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1 Cubic equations Acubicequationhastheform mc-ty-cubicequations ax 3 + bx 2 + cx + d = 0 where a 0 Allcubicequationshaveeitheronerealroot,orthreerealroots.Inthisunitweexplorewhythis isso. Thenwelookathowcubicequationscanbesolvedbyspottingfactorsandusingamethodcalled synthetic division. Finally we will see how graphs can help us locate solutions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. Afterreadingthistext,and/orviewingthevideotutorialonthistopic,youshouldbeableto: explainwhycubicequationspossesseitheronerealrootorthreerealroots usesyntheticdivisiontolocaterootswhenonerootisknown find approximate solutions by drawing a graph Contents. Introduction 2 2. Cubic equations and the nature of their roots 2 3. Solving cubic equations 5 4. Using graphs to solve cubic equations 0 c mathcentre 2009

2 . Introduction Inthisunitweexplainwhatismeantbyacubicequationandhowsuchanequationcanbe solved. The general strategy for solving a cubic equation is to reduce it to a quadratic equation, and then solve the quadratic by the usual means, either by factorising or using the formula. 2. Cubic equations and the nature of their roots Acubicequationhastheform ax 3 + bx 2 + cx + d = 0 Itmusthavethetermin x 3 oritwouldnotbecubic(andso a 0),butanyorallof b, cand d can be zero. For instance, are all cubic equations. x 3 6x 2 + x 6 = 0, 4x = 0, x 3 + 9x = 0 Justasaquadraticequationmayhavetworealroots,soacubicequationhaspossiblythree. Butunlikeaquadraticequationwhichmayhavenorealsolution,acubicequationalwayshasat leastonerealroot.wewillseewhythisisthecaselater.ifacubicdoeshavethreeroots,two or even all three of them may be repeated. This gives us four possibilities which are illustrated in the following examples. Suppose we wish to solve the equation This equation can be factorised to give x 3 6x 2 + x 6 = 0 (x )(x 2)(x 3) = 0 Thisequationhasthreerealroots,alldifferent-thesolutionsare x =, x = 2and x = 3. InFigureweshowthegraphof y = x 3 6x 2 + x 6. y x -2-3 Figure.Thegraphof y = x 3 6x 2 + x 6. Noticethatitstartslowdownontheleft,becauseas xgetslargeandnegativesodoes x 3 and itfinisheshighertotherightbecauseas xgetslargeandpositivesodoes x 3.Thecurvecrosses the x-axisthreetimes,oncewhere x =,oncewhere x = 2andoncewhere x = 3.Thisgives us our three separate solutions. 2 c mathcentre 2009

3 Supposewewishtosolvetheequation x 3 5x 2 + 8x 4 = 0. This equation can be factorised to give (x )(x 2) 2 = 0 Inthiscasewedohavethreerealrootsbuttwoofthemarethesamebecauseoftheterm (x 2) 2. Soweonlyhavetwodistinctsolutions.Figure2showsagraphof y = x 3 5x 2 + 8x 4. y x Figure2.Thegraphof y = x 3 5x 2 + 8x 4. Againthecurvestartslowtotheleftandgoeshightotheright.Itcrossesthe x-axisonceand thenjusttouchesitat x = 2.Sowehaveourtworoots x = and x = 2. Supposewewishtosolvetheequation x 3 3x 2 + 3x = 0. This equation can be factorised to give (x ) 3 = 0 Soalthoughtherearethreefactors,theyareallthesameandweonlyhaveasinglesolution x =.Thecorrespondingcurveis y = x 3 3x 2 + 3x andisshowninfigure3. y x Figure3.Thegraphof y = x 3 3x 2 + 3x. 3 c mathcentre 2009

