Partial fractions - mathcentre.ac.uk

1 Partial fractionsmc-TY-partialfractions-2009-1An algebraic fraction such as3x+ 52x2 5x 3can often be broken down into simpler parts calledpartial fractions . Specifically3x+ 52x2 5x 3=2x 3 12x+ 1In this unit we explain how this process is carried order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: explain the meaning of the terms proper fraction and improper fraction express an algebraic fraction as the sum of its Partial of adding and subtracting a fraction as the sum of its Partial where the denominator has a repeated in which the denominator has a quadratic with improper mathcentre 20091. IntroductionAnalgebraic fractionis a fraction in which the numerator and denominator are bothpolynomialexpressions.

2 Apolynomial expressionis one where every term is a multiple of a power ofx,such as5x4+ 6x3+ 7x+ 4 Thedegreeof a polynomial is the power of the highest term inx. So in this case the degree number in front ofxin each term is called itscoefficient. So, the coefficient ofx4is coefficient ofx3is consider the following algebraic fractions :xx2+ 2x3+ 3x4+x2+ 1In both cases the numerator is a polynomial of lower degree than the denominator. We call theseproper fractionsWith other fractions the polynomial may be of higher degree in the numerator or it may be ofthe same degree, for examplex4+x2+xx3+x+ 2x+ 4x+ 3and these are calledimproper PointIf the degree of the numerator is less than the degree of the denominator the fraction is said tobe aproper fractionIf the degree of the numerator is greater than or equal to the degree of the denominator thefraction is said to be animproper fraction2.

3 Revision of adding and subtracting fractionsWe now revise the process for adding and subtracting fractions . Consider2x 3 12x+ 1In order to add these two fractions together, we need to find the lowest common this particular case, it is(x 3)(2x+ 1). mathcentre 2009We write each fraction with this 3=2(2x+ 1)(x 3)(2x+ 1)and12x+ 1=x 3(x 3)(2x+ 1)So2x 3 12x+ 1=2(2x+ 1)(x 3)(2x+ 1) x 3(x 3)(2x+ 1)The denominators are now the same so we can simply subtract the numerators and divide theresult by the lowest common denominator to give2x 3 12x+ 1=4x+ 2 x+ 3(x 3)(2x+ 1)=3x+ 5(x 3)(2x+ 1)Sometimes in mathematics we need to do this operation in reverse. In calculus, for instance,or when dealing with the binomial theorem, we sometimes needto split a fraction up into itscomponent parts which are calledpartial fractions . We discuss how to do this in the 1 Use the rules for the addition and subtraction of fractions to simplifya)3x+ 1+2x+ 3b)5x 2 3x+ 2c)42x+ 1 2x+ 3d)13x 1 26x+ 93.

4 Expressing a fraction as the sum of its Partial fractionsIn the previous section we saw that2x 3 12x+ 1=3x+ 5(x 3)(2x+ 1)Suppose we start with3x+ 5(x 3)(2x+ 1). How can we get this back to its component parts ?By inspection of the denominator we see that the component parts must have denominators ofx 3and2x+ 1so we can write3x+ 5(x 3)(2x+ 1)=Ax 3+B2x+ 1whereAandBare involvexor powers ofxbecause otherwise theterms on the right would be improper next thing to do is to multiply both sides by the common denominator(x 3)(2x+ 1). Thisgives(3x+ 5)(x 3)(2x+ 1)(x 3)(2x+ 1)=A(x 3)(2x+ 1)x 3+B(x 3)(2x+ 1)2x+ 1 Then cancelling the common factors from the numerators and denominators of each term gives3x+ 5 =A(2x+ 1) +B(x 3)Now this is anidentity. This means that it is true for any values ofx, and because of this wecan substitute any values ofxwe choose into it. Observe that if we letx= 12the first termon the right will become zero and henceAwill disappear.

