Mathematics Revision Guides Partial Fractions Page 1 of MK HOME TUITION Mathematics Revision Guides Level: AS / A Level AQA : C4 Edecel: C4 OCR: C4 OCR MEI: C4 PARTIAL FRACTIONS Version : Date: 1-04-01 Eamples 1,, 5, 6 and 8 are copyrighted to their respective owners and used with their permission
Mathematics Revision Guides Partial Fractions Page of Partial Fractions The section on rational epressions included methods of addition and subtraction of algebraic fractions Thus an epression like 1 can be converted to a single fraction as follows: ( 1) ( ) ( )( 1) ( )( 1), to 4 ( )( 1) 7 ( )( 1) and finally Sometimes it might be useful to carry out the process in reverse, where we would convert an epression like 7 ( )( 1) into the form decomposing the epression into partial fractions A B 1 This process is known as Epanding A B 1 gives 1) B( ) ( )( 1) To find the values of A and B, we substitute convenient values for to make each term on the top line equal to zero Let -1 (to make +1) 0), then + 7 B(-), or 6 -B B - Let, (to make B(-) 0), then + 7 +1), or 9 A A 7 ( )( 1) 1 This method is also often illustrated by the cover-up rule Here we cover up each term in the denominator in turn and substitute the value of that makes the covered-up term equal to zero 7 ()( 1) 7 ( )() Covering up ( ) and substituting gives A 9 or 6 Covering up ( + 1) and substituting -1 gives B or - It must be stressed that the substitution method and the cover-up rule can only be used eclusively when we have distinct linear factors in the denominator In fact, when there are repeated or quadratic factors 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 epanded epression gives +1) + B(-) + 7 1) B( ) ( )( 1) From this, we can solve the simultaneous equations: A + B 1 (equating -coefficients) A B 7 (equating constants) Subtracting the second equation from the first gives B -6 and thus B - Substituting in the first equation gives A 1 and thus A
Mathematics Revision Guides Partial Fractions Page of Most eamination questions on partial fractions cover the following cases: Two distinct linear factors in the denominator Three distinct linear factors in the denominator Three linear factors in the denominator, but one repeated Two distinct linear factors in the denominator This case was covered in the previous eample here are two others Eample (1): Epress 4 9 ( )( ) in partial fractions (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 978019914590, Eercise 18A, Q1 ) The partial fraction will be in the form (Long working) Epand the partial fraction to the top line zero in turn A B ) B( ) ( )( ), and substitute values for to make each term in When, the term in A becomes zero, and thus 4-9 B(-), or B B When, the term in B becomes zero, and thus 4 9 -), or -1-1(A) A 1 4 9 ( )( ) 1 (Cover-up rule working) 4 9 ()( ) To find A, we cover up ( ) and substitute to give A 1 1 or 1 4 9 ( )() To find B, we cover up ( - ) and substitute to give B 1 or (Method of equating coefficients) The top line of the epanded epression Therefore -) + B(-) 4-9 Solving the simultaneous equations: ) B( ) ( )( ) is equivalent to 4 9 A + B 4 (equating -coefficients) -A B -9 (equating constants) A + B 8 ; -A B -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
