Partial fractions - University of Sheffield

1 Partial fractionsAn 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 fractions71c mathcentreAugust7,20031.

2 IntroductionAnalgebraic fractionis a fraction in which the numerator and denominator are both polyno-mial expressions. Apolynomial expressionis one where every term is a multiple of a powerofx, 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.

3 We calltheseproper 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. Revision of adding and subtracting fractionsWe now revise the process for adding and subtracting fractions .

4 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).c mathcentreAugust7,20032We 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.

5 In calculus, for instance,or when dealing with the binomial theorem, we sometimes need to 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. 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 ?

6 By inspection of the denominator we see that the component parts must have denominators ofx 3 and 2x+ 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).This gives(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.

7 Observe that if we letx= 12the first termon the right will become zero and henceAwill disappear. If we letx= 3 the second term onthe right will become zero and henceBwill mathcentreAugust7,2003 Ifx= 12 32+5 =B 12 3 72= 72 Bfrom whichB= 1 Now we want to try to 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+ 2 so we write3x(x 1)(x+2)

8 =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= 2 because 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=2c mathcentreAugust7,20034 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)

9 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. fractions where the denominator has a repeated factorConsider the following example in which the denominator has a 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 in the Partial fractions :x 1,x+ 2 andalso 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.

10 If we letx= 1then the firstand last terms on the right will be zero andAandCwill disappear. If we letx= 2 the firstand second terms will be zero andAandBwill 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. We could usex= 0. This will give us an equation inA, we already knowBandC, this would give mathcentreAugust7,2003 But here we shall demonstrate a different technique - one calledequatingcoefficients. We takeequation 1and 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.