C4 Partial Fractions?

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?

Reply 1

RDKGames

Study Forum Helper

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.

(edited 6 years ago)

Reply 2

AxSirlotl

Original post by RDKGames

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)

(edited 6 years ago)

Reply 3

moz4rt

Original post by AxSirlotl

The question was

Unparseable latex formula:

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) ??

Reply 4

AxSirlotl

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).

Reply 5

RDKGames

Study Forum Helper

Original post by AxSirlotl

The question was

Unparseable latex formula:

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{Bx+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.

(edited 6 years ago)

Reply 6

AxSirlotl

Original post by RDKGames

\displaystyle \frac{7x^2+2x-2}{x^2(x+1)}=\frac{A}{x}+\frac{Bx+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?

Reply 7

Notnek

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}

Reply 8

RDKGames

Study Forum Helper

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