTORING

316 (5-60) Chpter 5 Exponents nd Polynomils 5.9 SOLVING EQUATIONS BY FACTORING In this setion The Zero Ftor Property Applitions helpful hint Note tht the zero ftor property is our seond exmple of getting n equivlent eqution without doing the sme thing to eh side. Wht ws the first? E X A M P L E 1 The tehniques of ftoring n e used to solve equtions involving polynomils tht nnot e solved y the other methods tht you hve lerned. After you lern to solve equtions y ftoring, we will use this tehnique to solve some new pplied prolems in this setion nd in Chpters 6 nd 8. The Zero Ftor Property The eqution 0 indites tht the produt of two unknown numers is 0. But the produt of two rel numers is zero only when one or the other of the numers is 0. So even though we do not know extly the vlues of nd from 0, we do know tht 0 or 0. This ide is lled the zero ftor property. Zero Ftor Property The eqution 0 is equivlent to the ompound eqution 0 or 0. The next exmple shows how to use the zero ftor property to solve n eqution in one vrile. Using the zero ftor property Solve x 2 x 12 0. We ftor the left-hnd side of the eqution to get produt of two ftors tht re equl to 0. Then we write n equivlent eqution using the zero ftor property. x 2 x 12 0 (x 4)(x 3) 0 Ftor the left-hnd side. x 4 0 or x 3 0 Zero ftor property x 4 or x 3 Solve eh prt of the ompound eqution. Chek tht oth 4 nd 3 stisfy x 2 x 12 0. If x 4, we get ( 4) 2 ( 4) 12 16 4 12 0. If x 3, we get (3) 2 3 12 9 3 12 0. So the solution set is 4, 3. The zero ftor property is used only in solving polynomil equtions tht hve zero on one side nd polynomil tht n e ftored on the other side. The polynomils tht we ftored most often were the qudrti polynomils. The equtions tht we will solve most often using the zero ftor property will e qudrti equtions. Qudrti Eqution If,, nd re rel numers, with 0, then the eqution x 2 x 0 is lled qudrti eqution.

5.9 Solving Equtions y Ftoring (5-61) 317 M A T H A T W O R K Sems Merdo, professionl odyorder nd 1988 Ntionl Chmpion, hrges the wves off Hwii, Thiti, Indonesi, Mexio, nd Cliforni. In hoosing ord for ompetition nd for the mneuvers he wnts to perform, Merdo ftors in his height nd weight s well s the size, power, nd temperture of the wves he will e riding. In older wter softer, more flexile ord is used; in wrmer wter stiffer ord is hosen. When wves rsh on shore, the ride usully lsts 3 to 5 seonds, nd shorter ord with nrrow til is hosen for greter ontrol. When wves rek long snd r or reef, the ride n sometimes lst s long s 2 minutes, nd strighter ord with more surfe re is hosen so tht the ord will move fster nd llow the rider to pull off more mneuvers. Bsi mneuvers inlude ottom turns, erils, forwrd nd reverse 360 s, nd el rollos. As one of the top 10 odyorders in the world, Merdo helps to design the ords he uses. Performne levels re gretly inresed with fine-tuned equipment nd tehniques, he sys. In Exerise 63 of Setion 5.9 you will find the dimensions of given odyord. BODYBOARD DESIGNER In Chpter 8 we will study qudrti equtions further nd solve qudrti equtions tht nnot e solved y ftoring. Keep the following strtegy in mind when solving equtions y ftoring. study tip We re ll retures of hit. When you find ple in whih you study suessfully, stik with it. Using the sme ple for studying will help you to onentrte nd to ssoite the ple with good studying. Strtegy for Solving Equtions y Ftoring 1. Write the eqution with 0 on the right-hnd side. 2. Ftor the left-hnd side. 3. Use the zero ftor property to get simpler equtions. (Set eh ftor equl to 0.) 4. Solve the simpler equtions. 5. Chek the nswers in the originl eqution. E X A M P L E 2 Solving qudrti eqution y ftoring Solve eh eqution. ) 10x 2 5x ) 3x 6x 2 9 ) Use the steps in the strtegy for solving equtions y ftoring: 10x 2 5x Originl eqution 10x 2 5x 0 Rewrite with zero on the right-hnd side. 5x(2x 1) 0 Ftor the left-hnd side. 5x 0 or 2x 1 0 Zero ftor property x 0 or x 1 Solve for x. 2 The solution set is 0, 1 2. Chek eh solution in the originl eqution.

