Partial fractions An algebraic fraction such as partial fractions. Specifically mc-ty-partialfractions-009-3x + 5 x 5x 3 canoftenbebrokendownintosimplerpartscalled 3x + 5 x 5x 3 x 3 Inthisunitweexplainhowthisprocessiscarriedout. 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: explain the meaning of the terms proper fraction and improper fraction express an algebraic fraction as the sum of its partial fractions Contents. Introduction. Revision of adding and subtracting fractions 3. Expressing a fraction as the sum of its partial fractions 3 4. Fractions where the denominator has a repeated factor 5 5. Fractions in which the denominator has a quadratic term 6 6. Dealing with improper fractions 7 www.mathcentre.ac.uk c mathcentre 009
. Introduction An algebraic fraction is a fraction in which the numerator and denominator are both polynomial expressions. Apolynomialexpressionisonewhereeverytermisamultipleofapowerof x, such as 5x 4 + 6x 3 + 7x + 4 Thedegreeofapolynomialisthepowerofthehighesttermin x.sointhiscasethedegreeis 4. Thenumberinfrontof xineachtermiscalleditscoefficient. So,thecoefficientof x 4 is5. Thecoefficientof x 3 is6. Now consider the following algebraic fractions: x x + x 3 + 3 x 4 + x + Inbothcasesthenumeratorisapolynomialoflowerdegreethanthedenominator.Wecallthese proper fractions Withotherfractionsthepolynomialmaybeofhigherdegreeinthenumeratororitmaybeof the same degree, for example x 4 + x + x x + 4 x 3 + x + x + 3 and these are called improper fractions. Key Point Ifthedegreeofthenumeratorislessthanthedegreeofthedenominatorthefractionissaidto be a proper fraction Ifthedegreeofthenumeratorisgreaterthanorequaltothedegreeofthedenominatorthe fraction is said to be an improper fraction. Revision of adding and subtracting fractions We now revise the process for adding and subtracting fractions. Consider x 3 In order to add these two fractions together, we need to find the lowest common denominator. Inthisparticularcase,itis (x 3)(). www.mathcentre.ac.uk c mathcentre 009
We write each fraction with this denominator. x 3 () and (x 3)() So x 3 (x 3)() x 3 () (x 3)() x 3 (x 3)() Thedenominatorsarenowthesamesowecansimplysubtractthenumeratorsanddividethe result by the lowest common denominator to give x 3 4x + x + 3 (x 3)() 3x + 5 (x 3)() Sometimes in mathematics we need to do this operation in reverse. In calculus, for instance, orwhendealingwiththebinomialtheorem,wesometimesneedtosplitafractionupintoits component parts which are called partial fractions. We discuss how to do this in the following section. Exercises Use the rules for the addition and subtraction of fractions to simplify 3 x + + x + 3 5 x 3 x + 4 x + 3 d) 3x 6x + 9 3. Expressing a fraction as the sum of its partial fractions Intheprevioussectionwesawthat x 3 3x + 5 (x 3)() Suppose we start with 3x + 5 (x 3)().Howcanwegetthisbacktoitscomponentparts? By inspection of the denominator we see that the component parts must have denominators of x 3and sowecanwrite 3x + 5 A (x 3)() x 3 + B where Aand Barenumbers. Aand Bcannotinvolve xorpowersof xbecauseotherwisethe terms on the right would be improper fractions. Thenextthingtodoistomultiplybothsidesbythecommondenominator (x 3)(x+).This gives (3x + 5)(x 3)() A(x 3)() B(x 3)() + (x 3)() x 3 Then cancelling the common factors from the numerators and denominators of each term gives 3x + 5 A() + B(x 3) Nowthisisanidentity. Thismeansthatitistrueforanyvaluesof x,andbecauseofthiswe cansubstituteanyvaluesof xwechooseintoit. Observethatifwelet x thefirstterm ontherightwillbecomezeroandhence Awilldisappear. Ifwelet x 3thesecondtermon therightwillbecomezeroandhence Bwilldisappear. www.mathcentre.ac.uk 3 c mathcentre 009
