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Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiPage 1 of 23M.K. HOME TUITIONMathematics Revision GuidesLevel: AS / A LevelAQA : C4Edexcel: C4OCR: C4OCR MEI: C4PARTIAL FRACTIONSVersion : 3.2Date: 12-04-2013Examples 1, 3, 5, 6 and 8 are copyrighted to their respective owners and used with their permission.
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiPage 2 of 23Partial Fractions.The section on rational expressions included methods of addition and subtraction of algebraic fractions.Thus an expression like32can be converted to a single fraction as follows: x 2 x 13( x 1)2( x 2)3x 3 2 x 4x 7, toand finally .( x 2) ( x 1) ( x 2) ( x 1)( x 2) ( x 1)( x 2) ( x 1)Sometimes it might be useful to carry out the process in reverse, where we would convert anexpression likeABx 7into the form. This process is known as x 2 x 1( x 2) ( x 1)decomposing the expression into partial fractions.ExpandingABA( x 1) B( x 2)gives. x 2 x 1( x 2) ( x 1)To find the values of A and B, we substitute convenient values for x to make each term on the top lineequal to zero.Let x -1 (to make A(x 1) 0) , then x 7 B(x-2), or 6 -3B B -2.Let x 2, (to make B(x-2) 0), then x 7 A(x 1), or 9 3A A 3.Hence32x 7 . ( x 2) ( x 1) x 2 x 1This method is also often illustrated by the cover-up rule. Here we cover up each term in thedenominator in turn and substitute the value of x that makes the covered-up term equal to zero.x 7Covering up (x – 2) and substituting x 2 gives A 93 or 3.(.) ( x 1)x 7Covering up (x 1) and substituting x -1 gives B 63 or -2.( x 2) (.)It must be stressed that the substitution method and the cover-up rule can only be used exclusivelywhen we have distinct linear factors in the denominator. In fact, when there are repeated or quadraticfactors in the denominator, the cover-up method is ‘dodgy’ and should not be used (see later) !Another method of finding A and B is to equate the coefficients.Inspection of the numerators in the expanded expressionA( x 1) B( x 2)( x 2) ( x 1)gives A(x 1) B(x-2) x 7.From this, we can solve the simultaneous equations:A B 1 (equating x-coefficients)A – 2B 7 (equating constants).Subtracting the second equation from the first gives 3B -6 and thus B -2.Substituting in the first equation gives A – 2 1 and thus A 3.
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiPage 3 of 23Most examination questions on partial fractions cover the following cases:Two distinct linear factors in the denominatorThree distinct linear factors in the denominatorThree linear factors in the denominator, but one repeatedTwo distinct linear factors in the denominator.This case was covered in the previous example – here are two others.Example (1): Express4x 9in partial fractions.( x 2) ( x 3)(Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 9780199142590, Exercise 18A, Q.1 )The partial fraction will be in the form(Long working)Expand the partial fraction toAB. x 2 x 3A( x 3) B( x 2), and substitute values for x to make each term in( x 2) ( x 3)the top line zero in turn.When x 3, the term in A becomes zero, and thus 4x - 9 B(x-2), or 3 B B 3.When x 2, the term in B becomes zero, and thus 4x – 9 A(x-3), or -1 -1(A) A 1.Hence134x 9 . ( x 2) ( x 3) x 2 x 3(Cover-up rule working)4x 9To find A, we cover up (x – 2) and substitute x 2 to give A 11 or 1.(.) ( x 3)4x 9To find B, we cover up (x - 3) and substitute x 3 to give B 13 or 3.( x 2) (.)(Method of equating coefficients)The top line of the expanded expressionA( x 3) B( x 2)is equivalent to 4x – 9.( x 2) ( x 3)Therefore A(x-3) B(x-2) 4x - 9.Solving the simultaneous equations:A B 4-3A – 2B -9(equating x-coefficients)(equating constants).Hence 2A 2B 8 ; -3A – 2B -9.Adding the last two equations gives -A -1 and thus A 1.Substituting in the first equation gives 1 B 4 and thus B 3.