318 (5-62) Chpter 5 Exponents nd Polynomils ) First rewrite the eqution with 0 on the right-hnd side nd the left-hnd side in order of desending exponents: 3x 6x 2 9 Originl eqution 6x 2 3x 9 0 Add 9 to eh side. 2x 2 x 3 0 Divide eh side y 3. (2x 3)(x 1) 0 Ftor. 2x 3 0 or x 1 0 Zero ftor property x 3 or x 1 Solve for x. 2 The solution set is 1, 3 2. Chek eh solution in the originl eqution. CAUTION If we divide eh side of 10x 2 5x y 5x, we get 2x 1, or x 1 2. We do not get x 0. By dividing y 5x we hve lost one of the ftors nd one of the solutions. In the next exmple there re more thn two ftors, ut we n still write n equivlent eqution y setting eh ftor equl to 0. E X A M P L E 3 lultor lose-up To hek, use Y= to enter y 1 2x 3 3x 2 8x 12. Then use the vriles feture (VARS) to find y 1 ( 2), y 1 (3 2), nd y 1 (2). Solving ui eqution y ftoring Solve 2x 3 3x 2 8x 12 0. First notie tht the first two terms hve the ommon ftor x 2 nd the lst two terms hve the ommon ftor 4. x 2 (2x 3) 4(2x 3) 0 Ftor y grouping. (x 2 4)(2x 3) 0 Ftor out 2x 3. (x 2)(x 2)(2x 3) 0 Ftor ompletely. x 2 0 or x 2 0 or 2x 3 0 Set eh ftor equl to 0. x 2 or x 2 or x 3 2 The solution set is 2, 3 2, 2. Chek eh solution in the originl eqution. The eqution in the next exmple involves solute vlue. E X A M P L E 4 Solving n solute vlue eqution y ftoring Solve x 2 2x 16 8. First write n equivlent ompound eqution without solute vlue: x 2 2x 16 8 or x 2 2x 16 8 x 2 2x 24 0 or x 2 2x 8 0 (x 6)(x 4) 0 or (x 4)(x 2) 0 x 6 0 or x 4 0 or x 4 0 or x 2 0 x 6 or x 4 or x 4 or x 2 The solution set is 2, 4, 4, 6. Chek eh solution.

Applitions 5.9 Solving Equtions y Ftoring (5-63) 319 Mny pplied prolems n e solved y using equtions suh s those we hve een solving. E X A M P L E 5 helpful hint To prove the Pythgoren theorem, drw two squres with sides of length, nd prtition them s shown. Are of room Ronld s living room is 2 feet longer thn it is wide, nd its re is 168 squre feet. Wht re the dimensions of the room? Let x e the width nd x 2 e the length. See Figure 5.1. Beuse the re of retngle is the length times the width, we n write the eqution x(x 2) 168. 2 2 2 Erse the four tringles in eh piture. Sine we strted with equl res, we must hve equl res fter ersing the tringles: 2 2 2 x FIGURE 5.1 x 2 We solve the eqution y ftoring: x 2 2x 168 0 (x 12)(x 14) 0 x 12 0 or x 14 0 x 12 or x 14 Beuse the width of room is positive numer, we disregrd the solution x 14. We use x 12 nd get width of 12 feet nd length of 14 feet. Chek this nswer y multiplying 12 nd 14 to get 168. Applitions involving qudrti equtions often require theorem lled the Pythgoren theorem. This theorem sttes tht in ny right tringle the sum of the squres of the lengths of the legs is equl to the length of the hypotenuse squred. The Pythgoren Theorem The tringle shown is right tringle if nd only if 2 2 2. Hypotenuse Legs We use the Pythgoren theorem in the next exmple.