If x 3 + 5 B ( 3 ) from which 7 7 B B Nowwewanttotrytofind A. If x 3 sothat A. Putting these results together we have 4 7A 3x + 5 (x 3)() A x 3 + B x 3 whichisthesumthatwestartedwith, andwehavenowbrokenthefractionbackintoits component parts called partial fractions. Example Suppose we want to express 3x (x )(x + ) asthesumofitspartialfractions. Observethatthefactorsinthedenominatorare x and x + sowewrite where A and B are numbers. 3x (x )(x + ) A x + B x + Wemultiplybothsidesbythecommondenominator (x )(x + ): 3x A(x + ) + B(x ) Thistimethespecialvaluesthatweshallchooseare x becausethenthefirsttermonthe rightwillbecomezeroand Awilldisappear,and x becausethenthesecondtermonthe right will become zero and B will disappear. If x 6 3B B 6 3 B www.mathcentre.ac.uk 4 c mathcentre 009
If x Putting these results together we have 3 3A A 3x (x )(x + ) x + x + and we have expressed the given fraction in partial fractions. Sometimes the denominator is more awkward as we shall see in the following section. Exercises Express the following as a sum of partial fractions x x + 5 3 (x + )(x 3) (x )(x + ) (x )(x ) d) (x + 4)(x ) 4. Fractions where the denominator has a repeated factor Considerthefollowingexampleinwhichthedenominatorhasarepeatedfactor (x ). Example Suppose we want to express 3x + (x ) (x + ) asthesumofitspartialfractions. Thereareactuallythreepossibilitiesforadenominatorinthepartialfractions: x, x + and alsothepossibilityof (x ),sointhiscasewewrite where A, Band Carenumbers. 3x + (x ) (x + ) A (x ) + B (x ) + C (x + ) Asbeforewemultiplybothsidesbythedenominator (x ) (x + )togive 3x + A(x )(x + ) + B(x + ) + C(x ) () Againwelookforspecialvaluestosubstituteintothisidentity.Ifwelet x thenthefirstand lasttermsontherightwillbezeroand Aand Cwilldisappear. Ifwelet x thefirstand secondtermswillbezeroand Aand Bwilldisappear. If x If x 4 3B sothat B 4 3 5 9C sothat C 5 9 Wenowneedtofind A. Thereisnospecialvalueof xthatwilleliminate Band Ctogiveus A. Wecoulduseanyvalue. Wecoulduse x 0. Thiswillgiveusanequationin A, Band C. Sincewealreadyknow Band C,thiswouldgiveus A. www.mathcentre.ac.uk 5 c mathcentre 009
But here we shall demonstrate a different technique- one called equating coefficients. We take equation and multiply-out the right-hand side, and then collect up like terms. 3x + A(x )(x + ) + B(x + ) + C(x ) A(x + x ) + B(x + ) + C(x ) (A + C)x + (A + B C)x + ( A + B + C) Thisisanidentitywhichistrueforallvaluesof x. Ontheleft-handsidetherearenoterms involving x whereasontherightwehave (A + C)x.Theonlywaythiscanbetrueisif A + C 0 Thisiscalledequatingcoefficientsof x. Wealreadyknowthat C 5 9 sothismeansthat A 5 9.Wealsoalreadyknowthat B 4 3.Puttingtheseresultstogetherwehave 3x + (x ) (x + ) 5 9(x ) + 4 3(x ) 5 9(x + ) and the problem is solved. Exercises 3 Express the following as a sum of partial fractions 5x + 7x + 5 (x + ) (x + ) x (x 3) () x + (x ) (x + ) 5. Fractions in which the denominator has a quadratic term Sometimes we come across fractions in which the denominator has a quadratic term which cannot befactorised.wewillnowlearnhowtodealwithcaseslikethis. Example Suppose we want to express asthesumofitspartialfractions. 