Question

Resolve the following expression into its simplest partial fractions:

Answer 1

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!

Answer 2

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

Answer 3

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

Answer 4

@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 \[(x-1)^{2}\]

Answer 5

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

Answer 6

@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?

Answer 7

so D = -0.5 right?

Answer 8

@FireKat97

yes

Answer 9

okay thanks @Jadedry

Answer 10

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 \(x\))

Answer 11

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?

Answer 12

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)

Answer 13

would the value for C be 3/2?

Answer 14

no, 1/2?

Answer 15

@BAdhi

Ah that is clearer, thank you.

@FireKat97

No, I don't think those are the answers.

Answer 16

oh lol okay

Answer 17

@FireKat97

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

Answer 18

hmm maybe C is -1

Answer 19

@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

Answer 20

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

Answer 21

@FireKat97

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

Answer 22

soooo we have an E too now?

Answer 23

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

Answer 24

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

Answer 25

@FireKat97

actually

that look pretty good

Answer 26

lololol legit?

Answer 27

@FireKat97

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

Answer 28

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

Answer 29

@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

Answer 30

ummmmmmmm soo the E is really just a D?

Answer 31

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

Answer 32

@FireKat97

Yep

Answer 33

@FireKat97

it says the answer is:

\[\frac{ 1}{ x-1 } - \frac{ 1 }{ 2(x-1)^2} - \frac{2x+1}{2(x^2+1)}\]

Answer 34

so, D is still -0.5

a= -1

b =- 0.5

c = 1

right @BAdhi ?

Answer 35

yes

Answer 36

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!

Answer 37

Yup I think I got it too, thanks @Jadedry and @BAdhi :D

Answer 38

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.

Answer 39

ok cool.. u are welcome

Answer 40

:)

Answer 41

Okay, closing the board guys. Thanks again!