5 If we letx= 3the second term onthe right will become zero and henceBwill mathcentre 2009 Ifx= 12 32+ 5 =B( 12 3)72= 72 Bfrom whichB= 1 Now we want to try to 314 = 7 Aso thatA= these results together we have3x+ 5(x 3)(2x+ 1)=Ax 3+B2x+ 1=2x 3 12x+ 1which is the sum that we started with, and we have now broken the fraction back into itscomponent parts calledpartial we want to express3x(x 1)(x+ 2)as the sum of its Partial that the factors in the denominator arex 1andx+ 2so we write3x(x 1)(x+ 2)=Ax 1+Bx+ 2whereAandBare multiply both sides by the common denominator(x 1)(x+ 2):3x=A(x+ 2) +B(x 1)This time the special values that we shall choose arex= 2because then the first term on theright will become zero andAwill disappear, andx= 1because then the second term on theright will become zero andBwill 2 6 = 3BB= 6 3B= mathcentre 2009 Ifx= 13 = 3AA= 1 Putting these results together we have3x(x 1)(x+ 2)=1x 1+2x+ 2and we have expressed the given fraction in Partial the denominator is more awkward as we shall see in the following 2 Express the following as a sum of Partial fractionsa)2x 1(x+ 2)(x 3)b)2x+ 5(x 2)(x+ 1)c)3(x 1)(2x 1)d)1(x+ 4)(x 2)4.

6 fractions where the denominator has a repeated factorConsider the following example in which the denominator hasa repeated factor(x 1) we want to express3x+ 1(x 1)2(x+ 2)as the sum of its Partial are actually three possibilities for a denominator inthe Partial fractions :x 1,x+ 2andalso the possibility of(x 1)2, so in this case we write3x+ 1(x 1)2(x+ 2)=A(x 1)+B(x 1)2+C(x+ 2)whereA,BandCare before we multiply both sides by the denominator(x 1)2(x+ 2)to give3x+ 1 =A(x 1)(x+ 2) +B(x+ 2) +C(x 1)2(1)Again we look for special values to substitute into this identity. If we letx= 1then the first andlast terms on the right will be zero andAandCwill disappear. If we letx= 2the first andsecond terms will be zero andAandBwill 14 = 3 Bso thatB=43 Ifx= 2 5 = 9 Cso thatC= 59We now need to findA. There is no special value ofxthat will eliminateBandCto give usA. We could use any value.

7 We could usex= 0. This will give us an equation inA, we already knowBandC, this would give mathcentre 2009 But here we shall demonstrate a different technique - one calledequating coefficients. We takeequation 1 and multiply-out the right-hand side, and then collect up like + 1 =A(x 1)(x+ 2) +B(x+ 2) +C(x 1)2=A(x2+x 2) +B(x+ 2) +C(x2 2x+ 1)= (A+C)x2+ (A+B 2C)x+ ( 2A+ 2B+C)This is an identity which is true for all values ofx. On the left-hand side there are no termsinvolvingx2whereas on the right we have(A+C)x2. The only way this can be true is ifA+C= 0 This is calledequating coefficientsofx2. We already know thatC= 59so this means thatA=59. We also already know thatB=43. Putting these results together we have3x+ 1(x 1)2(x+ 2)=59(x 1)+43(x 1)2 59(x+ 2)and the problem is 3 Express the following as a sum of Partial fractionsa)5x2+ 17x+ 15(x+ 2)2(x+ 1)b)x(x 3)2(2x+ 1)c)x2+ 1(x 1)2(x+ 1)5.

8 fractions in which the denominator has a quadratic termSometimes we come across fractions in which the denominatorhas a quadratic term which cannotbe factorised. We will now learn how to deal with cases like we want to express5x(x2+x+ 1)(x 2)as the sum of its Partial that the two denominators of the Partial fractions willbe(x2+x+1)and(x 2). When thedenominator contains a quadratic factor we have to considerthe possibility that the numeratorcan contain a term inx. This is because if it did, the numerator would still be of lower degreethan the denominator - this would still be a proper we write5x(x2+x+ 1)(x 2)=Ax+Bx2+x+ 1+Cx 2As before we multiply both sides by the denominator(x2+x+ 1)(x 2)to give5x= (Ax+B)(x 2) +C(x2+x+ 1)One special value we could use isx= 2because this will make the first term on the right-handside zero and soAandBwill mathcentre 2009 Ifx= 210 = 7 Cand soC=107 Unfortunately there is no value we can substitute which willenable us to get rid ofCso insteadwe use the technique of equating coefficients.

