Partial Fractions (DP IB Analysis & Approaches (AA)): Revision Note

Amber

Written by: Amber

Reviewed by: Mark Curtis

Updated on

Did this video help you?

Partial Fractions

What are partial fractions?

  • Partial fractions are the reverse of adding or subtracting algebraic fractions

    • e.g. fraction numerator 5 x over denominator open parentheses x minus 2 close parentheses open parentheses x plus 1 close parentheses end fraction in partial fractions is fraction numerator 2 over denominator x minus 2 end fraction plus fraction numerator 3 over denominator x plus 1 end fraction

      • This reverses fraction numerator 2 over denominator x minus 2 end fraction plus fraction numerator 3 over denominator x plus 1 end fraction equals fraction numerator 5 x over denominator open parentheses x minus 2 close parentheses open parentheses x plus 1 close parentheses end fraction

How do I find partial fractions when the denominator is quadratic?

  • STEP 1
    Factorise the denominator into two linear factors (and simplify if necessary)

    • e.g.  fraction numerator 5 x plus 5 over denominator x squared blank plus x minus 6 end fraction equals fraction numerator 5 x plus 5 over denominator left parenthesis x plus 3 right parenthesis left parenthesis x minus 2 right parenthesis end fraction

  • STEP 2
    Split the fraction into the sum of two smaller fractions with linear denominators and unknown constant numerators

    • Use A and B to represent the unknown numerators

      • e.g.  blank fraction numerator 5 x plus 5 over denominator left parenthesis x plus 3 right parenthesis left parenthesis x minus 2 right parenthesis end fraction identical to fraction numerator A over denominator x plus 3 end fraction plus fraction numerator B over denominator x minus 2 end fraction

  • STEP 3
    Multiply both sides by the denominator to eliminate fractions

    • e.g. up diagonal strike left parenthesis x plus 3 right parenthesis end strike up diagonal strike left parenthesis x minus 2 right parenthesis end strike cross times fraction numerator 5 x plus 5 over denominator up diagonal strike left parenthesis x plus 3 right parenthesis end strike up diagonal strike left parenthesis x minus 2 right parenthesis end strike end fraction equals up diagonal strike left parenthesis x plus 3 right parenthesis end strike left parenthesis x minus 2 right parenthesis cross times fraction numerator A over denominator up diagonal strike x plus 3 end strike end fraction plus left parenthesis x plus 3 right parenthesis up diagonal strike left parenthesis x minus 2 right parenthesis end strike cross times fraction numerator B over denominator up diagonal strike x minus 2 end strike end fraction

    • which gives 5 x plus 5 equals A open parentheses x minus 2 close parentheses plus B left parenthesis x plus 3 right parenthesis

  • STEP 4
    Find the unknown constants by substituting different numerical values into both sides to form and solve simultaneous equations in A and B

    • The easiest equations come from substituting in the roots of each linear factor

      • e.g. let x equals 25 open parentheses 2 close parentheses plus 5 blank equals A open parentheses 2 minus 2 close parentheses plus B open parentheses 2 plus 3 close parentheses equals 5 B

      • so 15 equals 5 B giving B equals 3

      • Let x equals negative 3: 5 open parentheses negative 3 close parentheses plus 5 equals A open parentheses negative 3 minus 2 close parentheses plus B open parentheses negative 3 plus 3 close parentheses equals negative 5 A

      • so negative 10 equals negative 5 A giving A equals 2

    • An alternative method is comparing coefficients

      • e.g. write it as  5 x plus 5 blank equals open parentheses A plus B close parentheses x plus left parenthesis negative 2 A plus 3 B right parenthesis

      • The coefficient of x on both sides must match, so 5 equals A plus B

      • The constant term on both sides must match, so 5 equals negative 2 A plus 3 B

      • Then solve these simultaneously

  • STEP 5
    Write the original fraction in partial fractions

    • Substitute A and B back into the expression

      • e.g.  fraction numerator 5 x plus 5 over denominator x squared plus x minus 6 end fraction equals fraction numerator 2 over denominator x plus 3 end fraction plus fraction numerator 3 over denominator x minus 2 end fraction blank

  • In general, if the denominator factorises then

    • 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

How do I find partial fractions when the denominator is the square of a linear term?

  • A squared linear factor in the denominator must be split into two different partial fractions of the form fraction numerator A over denominator a x plus b end fraction plus blank B over open parentheses a x plus b close parentheses squared

    • This can be seen in reverse by trying to add, for example, fraction numerator 1 over denominator x minus 3 end fraction plus 4 over open parentheses x minus 3 close parentheses squared

      • The lowest common denominator is open parentheses x minus 3 close parentheses squared

      • fraction numerator 1 over denominator x minus 3 end fraction plus 4 over open parentheses x minus 3 close parentheses squared equals... equals fraction numerator x plus 1 over denominator open parentheses x minus 3 close parentheses squared end fraction

  • In general, c over open parentheses a x plus b close parentheses squared equals fraction numerator A over denominator a x plus b end fraction plus B over open parentheses a x plus b close parentheses squared

    • Then use the same method as above

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

  • If the numerator and denominator are both linear, the fraction can be split into a constant and a simpler fraction

    • 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

      • Then use the same method as above

  • e.g. writeblank fraction numerator 12 x minus 2 over denominator 3 x minus 1 end fraction as A plus fraction numerator B over denominator 3 x minus 1 end fraction to get 12 x minus 2 blank equals A open parentheses 3 x minus 1 close parentheses plus B

    • Substitute in x equals 1 third to get B equals 2

    • Substitute in, say, x equals 0 to get negative 2 equals negative A plus 2 so A equals 4

      • blank fraction numerator 12 x minus 2 over denominator 3 x minus 1 end fraction space equals space 4 plus fraction numerator space 2 over denominator 3 x minus 1 end fraction

Examiner Tips and Tricks

In the exam, you will often be given the form in which to split partial fractions.

Worked Example

a) Express  fraction numerator 2 x minus 13 over denominator x squared minus x minus 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 minus 13 right parenthesis over denominator open parentheses x plus 1 close parentheses open parentheses x minus 3 close parentheses squared blank end fraction in the form fraction numerator A over denominator open parentheses x plus 1 close parentheses end fraction plus fraction numerator B over denominator x minus 3 blank end fraction plus fraction numerator C over denominator open parentheses x minus 3 close parentheses squared blank end fraction.

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

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

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

the (exam) results speak for themselves:

Did this page help you?

Amber

Author: Amber

Expertise: Maths Content Creator

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.

Mark Curtis

Reviewer: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.