## Jadedry one year ago Resolve the following expression into its simplest partial fractions:

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!

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

3. FireKat97

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

@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 $(x-1)^{2}$

5. FireKat97

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

@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?

7. FireKat97

so D = -0.5 right?

@FireKat97 yes

9. FireKat97

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 $$x$$)

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?

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)

13. FireKat97

would the value for C be 3/2?

14. FireKat97

no, 1/2?

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

16. FireKat97

oh lol okay

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

18. FireKat97

hmm maybe C is -1

@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

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

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

22. FireKat97

soooo we have an E too now?

23. FireKat97

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

24. 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

@FireKat97 actually that look pretty good

26. FireKat97

lololol legit?

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

28. FireKat97

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

@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

30. FireKat97

ummmmmmmm soo the E is really just a D?

31. FireKat97

@FireKat97 Yep

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

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

yes

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!

37. FireKat97

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

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.

ok cool.. u are welcome

40. FireKat97

:)