Mathematics Revision Guides Partial Fractions Page 4 of Eample (): Epress 4 5 1 in partial factions Firstly, we need to factorise the denominator to ( 4 )( + ), giving a resulting partial fraction of the form A B 4 (Long working) Epand the partial fraction to, ) B( 4) ( 4)( ) and substitute values for as in the last eample Substitute values of and for, so that the corresponding term in the top line equals zero When -, the term in A becomes zero, and thus 4 + 5 B(-4), or -7-7(B) B 1 When 4, the term in B becomes zero, and thus 4 + 5 +), or 1 7(A) A 4 5 1 1 4 (Cover-up rule working) 4 5 ()( ) To find A, we cover up ( 4) and substitute 4 to give A 7 1 or 4 5 ( 4)() To find B, we cover up ( + ) and substitute - to give B 7 7 or 1 (Method of equating coefficients) The top line of the epanded epression, +) + B(-4) 4 + 5 Solving the simultaneous equations: A + B 4 A 4B 5 (equating -coefficients) (equating constants) 4A + 4B 16 ; A 4B 5 Adding the last two equations gives 7A 1 and thus A Substituting in the first equation gives + B 4 and thus B 1
Mathematics Revision Guides Partial Fractions Page 5 of Three distinct linear factors in the denominator The general method of solving partial fractions of this type is the same as the case with two distinct linear factors, though there is more work involved Eample (): Epress 17 1 ( )( )( ) in partial fractions (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 978019914590, Eercise 18A, Q6 ) The partial fraction will now be of the form A B C (Long working) The epanded partial fraction is now )( ) B( )( ) C( )( ) ( )( )( ) We need to carry out three substitutions for this time When -, the terms in B and C both become zero, and thus + 17 + 1 +)( - ) 8 4 + 1 1)(-5) 5A -5 and hence A 1 When -, the terms in A and C both become zero, and thus + 17 + 1 B( +)( - ) 18 51 + 1 B(-1)(-6) 6B -1 and hence B - When, the terms in A and B both become zero, and thus + 17 + 1 C( + )( + ) 18 + 51 + 1 C(5)(6) 0C 90 and hence C 17 1 ( )( )( ) 1 (Cover-up rule working) 17 1 ()( )( ) To find A, we cover up ( + ) and substitute - to give 841 5 A (1) ( 5) or 5 or 1 17 1 ( )()( ) To find B, we cover up ( + ) and substitute - to give 1851 1 B ( 1) ( 6) or 6 1 or - 17 1 ( )( )() To find C, we cover up ( - ) and substitute to give 1851 1 C 56 90 or 0 or
Mathematics Revision Guides Partial Fractions Page 6 of The method of equating coefficients can also be used when there are three linear factors in the denominator Eample (a): Use the method of equating coefficients to epress fractions (same as Eample ()) 17 1 ( )( )( ) in partial The partial fraction is of the form A B C, and )( ) B( )( ) C( )( ) ( )( )( ) in epanded form We need to epand all the quadratic products in the numerator: - 9) + B( - 6) + C( + 5 + 6) + 17 + 1 This gives us three simultaneous equations in three unknowns: A + B + C (equating coefficients) (1) 5C B 17 (equating coefficients) () 6C - 6B - 9A 1 (equating constants) () We can substitute B 5C - 17 into equations (1) and () to obtain A + 6C 19 (1) 5C B 17 () -4C 9A -81 () 4A + 4C 76 4 (1) -4C 9A -81 () Adding the two equations gives 5A -5 and thus A 1 Substituting back into the original equation (1) gives C Finally, substituting into the original equation () gives B - 17 1 ( )( )( ) 1 When we have three factors in the denominator, the method of equating coefficients can become increasingly difficult, but there are occasions when we do need to use it
Mathematics Revision Guides Partial Fractions Page 7 of Eample (4): Epress 5 18 ( )( ) in partial fractions The partial fraction will now be of the form (Long working) A B C The epanded partial fraction is now The appropriate substitutions for are 0, and )( ) B( ) C( ) ( )( ) When 0, only the term in A is non-zero, and thus - 5-18 - )( + ) 18 -)() 6A -18 A When, only the term in B is non-zero, and thus - 5-18 B( + ) 8-10 - 18 B()(5) 10B -0 B - When -, only the term in C is non-zero, and thus - 5-18 C( - ) 18 + 15-18 C(-)(-5) 15C 15 and hence C 1 5 18 ( )( ) 1 (Cover-up rule working) 5 18 ()( )( ) 18 18 A ( ) or 6 or 5 18 ()( ) 81018 B 5 0 or 10 or - To find A, cover up and substitute 0 to give To find B, cover up ( - ) and substitute to give 5 18 ( )() 181518 C ( ) ( 5) To find A, cover up ( + ) and substitute - to give 15 or 15 or 1
