Question

Partial Fraction Decomposition...x^(2)+1/x^(3)+x^(2)-5x+3

Answer 1

\[\frac{x^2+1}{x^3+x^2-5x+3}\]?

Answer 2

this is going to suck because
\[x^3+x^2-5x+3=(x-1)^2(x+3)\]

Answer 3

you have to write
\[\frac{x^2+1}{x^3+x^2-5x+3}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+3}\]and then find the coefficients up top

Answer 4

Okay! I was just wondering where to put x+3 I was putting it with the A previously. Thanks!!

Answer 5

it makes no difference what you label the numerators
that is not the problem
the problem is finding them

Answer 6

Oh okay! I hope I can figure it out. Haha!

Answer 7

i can help with that if you like

Answer 8

Sure, if you don't mind

Answer 9

when you add that mess on the left, the numerator will be
\[A(x-1)(x+3)+B(x+3)+C(x-1)^2\] so first set \[x^2+1=A(x-1)(x+3)+B(x+3)+C(x-1)^2\]

Answer 10

i meant on the right doh

Answer 11

we can find \(B\) instantly by replacing \(x\) by \(1\) and getting
\[2=4B\] so \[b=\frac{1}{2}\]

Answer 12

How come there isn't a (x-1) with the B??

Answer 13

because the denominator of B is \((x-1)^2\)already, so all you need to do to that term to put it over the common denominator \((x-1)^2(x+3)\) is multiply top and bottom by \(x+3\)

Answer 14

Okay that makes sense. It gets cancelled out basically?

Answer 15

? nothing is cancelled

Answer 16

So you don't even multiply it with x-1 ?

Answer 17

the expression i wrote on the right of \[x^2+1=A(x-1)(x+3)+B(x+3)+C(x-1)^2\] is the numerator if you were to add \[\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+3}\]

Answer 18

to add, you would have to multiply \(\frac{A}{x-1}\) top and bottom by \((x-1)(x+3)\) because the denominator will be \((x-1)^2(x+3)\)

Answer 19

you would have to multiply \(\frac{B}{(x-1)^2}\) top and bottom by \(x+3\) and you would have to multiply \(\frac{C}{x+3}\) top and bottom by \((x-1)^2\)
that would put each term over the right denominator

Answer 20

that is why i wrote \[x^2+1=A(x-1)(x+3)+B(x+3)+C(x-1)^2\]

Answer 21

so far so good?

Answer 22

yeah! thanks

Answer 23

ok now did you get how i got \(B=\frac{1}{2}\)?

Answer 24

Yeah!

Answer 25

now comes the hard part

Answer 26

oh maybe not
lets try making \(x=-3\)

Answer 27

then \[x^2+1=A(x-1)(x+3)+B(x+3)+C(x-1)^2\] turns in to
\[10=16C\] so
\[C=\frac{5}{8}\]

Answer 28

that ok too?

Answer 29

Yep!

Answer 30

ok now comes the think part, because i really don't want to multiply all that crap out, which is another way to do it, but i don't like doing that if i can avoid it, because i often make an algebra mistake

Answer 31

visualize the constant when you multiply this out \[A(x-1)(x+3)+B(x+3)+C(x-1)^2\]
it should be clear it is \[-3A+3B+C\] right?

Answer 32

Yes

Answer 33

sure? if not, you have to multiply all that crap out

Answer 34

I actually didnt get the +C?

Answer 35

but lets pretend it is more or less obvious
now we know \(B=\frac{1}{2}, C=\frac{5}{8}\)

Answer 36

Wait nevermind..got it!

Answer 37

k good
it is the constant when you square \((x-1)\)

Answer 38

on the left, the constant is 1
on the right, the constant is
\[-3A+3B+C\] i.e.
\[1=-3A+3\times \frac{1}{2}+\frac{5}{8}\] solve for A

Answer 39

"we can find B instantly by replacing x by 1 and ..."

Keep in mind that this may be confusing. Fundamentally, x = 1 is NOT in the Domain of the Original Function. This odd substitution makes little sense, excepting that it does help to find the solutions more quickly in many cases.

Answer 40

true, but i was not putting it in the original function
actually there is an even snappier way to do it but i am not sure i can draw it here

Answer 41

in any case if we did it right we should get \(A=\frac{3}{8}\) and that is the end

Answer 42

I was taught this weird way of doing with a system of equations and then you take out the x^2 and x. I ended up with the answers A=-1/2 B=-2/3 C=3/2 and wow what a difference! thanks how would I write what we got?

Answer 43

It doesn't matter where you put it. It NOT in the Domain. You should not be putting it anywhere. It still works, though.

Answer 44

Do you know which of the answers is correct? I am so confused!

Answer 45

You should not be confused. Just drag through it and set it up and solve it.

Final Answer: \(\dfrac{3/8}{x-1}+\dfrac{1/2}{(x-1)^{2}}+\dfrac{5/8}{x+3}\)