# How to demonestrate this!!!!

• February 5th, 2007, 05:25 AM
almirza
How to demonestrate this!!!!
Hi

I have a rule in my book about the partial fractions
which is:

(6x^2-29x-29)/(x+1)(x+3)^2 =

A/(x+1) + B/(x-3) + C/(x-3)^2

How to demonestrate this !!!!!!!!!

By the way this not my homework and I asked my teacher but she said something I did not understaned!!!!!!

ALMIRZA

MOd edit: I have changed 'ti' into 'to' in the title, it seemed more appropriate :wink:
Megabrain
• February 6th, 2007, 08:58 AM
river_rat
Demonstrate what exactly? The partial fraction decomposition?
• February 8th, 2007, 02:09 AM
almirza
Demostrate how the part b4 the equal sign equals the part after the equal sign.

Particularly:
How the Denominatorof the part b4 the equal sign could be split into the denominators of the parts after te equal sign in the first place???
• February 8th, 2007, 01:59 PM
Thisnamesucks
Hi

I had a problem with partial fractions myself recently. Fortunately it was similar to this equation. I could work it through, but didn't understand the use of the extra factor.

After asking my teacher he gave me an explanation, but it didn't feel like a suitable explanation. So I went about explaining partial fractions to myself and came up with what follows.

When splitting a fraction into partial fractions, you must find factors of the denominator and place letters over them, which you then find the values of algebraically, e.g.

1/x^2+1 = (A/x+1) + (B/x-1)

x^2 + 1 can be represented as (x+1)(x-1)

As such both (x+1) and (x-1) can be said to be factors of the denominator. These are the numbers used as the denominators on the right side of the equation (where the A, B and C are).

Hence: 1 = A(x-1) + B(x+1)

In the case you have presented, we have two factors that are the same, which are represented as a squared bracket ((x-3) and (x-3) again).

(6x^2-29x-29)/(x+1)(x+3)^2 = A/(x+1) + B/(x+3) + C/(x+3)^2

You'd think you could write this as "= A/(x+1) + B/(x+3) + C/(x+3)" giving (x+3)^2, but B and C would share the same denominator and hence could be written as
(B + C)/(x+3), which, when put back into a complex fraction with the denominator of (x+1), would give the combined denominator of (x+1)(x-3) instead of (x+1)(x+3)^2.

Now you might then ask, well why don't we just write it as = A/(x+1) + (B)/(x+3)^2. We could do this, but the accepted rule is to put a linear equation over a quadratic equation (e.g. (Bx + C)/(x+3)^2). Either way, you end up involving x in your answer to the numerator, so you might as well save time (and calculations) by putting it in to start with.

We introduce the non-squared term of (x+3) to split up the Bx and C that would usually be put over (x+3)^2. This way we don't have to separate the B and C by equating coefficients of x.

I will now split the fraction into partial fractions for you both with and without the added B/(x+3) fraction:

WITH B/(x+3):

(6x^2-29x-29)/(x+1)(x+3)^2 = A/(x+1) + B/(x+3) + C/(x+3)^2
6x^2 - 29x - 29 = A(x+3)^2 + B(x+1)(x+3) + C(x+1)

(Use values of x that make two brackets = 0)

Let x = -1
6*(-1^2) - 29(-1) - 29 = A(2)^2 + B(0)(2) + C(0)
6 + 29 - 29 = 4A + 0B + 0C

(we can get rid of B and C as they are multiplied by 0)

6 = 4A
A = 6/4 = 3/2

Let x = -3
6*(-3^2) - 29(-3) - 29 = A(0)^2 + B(-2)(0) + C(-2)
54 + 87 - 29 = 0A + 0B - 2C

(we can get rid of A and B as they are multiplied by 0)

112 = -2C
C = 112/-2 = -56

To find B, we equate the coefficients. Put simply, we look at how many x^2 there are on the left, and see what brackets give x^2 on the right.

