DP IB Maths: AA HL

Revision Notes

1.1.3 Partial Fractions

Test Yourself

Partial Fractions

What are partial fractions?

  • Partial fractions allow us to simplify rational expressions into the sum of two or more fractions with constant numerators and linear denominators
    • This allows for integration of rational functions
  • The method of partial fractions is essentially the reverse of adding or subtracting fractions
    • When adding fractions, a common denominator is required
    • In partial fractions the common denominator is split into parts (factors)
  • If we have a rational function with a quadratic on the denominator partial fractions can be used to rewrite it as the sum of two rational functions with linear denominators
    •  This works if the non-linear denominator can be factorised into two distinct factors
      • For example: fraction numerator a x plus b over denominator left parenthesis c x plus d right parenthesis left parenthesis e x plus f right parenthesis end fraction equals fraction numerator A over denominator c x plus d end fraction plus fraction numerator B over denominator e x plus f end fraction
  • If we have a rational function with a linear numerator and denominator partial fractions can be used to rewrite it as the sum of a constant and a fraction with a linear denominator
    • The linear denominator does not need to be factorised
    • For example: fraction numerator a x plus b over denominator c x plus d end fraction equals A plus fraction numerator B over denominator c x plus d end fraction

How do I find partial fractions if the denominator is a quadratic?

  • STEP 1
    Factorise the denominator into the product of two linear factors
    • Check the numerator and cancel out any common factors
      • e.g.  fraction numerator 5 x blank plus blank 5 over denominator x squared blank plus blank x blank minus blank 6 end fraction equals blank fraction numerator 5 x blank plus blank 5 over denominator left parenthesis x blank plus blank 3 right parenthesis left parenthesis x blank minus blank 2 right parenthesis end fraction
  • STEP 2
    Split the fraction into a sum of two fractions with single linear denominators each having unknown constant numerators
    • Use A and B to represent the unknown numerators
      • e.g.  blank fraction numerator 5 x blank plus blank 5 over denominator left parenthesis x blank plus blank 3 right parenthesis left parenthesis x blank minus blank 2 right parenthesis end fraction blank identical to blank fraction numerator A over denominator x blank plus blank 3 end fraction plus fraction numerator B over denominator x blank minus blank 2 end fraction
  • STEP 3
    Multiply through by the denominator to eliminate fractions
    • Eliminate fractions by cancelling all common expressions
      • e.g.  blank 5 x plus 5 blank identical to A open parentheses x minus 2 close parentheses plus B left parenthesis x plus 3 right parenthesis
  • STEP 4
    Substitute values into the identity and solve for the unknown constants
    • Use the root of each linear factor as a value of  to find the unknowns
      • e.g. Let x = 2:  5 open parentheses 2 close parentheses plus 5 blank identical to A open parentheses open parentheses 2 close parentheses minus 2 close parentheses plus B open parentheses open parentheses 2 close parentheses plus 3 close parentheses etc
    • An alternative method is comparing coefficients
      • e.g.  5 x plus 5 blank identical to open parentheses A plus B close parentheses x plus left parenthesis negative 2 A plus 3 B right parenthesis
  • STEP 5
    Write the original as partial fractions
    • Substitute the values you found for A and B into your expression from STEP 2
      • e.g.  fraction numerator 5 x blank plus blank 5 over denominator x squared blank plus blank x blank minus blank 6 end fraction equals fraction numerator 2 over denominator x blank plus blank 3 end fraction plus fraction numerator 3 over denominator x blank minus blank 2 end fraction blank

How do I find partial fractions if the numerator and denominator are both linear?

  • If the denominator is not a quadratic expression you will be given the form in which the partial fractions should be expressed
  • For example expressblank fraction numerator 12 x blank minus blank 2 over denominator 3 x blank minus blank 1 end fraction in the form A plus fraction numerator B over denominator 3 x blank minus blank 1 end fraction
  • STEP 1
    Multiply through by the denominator to eliminate fractions
    • e.g.  12 x minus 2 blank identical to A open parentheses 3 x minus 1 close parentheses plus B
  • STEP 2
    Expand the expression on the right-hand side and compare coefficients
    • Compare the coefficients of x and solve for the first unknown
      • e.g.  12x = 3Ax 
      • therefore A = 4
    • Compare the constant coefficients and solve for the second unknown
        • e.g. - 2 = - A + B = - 4 + B
        • therefore B = 2
  • STEP 3
    Write the original as partial fractions
    • blank fraction numerator 12 x blank minus blank 2 over denominator 3 x blank minus blank 1 end fraction space equals space 4 space plus space fraction numerator space 2 over denominator 3 x space minus space 1 end fraction

 

How do I find partial fractions if the denominator has a squared linear term?

  • A squared linear factor in the denominator actually represents two factors rather than one
  • This must be taken into account when the rational function is split into partial fractions
    • For the squared linear denominator (ax + b)2  there will be two factors: (ax + b) and (ax + b)2
    • So the rational expression p over open parentheses a x plus b close parentheses squared becomes  fraction numerator A over denominator a x plus b end fraction plus blank B over open parentheses a x plus b close parentheses squared  
  • In IB you will be given the form into which you should split the partial fractions
    • Put the rational expression equal to the given form and then continue with the steps above
  • There is more than one way of finding the missing values when working with partial fractions
    • Substituting values is usually quickest, however you should look at the number of times a bracket is repeated to help you decide which method to use

Exam Tip

  • An exam question will often have partial fractions as part (a) and then integration or using the binomial theorem as part (b)
    • Make sure you use your partial fractions found in part (a) to answer the next part of the question

Worked example

a)
Express  fraction numerator 2 x blank minus blank 13 over denominator x squared minus blank x blank minus blank 2 end fraction  in partial fractions.             

1-1-3-aa-hl-partial-fractions-we-solution-a

b)
Express fraction numerator x left parenthesis 3 x blank minus blank 13 right parenthesis over denominator open parentheses x blank plus blank 1 close parentheses open parentheses x blank minus blank 3 close parentheses squared blank end fraction in the form fraction numerator A over denominator open parentheses x blank plus blank 1 close parentheses end fraction plus blank fraction numerator B over denominator x blank minus blank 3 blank end fraction plus blank fraction numerator C over denominator open parentheses x blank minus blank 3 close parentheses squared blank end fraction.

1-1-3-aa-hl-partial-fractions-we-solution-b

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Amber

Author: Amber

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.