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample (2): ExpressPage 4 of 234x 5in partial factions.x x 122Firstly, we need to factorise the denominator to (x – 4 )(x 3), giving a resulting partial fraction of theformAB, x 4 x 3(Long working)Expand the partial fraction toA( x 3) B( x 4)and substitute values for x as in the last example.( x 4) ( x 3)Substitute values of 3 and 2 for x, so that the corresponding term in the top line equals zero.When x -3, the term in A becomes zero, and thus 4x 5 B(x-4), or -7 -7(B) B 1.When x 4, the term in B becomes zero, and thus 4x 5 A(x 3), or 21 7(A) A 3.Hence4x 531 . x x 12 x 4 x 32(Cover-up rule working)4x 5To find A, we cover up (x – 4) and substitute x 4 to give A 21or 3.7(.) ( x 3)4x 5To find B, we cover up (x 3) and substitute x -3 to give B 77 or 1.( x 4) (.)(Method of equating coefficients)The top line of the expanded expression, A(x 3) B(x-4) 4x 5.Solving the simultaneous equations:A B 43A – 4B 5(equating x-coefficients)(equating constants).Hence 4A 4B 16 ; 3A – 4B 5.Adding the last two equations gives 7A 21 and thus A 3.Substituting in the first equation gives 3 B 4 and thus B 1.
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiPage 5 of 23Three distinct linear factors in the denominator.The general method of solving partial fractions of this type is the same as the case with two distinctlinear factors, though there is more work involved.Example (3): Express2 x 2 17 x 21in partial fractions.( x 2) ( x 3) ( x 3)(Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 9780199142590, Exercise 18A, Q.6 )The partial fraction will now be of the form(Long working)The expanded partial fraction is nowABC. x 2 x 3 x 3A( x 3) ( x 3) B( x 2) ( x 3) C ( x 2) ( x 3).( x 2) ( x 3) ( x 3)We need to carry out three substitutions for x this time.When x -2, the terms in B and C both become zero, and thus 2x2 17x 21 A(x 3)(x - 3) 8 – 34 21 A(1)(-5) 5A -5 and hence A 1.When x -3, the terms in A and C both become zero, and thus 2x2 17x 21 B(x 2)(x - 3) 18 – 51 21 B(-1)(-6) 6B -12 and hence B -2.When x 3, the terms in A and B both become zero, and thus 2x2 17x 21 C(x 2)(x 3) 18 51 21 C(5)(6) 30C 90 and hence C 3.Hence2 x 2 17 x 21123 ( x 2) ( x 3) ( x 3) x 2 x 3 x 3(Cover-up rule working)2 x 2 17 x 21To find A, we cover up (x 2) and substitute x -2 to give(.) ( x 3) ( x 3) 58 34 21A (1) ( 5) or 5 or 1.2 x 2 17 x 21To find B, we cover up (x 3) and substitute x -3 to give( x 2) (.) ( x 3)18 51 21B ( 1) ( 6 ) or 126 or -2.2 x 2 17 x 21To find C, we cover up (x - 3) and substitute x 3 to give( x 2) ( x 3) (.)C 9018 51 21or 30 or 3.5 6
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiPage 6 of 23The method of equating coefficients can also be used when there are three linear factors in thedenominator.2 x 2 17 x 21Example (3a): Use the method of equating coefficients to expressin partial( x 2) ( x 3) ( x 3)fractions (same as Example (3)). .The partial fraction is of the formandABC, x 2 x 3 x 3A( x 3) ( x 3) B( x 2) ( x 3) C ( x 2) ( x 3)in expanded form .( x 2) ( x 3) ( x 3)We need to expand all the quadratic products in the numerator:A(x2 - 9) B(x2 - x – 6) C(x2 5x 6) 2x2 17x 21.This gives us three simultaneous equations in three unknowns:A B C 25C – B 176C - 6B - 9A 21(equating x2 coefficients) (1)(equating x coefficients) (2)(equating constants)(3)We can substitute B 5C - 17 into equations (1) and (3) to obtainA 6C 195C – B 17-24C – 9A -81(1)(2)(3)4A 24C 76-24C – 9A -814 (1)(3)Adding the two equations gives –5A -5 and thus A 1.Substituting back into the original equation (1) gives C 3.Finally, substituting into the original equation (2) gives B -2.Hence2 x 2 17 x 21123 ( x 2) ( x 3) ( x 3) x 2 x 3 x 3When we have three factors in the denominator, the method of equating coefficients can becomeincreasingly difficult, but there are occasions when we do need to use it.