Mathematics Revision Guides Partial Fractions Page 8 of Eample (4a): Epress coefficients 5 18 ( )( ) in partial fractions, using the method of equating The partial fraction will take the form A B C Epanding gives )( ) B( ) C( ) ( )( ), giving a numerator of - 6) + B( + ) + C( - ) - 5-18 The resulting simultaneous equations are as follows: A + B + C (equating coefficients) (1) -A + B -C -5 (equating coefficients) () - 6A -18 (equating constants) () Note how equation () has turned out easy to solve we can see at once that A Substituting A into equations (1) and (), we obtain B + C -1 (1) B C -8 () B + C - (1) B C -8 () Subtraction gives 5C 5, and thus C 1; substitution in () gives B - 5 18 ( )( ) 1
Mathematics Revision Guides Partial Fractions Page 9 of Eample (5): Epress 7 81 ( )( 7)( ) in partial fractions (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 978019914590, Eercise 18A, Q1 ) The resulting partial fraction will be of the form A B C 7 (Long working) The epanded partial fraction is 7)( ) B( )( ) C( )( 7) ( )( 7)( ) The appropriate substitutions for this time are, -7 and The fractional substitution is a little more messy than the others! When, only the term in A is non-zero, and thus 7-81 +7)( - ) 111 81 10)() 0A 0 A 1 When -7, only the term in B is non-zero, and thus 7-81 B(- )( - ) -59-81 B(-10)(-17) 170B -40 B - When, only the term in C is non-zero, and thus 7-81 C(- )(+ 7) 111 81 C( )( 17 ) 4 C(-)(17) 51C -10 C (multiplying both sides by 4 to get rid of the fractions) 7 81 ( )( 7)( ) 1 7 (Cover-up rule working) 7 81 ()( 7)( ) 111 81 A 10 0 or 0 or 1 7 81 ( )()( ) ( 59) 81 B ( 10) ( 17) To find A, cover up ( - ) and substitute to give To find B, cover up ( + 7) and substitute -7 to give 40 or 170 or - 7 81 ( )( 7)() To find C, cover up ( -) and substitute (a bit messy!) to give (111) 81 A 17 ( ) ( ) 4 10 or ( ) 17 or 51 or (Again we have multiplied top and bottom by 4 to get rid of the awkward fractions)
Mathematics Revision Guides Partial Fractions Page 10 of Eample (5a): Epress 7 81 ( )( 7)( ) in partial fractions by equating coefficients The partial fraction will take the form A B C 7 7)( ) B( )( ) C( )( 7) ( )( 7)( ) + 11 1) + B( - 9 + 9) + C( +4-1) 7-81 The following simultaneous equations result: A + B + C 0 (equating coefficients) (1) 11A - 9B + 4C 7 (equating coefficients) () 9B 1A -1C -81 (equating constants) (), epanding to, and giving a numerator of We can rewrite equation (1) by epressing C as (A + B), and then substituting in the other two equations 11A - 9B 8(A + B) 7 A - 17B 7 () 9B 1A +4(A + B) -81 1A + 51B -81 () 9A - 51B 111 () 1A +51B -81 () Adding the above gives 0A 0 and hence A 1 Substituting in () gives - 17B 7-17B 4 and B - Substituting in (1) gives 4 + C 0 C 7 81 ( )( 7)( ) 1 7
Mathematics Revision Guides Partial Fractions Page 11 of Three 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 and equating coefficients Eample(6) : Epress ( 1) in partial fractions (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 978019914590, Eercise 18A, Q15) Note that the factor ( 1) is repeated here The required partial fraction will be of the form A B C 1 ( 1) Both ( 1) and its square are included in the denominators The cover-up rule should not be used here, because it may lead the student to think that the solution is of the form A B ( 1), thus missing out the term with ( - 1) in the denominator (Substitution working) The epanded partial fraction is now 1) B( )( 1) C( ) ( 1) We have used the LCM of the denominators here, namely (-1), and not (-1)(-1) We first carry out two substitutions for, ie 0 and 1 When 0, the terms in B and C both become zero, and thus + - 1) 0 + -1) A When 1, the terms in A and B both become zero, and thus + C + 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, since the term in B has no constant, and equating - terms looks more messy Looking at the epanded form 1) to give A + B, and since A, B 1 B( )( 1) C( ) ( 1), we can equate the coefficients of ( 1) 1 5 1 ( 1)
