Resolve the following expression into its simplest partial fractions:

- Jadedry

Resolve the following expression into its simplest partial fractions:

- katieb

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- Jadedry

the equation in question is: \[(x-2) / ( (x^2 +1)*(x-1)^2)\]
I've managed to figure out that \[-0.5 / (x-1) ^2 \] is one of the partial fractions, but I'm having a lot of trouble with the others.
Thanks in advance!

- BAdhi

In this particular scenario you can decompose this to following partial fractions,
\[\frac{x-2}{(x^2+1)(x-1)^2} = \frac{Ax+B}{x^2+1}+\frac{C}{(x-1)}+\frac{D}{(x-1)^2}\]
So what you have to do now is to find the values of the A,B,C,D

- FireKat97

@BAdhi can you please explain why its Ax + B and not simply A or B alone...

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## More answers

- Jadedry

@Firecat97
Because \[(x^{2} + 1)\] is quadratic due to the fact that x is squared
in the case of \[(x-1)^{2}\] , the x is not quadratic, but you have to come up with a numerator for \[(x-1) \] as well as \[(x-1)^{2}\]

- FireKat97

ohhhhh okayyy thanks! we just started learning this stuff last lesson :)

- Jadedry

@BAdhi
I managed to get that far in my calculations, but I never come up with the right numerators? I think my method may be too complex, is there a simple way to find the numerator when a simple substitute for x will not cancel out all but one constant?

- FireKat97

so D = -0.5 right?

- Jadedry

@FireKat97
yes

- FireKat97

okay thanks @Jadedry

- BAdhi

There are mainly two ways to find the constant
First one is the substituting values to the x and comparing numerator
Second one is comparing the coefficients of each order of x
The numerator is,
\(x-2 = (Ax+B)(x^2-2x+1)+C(x^3-x^2+x-1)+D(x^2+1)\)
now you have to compare the coefficients of some of the orders of x (choose the easiest one so that only one unknown or less number of coefficient is left)
For example take \(x^3\)
\(0 = A+C \implies C = -A\)
from constant =>
\(-2 = B-C+D \implies B = -1.5+C\)
compare another single such occation and youll find the value for C and thus values to others from above equations (i recommend coeff of \(x\))

- Jadedry

Ah, I see you've expanded the equations I didn't do that. I don't understand how you got:
0= A +C
I understand the second one, but I'm sorry to say the first one escapes me?

- BAdhi

the coefficient of \(x^3\) on the left side is 0
the coefficient of the \(x^3\) on the right side comes only from
\((Ax+\cdots)(x^2-\cdots) +C(x^3+\cdots)+\cdots = (A+C)x^3 +\cdots\)
when equating the left side to right,
0 = (A+C)

- FireKat97

would the value for C be 3/2?

- FireKat97

no, 1/2?

- Jadedry

@BAdhi
Ah that is clearer, thank you.
@FireKat97
No, I don't think those are the answers.

- FireKat97

oh lol okay

- Jadedry

@FireKat97
I think I came up with those numbers myself, but the textbook says they are wrong, no idea why.

- FireKat97

hmm maybe C is -1

- BAdhi

@FireKat97
If we take the numerator of (x-1)^2 as Cx+D jst like for (x^2+1)
\[\frac{Cx+D}{(x-1)^2} = \frac{C(x-1)+(D+C)}{(x-1)^2}=\frac{C(x-1)}{(x-1)^2}+\frac{(C-D)}{(x-1)^2}= \frac{C}{(x-1)}+\frac{E}{(x-1)^2}\]
So you can see it can furthur simplfied to that but you cant do it to x^2+1

- BAdhi

there is a square at for the denominator of E its not visible :(

- Jadedry

@FireKat97
Still not right, my book says the answer is 1??

- FireKat97

soooo we have an E too now?

- FireKat97

wait I think I see why C is 1 but I maybe wrong...

- FireKat97

so do we get A -2B + C -1 = 0
then -C -2(-1.5 + C) + C -1 = 0
-C + 3 - 2C +C - 1 = 0
-2C = -2
C= 1?
idkkkk

- Jadedry

@FireKat97
actually
that look pretty good

- FireKat97

lololol legit?

- Jadedry

@FireKat97
where did you get the eq ( A -2B + C -1 = 0 ) from though?

- FireKat97

I did the same thing that @BAdhi was doing earlier, but I did it to the coefficients for x as he had suggested.

- BAdhi

@FireKat97
Yes there is a E but also note that D has disappeard :P. so if you want you can replace E with D and you'd end up with the expression that i've used on the original question

- FireKat97

ummmmmmmm soo the E is really just a D?

- FireKat97

@Jadedry what does your textbook say the whole answer is btw?

- BAdhi

@FireKat97
Yep

- Jadedry

@FireKat97
it says the answer is:
\[\frac{ 1}{ x-1 } - \frac{ 1 }{ 2(x-1)^2} - \frac{2x+1}{2(x^2+1)}\]

- Jadedry

so, D is still -0.5
a= -1
b =- 0.5
c = 1
right @BAdhi ?

- BAdhi

yes

- Jadedry

ah i think I have it.
So if you substitute 2 for x
you get:
2a + b+ 5c +5d
if you substitute c for a
and -1.5 + c for b
you have:
-2c + c -1.5 +5c - 2.5 = 0
=4c - 4 = 0 -
therefore c = 1 as the equation requires
therefore,
b= -1.5 +c
b = -0.5
a = -c
a = -1
and d remains -0.5
Yeah! got it!

- Jadedry

Thanks @BAdhi , never would have got it without that a = -c , you also gave pretty solid advice.
Thanks @FireKat97 you managed to point me in the right direction
Never would have done it without you guys, this problem has been bugging me for days.

- BAdhi

ok cool.. u are welcome

- FireKat97

:)

- Jadedry

Okay, closing the board guys. Thanks again!

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