4 Aswithallthecubicswehaveseensofar,itstartslowdownontheleftandgoeshighupto theright. Noticethatthecurvedoescrossthe x-axisatthepoint x = butthe x-axisisalso atangenttothecurveatthispoint.thisisindicativeofthefactthattherearethreerepeated roots. Supposewewishtosolvetheequation x 3 + x 2 + x 3 = 0. This equation can be factorised to give (x )(x 2 + 2x + 3) = 0 Thequadratic x 2 + 2x + 3 = 0hasnorealsolutions,sotheonlysolutiontothecubicequation isobtainedbyputting x = 0,givingthesinglerealsolution x =. Thegraph y = x 3 + x 2 + x 3isshowninFigure4. y x -4 Figure4.Thegraphof y = x 3 + x 2 + x 3. Youcanseethatthegraphcrossesthe x-axisinoneplaceonly. Fromthesegraphsyoucanseewhyacubicequationalwayshasatleastonerealroot. The grapheitherstartslargeandnegativeandfinisheslargeandpositive(whenthecoefficientof x 3 ispositive),oritwillstartlargeandpositiveandfinishofflargeandnegative(whenthecoefficient of x 3 isnegative). Thegraphofacubicmustcrossthe x-axisatleastoncegivingyouatleastonerealroot.so, any problem you get that involves solving a cubic equation will have a real solution. Exercise Determinetherealrootsofthefollowingcubicequations-ifarootisrepeatedsayhowmany times. a) (x + )(x 2)(x 3) = 0 b) (x + )(x 2 2x + 20) = 0 c) (x 3) 2 (x + 4) = 0 d) (x + 2)(x 2 + 6x + 0) = 0 e) (x + 2)(x 2 + 7x + 0) = 0 f) (x 5)(x 2 0x + 25) = c mathcentre 2009

5 3. Solving cubic equations Nowletusmoveontothesolutionofcubicequations. Likeaquadratic,acubicshouldalways be re-arranged into its standard form, in this case The equation ax 3 + bx 2 + cx + d = 0 x 2 + 4x = 6 x isacubic,thoughitisnotwritteninthestandardform. Weneedtomultiplythroughby x, giving us x 3 + 4x 2 x = 6 andthenwesubtract6frombothsides,givingus Thisisnowinthestandardform x 3 + 4x 2 x 6 = 0 Whensolvingcubicsithelpsifyouknowoneroottostartwith. Supposewewishtosolve giventhat x = 2isasolution. x 3 5x 2 2x + 24 = 0 ThereisatheoremcalledtheFactorTheoremwhichwedonotprovehere. Itstatesthatif x = 2isasolutionofthisequation,then x+2isafactorofthiswholeexpression.thismeans that x 3 5x 2 2x + 24 = 0canbewrittenintheform where a and b are numbers. (x + 2)(x 2 + ax + b) = 0 Ourtasknowistofind aand b,andwedothisbyaprocesscalledsyntheticdivision. This involveslookingatthecoefficientsoftheoriginalcubicequation,whichare, 5, 2and24. Thesearewrittendowninthefirstrowofatable,thestartinglayoutforwhichis x = 2 Noticethattotherightoftheverticallinewewritedowntheknownroot x = 2. Wehave leftablanklinewhichwillbefilledinshortly. Inthefirstpositiononthebottomrowwehave broughtdownthenumberfromthefirstrow. Thenextstepistomultiplythenumber,justbroughtdown,bytheknownroot, 2,andwrite theresult, 2,intheblankrowinthepositionshown x = c mathcentre 2009

