# 3.6. Partial Fractions. Introduction. Prerequisites. Learning Outcomes

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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. For 4x + 7 example it can be shown that x 2 + 3x + 2 has the same value as x for any value of x. We say that 4x + 7 x 2 + 3x + 2 and that the partial fractions of is identically equal to x x + 7 x 2 + 3x + 2 are x + 2 and 3. The ability to express a fraction as its partial fractions is particularly useful in the study of Laplace transforms, Z-transforms, Control Theory and Integration. In this Section we explain how partial fractions are found. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... be familiar with addition, subtraction, multiplication and division of algebraic fractions distinguish between proper and improper fractions express an algebraic fraction as the sum of its partial fractions 60 HELM (2008): Workbook 3: Equations, Inequalities & 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 3 + x 2 + 3x + 7, x 2 + 3x 2 2x + 5 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. Task For each of the following fractions state the degree of the numerator (= n) and the degree of the denominator (= d). Hence classify the fractions as proper or improper. (a) x3 + x 2 + 3x + 7, (b) 3x2 2x + 5 x 2 + x 2 7x + 2, (c) x x 4 +, (d) s 2 + 4s + 5 (s 2 + 2s + 4)(s + 3) (a) Find the degree of denominator and numerator and hence classify (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) Here n = 2 and d = 2. State whether (b) is proper or improper: d n; the fraction is improper. (c) Noting that x = x, state whether (c) is proper or improper: d > n; the fraction is proper. HELM (2008): Section 3.6: Partial Fractions 6

3 (d) Find the degree of the numerator and denominator of (d): Removing the brackets in the denominator we see that d = 3. s n = 2 this fraction is proper. Exercise For each fraction state the degrees of the numerator and denominator, and hence determine which are proper and which are improper. (a) x, (b) x 2 (x )(x 2)(x 3), (c) x 3 x x 5 s (a) n =, d =, improper, (b) n = 2, d = 3, proper, (c) n = 3, d =, improper. The denominator of an algebraic fraction can often be factorised into a product of linear and/or quadratic factors. Before we can separate algebraic fractions into simpler (partial) fractions we need to 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. Irreducible quadratic factors are those of the form ax 2 + bx + c such as x 2 +, 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. factor such as 3x + 2 will produce a partial fraction of the form where is an unknown constant. In general a linear factor 3x + 2 ax + b will produce a partial fraction. The unknown constants for each partial ax + b fraction may be different and so we will call them, 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 to the original algebraic fraction for all values of x. Key Point 4 linear factor ax + b in the denominator gives rise to a single partial fraction of the form ax + b 62 HELM (2008): Workbook 3: Equations, Inequalities & Partial Fractions

4 The steps involved in expressing a proper fraction as partial fractions are illustrated in the following Example. Example 4 70 Express in terms of partial fractions. 2x 2 + 5x + 3 Solution Note that this fraction is proper. The denominator is factorised to give (2x + 3)(). Each of the linear factors produces a partial fraction. The factor 2x + 3 produces a partial fraction of the form B and the factor produces a partial fraction, where and B are constants 2x + 3 which we need to find. We write 70 (2x + 3)() = 2x B By multiplying both sides by (2x + 3)() we obtain 70 = () + B(2x + 3)... (*) 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. We find 7( ) + 0 = (0) + B( 2 + 3) 3 = B so that the constant B must equal 3. The constant can be found either by substituting some other value for x or alternatively by equating coefficients. Observe that, by rearranging the right-hand side, Equation (*) can be written as 70 = ( + 2B)x + ( + 3B) Comparing the coefficients of x on both sides we see that 7 = + 2B. We already know B = 3 and so 7 = + 2(3) = + 6 from which =. We can therefore write 70 2x 2 + 5x + 3 = 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. HELM (2008): Section 3.6: Partial Fractions 63

