Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    14
    ReputationRep:
    So there are 2 methods to to partial fractions: substitution and equating coefficients. Will there be certain questions where you literally won't be able to get the answer by using solely one method?
    I'm asking because I was doing a question and I couldn't do it by substitution, so I had to use equating coefficients and got the right answer. Was this down to me making a mistake in my method or is this a normal thing?
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (Original post by AxSirlotl)
    So there are 2 methods to to partial fractions: substitution and equating coefficients. Will there be certain questions where you literally won't be able to get the answer by using solely one method?
    No.

    What question did you fail to do by substitution?

    EDIT: Substitution doesn't always leave you with a single variable, sometimes you are left with 2 or more, and have to use simultaneously equations to solve for them after the substitutions.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by RDKGames)
    No.

    What question did you fail to do by substitution?

    EDIT: Substitution doesn't always leave you with a single variable, sometimes you are left with 2 or more, and have to use simultaneously equations to solve for them after the substitutions.
    The question was  7x^2+2x-2/ x^2(x+1)
    I managed to get my numerators as  Ax(x+1) + B(x+1) + Cx^2
    I don't know how to solve that using substitution (if it's right, which it probably isn't heh)
    Offline

    8
    ReputationRep:
    (Original post by AxSirlotl)
    The question was  7x^2+2x-2\fracx^2(x+1)
    I managed to get my numerators as  Ax(x+1) + B(x+1) + Cx^2
    I don't know how to solve that using substitution (if it's right, which it probably isn't heh)
    Would you not want the denominators to be x, x^2, and (x+1) ??
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by moz4rt)
    Would you not want the denominators to be x, x^2, and (x+1) ??
    Those are the ones I used, I kept editing my reply because I couldn't get it to display as a fraction, that's why the denominator isn't shown as what you said it should be (I think).
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (Original post by AxSirlotl)
    The question was  7x^2+2x-2\fracx^2(x+1)
    I managed to get my numerators as  Ax(x+1) + B(x+1) + Cx^2
    I don't know how to solve that using substitution (if it's right, which it probably isn't heh)
    \displaystyle \frac{7x^2+2x-2}{x^2(x+1)}=\frac{A}{x}+\frac{B  x+C}{x^2}+\frac{D}{x+1}

    \displaystyle \Rightarrow 7x^2+2x-2=Ax(x+1)+(Bx+C)(x+1)+Dx^2

    x=0 : -2=C

    x=-1 : 3 = D

    Then just pick any other x

    x=1 : 7 = 2A+2(B-2)+3

    x=2 : 30 = 6A + 3(2B-2) + 12


    And as per my note; solve the last two simultaneously.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by RDKGames)
    \displaystyle \frac{7x^2+2x-2}{x^2(x+1)}=\frac{A}{x}+\frac{B  x+C}{x^2}+\frac{D}{x+1}

    \displaystyle \Rightarrow 7x^2+2x-2=Ax(x+1)+(Bx+C)(x+1)+Dx^2

    x=0 : -2=C

    x=-1 : 3 = D

    Then just pick any other x

    x=1 : 7 = 2A+2(B-2)+3

    x=2 : 30 = 6A + 3(2B-2) + 12


    And as per my note; solve the last two simultaneously.
    Okidoki thanks, I think that makes sense. Quick question, why is it Bx+C and not just B?
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by AxSirlotl)
    Okidoki thanks, I think that makes sense. Quick question, why is it Bx+C and not just B?
    You don't need Bx+C, a constant term is fine instead. If you have a factor of the form (ax+b)^2 in the denominator then the corresponding partial fraction terms have the form

    \displaystyle \frac{A}{ax+b} + \frac{B}{(ax+b)^2}

    So in this case a factor of x^2 leads to

    \displaystyle \frac{A}{x} + \frac{B}{x^2}
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (Original post by AxSirlotl)
    Okidoki thanks, I think that makes sense. Quick question, why is it Bx+C and not just B?
    Sorry got a bit derailed and didn't mean to use Bx+C. In this situation, as said above, you just need B. Generally when you're setting these things up you use a polynomial of one order less than the denominator for the numerator, hence my mistake.

    Anyway, x=0 and x=-1 give two of the variables, then only one other remains. Plug in x=1 or some other x and rearrange for it.


    Alternatively, and this is what I had in mind while responding to you, is that \frac{7x^2+2x-2}{x^2(x+1)}=\frac{Ax+B}{x^2}+ \frac{C}{x+1} works too (and follows the rule I had earlier) to give \frac{4x-2}{x^2}+\frac{3}{x+1} which split to give the answer in the terms you're used to. Though you dont use this method at A-Level.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Will you be richer or poorer than your parents?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.