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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- Jan 16th 2010, 12:44 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? - Jan 16th 2010, 12:50 AMProve It
- Jan 16th 2010, 12:52 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? - Jan 16th 2010, 12:54 AMProve It
- Jan 16th 2010, 12:56 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? - Jan 16th 2010, 12:57 AMProve It
- Jan 16th 2010, 01:00 AMPunch
Express in partial fraction $\displaystyle \frac{17x^2+23x+12}{(3x+4)(x^2+4)}$

- Jan 16th 2010, 01:13 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

$\displaystyle \frac{A}{3x + 4} + \frac{Bx + C}{x^2 + 4} = \frac{17x^2 + 23x + 12}{(3x + 4)(x^2 + 4)}$.

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

Getting the common denominator:

$\displaystyle \frac{A(x^2 + 4) + (Bx + C)(3x + 4)}{(3x + 4)(x^2 + 4)} = \frac{17x^2 + 23x + 12}{(3x + 4)(x^2 + 4)}$.

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

So $\displaystyle A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12$.

In this case, there does not exist an $\displaystyle x$ that you can use to eliminate the $\displaystyle A$.

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

$\displaystyle Ax^2 + 4A + 3Bx^2 + 4Bx + 3Cx + 4C = 17x^2 + 23x + 12$

Equating like powers of $\displaystyle x$, we get

$\displaystyle (A + 3B)x^2 + (4B + 3C)x + 4A + 4C = 17x^2 + 23x + 12$.

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

$\displaystyle A + 3B = 17$

$\displaystyle 4B + 3C = 23$

$\displaystyle 4A + 4C = 12$.

Now you have three equations in three unknowns that you can solve simultaneously. Can you go from here? - Jan 16th 2010, 01:18 AMPunch
Well, one of the other method is to substitute x= -4/3?

- Jan 16th 2010, 01:24 AMProve It
Yes, if you substitute $\displaystyle -\frac{4}{3}$ you can eliminate $\displaystyle B$ and $\displaystyle C$ to find $\displaystyle A$.

But there does not exist any number that will be able to eliminate $\displaystyle A$ to find $\displaystyle B$ and $\displaystyle C$.

Using your method...

$\displaystyle A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12$

$\displaystyle A\left[\left(-\frac{4}{3}\right)^2 + 4\right] + (Bx + C)\left[3\left(-\frac{4}{3}\right) + 4\right] = 17\left(-\frac{4}{3}\right)^2 + 23\left(-\frac{4}{3}\right) + 12$

$\displaystyle \frac{52}{9}A = \frac{272}{9} - \frac{92}{3} + 12$

$\displaystyle \frac{52}{9}A = \frac{272}{9} - \frac{276}{9} + \frac{108}{9}$

$\displaystyle \frac{52}{9}A = \frac{104}{9}$

$\displaystyle A = 2$.

Now that you have $\displaystyle A$ 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 $\displaystyle x$. - Jan 16th 2010, 01:29 AMPunch
However I have seen some examples in which they substitute x=0 to find other unknowns

- Jan 16th 2010, 01:34 AMProve It
You substitute whatever will make the coefficients of A, B, C, etc = 0.

So in your case, we wanted the coefficient of $\displaystyle Bx + C$ to become 0.

So we let $\displaystyle 3x + 4 = 0$

$\displaystyle 3x = -4$

$\displaystyle x = -\frac{4}{3}$.

Notice that the coefficient of A is $\displaystyle x^2 + 4$. To try to eliminate A, we would need to let $\displaystyle x^2 + 4 = 0$

$\displaystyle x^2 = -4$.

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

You would only substitute $\displaystyle x = 0$ if one of your coefficients happened to be $\displaystyle x$. - Jan 16th 2010, 01:39 AMPunch
Substituting 0 means finding c and so I need to substitute 1 after 0?

- Jan 16th 2010, 01:41 AMProve It
You would not be able to find C by substituting $\displaystyle x = 0$ or $\displaystyle x = 1$. You will still have A in the equation.

Edit: Actually, if you have already found A, then yes, you can substitute $\displaystyle x = 0$.

You have

$\displaystyle A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12$.

Substituting $\displaystyle A = 2$ and $\displaystyle x = 0$ we find

$\displaystyle 2(0^2 + 4) + [B(0) + C][3(0) + 4] = 17(0)^2 + 23(0) + 12$

$\displaystyle 8 + 4C = 12$

$\displaystyle 4C = 4$

$\displaystyle C = 1$. - Jan 16th 2010, 01:42 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.