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1 Partial fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as for any value of x. x 2 +3x+2 x+2 x+1 We say that 4x +7 1 is identically equal to x 2 +3x +2 x x +1 4x+7 and that the partial fractions of are 1 and 3. The ability to express a fraction as x 2 +3x+2 x+2 x+1 its partial fractions is particularly useful in the study of Laplace transforms, of z-transforms, in Control Theory and in integration. In this Block we explain how partial fractions are found. Prerequisites Before starting this Block you should... Learning Outcomes After completing this Block you should be able to... understand the distinction between proper and improper fractions express an algebraic fraction as the sum of its partial fractions be familiar with addition, subtraction, multiplication and division of algebraic fractions Learning Style To achieve what is expected of you... allocate sufficient study time briefly revise the prerequisite material attempt every guided exercise and most of the other exercises

2 1. Proper and improper fractions Frequently we find that an algebraic fraction appears in the form algebraic fraction = numerator denominator where both numerator and denominator are polynomials. For example x 3 + x 2 +3x +7, x x 2 2x +5 x 2 7x +2, and x x 4 +1, The degree of the numerator, n say, is the highest power occurring in the numerator. The degree of the denominator, d say, is the highest power occurring in the denominator. If d > n the fraction is said to be proper; the third expression above is such an example. If d n the fraction is said to be improper; the first and second expressions above are examples of this type. Before calculating the partial fractions of an algebraic fraction it is important to decide whether the fraction is proper or improper. Now do this exercise For each of the following fractions state the degrees of the numerator and denominator. Hence classify the fractions as proper or improper. a) x3 +x 2 +3x+7, x 2 +1 b) 3x2 2x+5, Here n = 2 and d =2. x 2 7x+2 x c), Noting that x = x 4 +1 x1 we see that n = 1 and d =4. s d) 2 +4s+5 Find the degree of the numerator and denominator. (s 2 +2s+4)(s+3) More exercises for you to try 1. For each fraction state the degrees of the numerator and denominator, and hence determine which are proper and which are improper. a) x+1 x, b) x 2 x 3 x, (x 1)(x 2)(x 3) c) x 5 The denominator of an algebraic fraction can often be factorised into a product of linear and quadratic factors. Before we can separate algebraic fractions into simpler (partial) fractions we will completely factorise the denominators into linear and quadratic factors. Linear factors are those of the form ax + b; for example 2x +7,3x 2 and 4 x. Quadratic factors are those of the form ax 2 + bx + c such as x 2 + x + 1, and 4x 2 2x + 3, which cannot be factorised into linear factors (these are quadratics with complex roots). 2. Proper fractions with linear factors Firstly we describe how to calculate partial fractions for proper fractions where the denominator may be written as a product of linear factors. The steps are as follows: Factorise the denominator. Each factor will produce a partial fraction. A factor such as 3x + 2 will produce a partial Engineering Mathematics: Open Learning Unit Level 0 2

3 fraction of the form A where A is an unknown constant. In general a linear factor 3x+2 ax + b will produce a partial fraction A. The unknown constants for each partial ax+b fraction may be different and so we will call them A, B, C and so on. Evaluate the unknown constants by equating coefficients or using specific values of x. The sum of the partial fractions is identical in value to the original algebraic fraction for any value of x. Key Point A linear factor ax + b in the denominator gives rise to a single partial fraction of the form A ax + b. The steps involved are illustrated in the following example. Example Express 7x+10 2x 2 +5x+3 in terms of partial fractions. Solution Note that this fraction is proper. The denominator is factorised to give (2x + 3)(x + 1). Each of the linear factors produces a partial fraction. The factor 2x + 3 produces a partial fraction of the form A B. The factor x + 1 produces a partial fraction, where A and B are constants 2x+3 x+1 which we now try to find. We write 7x +10 (2x + 3)(x +1) = A 2x +3 + B x +1 By multiplying both sides by (2x + 3)(x + 1) we obtain 7x +10 = A(x +1)+B(2x + 3) (1) We may now let x take any value we choose. By an appropriate choice we can simplify the right-hand side. Let x = 1 because this choice eliminates A. We find 7( 1)+ 10 = A(0) + B( 2 + 3) 3 = B so that the constant B must equal 3. The constant A can be found by substituting other values for x or alternatively by equating coefficients. 3 Engineering Mathematics: Open Learning Unit Level 0

