anonymous 5 years ago Find the partial fraction decomposition of the rational function. 9x^2 − 3x − 9 x^4 + 3x3

1. anonymous

are you looking for an answer or the process?

2. anonymous

both

3. anonymous

well first your need to set up the equivalence $\frac{9x^2-3x-9}{x^3(x+3)}=$$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+3}$

4. anonymous

ok

5. anonymous

Thats whats confusing me

6. anonymous

what is confusing me is that this interface is buggy

7. anonymous

now it works...

8. anonymous

yea

9. anonymous

then rewrite that as $9x^2-3x-0=$$Ax^2(x+3)+Bx(x+3)$$+C(x+3)+Dx^3$

10. anonymous

that should be a 9 where that 0 is

11. anonymous

k

12. anonymous

then rewrite that as $9x^2-3x-9=$$Ax^2(x+3)+Bx(x+3)$$+C(x+3)+Dx^3$

13. anonymous

so are we good to this point?

14. anonymous

yes

15. anonymous

now we can choose x=0 and determine that C=-3

16. anonymous

ok

17. anonymous

and next let x=-3 to determine that D=-3

18. anonymous

are you good with what I am doing?

19. anonymous

yea i get it

20. anonymous

now however we are going to need to choose two different values of x and get a system of two equations in two variable tosolve for A and B

21. anonymous

alright

22. anonymous

so I chose x=1 and got 3=A+B

23. anonymous

Ok

24. anonymous

next I chose x=-1 and got 3=A-B

25. anonymous

adding those equations together I get 6=2A which says A=3

26. anonymous

then using A=3 in either of the other two either of the previous two equations I determine that B=0

27. anonymous

Now finally substituting those values for A,B,C,D we get$\frac{9x^2-3x-3}{x^3(x+3)}=$$\frac{3}{x}-\frac{3}{x^3}-\frac{3}{x+3}$

28. anonymous

ok i see

29. anonymous

Thats alot!

30. anonymous