When the brackets are multiplied out:

[6x^2] - 29x - 29 = [Ax^2] + 6Ax + 9A + [Bx^2] + 4Bx + 3B + Cx + C

There are 6x^2 on the left
There are Ax^2 + Bx^2 on the right (see x^2 terms in square brackets)

Therefore we can say that:
Ax^2 + Bx^2 = 6x^2
By cancelling x^2: A + B = 6

We know A = 3/2, so: 3/2 + B = 6
B = 6 - 3/2
= 4 and 1/2
(or 9/2 as a top heavy fraction)

We can now substitute these values into the original equation:

6x^2 - 29x - 29/(x+1)(x+3)^2 = 3/(2(x+1)) + 9/(2(x+3)) - 56/(x+3)^2

WITHOUT B/(x+3):

(6x^2-29x-29)/(x+1)(x+3)^2 = A/(x+1) + (Bx + C)/(x+3)^2
6x^2 - 29x - 29 = A(x+3)^2 + (Bx+C)(x+1)

(Use values of x that make one bracket = 0)

Let x = -1
6*(-1^2) - 29(-1) - 29 = A(2)^2 + (-B + C)(0)
6 + 29 - 29 = 4A + 0(-B + C)

(we can get rid of (-B + C) as it is multiplied by 0)

6 = 4A
A = 6/4 = 3/2

Let x = -3
6*(-3^2) - 29(-3) - 29 = A(0)^2 + (-3B + C)(-2)
54 + 87 - 29 = 0A - 2(-3B + C)

(we can get rid of A as it is multiplied by 0)

112 = -2(-3B + C)
(-3B + C) = -56

The problem now is that we have no way to identify what B and C equal as there are no ways of separating them.

Without the extra factor, we have to resort to equating the coefficients:

6x^2 - 29x - 29 = A(x+3)^2 + (Bx+C)(x+1)

Equating... x^2: 6 = A + B [1]
x: -29 = 6A + B + C [2]
Numbers: -29 = 9A + C [3]

There are now a number of ways to find what A, B and C equal. The simplest is to substitute the value of A into equation [1].

6 = 3/2 + B
B = 9/2 (as we found before)

For C, substitute into [3]
-29 = 9*(3/2 ) + C
C = -29 - (27/2 )
= -42 and 1/2
(or -75/2 )

Note how C now equals a different number to when we used the extra fraction?

We could also have treated these as simultaneous equations:

[3] - [2]: (9A + C) - (6A + B + C) = -29 - (-29)
3A - B = 0 [4]

[4] - 3*[1]: 3A - B - (3A + 3B) = 0 - 3*6
-4B = -18
B = 9/2

[2] - [1]: (6A + B + C) - (A + B) = -29 - 6
5A + C = -35 [5]

Notice how the numbers now get really ugly...

[5] - (5/9)*[3]: (5A + C) - (5A + (5/9)*C) = -35 + (5/9)*29
(4/9)*C = -170/9
C = -75/2

We get the same answer when we don't use the extra fraction, but the numbers are more awkward and there's a lot more working to be done. I know it's hard to see why it works, but hopefully some of this will help you see how it simplifies what could be a very ugly set of calculations. You have to agree, C looks a lot nicer when it equals -56 instead of -75/2.

If you have any queries, feel free to e-mail me.

- TNS
• February 9th, 2007, 08:17 AM
river_rat
Um, what exactly are you doing thisnamesucks?

Ok almirza - let me walk you through this. First off, your partial fraction decomposition is wrong. If you want a fraction to be the sum of simpler fractions (i.e. a linear combination of simple fractions) then the denominators in your partial fraction decomposition must be the factors of the denominator you wish to decompose.

So (6x^2-29x-29)/(x+1)(x+3)^2 = A/(x+1) + B/(x+3) + C/(x+3)^2 (notice your first mistake). Now we clear denominators and try to solve for the constants A, B and C. To do that we equate coefficients on each side of the equation. After doing the tedious algebra you arrive at the following set of equations:

1. A+B = 6
2. 6A+4B+C = -29
3. 9A+3B+C = -29

Which you can solve simultaneously to find that A = 3/2, B=9/2 and C=-56

Any questions?