Mathematics Revision Guides – Partial FractionsAuthor: Mark Kudlowski2 x 2 5 x 18Example (4): Expressin partial fractions.x( x 2) ( x 3)The partial fraction will now be of the formABC. x x 2 x 3(Long working)The expanded partial fraction is nowA( x 2) ( x 3) Bx ( x 3) Cx( x 2).x( x 2) ( x 3)The appropriate substitutions for x are 0, 2 and –3.When x 0, only the term in A is non-zero, and thus 2x2 - 5x - 18 A(x - 2)(x 3) – 18 A(-2)(3) 6A -18 A 3.When x 2, only the term in B is non-zero, and thus 2x2 - 5x - 18 Bx(x 3) 8 - 10 - 18 B(2)(5) 10B -20 B -2.When x -3, only the term in C is non-zero, and thus 2x2 - 5x - 18 Cx(x - 2) 18 15 - 18 C(-3)(-5) 15C 15 and hence C 1.Hence2 x 2 5 x 18321 . x( x 2) ( x 3) x x 2 x 3(Cover-up rule working)2 x 2 5 x 18To find A, cover up x and substitute x 0 to give(.) ( x 2) ( x 3) 18 18A ( 2 ) 3 or 6 or 3.2 x 2 5 x 18x(.) ( x 3)B To find B, cover up (x - 2) and substitute x 2 to give 208 10 182 5 or 10 or -2.2 x 2 5 x 18To find A, cover up (x 3) and substitute x -3 to givex( x 2) (.)1518 15 18C ( 3) ( 5) or 15 or 1.Page 7 of 23
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample (4a): Express2 x 2 5 x 18in partial fractions, using the method of equatingx( x 2) ( x 3)coefficients.The partial fraction will take the formExpanding givesABC. x x 2 x 3A( x 2) ( x 3) Bx ( x 3) Cx( x 2), giving a numerator ofx( x 2) ( x 3)A(x2 - x – 6) B(x2 3x) C(x2 - 2x) 2x2 - 5x - 18.The resulting simultaneous equations are as follows:A B C 2-A 3B -2C -5- 6A -18(equating x2 coefficients) (1)(equating x coefficients) (2)(equating constants)(3)Note how equation (3) has turned out easy to solve – we can see at once that A 3.Substituting A 3 into equations (1) and (2), we obtainB C -13B – 2C -8(1)(2)3B 3C -33B – 2C -83 (1)(2)Subtraction gives 5C 5, and thus C 1; substitution in (2) gives B -2.HencePage 8 of 232 x 2 5 x 18321 . x( x 2) ( x 3) x x 2 x 3
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample (5): ExpressPage 9 of 2337 x 81in partial fractions.( x 3) ( x 7) (2 x 3)(Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 9780199142590, Exercise 18A, Q.31 )The resulting partial fraction will be of the form(Long working)The expanded partial fraction isABC. x 3 x 7 2x 3A( x 7) (2 x 3) B( x 3) (2 x 3) C ( x 3) ( x 7).( x 3) ( x 7) (2 x 3)The appropriate substitutions for x this time are 3, -7 and32. The fractional substitution is a little moremessy than the others !When x 3, only the term in A is non-zero, and thus 37x - 81 A(x 7)(2x - 3) 111 – 81 A(10)(3) 30A 30 A 1.When x -7, only the term in B is non-zero, and thus 37x - 81 B(x- 3)(2x - 3) -259 - 81 B(-10)(-17) 170B -340 B -2.When x 32, only the term in C is non-zero, and thus 37x - 81 C(x- 3)(x 7) 317 1112 – 81 C( 2 )( 2 ) 222 – 324 C(-3)(17) (multiplying both sides by 4 to get rid of the fractions) 51C -102 C 2.Hence12237 x 81 . ( x 3) ( x 7) (2 x 3) x 3 x 7 2 x 3(Cover-up rule working)37 x 81To find A, cover up (x - 3) and substitute x 3 to give(.) ( x 7) (2 x 3)30111 81A 10 3 or 30 or 1.37 x 81To find B, cover up (x 7) and substitute x -7 to give( x 3) (.) (2 x 3)( 259) 81 340B ( 10) ( 17) or 170 or -2.37 x 81To find C, cover up (2x -3) and substitute x ( x 3) ( x 7) (.)A (111) 812( 32 ) ( 17)232(a bit messy!) to give 102222 324or ( 3) 17 or 51 or 2.(Again we have multiplied top and bottom by 4 to get rid of the awkward fractions.)
