1. 43: Partial Fractions © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

2. Partial Fractions Module C4 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

3. Partial Fractions In the presentation on algebraic fractions we saw how to add 2 algebraic fractions. To find partial fractions for an expression, we need to reverse the process of adding fractions. We will also develop a method for reducing a fraction to 3 partial fractions.

4. Partial Fractions We’ll start by adding 2 fractions. e.g. The partial fractions for are

5. Partial Fractions The expressions are equal for all values of x so we have an identity. The identity will be important for finding the values of A and B. To find the partial fractions, we start with

6. Partial Fractions Multiply by the denominator of the l.h.s. So, If we understand the cancelling, we can in future go straight to this line from the 1 st line. To find the partial fractions, we start with

7. Partial Fractions This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. Why should I choose x = 2 ? ANS: x = 2 means the coefficient of B is zero, so B disappears and we can solve for A.

8. Partial Fractions This is where the identity is important. What value would you substitute next ? ANS: Any value would do but x =  1 is good. The expressions are equal for all values of x, so I can choose to let x = 2.

9. Partial Fractions This is where the identity is important. So, The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

10. Partial Fractions This is where the identity is important. So, The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

11. Partial Fractions This is where the identity is important. So, The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

12. Partial Fractions Solution: Let Multiply by : It’s very important to write this each time e.g. 2 Express the following as 2 partial fractions.

13. Partial Fractions We never leave fractions piled up like this, so The “halves” are written in the denominators ( as 2s ) and the minus sign is moved to the front of the 2 nd fraction. So, Finally, we need to check the answer. A thorough check would be to reverse the process and put the fractions together over a common denominator.

14. Partial Fractions Another check is to use the “cover-up” method: We get To check B, substitute x =  in the l.h.s. but cover- up Cover-up on the l.h.s. and substitute x = 3 into the l.h.s. only To check A, find the value of x that makes the factor under A equal to zero ( x = 3 )

15. Partial Fractions The method we’ve used finds partial fractions for expressions I’ll call Type 1 where, the denominator has 2 linear factors, e.g.

16. Partial Fractions where, the denominator has 2 linear factors, The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. ( we may have to factorise to find them )

17. Partial Fractions and the numerator is a polynomial of lower degree than the denominator The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors,

18. Partial Fractions The degree of a polynomial is given by the highest power of x. Here the numerator is of degree The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. 1 and the denominator of degree and the numerator is a polynomial of lower degree than the denominator where, the denominator has 2 linear factors,

19. Partial Fractions where, the denominator has 2 linear factors, The degree of a polynomial is given by the highest power of x. and the numerator is a polynomial of lower degree then the denominator The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. and the denominator of degree 2 Here the numerator is of degree 1

20. Partial Fractions where, the denominator has 2 linear factors, The degree of a polynomial is given by the highest power of x. and the numerator is a polynomial of lower degree then the denominator The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. and the denominator of degree 2 Here the numerator is of degree 1

21. Partial Fractions SUMMARY To find partial fractions for expressions like Let Multiply by the denominator of the l.h.s. Substitute a value of x that makes the coefficient of B equal to zero and solve for A. Substitute a value of x that makes the coefficient of A equal to zero and solve for B. Check the result by reversing the method or using the “cover-up” method.

22. Partial Fractions Express each of the following in partial fractions. 1. Exercises 2. 3.4.

23. Partial Fractions Solutions: 1. Multiply by : So, Check:

24. Partial Fractions Solutions: 1. Multiply by : So, Check:

25. Partial Fractions Solutions: 1. Multiply by : So, Check: ( you don’t need to write out the check in full )

26. Partial Fractions Solutions: Multiply by So, 2. ( I won’t write out any more checks but it is important to do them. )

27. Partial Fractions Solutions: Multiply by So, 3.

28. Partial Fractions Solutions: Multiply by 4.So,

29. Partial Fractions If the denominator has 3 factors, we just extend the method. e.g. Solution: Multiply by So,

30. Partial Fractions The next type of fraction we will consider has a repeated linear factor in the denominator. e.g. 1 We would expect the partial fractions to be oreither This is wrong because the first 2 fractions just give, which is the same as having only one constant. We will try this to see why it is also wrong.

31. Partial Fractions Suppose Multiply by : However, Substituting B = 3 gives A = 1, an inconsistent result We need 3 constants if the degree of the denominator is 3.

32. Partial Fractions So, for we need It would also be correct to write but the fractions are not then reduced to the simplest form

33. Partial Fractions Using Multiply by :

34. Partial Fractions Using Multiply by :

35. Partial Fractions Using Multiply by : There is no other obvious value of x to use so we can choose any value. e.g. Subst. for A and C :

36. Partial Fractions There is however, a neater way of finding B. Since this is an identity, the terms on each side must be the same. For example, we have on the l.h.s. so there must be on the r.h.s. We had

37. Partial Fractions There is however, a neater way of finding B. Since this is an identity, the terms on each side must be the same. So, equating the coefficients of : For example, we have on the l.h.s. so there must be on the r.h.s. We had

38. Partial Fractions There is however, a neater way of finding B. Since this is an identity, the terms on each side must be the same. So, equating the coefficients of : For example, we have on the l.h.s. so there must be on the r.h.s. We had Since We could also equate the coefficients of x ( but these are harder to pick out ) or the constant terms ( equivalent to putting x = 0 ).