5x (x + x + )(x ) Notethatthetwodenominatorsofthepartialfractionswillbe (x +x+)and (x ).Whenthe denominator contains a quadratic factor we have to consider the possibility that the numerator cancontainatermin x. Thisisbecauseifitdid,thenumeratorwouldstillbeoflowerdegree thanthedenominator-thiswouldstillbeaproperfraction.sowewrite 5x (x + x + )(x ) Ax + B x + x + + C x Asbeforewemultiplybothsidesbythedenominator (x + x + )(x )togive 5x (Ax + B)(x ) + C(x + x + ) Onespecialvaluewecoulduseis x becausethiswillmakethefirsttermontheright-hand sidezeroandso Aand Bwilldisappear. www.mathcentre.ac.uk 6 c mathcentre 009
If x 0 7C andso C 0 7 Unfortunatelythereisnovaluewecansubstitutewhichwillenableustogetridof Csoinstead we use the technique of equating coefficients. We have 5x (Ax + B)(x ) + C(x + x + ) Ax Ax + Bx B + Cx + Cx + C (A + C)x + ( A + B + C)x + ( B + C) Westillneedtofind Aand B. Thereisnoterminvolving x ontheleftandsowecanstate that A + C 0 Since C 0 wehave A 0 7 7. Theleft-handsidehasnoconstanttermandso B + C 0 sothat B C Butsince C 0 7 then B 5 7.Puttingalltheseresultstogetherwehave 5x (x + x + )(x ) 0x + 5 0 7 7 x + x + + 7 x 0x + 5 7(x + x + ) + 0 7(x ) 5( ) 7(x + x + ) + 0 7(x ) Exercises 4 Express the following as a sum of partial fractions x 3x 7 (x + x + )(x ) 3 (x + 3)(x + ) x (x x + )(3x ) 6. Dealing with improper fractions Sofarwehaveonlydealtwithproperfractions,forwhichthenumeratorisoflowerdegreethan the denominator. We now look at how to deal with improper fractions. Consider the following example. Example Supposewewishtoexpress 4x3 + 0x + 4 x() in partial fractions. www.mathcentre.ac.uk 7 c mathcentre 009
Thenumeratorisofdegree3. Thedenominatorisofdegree. Sothisfractionisimproper. Thismeansthatifwearegoingtodividethenumeratorbythedenominatorwearegoingto divideatermin x 3 byonein x,whichgivesrisetoatermin x. Consequentlyweexpressthe partial fractions in the form: 4x 3 + 0x + 4 x() Ax + B + C x + D Multiplying both sides by the denominator x() gives 4x 3 + 0x + 4 Ax () + Bx() + C() + Dx Notethatbysubstitutingthespecialvalue x 0,alltermsontherightexceptthethirdwillbe zero.ifweusethespecialvalue x alltermsontherightexceptthelastonewillbezero. If x 0 4 C If x 4 8 0 + 4 D 5 + 4 D D D 3 Specialvalueswillnotgive Aor Bsoweshallhavetoequatecoefficients. 4x 3 + 0x + 4 Ax () + Bx() + C() + Dx Ax 3 + Ax + Bx + Bx + Cx + C + Dx Ax 3 + (A + B)x + (B + C + D)x + C Nowlookatthetermin x 3. A 4 sothat A Nowlookatthetermin x.thereisnosuchtermontheleft.so A + B 0 sothat A B sothat B Putting all these results together gives and the problem is solved. Exercise 5 4x 3 + 0x + 4 x() x + 4 x + 3 Expressthefollowingasasumofpowersof xandpartialfractions x 3 + x + x 4 + 3x + x + 3x + 7x x + 3 www.mathcentre.ac.uk 8 c mathcentre 009
Answers Exercise 5x + (x + )(x + 3) Exercise x + + x 3 Exercise 3 x + (x + ) + 3 x + 6 (x )(x + ) 3 x x + (x ) + (x ) + (x + ) Exercise 4 x + x + 3 x Exercise 5 x + x + x + 0 ()(x + 3) d) (3x )(6x + 9) 3 x 6 x d) 6(x ) 6(x + 4) 49(x 3) + 3 7(x 3) 49() 4 x + 3 x 3 x + x 6x + 7 + 6 x + 45 x + x + 3 7(x x + ) + 6 7(3x ) 7x + 6 x + 3 www.mathcentre.ac.uk 9 c mathcentre 009