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# C4 Partial Fractions? watch

1. 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?
2. (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.
3. (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
I managed to get my numerators as
I don't know how to solve that using substitution (if it's right, which it probably isn't heh)
4. (Original post by AxSirlotl)
The question was
I managed to get my numerators as
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) ??
5. (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).
6. (Original post by AxSirlotl)
The question was
I managed to get my numerators as
I don't know how to solve that using substitution (if it's right, which it probably isn't heh)

Then just pick any other

And as per my note; solve the last two simultaneously.
7. (Original post by RDKGames)

Then just pick any other

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?
8. (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 , a constant term is fine instead. If you have a factor of the form in the denominator then the corresponding partial fraction terms have the form

So in this case a factor of leads to

9. (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 . 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 works too (and follows the rule I had earlier) to give which split to give the answer in the terms you're used to. Though you dont use this method at A-Level.

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