Mathematics Revision Guides Partial Fractions Page 1 of Eample(6a) : Epress throughout ( 1) in partial fractions using the method of equating coefficients Note that the factor ( 1) is repeated here The required partial fraction will be of the form A B C 1 ( 1) Both ( 1) and its square are included in the denominators The epanded partial fraction is 1) B( )( 1) C( ) ( 1) We have used the LCM of the denominators here, namely (-1), and not (-1)(-1) Thus, - + 1) + B( - ) + C + The following simultaneous equations result: A + B (equating coefficients) (1) -A - B + C 0 (equating coefficients) () A (equating constants) () From the above, we can deduce B 1 and (-4 1 + C) 0 and thus C 5 ( 1) 1 5 1 ( 1)
Mathematics Revision Guides Partial Fractions Page 1 of Eample(7) : Epress 4 5 ( 1)( ) in partial fractions The required partial fraction will be of the form A B C 1 ( ) Both ( + ) and its square are included in the denominators (Substitution working) Epanding the partial fraction we have ) B( 1)( ) C( 1) ( 1)( ) Substitute values of 1 and - for here When 1, the terms in B and C both become zero, and thus 4 + 5 + ) 4 + 5 ) 9A 9 A 1 When -, the terms in A and B both become zero, and thus 4 + 5 C(-1) -8 + 5 (--1)C - -C C 1 Final step finding B by equating the coefficients of We still need to find B Looking at the epanded form ) B( 1)( ) C( 1) ( 1)( ), we can equate the coefficients of to give A + B 0, and since A 1, B -1 (The numerator of the top line, 4 + 5, has no term in and thus its coefficient is 0) Therefore 4 5 ( 1)( ) 1 1 1 1 ( )
Mathematics Revision Guides Partial Fractions Page 14 of Eample(7a) : Epress coefficients throughout 4 5 ( 1)( ) in partial fractions, using the method of equating The required partial fraction will be of the form A B C 1 ( ) Both ( + ) and its square are included in the denominators Epanding the partial fraction we have ) B( 1)( ) C( 1) ( 1)( ) Thus, + 4 + 4) + B( + - ) + C(-1) 4 + 5 The following simultaneous equations result: A + B 0 (equating coefficients) (1) 4A + B + C 4 (equating coefficients) () 4A - B - C 5 (equating constants) () From Equation (1) we can substitute A for B in Equations and A + C 4 (equating coefficients) () 6A - C 5 (equating constants) () Adding gives 9A 9 and hence A 1 and B -1 Substituting in Eqn gives + C 4 and hence C 1 From the above, we can deduce B 1 and (-4 1 + C) 0 and thus C 5 Therefore 4 5 ( 1)( ) 1 1 1 1 ( )
Mathematics Revision Guides Partial Fractions Page 15 of Eample(8) : Epress 5 6 1 ( 4) ( ) in partial fractions (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 978019914590, Eercise 18A, Q18 ) The required partial fraction will be of the form A B C 4 ( 4) Both ( 4) and its square are included in the denominators (Substitution working) The epanded partial fraction is now 4) Again, there are messy fractions involved here take care! Substitute values of and 4 for here B( )( 4) C( ) ( )( 4) When, the terms in B and C both become zero, and thus 5 6-1 - 4) 45 4-9 - 1-4) 75 ( ) - 5 4 ) 75 ( ) 5 4 4 ) A - When 4, the terms in A and B both become zero, and thus 5 6-1 C(-) 80 4-1 (8-)C 5 5C C 7 Final step finding B by equating the coefficients of We still need to find B Looking at the epanded form 4) B( )( 4) C( ) ( )( 4) coefficients of to give A + B 5, and since A -, B 4, we can equate the 5 6 1 ( 4) ( ) 4 7 4 ( 4)