5 Task Express 9 4x 3x 2 x 2 in partial fractions. First factorise the denominator: 3x 2 x 2 = (3x + 2)(x ) Because there are two linear factors we write 9 4x 3x 2 x 2 = 3x B x Multiply both sides by (3x + 2)(x ) to obtain the equation from which to find and B: 9 4x = 9 4x = (x ) + B(3x + 2) Substitute an appropriate value for x to obtain B: Substitute x = and get B = Equating coefficients of x to obtain the value of : 4 = + 3B, = 7 since B = Finally, write down the partial fractions: 9 4x 3x 2 x 2 = 7 3x x 64 HELM (2008): Workbook 3: Equations, Inequalities & Partial Fractions

6 . Find the partial fractions of (a) Exercises Check by adding the partial fractions together again. 2. Express each of the following as the sum of partial fractions: (a) 5x ()(x 2), (b) 7x + 25 (x + 4)(x + 3), (c) (x )(2). 3 ()(x + 2), (b) 5 x , (c) 3 (2)(x 3), s 2 (a) + 3 x 2, (b) 3 x x + 3 2(a) (c) 4 x + 3 2, 3 3 x + 2, 2(b) 5 x x + 4, 2(c) 6 7(2) 3 7(x 3). 3. Proper fractions with repeated linear factors Sometimes a linear factor appears more than once. For example in x = ()() which equals () 2 the factor () occurs twice. We call it a repeated linear factor. The repeated linear factor () 2 produces two partial fractions of the form + B. In general, a repeated linear () 2 factor of the form (ax + b) 2 generates two partial fractions of the form ax + b + B (ax + b) 2 This is reasonable since the sum of two such fractions always gives rise to a proper fraction: ax + b + B (ax + b) = (ax + b) 2 (ax + b) + B x(a) + b + B = 2 (ax + b) 2 (ax + b) 2 Key Point 5 repeated linear factor (ax + b) 2 in the denominator produces two partial fractions: ax + b + B (ax + b) 2 Once again the unknown constants are found by either equating coefficients and/or substituting specific values for x. HELM (2008): Section 3.6: Partial Fractions 65

7 Task Express 08 4x 2 + 2x + 9 in partial fractions. First factorise the denominator: 4x 2 + 2x + 9 = (2x + 3)(2x + 3) = (2x + 3) 2 There is a repeated linear factor (2x + 3) which gives rise to two partial fractions of the form 08 (2x + 3) = 2 2x B (2x + 3) 2 Multiply both sides through by (2x + 3) 2 to obtain the equation to be solved to find and B: 08 = (2x + 3) + B Now evaluate the constants and B by equating coefficients: Equating the x coefficients gives 0 = 2 so = 5. Equating constant terms gives 8 = 3 + B from which B = 3. Finally express the answer in partial fractions: 08 (2x + 3) 2 = 5 2x (2x + 3) 2 66 HELM (2008): Workbook 3: Equations, Inequalities & Partial Fractions

8 Express the following in partial fractions. (a) (d) Exercises 3 x 7x 5, (b) (c) x 2 2 (x ) 2 58 (x + 4) 2 (e) 34 x x 2 (f) 5x2 + 23x + 24 ()(x ) 2 (2x + 3)(x + 2) 2 (g) 6x2 30x + 25 (3x 2) 2 (x + 7) (h) s + 2 (s + ) 2 (i) 2s + 3 s 2. s (a) x + 2 (x ) 2 (b) 7 x + 8 (x ) 2 (c) (d) 5 x (x + 4) 2 (e) + x + (f) (x ) 2 3 x (x + 4) 2 3 2x x (x + 2) 2 (g) 3x 2 + (3x 2) 2 + x + 7 (h) s + + (i) 2 (s + ) 2 s + 3 s Proper fractions with quadratic factors Sometimes when a denominator is factorised it produces 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 x + B x 2 +. In general a quadratic factor of the form ax2 + bx + c produces a single partial fraction of the form x + B ax 2 + bx + c. Key Point 6 quadratic factor ax 2 + bx + c in the denominator produces a partial fraction of the form x + B ax 2 + bx + c HELM (2008): Section 3.6: Partial Fractions 67