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample (5a): ExpressPage 10 of 2337 x 81in partial fractions by equating coefficients.( x 3) ( x 7) (2 x 3)The partial fraction will take the formABC, expanding to x 3 x 7 2x 3A( x 7) (2 x 3) B( x 3) (2 x 3) C ( x 3) ( x 7), and giving a numerator of( x 3) ( x 7) (2 x 3)A(2x2 11x – 21) B(2x2 - 9x 9) C(x2 4x - 21) 37x - 81.The following simultaneous equations result:2A 2B C 011A - 9B 4C 379B – 21A -21C -81(equating x2 coefficients) (1)(equating x coefficients) (2)(equating constants)(3)We can rewrite equation (1) by expressing C as –2(A B), and then substituting in the other twoequations.11A - 9B –8(A B) 37 9B – 21A 42(A B) -81 9A - 51B 11121A 51B -813A - 17B 37 (2)21A 51B -81 (3)3 (2)(3)Adding the above gives 30A 30 and hence A 1Substituting in (2) gives 3 - 17B 37 -17B 34 and B -2.Substituting in (1) gives 2 – 4 C 0 C 2.Hence12237 x 81 . x 3 x 7 2x 3( x 3) ( x 7) (2 x 3)
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiPage 11 of 23Three linear factors in the denominator, but with one repeated pair.The method here is again slightly different: here we have to use a combination of substitution andequating coefficients.3x 2 2in partial fractions.x( x 1) 2Example(6) : Express(Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 9780199142590, Exercise 18A, Q.15)Note that the factor (x – 1) is repeated here. The required partial fraction will be of the formABC. Both (x – 1) and its square are included in the denominators. x x 1 ( x 1) 2The cover-up rule should not be used here, because it may lead the student to think that the solutionis of the formAB, thus missing out the term with (x - 1) in the denominator. x ( x 1) 2(Substitution working)The expanded partial fraction is nowA( x 1) 2 B( x) ( x 1) C ( x).x( x 1) 2We have used the L.C.M. of the denominators here, namely x(x-1)2, and not x(x-1)(x-1)2.We first carry out two substitutions for x , i.e. 0 and 1.When x 0, the terms in B and C both become zero, and thus 3x2 2 A(x - 1)2 0 2 A(-1)2 A 2.When x 1, the terms in A and B both become zero, and thus 3x2 2 Cx 3 2 C(1) C 5.Note that we have not been able to find B at this stage.Final step – finding B by equating coefficients .The best option is to equate the coefficients of x2 , since the term in B has no constant, and equating xterms looks more messy.Looking at the expanded formA( x 1) 2 B( x)( x 1) C ( x), we can equate the coefficients ofx( x 1) 2x2 to give A B 3, and since A 2, B 1.Hence3x 2 2x( x 1) 2 215. x x 1 ( x 1) 2
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample(6a) : ExpressPage 12 of 233x 2 2in partial fractions using the method of equating coefficientsx( x 1) 2throughout.Note that the factor (x – 1) is repeated here. The required partial fraction will be of the formABC. Both (x – 1) and its square are included in the denominators. x x 1 ( x 1) 2A( x 1) 2 B( x) ( x 1) C ( x)The expanded partial fraction is.x( x 1) 2We have used the L.C.M. of the denominators here, namely x(x-1)2, and not x(x-1)(x-1)2.Thus, A(x2 - 2x 1) B(x2 - x) Cx 3x2 2.The following simultaneous equations result:(equating x2 coefficients) (1)(equating x coefficients) (2)(equating constants)(3)A B 3-2A - B C 0A 2From the above, we can deduce B 1 and (-4 – 1 C) 0 and thus C 5.Hence3x 2 2x( x 1) 2 215. x x 1 ( x 1) 2