4 Solution Observe that, by rearranging the right-hand side, Equation (1) can be written as 7x +10=(A +2B)x +(A +3B) Comparing the coefficients of x on both sides we see that 7 = A +2B. We already know B =3 and so 7 = A + 2(3) = A +6 from which A = 1. We can therefore write 7x +10 2x 2 +5x +3 = 1 2x x +1 We have succeeded in expressing the given fraction as the sum of its partial fractions. The result can always be checked by adding the fractions on the right. Try each part of this exercise Express in partial fractions. 9 4x 3x 2 x 2 Part (a) First factorise the denominator: Because there are two linear factors we write 9 4x 3x 2 x 2 = 9 4x (3x + 2)(x 1) = A 3x +2 + B x 1 Part (b) Multiply both sides by (3x + 2)(x 1) to obtain the equation from which we can find values for A and B. Part (c) By substituting an appropriate value for x obtain B. Part (d) Finally by equating coefficients of x obtain the value of A. Part (e) Finally, write down the partial fractions: Engineering Mathematics: Open Learning Unit Level 0 4

5 More exercises for you to try 5x 1 1. (a) Find the partial fractions of. (x+1)(x 2) (b) Check your answer by adding the partial fractions together again. 7x+25 (c) Express in partial fractions. (x+4)(x+3) (d) Check your answer by adding the partial fractions together again. 11x+1 2. Find the partial fractions of. (x 1)(2x+1) Express each of the following as the sum of partial fractions: 3. (a) 3 (x+1)(x+2), (b) 5 x 2 +7x+12, (c) 3 (2x+1)(x 3), 3. Proper fractions with repeated linear factors Sometimes a linear factor appears more than once. For example in 1 x 2 +2x +1 = 1 (x + 1)(x +1) which equals 1 (x +1) 2 the factor (x+1) occurs twice. We call it a repeated linear factor. The repeated linear factor (x +1) 2 produces two partial fractions of the form A + B. In general, a repeated linear x+1 (x+1) 2 factor of the form (ax + b) 2 generates two partial fractions of the form A ax + b + B (ax + b) 2 This is reasonable since the sum of two such fractions always gives rise to a proper fraction: A ax + b + B A(ax + b) = (ax + b) 2 (ax + b) + B 2 (ax + b) = x(aa)+ab + B 2 (ax + b) 2 Key Point A repeated linear factor (ax + b) 2 in the denominator produces two partial fractions: A ax + b + B (ax + b) 2 Once again the unknown constants are found by either equating coefficients and/or substituting specific values for x. Try each part of this exercise 10x +18 Express in partial fractions. 4x 2 +12x +9 5 Engineering Mathematics: Open Learning Unit Level 0

6 Part (a) First the denominator is factorised. We have found a repeated linear factor. The repeated linear factor (2x + 3) gives rise to two partial fractions of the form 10x +18 (2x +3) = A 2 2x +3 + B (2x +3) 2 Part (b) Multiply both sides through by (2x +3) 2 to obtain the equation which must be solved to find A and B. Part (c) Now evaluate the constants A and B by equating coefficients. Equating coefficients of x gives Equating constant terms gives 18 = 3A + B from which B = 3. So, finally, we may write More exercises for you to try Express the following in partial fractions. (a) 3 x 7x 15, (b) (c) x 2 2x +1 (x 1) 2 (f) 5x2 +23x +24 (2x + 3)(x +2) 2 (g) 6x2 30x +25 (3x 2) 2 (x +7) 10x +18 (2x +3) 2 = 5 2x (2x +3) 2 3x +14 x 2 +8x +16 (h) (d) s +2 (s +1) 2 (i) 5x +18 (x +4) 2 (e) 2s +3 s 2. 2x 2 x +1 (x + 1)(x 1) 2 4. Proper fractions with quadratic factors Sometimes a denominator is factorised producing a quadratic term which cannot be factorised into linear factors. One such quadratic factor is x 2 + x + 1. This factor produces a partial fraction of the form Ax+B. In general a quadratic factor of the form x 2 +x+1 ax2 + bx + c produces a single partial fraction of the form Ax+B ax 2 +bx+c. Key Point A quadratic factor ax 2 + bx + c in the denominator produces a partial fraction of the form Ax + B ax 2 + bx + c Engineering Mathematics: Open Learning Unit Level 0 6