9 Task Express as partial fractions 3 (x 2 + 0)(x ) Note that the quadratic factor cannot be factorised further. We have 3 (x 2 + 0)(x ) = x + B x C x First multiply both sides by (x 2 + 0)(x ): 3 = (x + B)(x ) + C(x 2 + 0) Evaluate C by letting x = : 4 = 2C so that C = 3 Equate coefficients of x 2 and hence find, and then substitute any other value for x (or equate coefficients of x) to find B: 3, 7 3. = B = Finally express in partial fractions: 3 (x 2 + 0)(x ) = x x x = 7 x 3(x 2 + 0) + 3(x ) 68 HELM (2008): Workbook 3: Equations, Inequalities & Partial Fractions

10 Engineering Example 3 dmittance dmittance, Y, is a quantity which is used in analysing electronic circuits. 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. The fraction is proper. The denominator contains an irreducible quadratic factor, which cannot be factorised further, and also a linear factor. Thus s 2 + 4s + 5 (s 2 + 2s + 4)(s + 3) = s + B s 2 + 2s C s + 3 Multiplying both sides of Equation () by (s 2 + 2s + 4)(s + 3) we obtain s 2 + 4s + 5 = (s + B)(s + 3) + C(s 2 + 2s + 4) (2) To find the constant C we let s = 3 in Equation (2) to eliminate and B. Thus so that ( 3) 2 + 4( 3) + 5 = C(( 3) 2 + 2( 3) + 4) () 2 = 7C and so C = 2 7 Equating coefficients of s 2 in Equation (2) we find = + C so that = C = 2 7 = 5 7. Equating constant terms in Equation (2) gives ( ) 2 so that 3B = 5 4C = = 3B + 4C = 27 7 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) HELM (2008): Section 3.6: Partial Fractions 69

11 Exercise Express each of the following as the sum of its partial fractions. (a) 3 (x 2 + )(x 2), (b) 27x 2 4x + 5 (6x 2 + x + 2)(x 3), (c) 2x + 4 4x 2 + 2x + 9, (d) 6x 2 + 3x + 2 (x 2 + 5)(x ) s 3 3(x + 3) (a) 7(x 2) 7(x 2 + ) (d) 3 x x. (b) 3 6x 2 + x x 3 (c) 2x (2x + 3) 2 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 n is the degree of the numerator and d is the degree of the denominator. Recall that a polynomial of degree 0 is a constant, say, a polynomial of degree has the form x + B, a polynomial of degree 2 has the form x 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 x 2 + Bx + C. Key Point 7 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. 70 HELM (2008): Workbook 3: Equations, Inequalities & Partial Fractions

12 Example 42 Express as partial fractions 2x 2 x 2 Solution The fraction is improper because n = 2, d = and so d n. Here n d =, so we need to include as 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. So altogther we have 2x 2 x 2 = + (Bx + C) Multiplying both sides by we find 2x 2 x 2 = + (Bx + C)() = Bx 2 + (C + B)x + (C + ) Equating coefficients of x 2 gives B = 2. Equating coefficients of x gives = C + B and so C = B = 3. Equating the constant terms gives 2 = C + and so = 2 C = 2 ( 3) =. Finally, we have 2x 2 x 2 = + 2x 3 Exercise 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, (d) 2x2 + 7x + 7 x + 2 (e) 3x5 + 4x 4 2x 3 40x 2 24x 29, (f) 4x5 + 8x x x x + 9 (x + 2) 2 (x 3) (x 2 + )(2) s (a) + x + 2 (b) x 3, (c) + (e) (d) 2x x + 2, (x + 2) + 2 x x 3 + 3x2 + x + 2, (f) 2x 2 + x x 2 + HELM (2008): Section 3.6: Partial Fractions 7

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