Mathematics Revision Guides Partial Fractions Page 16 of Eample(8a) : Epress coefficients throughout 5 6 1 ( 4) ( ) in partial fractions, using the method of equating The required partial fraction will be of the form A B C 4 ( 4) The epanded partial fraction is now 4) B( ) ( 4) C( ) ( ) ( 4) Thus, - 8 + 16) + B( - 11 + 1) + C(-) 5-6 1 The following simultaneous equations result: A + B 5 (equating coefficients) (1) -8A - 11B + C -6 (equating coefficients) () 16A + 1B - C -1 (equating constants) () The equations are more messy than in the last one, but we can substitute 5 B for A in Equations and -8(5 B) - 11B + C -6 0+ 16B 11B + C -6 40 +5B + C -6 B + C 4 () 16(5 B) + 1B - C -1 80 B + 1B -C -1 80 0B - C -1 0B - C -101 () Quadrupling Eqn () gives 0B + 8C 16-0B C -101 Adding the two equations gives 5C 5 and hence C 7 Substituting in 0B + 8C 16 gives 0B 80 and hence B 4 Finally, recalling A 5 B gives A - 5 6 1 ( 4) ( ) 4 7 4 ( 4)
Mathematics Revision Guides Partial Fractions Page 17 of Improper Partial Fractions The rational epression to be reduced into partial fractions must have the numerator of lower degree than the denominator If this is not the case, we have an improper epression corresponding to a topheavy or improper fraction in arithmetic To convert an improper epression into the required form, we can either try factorising the numerator, or using long division Eample (9) Epress 4 5 1 in partial fractions Epress the result in the form A B C 4 D This is an improper epression because the degree of the numerator is not less than that of the denominator Substituting 1 in the numerator gives a value of 0, therefore ( - 1) is a factor of the numerator by the factor theorem The epression therefore factorises into ( 1) (4 5) ( 4)( ), leaving the fractional part in the correct form The full working is given here in Eample (), and the solution is 4 1 1 Eample (10) Epress 11 9 ( )( ) in partial fractions The numerator in the epression cannot be simplified by factorisation, so we need to use long division We must also epand the denominator to give - 5+ 6 (Quotient) - 5+ 6-11 +9-15 +18 4-9 (Remainder) The quotient is 4 9 ( )( ) working is also identical to that in Eample (1), again leaving the fractional part in the correct form The
Mathematics Revision Guides Partial Fractions Page 18 of Occasionally, the following cases turn up in eams, depending on the board and syllabus: One linear factor and one quadratic factor that cannot be factorised Two factors in the denominator, but repeated Three linear factors in the denominator, all repeated The cover-up rule should not be used in any of those cases One linear factor and one quadratic factor that cannot be factorised Eample (11): Epress 5 1 ( 1)( 1) in partial fractions (Copyright OUP, Understanding Pure Mathematics, Sadler & Thorning, ISBN 978019914590, Chapter 18, Eample ) The quadratic cannot be factorised in fact it does not even have real solutions The partial fraction form of the epression is therefore A B C 1 1 (Long working) The epanded form of the partial fraction is 1) ( B C)( 1) ( 1)( 1) When -1, the term in B and C becomes zero, and thus 5 - - 1 + 1) 6 A A Having found A, we can substitute 0 to find C Thus 5 - - 1 (0 + 1) + (0 + C) (0 +1) -1 + C C -4 We find B by equating the terms in on each side of the epression : 5 ( + B) 5 + B B 5 1 ( 1)( 1) 4 1 1