7 Try each part of this exercise 3x +1 Express as partial fractions (x 2 + x + 10)(x 1) Note that the quadratic factor cannot be factorised further. We have 3x +1 (x 2 + x + 10)(x 1) = Ax + B x 2 + x C x 1 Part (a) Multiply both sides by (x 2 + x + 10)(x 1) To evaluate C we can let x = 1 which eliminates the first term on the right. This gives 4=12C so that C = 1 3 Equate coefficients of x 2 and hence find A. Part (b) Using the result that A = 1/3 substitute any other value for x or equate coefficients of x to find B. Finally 3x +1 (x 2 + x + 10)(x 1) = 1x x 2 + x x 1 = 7 x 3(x 2 + x + 10) + 1 3(x 1) Example Admittance Admittance, Y, is a quantity which is used in analysing electronic circuits. A typical expression for admittance is Y (s) = s 2 +4s +5 (s 2 +2s + 4)(s +3) where s can be thought of as representing frequency. To predict the behaviour of the circuit it is often necessary to express the admittance as the sum of its partial fractions and find the effect of each part separately. Express Y (s) in partial fractions. 7 Engineering Mathematics: Open Learning Unit Level 0

8 Solution The fraction is proper. The denominator contains a quadratic factor which cannot be factorised further, and also a linear factor. Thus s 2 +4s +5 (s 2 +2s + 4)(s +3) = As + B s 2 +2s +4 + C s +3 Multiplying both sides by (s 2 +2s + 4)(s + 3) we obtain s 2 +4s +5=(As + B)(s +3)+C(s 2 +2s +4) To find the constant C we can let s = 3 to eliminate A and B. Thus so that Equating coefficients of s 2 we find so that A =1 C =1 2 7 = 5 7. ( 3) 2 +4( 3) + 5 = C(( 3) 2 +2( 3) + 4) 2=7C and so C = 2 7 1=A + C Equating constant terms gives 5 = 3B + 4C ( ) 2 so that 3B =5 4C =5 4 = so B = 9 7 Finally Y (s) = which can be written as Y (s) = s s +5 (s 2 +2s + 4)(s +3) = s s 2 +2s s +3 5s +9 7(s 2 +2s +4) + 2 7(s +3) More exercises for you to try Express each of the following as the sum of its partial fractions. (a) 3 (x 2 + x + 1)(x 2), (b) 27x 2 4x +5 (6x 2 + x + 2)(x 3), (c) 2x +4 4x 2 +12x +9, (d) 6x 2 +13x +2 (x 2 +5x + 1)(x 1) Engineering Mathematics: Open Learning Unit Level 0 8

9 5. Improper fractions When calculating the partial fractions of improper fractions an extra polynomial is added to any partial fractions that would normally arise. The added polynomial has degree n d where d is the degree of the denominator and n is the degree of the numerator. Recall that a polynomial of degree 0 is a constant, A say, a polynomial of degree 1 has the form Ax + B, a polynomial of degree 2 has the form Ax 2 + Bx + C, and so on. If, for example, the improper fraction is such that the numerator has degree 5 and the denominator has degree 3, then n d = 2, and we need to add a polynomial of the form Ax 2 + Bx+ C. Key Point If a fraction is improper an additional term is included taking the form of a polynomial of degree n d, where n is the degree of the numerator and d is the degree of the denominator. Example Express as partial fractions 2x 2 x 2 x +1 Solution The fraction is improper because n =2,d = 1 and so d n. Further, note that n d =1. We therefore need to include an extra term: a polynomial of the form Bx + C, in addition to the usual partial fractions. The linear term in the denominator gives rise to a partial fraction A. x+1 So 2x 2 x 2 = A x +1 x +1 + Bx + C Multiplying both sides by x + 1 we find 2x 2 x 2=A +(Bx + C)(x +1)=Bx 2 +(C + B)x +(C + A) Equating coefficients of x 2 gives B =2. Equating coefficients of x gives 1 =C + B and so C = 1 B = 3. Equating the constant terms gives 2 =C + A and so A = 2 C = 2 ( 3)=1. Finally we have 2x 2 x 2 x +1 = 1 x +1 +2x 3 9 Engineering Mathematics: Open Learning Unit Level 0

