Question

Need help doing partial fraction decomposition step by step...
4/x(x-1) can someone please help

Answer 1

We have two linear factors in our denominator,
so our setup will look like this:
\[\Large\rm \frac{4}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\]Understand that part? :U

Answer 2

yes.

Answer 3

thanks for replying, my bad I left my computer unattended

Answer 4

@LALFAN4E, do you have it from there?

Answer 5

You still have to find the values of A and B so it's not done yet.

Answer 6

ok

Answer 7

i solved that one

Answer 8

i moved on to a harder question can you help me?

Answer 9

1/[^{3}\]/\[^{3}-8\]

Answer 10

disregard that
it's 1/x^3-8

Answer 11

Im at 1= (A+B)x^2 + (2A-2B+C)X + (4A-2C)

Answer 12

You should learn how to post fractions properly. What you posted could be interpreted as

\(\dfrac{1}{x^3} - 8\) rather than \(\dfrac{1}{x^3 - 8}\)

Answer 13

yea my bad, how do you do that.. I'm pretty new

Answer 14

To avoid the confusion, use parentheses to emphasize that x^3 - 8 is in the denominator:

1/(x^3 - 8)

Answer 15

right

Answer 16

ok I expanded the expression out... but as to solving for A B and C I'm confused... given that the total expression =1, what do I set A+B = ? then, 2A-2B+C and 4A-2C

Answer 17

the examples in my book show expressions where x-3 = ...

Answer 18

in other words... how do I form the system of equations to answer A B & C

Answer 19

1/(x^3-8)?
Hmm I don't remember how to take that one apart...
I guess we can use the difference of cubes formula:\[\Large\rm =\frac{1}{(x-2)(x^2+2x+4)}\]

Answer 20

got that...

Answer 21

then expanded it for A B C

Answer 22

\[\Large\rm =\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}\]Like that?

Answer 23

yea

Answer 24

then cross multiplied

Answer 25

\[\Large\rm 1=A(x^2+2x+4)+(Bx+C)(x-2)\]k

Answer 26

yup

Answer 27

expand...

Answer 28

Nahhh don't expand :\
I mean if you prefer to do it that way.. then fine..

Answer 29

ok without expanding can you show me and explain?

Answer 30

It's always nice to try and solve it without expanding if you're able.
Just plug in key values for x.

So in this case, x=2 would cancel out that entire B/C block of stuff, yes?
So plugging in x=2 gives us,\[\Large\rm 1=A(2^2+2\cdot2+4)+0\]

Answer 31

well [B(2)+C)] how is that gone?

Answer 32

[B(2)+C)](2-2)

Because it's multiplying the 2-2

Answer 33

oh ... I see my bad. good

Answer 34

So that gets us our A value nice and easy,\[\Large\rm A=\frac{1}{12}\]

Answer 35

Let's plug it back into our setup:\[\Large\rm 1=\frac{1}{12}(x^2+2x+4)+(Bx+C)(x-2)\]

Answer 36

Now our A is out of the way.
Can you think of any number that might let us solve for B or C? :o
We have another good one we can work with.

Answer 37

gimme a sec, writing and trying to understand.

Answer 38

0?

Answer 39

x=0

Answer 40

or am I completely off?

Answer 41

Ah yes good :) x=0 will zero out the B unknown for us, allowing us to solve for C!

Answer 42

x=0:
\[\Large\rm 1=\frac{1}{12}(4)+(C)(-2)\]Mmm k what'd we get for C?

Answer 43

got it!

Answer 44

C= -1/3

Answer 45

Ok good.

Plug it back in,\[\Large\rm 1=\frac{1}{12}(x^2+2x+4)+\left(Bx-\frac{1}{3}\right)(x-2)\]Then to solve for B, plug in any other value.
Preferably something that's easy to work with, like x=1.

Answer 46

B=1/12 correct?

Answer 47

Hmm I came up with -1/12.
lemme check my math real quick.

Answer 48

because B x -1 would be -B.... thus i get -B = -1/12

Answer 49

So when you multiply everything out you should get,\[\Large\rm 1=\frac{7}{12}-B+\frac{1}{3}\]That look right?
You distribute the -1 to the 1/3 also?

Answer 50

the -1 to the -1/3 i mean*

Answer 51

my bad.. i messed up... i had 9/12 -___- did the algebra inside wrong

Answer 52

ah :3

Answer 53

Cool I think we got it!

Sheesh that one was tricky! :U

Answer 54

it was man... really appreciate the help. this site is too damn good

Answer 55

hey but wouldn't the answer come out to 10/12?

Answer 56

oh nm.... dumb. I see my mistake again... -___-

Answer 57

i'm stressed out, missed class lecture for the hardest section of my chapter test lol, but I think I got it from here on out

Answer 58

Hey lastly man, this is the answer in the book...

Answer 59

for B, why is there a (x+4) ?

Answer 60

They just factored out the -1/12 from each term,
if you distribute it back in, you can see what's going on,
\[\Large\rm -\frac{1}{12}(x+4)=\color{orangered}{-\frac{1}{12}}x\color{royalblue}{-\frac{1}{3}}\]
That matches what we determined for B and C, yes?
\[\Large\rm \color{orangered}{B=-\frac{1}{12}}, \qquad \color{royalblue}{C=-\frac{1}{3}}\]