Mathematics Revision Guides Partial Fractions Page 19 of Eample (11a): Epress coefficients throughout 5 1 ( 1)( 1) in partial fractions, using the method of equating The partial fraction form of the epression is A B C 1 1 The epanded form of the partial fraction is Thus, + 1) + B(+ 1) + C(+1) 5-1 The following simultaneous equations result: A + B 5 (equating coefficients) (1) B + C - (equating coefficients) () A + C -1 (equating constants) () 1) ( B C)( 1) ( 1)( Substituting B 5-A in equation gives 5-A+C - or A+C -7 -A + C -7 (equating coefficients) () A + C -1 (equating constants) () C -8 and C -4; from Eq, A and from Eq1, B 1) 5 1 ( 1)( 1) 4 1 1
Mathematics Revision Guides Partial Fractions Page 0 of Eample (1): Epress 10 11 15 ( 1)( 5) in partial fractions The quadratic cannot be factorised, although it does have two real solutions of 5 The partial fraction form of the epression is therefore A B C 1 5 (Long working) The epanded form of the partial fraction is 5) ( B C)( 1) ( 1)( 5) When 1, the term in (B + C)( 1) becomes zero, and thus 10-11 - 15-5) -16-4A A 4 Again, the cover-up rule can only be used to find A 10 11 15 ()( 5) 16 Cover up - 1 and substitute 1 to give A 4 or 4 Having found A using either method, we can substitute 0 to find C Thus 10-11 - 15 4(0-5) + (0 + C)(0-1) -0-15 + C C -5 We find B by equating the terms in on each side of the epression : 10 (4 + B) 10 4 + B B 6 10 ( 1) ( 11 15 5) 4 6 5 1 5
Mathematics Revision Guides Partial Fractions Page 1 of Eample (1a): Epress coefficients throughout 10 11 15 ( 1)( 5) in partial fractions, using the method of equating The partial fraction form of the epression is therefore A B C 1 5 The epanded form of the partial fraction is Thus, - 5) + B(-1) + C(-1) 10-11 15 The following simultaneous equations result: A + B 10 (equating coefficients) (1) -B + C -11 (equating coefficients) () -5A - C -15 (equating constants) () 5) ( B C)( 1) ( 1)( 5) Substituting B 10-A and hence B A 10 in equation gives A-10+C -11 or A+C -1 A + C -1 Eqn() -5A - C -15 Eqn() 5A + 5C -5 5 Eqn() -5A - C -15 Eqn () Adding the equations above gives 4C -0 and C -5 From Eqn(), A-5-1, so A 4 Finally, from Eqn(1), B 6 10 11 15 ( 1)( 5) 4 6 5 1 5
Mathematics Revision Guides Partial Fractions Page of Two factors in the denominator, but repeated Here, the method of equating coefficients is the only feasible one Eample (1): Epress 4 ( 1) in partial fractions The partial fraction form of the epression will be The epanded form is therefore A B ( 1) ( 1) (compare Eample ()) 4 ( 1) A B ( 1) ( 1) Multiplying both sides by ( - 1) gives 4 + 1) + B When 1, 4 + B B 7 Equating the terms in gives A 4 4 ( 1) 4 7 ( 1) ( 1) Eample (14): Epress 1 ( ) in partial fractions Epanding, we have 1 ( ) A B ( ) ( ) Multiplying both sides by ( - ) gives - 1 ) + B When, 6-1 B B 5 Equating the terms in gives A 1 ( ) 5 ( ) ( )
Mathematics Revision Guides Partial Fractions Page of Three factors in the denominator, but repeated Here again, the method of equating coefficients is the only feasible one Eample (15): Epress ( 1) in partial fractions The partial fraction form of the epression will be The epanded form is therefore A B C 1) ( 1) ( 1) ( (see Eample (6)) ( 1) A B ( 1) ( 1) ( C 1) Multiplying both sides by ( + 1) gives + + + 1) + B( + 1) + C + + 1) + B( + 1) + C When -1, + (-1) + 0) + B(0) + C C 4 A and B need to be found by equating coefficients + 1) + B( + 1) + 4 + + 1) + B( + 1) + 4 A + (B + A) + (A + B + 4), (constant terms included for completeness only) Equating the terms in gives A Equating the terms in gives B + A 1 B - ( 1) ( 1) ( 1) 4 ( 1) (Again, the cover-up rule is no help here!) Eample (16): Epress 4 17 1 ( ) in partial fractions 4 17 1 ( ) A B ( ) ( ) ( C ) Multiplying both sides by ( - ) gives 4 17 + 1 ) + B( ) + C 4 + 4) + B( ) + C When, 16 4 + 1 0) + B(0) + C C ) + B( ) + 4 + 4) + B( ) + Equating the terms in gives A 4 Equating the terms in gives B - 4A -17 B -1 4 17 1 ( ) 4 1 ( ) ( ) ( )