39. Partial Fractions So,

40. Partial Fractions So, We get The “cover-up” method can only be used to check A and C so for a proper check we need to put the r.h.s. back over a common denominator. So the numerator gives:

41. Partial Fractions SUMMARY Let To find partial fractions for expressions with repeated factors, e.g. Check the answer by using a common denominator for the right-hand side. Work in the same way as for type 1 fractions, using the two obvious values of x and either any other value or the coefficients of. N.B. B can sometimes be zero.

42. Partial Fractions Exercises Express each of the following in partial fractions. 1.2.

43. Partial Fractions Solutions: Multiply by : Coefficient of : So, 1. Let

44. Partial Fractions Solutions: Multiply by : Coefficient of : So, 2. Let

45. Partial Fractions If you are taking the OCR/MEI spec you can skip the next section. SKIP

46. Partial Fractions You may meet a question that combines algebraic division and partial fractions. e.g. Find partial fractions for The degree of the denominator is equal to the degree of the numerator. Both are degree 2. This is called an improper fraction. If the degree of the numerator is higher than the denominator the fraction is also improper. In an exam you are likely to be given the form of the partial fractions.

47. Partial Fractions e.g. 1 Find the values of A, B and C such that Solution:We don’t need to change our method Multiply by : So,

48. Partial Fractions e.g. 2 Find the values of A, B and C such that Solution: Multiply by : So, If you notice at the start, by looking at the terms on the l.h.s., that A = 2, the solution will be shorter as you can start with x = 0 and find C, then B.

49. Partial Fractions You will need to divide out but you will probably only need one stage of division so it will be easy. If you aren’t given the form of the partial fractions, you just need to watch out for an improper fraction.

50. Partial Fractions e.g. 2

51. Partial Fractions e.g. 2

52. Partial Fractions So, We can now find partial fractions for We get So, e.g. 2

53. Partial Fractions 1. Express the following in partial fractions: Exercise Solution: Dividing out:

54. Partial Fractions Partial Fractions: Let Multiply by : So,

55. Partial Fractions The next ( final ) section is only for those of you doing the OCR/MEI spec.

56. Partial Fractions The 3 rd type of partial fractions has a quadratic factor in the denominator that will not factorise. e.g. The partial fractions are of the form The method is no different but the easiest way to find A, B and C is to use the obvious value of x but then equate coefficients of the term and equate constants.

57. Partial Fractions 1. Express the following in partial fractions: Exercise Solution: Multiply by :Coefficient of Constants:

58. Partial Fractions So,

59. Partial Fractions

60. The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

61. Partial Fractions e.g. Express the following as 2 partial fractions. Solution: Let Multiply by : It’s very important to write this each time

62. Partial Fractions We never leave fractions piled up like this, so The “halves” are written in the denominators ( as 2s ) and the minus sign is moved to the front of the 2 nd fraction. So, Finally, we need to check the answer. A thorough check would be to reverse the process and put the fractions together over a common denominator.

63. Partial Fractions Another check is to use the “cover-up” method: We get To check B, substitute x =  in the l.h.s. but cover- up Cover-up on the l.h.s. and substitute x = 3 into the l.h.s. only To check A, find the value of x that makes the factor under A equal to zero ( x = 3 )

64. Partial Fractions SUMMARY To find partial fractions for expressions like Let Multiply by the denominator of the l.h.s. Substitute a value of x that makes the coefficient of B equal to zero and solve for A. Substitute a value of x that makes the coefficient of A equal to zero and solve for B. Check the result by reversing the method or using the “cover-up” method.

65. Partial Fractions The next type of fraction we will consider has a repeated linear factor in the denominator. e.g. 1 We would expect the partial fractions to be or either This is wrong because the first 2 fractions just give, which is the same as having only one constant. We will try this to see why it is also wrong.

66. Partial Fractions Suppose Multiply by : However, Substituting A = 2 gives B = 1, an inconsistent result We need 3 constants if the degree of the denominator is 3.

67. Partial Fractions Using Multiply by : There is no other obvious value of x to use so we can choose any value. e.g. Subst. for A and C :

68. Partial Fractions There is however, a neater way of finding B. Since this is an identity, the terms on each side must be the same. So, equating the coefficients of : Since For example, we have on the l.h.s. so there must be on the r.h.s. We had We could also equate the coefficients of x ( but these are harder to pick out ) or the constant terms.

69. Partial Fractions SUMMARY Let To find partial fractions for expressions with repeated factors, e.g. Check the answer by using using a common denominator for the right-hand side. Work in the same way as for type 1 fractions, using the two obvious values of x and either any other value or the coefficients of. N.B. B can sometimes be zero.

70. Partial Fractions You may meet a question that combines algebraic division and partial fractions. e.g. Find partial fractions for The degree of the denominator is equal to the degree of the numerator. Both are degree 2. This is called an improper fraction. If the degree of the numerator is higher than the denominator the fraction is also improper. In an exam you are likely to be given the form of the partial fractions.

71. Partial Fractions e.g. 1 Find the values of A, B and C such that Solution:We don’t need to change our method Multiply by : So,

72. Partial Fractions You will need to divide out but you will probably only need one stage of division so it will be easy. If you aren’t given the form of the partial fractions, you just need to watch out for an improper fraction.

73. Partial Fractions So, We can now find partial fractions for We get So, e.g. 2

74. Partial Fractions OCR/MEI only The 3 rd type of partial fractions has a quadratic factor in the denominator that will not factorise. e.g. The partial fractions are of the form The method is no different but the easiest way to find A, B and C is to use the obvious value of x but then equate coefficients of the term and equate constants.