Okay, for example, I have 1 value which I can sub x into. And the value is 2. So now I sub x=2 and I have to find unknowns A, B and C.

After that I found A=5 and have unknown B. What should I sub x now?

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- January 16th 2010, 01:44 AMPunchA little minor question on Partial Fraction
Okay, for example, I have 1 value which I can sub x into. And the value is 2. So now I sub x=2 and I have to find unknowns A, B and C.

After that I found A=5 and have unknown B. What should I sub x now? - January 16th 2010, 01:50 AMProve It
- January 16th 2010, 01:52 AMPunch
Erm, not asking a question but instead clarifying something. Okay probably if i explain it this way it would be easier to understand.

What I am asking is, what should I substitute x as when I run out of values to substitute x as? - January 16th 2010, 01:54 AMProve It
- January 16th 2010, 01:56 AMPunch
Oh, so this is the case...

Alright then this poses a new problem and a new question in my head, how do I know what to substitute x as when I run out of values to substitute x as? - January 16th 2010, 01:57 AMProve It
- January 16th 2010, 02:00 AMPunch
Express in partial fraction

- January 16th 2010, 02:13 AMProve It
Checklist:

1. The numerator has a degree less than the denominator. (Tick)

2. The denominator can be factorised. (Tick)

Now, we require a partial fraction of the form

.

The numerator is always of one less degree than the denominator.

Getting the common denominator:

.

Can you see that since the denominators are equal, so must the numerators...

So .

In this case, there does not exist an that you can use to eliminate the .

So we will use another method. Expand all the brackets...

Equating like powers of , we get

.

Can you see that for this equation to be true, then

.

Now you have three equations in three unknowns that you can solve simultaneously. Can you go from here? - January 16th 2010, 02:18 AMPunch
Well, one of the other method is to substitute x= -4/3?

- January 16th 2010, 02:24 AMProve It
Yes, if you substitute you can eliminate and to find .

But there does not exist any number that will be able to eliminate to find and .

Using your method...

.

Now that you have you should be able to solve the simultaneous equations I gave you easier.

But like I said, in this case, the ONLY method that will get you all three unknowns is to equate like powers of . - January 16th 2010, 02:29 AMPunch
However I have seen some examples in which they substitute x=0 to find other unknowns

- January 16th 2010, 02:34 AMProve It
You substitute whatever will make the coefficients of A, B, C, etc = 0.

So in your case, we wanted the coefficient of to become 0.

So we let

.

Notice that the coefficient of A is . To try to eliminate A, we would need to let

.

Is it possible to square a number and end up with something negative?

You would only substitute if one of your coefficients happened to be . - January 16th 2010, 02:39 AMPunch
Substituting 0 means finding c and so I need to substitute 1 after 0?

- January 16th 2010, 02:41 AMProve It
You would not be able to find C by substituting or . You will still have A in the equation.

Edit: Actually, if you have already found A, then yes, you can substitute .

You have

.

Substituting and we find

. - January 16th 2010, 02:42 AMPunch
Look you have already found a to be 2 so the a part will make a value. which will help in deducing what c is... and after finding what c is, we substitute x=1 to find b.