Question

Can anyone explain to me how a partial fraction decomposition work?

I got a question of


18
--------
x(x^2+6)

Answer 1

So for partial fractions you essentially factor your denominator and create two fractions from what was before, 1. So in general, it is of the form
\[\frac{f(x)}{some factor-able function}=\frac{A}{1st Factor}+\frac{B}{2nd Factor}\]
Now when you have a factor that is a polynomial, you must include one less than that above. So with your quadratic, the B will become Bx+C.

Answer 2

So, if we apply this to your problem. The bottom is already factored for us. So we set it up as such:\[\frac{18}{x(x^2+6)}=\frac{A}{x}+\frac{Bx+C}{x^2+6}\]

Answer 3

So now, what do we have to do in order to add two fractions?

Answer 4

multiply both sides, right?

Answer 5

Not quite, think back to when you first learned how to add \[\frac{1}{3}+\frac{1}{2}\] What did you have to do?

Answer 6

oh make sure the other side is equal in the denominator

1/3 + 1/2 -> 2/6 + 3/6 = 5/6

Answer 7

There we go. So we had to find a common denominator. To do that we had to multiply by something that reduced to 1 right?

Answer 8

In the example it was 2/2 or 3/3 yea?

Answer 9

I'm sorry i'm confused; i don't know what to do for the numerator

Answer 10

ok, so how did you turn 1/3 into 2/6?

Answer 11

just multiplied by 2

Answer 12

Wel, you multiplied by \(\large \frac{2}{2}\) right?

Answer 13

right

Answer 14

if you only multiplied by 2, you would have 2/3 yea?

Answer 15

yes

Answer 16

So, the same principle applies here. Your denominator is x, so what do you have to multiply the other fraction by in order to create a common denominator?

Answer 17

\[\frac{18}{x(x^2+6)}=\frac{A}{x}(\frac{?}{?})+\frac{Bx+C}{x^2+6}(\frac{?}{?})\]
We know our common denominator is \(x(x^2+6)\). We just need to turn the right into one fraction.

Answer 18

A(x^2+6)+(Bx+C)
===========
x(x^2+c)

Answer 19

Close, you forgot to multiply the second fraction by something

Answer 20

well, reallly the second numerator

Answer 21

And also watch that 'c' in the denominator should be 6 :)

Answer 22

I'm also assuming the c on the bottom is a type-o

Answer 23

A(x^2+6)+(Bx+C)x
===========
x(x^2+6) ??

Answer 24

There we go.

Answer 25

So now get rid of all of your parentheses and group the like terms

Answer 26

including the numerator?

Answer 27

Sorry, I should have specified, only the numerator

Answer 28

Don't worry about the denominator at the moment

Answer 29

Ax^2+6A + Bx^2 +Cx
-------------------
x(x^2+6)

Answer 30

Good, so now we have :
\[\frac{18}{x(x^2+6)}=\frac{Ax^2+6A+Bx^2+Cx}{x(x^2+6)}\]

Answer 31

So, notice how the denominators are the same on both sides?

Answer 32

That means that the tops must be the same thing

Answer 33

so, we don't have an x^2 term on the left right?

Answer 34

or an x for that matter, yea?

Answer 35

*In terms of the numerator...remember the denominators are now gone

\[\large \frac{18}{\cancel{x(x^2+6)}}=\frac{Ax^2+6A+Bx^2+Cx}{\cancel{x(x^2+6)}}\]
\[\large 18 = Ax^2 + 6A + Bx^2 + Cx\]

Answer 36

hmm ok but does the letter A nad B stay?

Answer 37

So, here is where we use logic

Answer 38

since you don't have an x^2 or an x term on the left, what must their coefficients be?

Answer 39

3?

Answer 40

If you multiply x^2 by 3 it will go away?

Answer 41

What can you multiply something by to make it disappear?

Answer 42

0?

Answer 43

yea, so we really have
\[0x^2+0x+18=Ax^2+6A+Bx^2+Cx\]

Answer 44

I'm not sure this is going the right direction, since my answer got

3 -3x
- + -----
x x^2 + 6

Answer 45

That is entirely correct

Answer 46

so how do we reach that?

Answer 47

I thought you reached that yourself?

Answer 48

No, in my test, I'm just practicing for a final and wanted to know how i can do that

Answer 49

oh, ok, so we are here:\[0x^2+0x+18=Ax^2+6A+Bx^2+Cx\]

Answer 50

Now, we can set the like terms equal to each other and solve. So for example, we have: \[0x=Cx\] When we solve for C, we get C=0. Right?

Answer 51

right

Answer 52

We can do the same for each type of term. So, you try setting it up for the constant term

Answer 53

Don't solve, just set it up first

Answer 54

set what up?

Answer 55

Set up the equation for the constant terms

Answer 56

Like I did for the x terms

Answer 57

You mean A + Bx+C
--- ---------
x^2+6

Answer 58

Not quite. So above we had:

"Now, we can set the like terms equal to each other and solve. So for example, we have:
0x=Cx
When we solve for C, we get C=0. Right?"

Answer 59

We set the x terms equal and solved for C, I would like you to do this for the constant terms(the ones without an x).

Answer 60

So, if you are confused, start by telling me which terms are constant.

Answer 61

\[0x^2+0x+18=Ax^2+6A+Bx^2+Cx\]

*Set the \(\large x^2\) terms on the right side equal to the \(\large x^2\) terms on the left side* etc...

Answer 62

You still there? @umulas

Answer 63

Just in case I am not on when you come back, you only have like 3 more steps. Here is a good resource for you: http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx