# is identically equal to x 2 +3x +2

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1 Partial fractions.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 +x+2 x+2 x+ We say that 4x +7 is identically equal to x 2 +x +2 x x+7 and that the partial fractions of are and. The ability to express a fraction as x 2 +x+2 x+2 x+ its partial fractions is particularly useful in the study of Laplace transforms, of z-transforms, in Control Theory and in integration. In this section we explain how partial fractions are found. Prerequisites Before starting this Section you should... Learning Outcomes After completing this Section you should be able to... be familiar with addition, subtraction, multiplication and division of algebraic fractions understand the distinction between proper and improper fractions express an algebraic fraction as the sum of its partial fractions

2 . 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 + x 2 +x +7 x 2 2x +5, x 2 + x 2 7x +2, and x x 4 +, 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. For each of the following fractions state the degrees of the numerator and denominator. Hence classify the fractions as proper or improper. a) x + x 2 +x +7, b) x2 2x +5 x 2 + x 2 7x +2, c) x x 4 +, d) s 2 +4s +5 (s 2 +2s + 4)(s +) a) The degree of the numerator, n, is. The degree of the denominator, d, is2. Because d n the fraction is improper. b) Here n =2and d =2. State whether this fraction is proper or improper. d n; the fraction is improper. c) Noting that x = x we see that n =and d =4. State whether this fraction is proper or improper. d) Find the degree of the numerator and denominator. d>n; the fraction is proper. Removing the brackets in the denominator we see that it has degree. The degree of the numerator is 2 and so this fraction is proper. HELM (VERSION : March 8, 2004): Workbook Level 0 2

3 Exercises. For each fraction state the degrees of the numerator and denominator, and hence determine which are proper and which are improper. a) x+ x, b) x 2 x x, (x )(x 2)(x ) c) x 5 Answers. a) n =,d=,improper, b) n =2,d=,proper, c) n =,d=,improper. 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,x 2 and 4 x. Quadratic factors are those of the form ax 2 + bx + c such as x 2 +, and 4x 2 2x +, 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 x +2 will produce a partial fraction of the form A where A is an unknown constant. In general a linear factor x+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. HELM (VERSION : March 8, 2004): Workbook Level 0

4 Example Express 7x+0 2x 2 +5x+ in terms of partial fractions. Solution Note that this fraction is proper. The denominator is factorised to give (2x + )(x + ). Each of the linear factors produces a partial fraction. The factor 2x + produces a partial fraction of the form A B. The factor produces a partial fraction, where A and B are constants 2x+ x+ which we now try to find. We write 70 (2x + )() = A 2x + + B By multiplying both sides by (2x + )()weobtain 70 = A()+B(2x +) () We may now let x take any value we choose. By an appropriate choice we can simplify the right-hand side. Let x = because this choice eliminates A. We find 7( )+ 0 = A(0) + B( 2 + ) = B so that the constant B must equal. The constant A can be found by substituting other values for x or alternatively by equating coefficients. Observe that, by rearranging the right-hand side, Equation () can be written as 70=(A +2B)x +(A +B) Comparing the coefficients of x on both sides we see that 7 = A +2B. Wealready know B = and so 7 = A + 2() = A +6 from which A =.Wecan therefore write 70 2x 2 +5x + = 2x + + 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. Express 9 4x x 2 x 2 in partial fractions. HELM (VERSION : March 8, 2004): Workbook Level 0 4

5 First factorise the denominator: x 2 x 2= (x + 2)(x ) Because there are two linear factors we write 9 4x x 2 x 2 = 9 4x (x + 2)(x ) = A x +2 + B x Multiply both sides by (x + 2)(x ) to obtain the equation from which we can find values for A and B. 9 4x = 9 4x = A(x ) + B(x +2) By substituting an appropriate value for x obtain B. substitute x =and get B = Finally by equating coefficients of x obtain the value of A. 4 =A +B, A = 7 since B = Finally, write down the partial fractions: 9 4x x 2 x 2 = 7 + x+2 x 5 HELM (VERSION : March 8, 2004): Workbook Level 0

6 Exercises 5x. (a) Find the partial fractions of. (x+)(x 2) (b) Check your answer by adding the partial fractions together again. 7x+25 (c) Express in partial fractions. (x+4)(x+) (d) Check your answer by adding the partial fractions together again. x+ 2. Find the partial fractions of. (x )(2x+) Express each of the following as the sum of partial fractions:. (a) (x+)(x+2), (b) 5 x 2 +7x+2, (c) (2x+)(x ), Answers 2. +, c) + 4 x+ x 2 x+4 x , x 2x+. (a), (b) 5 5, (c) 6. x+ x+2 x+ x+4 7(2x+) 7(x ). Proper fractions with repeated linear factors Sometimes a linear factor appears more than once. For example in x 2 +2 = (x + )() which equals () 2 the factor (x+)occurs twice. We call it a repeated linear factor. The repeated linear factor () 2 produces two partial fractions of the form A + B. In general, a repeated linear x+ (x+) 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 HELM (VERSION : March 8, 2004): Workbook Level 0 6