320 (5-64) Chpter 5 Exponents nd Polynomils E X A M P L E 6 5 x 7 x FIGURE 5.2 Using the Pythgoren theorem Shirley used 14 meters of fening to enlose retngulr region. To e sure tht the region ws retngle, she mesured the digonls nd found tht they were 5 meters eh. (If the opposite sides of qudrilterl re equl nd the digonls re equl, then the qudrilterl is retngle.) Wht re the length nd width of the retngle? The perimeter of retngle is twie the length plus twie the width, P 2L 2W. Beuse the perimeter is 14 meters, the sum of one length nd one width is 7 meters. If we let x represent the width, then 7 x is the length. We use the Pythgoren theorem to get reltionship mong the length, width, nd digonl. See Figure 5.2. x 2 (7 x) 2 5 2 x 2 49 14x x 2 25 2x 2 14x 24 0 Pythgoren theorem Simplify. Simplify. x 2 7x 12 0 Divide eh side y 2. (x 3)(x 4) 0 Ftor the left-hnd side. x 3 0 or x 4 0 Zero ftor property x 3 or x 4 7 x 4 or 7 x 3 Solving the eqution gives two possile retngles: 3 y 4 retngle or 4 y 3 retngle. However, those re identil retngles. The retngle is 3 meters y 4 meters. WARM-UPS True or flse? Explin your nswer. 1. The eqution (x 1)(x 3) 12 is equivlent to x 1 3 or x 3 4. Flse 2. Equtions solved y ftoring my hve two solutions. True 3. The eqution d 0 is equivlent to 0 or d 0. True 4. The eqution x 2 4 5 is equivlent to the ompound eqution x 2 4 5 or x 2 4 5. Flse 5. The solution set to the eqution (2x 1)(3x 4) 0 is 1 2, 4 3. True 6. The Pythgoren theorem sttes tht the sum of the squres of ny two sides of ny tringle is equl to the squre of the third side. Flse 7. If the perimeter of retngulr room is 38 feet, then the sum of the length nd width is 19 feet. True 8. Two numers tht hve sum of 8 n e represented y x nd 8 x. True 9. The solution set to the eqution x(x 1)(x 2) 0is 1, 2. Flse 10. The solution set to the eqution 3(x 2)(x 5) 0 is 3, 2, 5. Flse

5.9 Solving Equtions y Ftoring (5-65) 321 5. 9 EXERCISES Reding nd Writing After reding this setion, write out the nswers to these questions. Use omplete sentenes. 1. Wht is the zero ftor property? The zero ftor property sys tht if 0 then either 0 or 0. 2. Wht is qudrti eqution? A qudrti eqution is n eqution of the form x 2 x 0 with 0. 3. Where is the hypotenuse in right tringle? The hypotenuse of right tringle is the side opposite the right ngle. 4. Where re the legs in right tringle? The legs of right tringle re the sides tht form the right ngle. 5. Wht is the Pythgoren theorem? The Pythgoren theorem sys tht tringle is right tringle if nd only if the sum of the squres of the legs is equl to the squre of the hypotenuse. 6. Where is the digonl of retngle? The digonl of retngle is the line segment tht joins two opposite verties. Solve eh eqution. See Exmples 1 3. 7. (x 5)(x 4) 0 4, 5 8. ( 6)( 5) 0 5, 6 9. (2x 5)(3x 4) 0 5 2, 4 3 10. (3k 8)(4k 3) 0 8 3, 3 4 11. w 2 5w 14 0 7, 2 12. t 2 6t 27 0 3, 9 13. m 2 7m 0 0, 7 14. h 2 5h 0 0, 5 15. 2 20 4, 5 16. p 2 p 42 6, 7 17. 3x 2 3x 36 0 3, 4 18. 2x 2 16x 24 0 6, 2 19. z 2 3 2 z 10 4, 5 2 20. m 2 1 1 m 2 3 3, 2 3 21. x 3 4x 0 2, 0, 2 22. 16x x 3 0 4, 0, 4 23. w 3 4w 2 25w 100 0 5, 4, 5 24. 3 2 2 16 32 0 4, 2, 4 25. n 3 2n 2 n 2 0 1, 1, 2 26. w 3 w 2 25w 25 0 5, 1, 5 Solve eh eqution. See Exmple 4. 27. x 2 5 4 3, 1, 1, 3 28. x 2 17 8 5, 3, 3, 5 29. x 2 2x 36 12 8, 6, 4, 6 30. x 2 2x 19 16 7, 3, 1, 5 31. x 2 4x 2 2 4, 2, 0 32. x 2 8x 8 8 8, 4, 0 33. x 2 6x 1 8 7, 3, 1 34. x 2 x 21 9 5, 3, 4, 6 Solve eh eqution. 