6 Thenumbersinthesecondcolumnarethenadded, 5+ 2 = 7,andtheresultwritteninthe bottom row as shown x = Then,thenumberjustwrittendown, 7,ismultipliedbytheknownroot, 2,andwewritethe result, 4,intheblankrowinthepositionshown. Thenthenumbersinthiscolumnareadded: The process continues: x = x = x = Notethatthefinalnumberinthebottomrow(obtainedbyadding24and 24)iszero.Thisis confirmationthat x = 2isarootoftheoriginalcubic. Ifthisvalueturnsouttobenon-zero thenwedonothavearoot. Atthisstagethecoefficientsinthequadraticthatwearelookingforarethefirstthreenumbers inthebottomrow.sothequadraticis Sowehavereducedourcubicto The quadratic term can be factorised to give givingusthesolutions x = 2, 3 or4 x 2 7x + 2 (x + 2)(x 2 7x + 2) = 0 (x + 2)(x 3)(x 4) = 0 Inthepreviousweweregivenoneoftheroots.Ifarootisnotknownit salwaysworth trying a few simple values. 6 c mathcentre 2009

7 Supposewewishtosolve x 3 7x 6 = 0 Averysimplevaluewemighttryis x =.Substituting x = intheleft-handsidewefind 3 7() 6 whichequals 2,sothisvalueisclearlynotasolution.Supposewetrythevalue x = : ( ) 3 7( ) 6 whichdoesequalzero,so x = isasolution.thismeansthat x + isafactorandthecubic canbewrittenintheform (x + )(x 2 + ax + b) = 0 We can perform synthetic division to find the other factors. Asbefore,wetakethecoefficientsoftheoriginalcubicequation,whichare, 0, 7and 6. Thesearewrittendowninthefirstrowofatable,thestartinglayoutforwhichis x = Totherightoftheverticallinewewritedowntheknownroot x =. Inthefirstpositionon thebottomrowwehavebroughtdownthenumberfromthefirstrow. Thenextstepistomultiplythenumber,justbroughtdown,bytheknownroot,,andwrite theresult,,intheblankrowinthepositionshown x = Thenumbersinthesecondcolumnarethenadded, 0 + =,andtheresultwritteninthe bottom row as shown x = Then,thenumberjustwrittendown,,ismultipliedbytheknownroot,,andwewritethe result,,intheblankrowinthepositionshown. Thenthenumbersinthiscolumnareadded: x = x = c mathcentre 2009

8 The process continues: x = Atthisstagethecoefficientsinthequadraticthatwearelookingforarethefirstthreenumbers inthebottomrow.sothequadraticis Sowenowneedtosolvetheequation x 2 x 6 (x + )(x 2 x 6) = 0 which factorises to give (x + )(x 3)(x + 2) = 0 andthethreesolutionstothecubicequationare x = 2, or3. Sometimesyoumaybeabletospotafactorasshowninthefollowing. Supposewewishtosolvetheequation x 3 4x 2 9x + 36 = 0. Observethatthefirsttwotermsinthecubiccanbefactorisedas x 3 4x 2 = x 2 (x 4). The secondpairoftermscanbefactorisedas 9x + 36 = 9(x 4). Thissortofobservationcan only be made when you have had sufficient practice and experience of handling expressions like this. However, the observation enables us to proceed as follows: x 3 4x 2 9x + 36 = 0 x 2 (x 4) 9(x 4) = 0 Thecommonfactorof (x 4)canbeextractedtogive (x 2 9)(x 4) 0 and the difference of two squares can be factorised giving givingussolutions x = 3, 3 or4 (x + 3)(x 3)(x 4) = 0 () Youmayhavenoticedthatineachexamplewehavedone,everyrootisafactoroftheconstant termintheequation.inthelast, 3, 3and 4alldivideintotheconstantterm36.As longasthecoefficientof x 3 inthecubicequationisthismustbethecase. Thisisbecause, referring to equation() for example, the constant term arises from multiplying the numbers 3, 3 and 4. This gives us another possible approach. 8 c mathcentre 2009