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample(7) : ExpressPage 13 of 234x 5in partial fractions.( x 1) ( x 2) 2The required partial fraction will be of the formABC. Both (x 2) and its square are included in the denominators. x 1 x 2 ( x 2) 2(Substitution working)Expanding the partial fraction we haveA( x 2) 2 B( x 1) ( x 2) C ( x 1).( x 1) ( x 2) 2Substitute values of 1 and -2 for x here.When x 1 , the terms in B and C both become zero, and thus 4x 5 A(x 2)2 4 5 A(32 ) 9A 9 A 1.When x -2, the terms in A and B both become zero, and thus 4x 5 C(x-1) -8 5 (-2-1)C -3 -3C C 1.Final step – finding B by equating the coefficients of x2.We still need to find B.Looking at the expanded formA( x 2) 2 B( x 1) ( x 2) C ( x 1), we can equate the( x 1) ( x 2) 2coefficients of x2 to give A B 0, and since A 1, B -1. (The numerator of the top line, 4x 5,has no term in x2 and thus its coefficient is 0.)Therefore4x 5111 2x 1 x 2 ( x 2) 2( x 1) ( x 2)
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample(7a) : ExpressPage 14 of 234x 5in partial fractions, using the method of equating( x 1) ( x 2) 2coefficients throughout.The required partial fraction will be of the formABC. Both (x 2) and its square are included in the denominators. x 1 x 2 ( x 2) 2Expanding the partial fraction we haveA( x 2) 2 B( x 1) ( x 2) C ( x 1).( x 1) ( x 2) 2Thus, A(x2 4x 4) B(x2 x - 2) C(x-1) 4x 5.The following simultaneous equations result:A B 04A B C 44A - 2B - C 5(equating x2 coefficients) (1)(equating x coefficients) (2)(equating constants)(3)From Equation (1) we can substitute –A for B in Equations 2 and 3.3A C 46A - C 5(equating x coefficients)(equating constants)(2)(3)Adding gives 9A 9 and hence A 1 and B -1.Substituting in Eqn.2 gives 3 C 4 and hence C 1.From the above, we can deduce B 1 and (-4 – 1 C) 0 and thus C 5.Therefore4x 5111 2x 1 x 2 ( x 2) 2( x 1) ( x 2)
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample(8) : ExpressPage 15 of 235 x 2 6 x 21in partial fractions.( x 4) 2 (2 x 3)(Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 9780199142590, Exercise 18A, Q.18 )The required partial fraction will be of the formABC. 2 x 3 x 4 ( x 4) 2Both (x – 4) and its square are included in the denominators.(Substitution working)The expanded partial fraction is nowA( x 4) 2 B(2 x 3)( x 4) C (2 x 3).(2 x 3)( x 4) 2Again, there are messy fractions involved here – take care !Substitute values of 32 and 4 for x here.When x 32, the terms in B and C both become zero, and thus 5x2 – 6x - 21 A(x - 4)2 454 - 9 - 21 A( 32 -4)2 ( 75) A(- 52 )24 ( 75) A( 25)44 A -3.When x 4, the terms in A and B both become zero, and thus 5x2 – 6x - 21 C(2x-3) 80 –24 - 21 (8-3)C 35 5C C 7.Final step – finding B by equating the coefficients of x2.We still need to find B.A( x 4) 2 B(2 x 3) ( x 4) C (2 x 3)Looking at the expanded form, we can equate the(2 x 3) ( x 4) 2coefficients of x2 to give A 2B 5, and since A -3, B 4.Hence5 x 2 6 x 21 347 . 22 x 3 x 4 ( x 4) 2( x 4) (2 x 3)
Mathematics Revision Guides – Partial FractionsAuthor: Mark KudlowskiExample(8a) : ExpressPage 16 of 235 x 2 6 x 21in partial fractions, using the method of equating( x 4) 2 (2 x 3)coefficients throughout.The required partial fraction will be of the formThe expanded partial fraction is nowABC. 2 x 3 x 4 ( x 4) 2A( x 4)2 B(2 x 3) ( x 4) C (2 x 3).(2 x 3) ( x 4)2Thus, A(x2 - 8x 16) B(2x2 - 11x 12) C(2x-3) 5x2 - 6x – 21.The following simultaneous equations result:A 2B 5-8A - 11B 2C -616A 12B - 3C -21(equating x2 coefficients) (1)(equating x coefficients) (2)(equating constants)(3)The equations are more messy than in the last
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