10 More exercises for you to try Express each of the following improper fractions in terms of partial fractions. (a) x +3 3x 7, (b) x +2 x 3, (c) x2 +2x +2 x +1, (d) 2x2 +7x +7 x +2 (e) 3x5 +4x 4 21x 3 40x 2 24x 29, (f) 4x5 +8x 4 +23x 3 +27x 2 +25x +9 (x +2) 2 (x 3) (x 2 + x + 1)(2x +1) Engineering Mathematics: Open Learning Unit Level 0 10

11 6. Computer Exercise or Activity For this exercise it will be necessary for you to access the computer package DERIVE. DERIVE can be used to obtain the partial fraction expansion of both proper and improper fractions. For example to find the partial fraction version of (an improper fraction) x 3 3x 2 +1 (x 1)(x +2) simply key Author:Expression (x 3 3x 2+1)/((x 1)(x + 2)). DERIVE responds x 3 3 x 2 +1 (x 1) (x +2) Now hit Simplify:Expand to obtain the Expand Expression box. Choose x as the Expansion Variable and Rational in the Amount box. Hit the Expand button and DERIVE responds 19 3 (x +2) 1 3 (x 1) + x 4 It would be a useful exercise to check the partial fraction expansions of the many algebraic fractions presented in this Block. 11 Engineering Mathematics: Open Learning Unit Level 0

12 End of Block 3.6 Engineering Mathematics: Open Learning Unit Level 0 12

13 (a) The degree of the numerator, n, is 3. The degree of the denominator, d, is 2. Because d n the fraction is improper. (b) d n; the fraction is improper. (c) d>n; the fraction is proper. (d) Removing the brackets in the denominator we see that it has degree 3. The degree of the numerator is 2 and so this fraction is proper. 13 Engineering Mathematics: Open Learning Unit Level 0

14 1. a) n =1,d= 1, improper, b) n =2,d= 3, proper, c) n =3,d= 1, improper. Engineering Mathematics: Open Learning Unit Level 0 14

15 (3x + 2)(x 1) 15 Engineering Mathematics: Open Learning Unit Level 0

16 9 4x = A(x 1) + B(3x +2) Engineering Mathematics: Open Learning Unit Level 0 16

17 substitute x = 1 and get B =1 17 Engineering Mathematics: Open Learning Unit Level 0

18 4 =A +3B, A = 7 since B =1 Engineering Mathematics: Open Learning Unit Level 0 18

19 7 3x x 1 19 Engineering Mathematics: Open Learning Unit Level 0

20 x x 2, c) 3 x x +3 4 x x +1, 3. (a) 3 x +1 3 x +2, (b) 5 x +3 5 x +4, (c) 6 7(2x +1) 3 7(x 3). Engineering Mathematics: Open Learning Unit Level 0 20

21 (2x + 3)(2x +3)=(2x +3) 2 21 Engineering Mathematics: Open Learning Unit Level 0

22 10x +18=A(2x +3)+B Engineering Mathematics: Open Learning Unit Level 0 22

23 10=2A so A =5. 23 Engineering Mathematics: Open Learning Unit Level 0

24 (a) 1 x (x 1) 2 (b) 7 x (x 1) 2 (c) (d) 5 x +4 2 (x +4) 2 (e) (g) 1 3x (3x 2) x +7 1 x x (x 1) 2 (f) (h) 3 x (x +4) 2 1 s (i) 2 (s +1) 2 s + 3 s x x (x +2) 2 Engineering Mathematics: Open Learning Unit Level 0 24

25 3x +1=(Ax + B)(x 1) + C(x 2 + x + 10) 25 Engineering Mathematics: Open Learning Unit Level 0

26 7 3. Engineering Mathematics: Open Learning Unit Level 0 26

27 (a) (c) 3 3(x +3) 7(x 2) 7(x 2 + x +1) 1 2x (2x +3) 2 (d) (b) 3x +1 6x 2 + x x 3 3x +1 x 2 +5x x Engineering Mathematics: Open Learning Unit Level 0

28 (a) x +2 (b) x 3, (c) 1 + x + 1 x +1 (e) (d) 2x x +2, 1 (x +2) x x 3 +3x2 + x +2, (f)2x 2 + x x x 2 + x +1 Engineering Mathematics: Open Learning Unit Level 0 28

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