7 Once again the unknown constants are found by either equating coefficients and/or substituting specific values for x. Express 08 4x 2 +2x +9 in partial fractions. First the denominator is factorised. 4x 2 +2x +9= (2x + )(2x +)=(2x +) 2 We have found a repeated linear factor. The repeated linear factor (2x +)gives rise to two partial fractions of the form 08 (2x +) = A 2 2x + + B (2x +) 2 Multiply both sides through by (2x +) 2 to obtain the equation which must be solved to find A and B. 08=A(2x +)+B Now evaluate the constants A and B by equating coefficients. Equating coefficients of x gives 0 =2A so A =5. Equating constant terms gives 8 = A + B from which B =. So, finally, we may write 08 (2x +) = 5 2 2x + + (2x +) 2 Express the following in partial fractions. Exercises (a) x 7x 5, (b) (c) x 2 2 (x ) 2 4 x (d) 58 (x +4) 2 (e) 2x 2 (x + )(x ) 2 (f) 5x2 +2x +24 (2x + )(x +2) 2 (g) 6x2 0x +25 (x 2) 2 (x +7) (h) s +2 (s +) 2 (i) 2s + s 2. 7 HELM (VERSION : March 8, 2004): Workbook Level 0

8 Answers (a) x + 2 (b) 7 (x ) 2 x + 8 (c) (x ) 2 x (x +4) 2 (d) 5 x +4 2 (e) (x +4) 2 + x + (f) (x ) 2 (g) x 2 + (x 2) + 2 x +7 (h) s + + (s +) (i) 2 2 s + s. 2 2x + + x (x +2) 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 +. This factor produces a partial fraction of the form Ax+B. In general a quadratic factor of the form x 2 +x+ 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 Express as partial fractions (x 2 + x + 0)(x ) Note that the quadratic factor cannot be factorised further. We have (x 2 + x + 0)(x ) = Ax + B x C x Multiplying both sides by (x 2 + x + 0)(x ) gives = (Ax + B)(x ) + C(x 2 + x + 0) HELM (VERSION : March 8, 2004): Workbook Level 0 8

9 To evaluate C we can let x =which eliminates the first term on the right. This gives 4=2C so that C = Equate coefficients of x 2 and hence find A. Finally substitute any other value for x or equate coefficients of x to find B. A = B =, 7. Finally (x 2 + x + 0)(x ) = x + 7 x x = 7 x (x 2 + x + 0) + (x ) Example 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 +) 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. 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 +) = As + B s 2 +2s +4 + C s + Multiplying both sides by (s 2 +2s + 4)(s +)weobtain s 2 +4s + 5=(As + B)(s +)+C(s 2 +2s +4) To find the constant C we can let s = toeliminate A and B. Thus so that ( ) 2 +4( ) + 5 = C(( ) 2 +2( ) + 4) 2=7C and so C = 2 7 Equating coefficients of s 2 we find =A + C so that A = C = 2 7 = HELM (VERSION : March 8, 2004): Workbook Level 0

10 Solution contd. Equating constant terms gives 5 = B + 4C ( ) 2 so that B =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 +) = s s 2 +2s s + 5s +9 7(s 2 +2s +4) + 2 7(s +) Exercises Express each of the following as the sum of its partial fractions. (a) (x 2 + x + )(x 2), (b) 27x 2 4x +5 (6x 2 + x + 2)(x ), (c) 2x +4 4x 2 +2x +9, (d) 6x 2 +x +2 (x 2 +5x + )(x ) Answers (a) (x+) x+ (b) + 4 (c) + x+ (d) +. 7(x 2) 7(x 2 +x+) 6x 2 +x+2 x 2x+ (2x+) 2 x 2 +5x+ x 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 apolynomial of degree 0 is a constant, A say, apolynomial of degree has the form Ax + B, apolynomial 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, then n d =2,and we need to add a polynomial of the form Ax 2 + Bx+ C. HELM (VERSION : March 8, 2004): Workbook Level 0 0

11 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 Solution The fraction is improper because n =2,d =and so d n. Further, note that n d =. We therefore need to include an extra term: a polynomial of the form Bx + C, inaddition to the usual partial fractions. The linear term in the denominator gives rise to a partial fraction A. x+ So 2x 2 x 2 = A + Bx + C Multiplying both sides by x + wefind 2x 2 x 2=A +(Bx + C)()=Bx 2 +(C + B)x +(C + A) Equating coefficients of x 2 gives B =2. Equating coefficients of x gives =C + B and so C = B =. Equating the constant terms gives 2 =C + A and so A = 2 C = 2 ( ) =. Finally we have 2x 2 x 2 = +2x Exercises Express each of the following improper fractions in terms of partial fractions. (a) x + x 7, (b) x +2 x, (c) x2 +2x +2, (d) 2x2 +7x +7 x +2 (e) x5 +4x 4 2x 40x 2 24x 29, (f) 4x5 +8x 4 +2x +27x 2 +25x +9 (x +2) 2 (x ) (x 2 + x + )(2) Answers (a) + (b) + 2, (c) + x + (d) 2x ++, x+2 x x+ x+2 (e) + + (x+2) 2 x+2 x +x2 + x +2, (f)2x 2 + x x+ x 2 +x+ HELM (VERSION : March 8, 2004): Workbook Level 0

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