# A little minor question on Partial Fraction

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• Jan 16th 2010, 01:44 AM
Punch
A 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, 01:50 AM
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Quote:

Originally Posted by Punch
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?

It would help if you post the entire question...
• Jan 16th 2010, 01:52 AM
Punch
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, 01:54 AM
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Quote:

Originally Posted by Punch
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?

The value of x that you substitute is different for every problem.

This is why I need to see the problem.
• Jan 16th 2010, 01:56 AM
Punch
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, 01:57 AM
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Quote:

Originally Posted by Punch
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?

All of these questions will be answered when you POST A PROBLEM so that I CAN DEMONSTRATE HOW TO SOLVE IT.
• Jan 16th 2010, 02:00 AM
Punch
Express in partial fraction $\frac{17x^2+23x+12}{(3x+4)(x^2+4)}$
• Jan 16th 2010, 02:13 AM
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Quote:

Originally Posted by Punch
Express in partial fraction $\frac{17x^2+23x+12}{(3x+4)(x^2+4)}$

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

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

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

Getting the common denominator:

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

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

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

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

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

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

Equating like powers of $x$, we get

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

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

$A + 3B = 17$

$4B + 3C = 23$

$4A + 4C = 12$.

Now you have three equations in three unknowns that you can solve simultaneously. Can you go from here?
• Jan 16th 2010, 02:18 AM
Punch
Well, one of the other method is to substitute x= -4/3?
• Jan 16th 2010, 02:24 AM
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Quote:

Originally Posted by Punch
Well, one of the other method is to substitute x= -4/3?

Yes, if you substitute $-\frac{4}{3}$ you can eliminate $B$ and $C$ to find $A$.

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

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

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

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

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

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

$A = 2$.

Now that you have $A$ 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 $x$.
• Jan 16th 2010, 02:29 AM
Punch
However I have seen some examples in which they substitute x=0 to find other unknowns
• Jan 16th 2010, 02:34 AM
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You substitute whatever will make the coefficients of A, B, C, etc = 0.

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

So we let $3x + 4 = 0$

$3x = -4$

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

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

$x^2 = -4$.

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

You would only substitute $x = 0$ if one of your coefficients happened to be $x$.
• Jan 16th 2010, 02:39 AM
Punch
Substituting 0 means finding c and so I need to substitute 1 after 0?
• Jan 16th 2010, 02:41 AM
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You would not be able to find C by substituting $x = 0$ or $x = 1$. You will still have A in the equation.

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

You have

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

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

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

$8 + 4C = 12$

$4C = 4$

$C = 1$.
• Jan 16th 2010, 02:42 AM
Punch
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.
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