9 Suppose we wish to solve the equation x 3 6x 2 6x 7 = 0 Ifthereistobeasolution,thenbecausethecoefficientof x 3 is,itisgoingtobeaninteger andit sgoingtobeafactorof7. Thisleavesuswithonlyfourpossibilities:,, 7,and 7. Sowecantryeachoftheminturn. Youcanseefairlyquicklythatand don twork. So wewilltry7. Ratherthansubstitute x = 7intothecubicequationweshallimmediatelytryto synthetically divide this expression by x = Becausethefourthnumberinthelastrowiszeroitfollowsthat x = 7isindeedaroot. Had thisnumbernotbeenzerothen x = 7wouldnothavebeenaroot. So7isindeedaroot,and the resulting quadratic is x 2 + x + = 0 Nowyouwillfindthatifyoutrytosolveitthatthequadraticequation x 2 + x + = 0hasno realsolutions,sotheonlypossiblesolutiontothiscubicis x = 7. Certain basic identities which you may wish to learn can help in factorising both cubic and quadratic equations. Supposewewantedtosolvetheequation x 3 + 3x 2 + 3x + = 0. Thishasthewidely-knownfactorisation (x + ) 3 = 0fromwhichwehavetheroot x = repeated three times. Exercise 2.Foreachofthefollowingcubicequationsonerootisgiven.Determinetheotherrootsof each cubic. a) x 3 + 3x 2 6x 8 = 0hasarootat x = 2. b) x 3 + 2x 2 2x + 8 = 0hasarootat x = 3. c) x 3 + 4x 2 + 7x + 6 = 0hasarootat x = 2. d) 2x 3 + 9x 2 + 3x 4 = 0hasarootat x = For each of the following cubic equations use synthetic division to determine if the given valueof xisarootoftheequation.whereitis,determinetheotherrootsoftheequation. a)is x = 2arootoftheequation x 3 + 9x x + 24 = 0? 9 c mathcentre 2009

10 b)is x = 4arootoftheequation x 3 6x 2 + 9x + = 0? c)is x = arootoftheequation x 3 + 6x 2 + 3x 5 = 0? d)is x = 2arootoftheequation x 3 + 2x 2 20x + 24 = 0? 4. Using graphs to solve cubic equations Ifyoucannotfindasolutionbythesemethodsthendrawanaccurategraphofthecubicexpression.Thepointswhereitcrossesthe x-axiswillgiveyouthesolutionstotheequationbuttheir accuracy will be limited to the accuracy of your graph. Youmayindeedfindthatthegraphcrossesthe x-axisatapointthatwouldsuggestafactor, forinstance,ifyoudrawagraphthatappearstocrossthe x-axisat,say, x = ( 2,thenit sworth trying to find out whether x ) is indeed a factor. 2 Supposewewishedtosolve x 3 + 4x 2 + x 5 = 0. Nowthisequationwillnotyieldafactorbyanyofthemethodsthatwehavediscussed. Soa graphof y = x 3 + 4x 2 + x 5hasbeendrawnasshowninFigure5. y x Figure5.Agraphof y = x 3 + 4x 2 + x 5 Itcrossesthe x-axisatthreeplacesandhencetherearethreerealroots.aswestatedbeforetheir accuracywillbelimitedtotheaccuracyofthegraph. Fromthegraphwefindtheapproximate solutions x 3.2,.7, 0.9. Exercise 3 Use graphs to solve, as accurately as you can, the following cubic equations. a) x 3 4x 2 6x + 5 = 0 b) 2x 3 3x 2 4x 35 = 0 c) x 3 3x 2 x + = c mathcentre 2009

11 Answers Exercise a),2,3 b),2,0 c) 4,3(twice) d) 2 e) 5, 2(twice) f) 5(three times) Exercise 2.a) x = and x = 4 b) x = 6and x = c)nootherroots d) x = and x = a) x = 2isaroot.Otherrootsare x = 3and x = 4 b) x = 4isnotaroot c) x = isnotaroot d) x = 2isaroot.Otherrootsare x = 6and x = 2 Exercise 3 All answers are correct to 2 significant figures. a).6,0.62,5 b)3.5,nootherrealroots c) 0.68,0.46,3.2 c mathcentre 2009

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