Hey there! Sign in to join this conversationNew here? Join for free
Turn on thread page Beta
    • Thread Starter
    Offline

    16
    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
    Offline

    20
    ReputationRep:
    Community Assistant
    (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

    16
    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

    10
    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

    16
    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
    Offline

    20
    ReputationRep:
    Community Assistant
    (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

    16
    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?
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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
    Offline

    20
    ReputationRep:
    Community Assistant
    (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.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 11, 2017
Poll
“Yanny” or “Laurel”
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

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

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