35. 2x 2 x 6 3 2, 2 36. 3x 2 14x 5 5, 1 3 37. x 2 5x 6 6, 3, 2, 1 38. x 2 6x 4 12 8, 4, 2, 2 39. x 2 5x 6 6, 1 40. x 5x 6 1 41. (x 2)(x 1) 12 5, 2 42. (x 2)(x 3) 20 7, 2 43. y 3 9y 2 20y 0 5, 4, 0 44. m 3 2m 2 3m 0 1, 0, 3 45. 5 3 45 3, 0, 3 46. 5x 3 125x 5, 0, 5 47. (2x 1)(x 2 9) 0 3, 1 2,3 48. (x 1)(x 3)(x 9) 0 3, 1, 9 49. 4x 2 12x 9 0 3 2 50. 16x 2 8x 1 0 1 4 Solve eh eqution for y. Assume nd re positive numers. 51. y 2 y 0 0, 52. y 2 y y 0, 53. 2 y 2 2 0, 54. 9y 2 6y 2 0 3 55. 4y 2 4y 2 0 2 56. y 2 2 0, 57. y 2 3y y 3 3, 1 58. 2 y 2 2y 2 0 Solve eh prolem. See Exmples 5 nd 6. 59. Color print. The length of new super size olor print is 2 inhes more thn the width. If the re is 24 squre inhes, wht re the length nd width? Width 4 inhes, length 6 inhes

322 (5-66) Chpter 5 Exponents nd Polynomils 60. Tennis ourt dimensions. In singles ompetition, eh plyer plys on retngulr re of 117 squre yrds. Given tht the length of tht re is 4 yrds greter thn its width, find the length nd width. Width 9 yrds, length 13 yrds 61. Missing numers. The sum of two numers is 13 nd their produt is 36. Find the numers. 4 nd 9 62. More missing numers. The sum of two numers is 6.5, nd their produt is 9. Find the numers. 2 nd 4.5 63. Bodyording. The Sems Chnnel pro odyord shown in the figure hs length tht is 21 inhes greter thn its width. Any rider weighing up to 200 pounds n use it euse its surfe re is 946 squre inhes. Find the length nd width. Length 43 inhes, width 22 inhes x 21 in. Height (feet) ) Use the ompnying grph to estimte the mximum height rehed y the rrow. 64 feet d) At wht time does the rrow reh its mximum height? 2 seonds 70 60 50 40 30 20 10 0 0 1 2 3 4 5 Time (seonds) FIGURE FOR EXERCISE 65 66. Time until impt. If n ojet is dropped from height of s 0 feet, then its ltitude fter t seonds is given y the formul S 16t 2 s 0. If pk of emergeny supplies is dropped from n irplne t height of 1600 feet, then how long does it tke for it to reh the ground? 10 seonds 67. Yolnd s loset. The length of Yolnd s loset is 2 feet longer thn twie its width. If the digonl mesures 13 feet, then wht re the length nd width? Width 5 feet, length 12 feet x in. FIGURE FOR EXERCISE 63 64. New dimensions in grdening. Mry Gold hs retngulr flower ed tht mesures 4 feet y 6 feet. If she wnts to inrese the length nd width y the sme mount to hve flower ed of 48 squre feet, then wht will e the new dimensions? 6 feet y 8 feet 2x 2 ft 13 ft x ft 4 ft 6 ft x ft FIGURE FOR EXERCISE 67 68. Ski jump. The se of ski rmp forms right tringle. One leg of the tringle is 2 meters longer thn the other. If the hypotenuse is 10 meters, then wht re the lengths of the legs? 6 feet nd 8 feet x ft FIGURE FOR EXERCISE 64 65. Shooting rrows. An rher shoots n rrow stright upwrd t 64 feet per seond. The height of the rrow h(t) (in feet) t time t seonds is given y the funtion h(t) 16t 2 64t. ) Use the ompnying grph to estimte the mount of time tht the rrow is in the ir. 4 seonds ) Algerilly find the mount of time tht the rrow is in the ir. 4 seonds 10 m x + 2 m x m FIGURE FOR EXERCISE 68 69. Trimming gte. A totl of 34 feet of 1 4 lumer is used round the perimeter of the gte shown in the figure on the next pge. If the digonl re is 13 feet long, then wht re the length nd width of the gte? Width 5 feet, length 12 feet