9 We have5x= (Ax+B)(x 2) +C(x2+x+ 1)=Ax2 2Ax+Bx 2B+Cx2+Cx+C= (A+C)x2+ ( 2A+B+C)x+ ( 2B+C)We still need to findAandB. There is no term involvingx2on the left and so we can statethatA+C= 0 SinceC=107we haveA= left-hand side has no constant term and so 2B+C= 0so thatB=C2 But sinceC=107thenB=57. Putting all these results together we have5x(x2+x+ 1)(x 2)= 107x+57x2+x+ 1+107x 2= 10x+ 57(x2+x+ 1)+107(x 2)=5( 2x+ 1)7(x2+x+ 1)+107(x 2)Exercises 4 Express the following as a sum of Partial fractionsa)x2 3x 7(x2+x+ 2)(2x 1)b)13(2x+ 3)(x2+ 1)c)x(x2 x+ 1)(3x 2)6. Dealing with improper fractionsSo far we have only dealt with proper fractions , for which thenumerator is of lower degree thanthe denominator. We now look at how to deal with improper the following we wish to express4x3+ 10x+ 4x(2x+ 1)in Partial mathcentre 2009 The numerator is of degree 3. The denominator is of degree 2.

10 So this fraction is means that if we are going to divide the numerator by the denominator we are going todivide a term inx3by one inx2, which gives rise to a term inx. Consequently we express thepartial fractions in the form:4x3+ 10x+ 4x(2x+ 1)=Ax+B+Cx+D2x+ 1 Multiplying both sides by the denominatorx(2x+ 1)gives4x3+ 10x+ 4 =Ax2(2x+ 1) +Bx(2x+ 1) +C(2x+ 1) +DxNote that by substituting the special valuex= 0, all terms on the right except the third will bezero. If we use the special valuex= 12all terms on the right except the last one will be 04 =CIfx= 12 48 102+ 4 = 12D 12 5 + 4 = 12D 112= 12DD= 3 Special values will not giveAorBso we shall have to equate + 10x+ 4 =Ax2(2x+ 1) +Bx(2x+ 1) +C(2x+ 1) +Dx= 2Ax3+Ax2+ 2Bx2+Bx+ 2Cx+C+Dx= 2Ax3+ (A+ 2B)x2+ (B+ 2C+D)x+CNow look at the term 4so thatA= 2 Now look at the term inx2. There is no such term on the left. SoA+ 2B= 0so thatA= 2 Bso thatB= 22= 1 Putting all these results together gives4x3+ 10x+ 4x(2x+ 1)= 2x 1 +4x+32x+ 1and the problem is 5 Express the following as a sum of powers ofxand Partial fractionsa)x3+ 1x2+ 1b)2x4+ 3x2+ 1x2+ 3x+ 2c)7x2 1x+ mathcentre 2009 AnswersExercise 1a)5x+ 11(x+ 1)(x+ 3)b)2x+ 16(x 2)(x+ 2)c)10(2x+ 1)(x+ 3)d)11(3x 1)(6x+ 9)Exercise 2a)1x+ 2+1x 3b)3x 2 1x+ 1c)3x 1 62x 1d)16(x 2) 16(x+ 4)Exercise 3a)2x+ 2 1(x+ 2)2+3x+ 1b)149(x 3)+37(x 3)2 249(2x+ 1)c)12(x 1)+1(x 1)2+12(x+ 1)Exercise 4a)2x+ 1x2+x+ 2 32x 1b)42x+ 3 2x 3x2+ 1c) 2x+ 37(x2 x+ 1)+67(3x 2)Exercise 5a)x+ x+ 1x2+ 1b)2x2 6x+ 17 +6x+ 1 45x+ 2c)7x 21 +62x+ mathcentre 2009