Chpter 5 Collortive Ativities (5-67) 323 13 ft 74. Arrnging the rows. Mr. Converse hs 112 students in his lger lss with n equl numer in eh row. If he rrnges the desks so tht he hs one fewer rows, he will hve two more students in eh row. How mny rows did he hve originlly? 8 GETTING MORE INVOLVED FIGURE FOR EXERCISE 69 70. Perimeter of retngle. The perimeter of retngle is 28 inhes, nd the digonl mesures 10 inhes. Wht re the length nd width of the retngle? Length 8 inhes, width 6 inhes 71. Conseutive integers. The sum of the squres of two onseutive integers is 25. Find the integers. 3 nd 4, or 4 nd 3 72. Pete s grden. Eh row in Pete s grden is 3 feet wide. If the rows run north nd south, he n hve two more rows thn if they run est nd west. If the re of Pete s grden is 135 squre feet, then wht re the length nd width? Length 15 feet, width 9 feet 73. House plns. In the plns for their drem house the Bileys hve mster edroom tht is 240 squre feet in re. If they inrese the width y 3 feet, they must derese the length y 4 feet to keep the originl re. Wht re the originl dimensions of the edroom? Length 20 feet, width 12 feet 75. Writing. If you divide eh side of x 2 x y x, you get x 1. If you sutrt x from eh side of x 2 x, you get x 2 x 0, whih hs two solutions. Whih method is orret? Explin. 76. Coopertive lerning. Work with group to exmine the following solution to x 2 2x 1: x(x 2) 1 x 1 or x 2 1 x 1 or x 1 Is this method orret? Explin. 77. Coopertive lerning. Work with group to exmine the following steps in the solution to 5x 2 5 0 5(x 2 1) 0 5(x 1)(x 1) 0 x 1 0 or x 1 0 x 1 or x 1 Wht hppened to the 5? Explin. COLLABORATIVE ACTIVITIES Mgi Triks Jim nd Sdr re tlking one dy fter lss. Sdr: Jim, I hve trik for you. Think of numer etween 1 nd 10. I will sk you to do some things to this numer. Then t the end tell me your result, nd I will tell you your numer. Jim: Oh, yeh you proly rig it so the result is my numer. Sdr: Come on Jim, give it try nd see. Jim: Oky, oky, I thought of numer. Sdr: Good, now write it down, nd don t let me see your pper. Now dd x. Got tht? Now multiply everything y 2. Jim: Hey, I didn t know you were going to mke me think! This is lger! Sdr: I know, now just do it. Oky, now squre the polynomil. Got tht? Now sutrt 4x 2. Jim: How did you know I hd 4x 2? I told you this ws rigged! Sdr: Of ourse it s rigged, or it wouldn t work. Do you wnt to finish or not? Jim: Yeh, I guess so. Go hed, wht do I do next? Sdr: Divide y 4. Oky, now sutrt the x-term. Jim: Just ny old x-term? Got ny prtiulr oeffiient in mind? Grouping: Two students per group Topi: Prtie with exponent rules, multiplying polynomils Sdr: Now stop tesing me. I know you only hve one x-term left, so sutrt it. Jim: H, h, I ould give you hint out the oeffiient, ut tht wouldn t e fir, would it? Sdr: Well you ould, nd then I ould tell you your numer, or you ould just tell me the numer you hve left fter sutrting. Jim: Oky, the numer I hd left t the end ws 25. Let s see if you n tell me wht the oeffiient of the x-term I sutrted is. Sdr: Ah, then the numer you hose t the eginning ws 5, nd the oeffiient ws 10! Jim: Hey, you re right! How did you do tht? In your group, follow Sdr s instrutions nd determine why she knew Jim s numer. Mke up nother set of instrutions to use s mgi trik. Be sure to use vriles nd some of the exponent rules or rules for multiplying polynomils tht you lerned in this hpter. Exhnge instrutions with nother group nd see whether you n figure out how their trik works.