## anonymous 5 years ago Find the partial fraction decomposition of the rational function. 5x^2 + 8/x3 + x2

1. anonymous

Is this $$\frac{5x^2+8}{x^3+x^2}$$?

2. anonymous

yes

3. anonymous

got a 5x^2 + 8 = A/x^2 + B/x + 1 A(x+1) + B(x^2) (Ax+A) + (Bx^2) (Ax + Bx^2) + A 5x^2 + 8 = (A + B) x + (A)x^2 5x^2+x+8=(A+B) + (A)

4. anonymous

but i think im wrong

5. anonymous

the x^2 is throwing me off

6. anonymous

So the problem is in how you split it up in the first step. First you've got to realize that the denominator splits into a factor that gets taken into a power. Namely, $$x^3+x^2$$ becomes $$x^2(x+1)$$, where $$x$$ is a factor taken to the second power. The split up thus goes as follows: \begin{align*} \frac{5x^2+8}{x^3+x^2} &= \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}\\ 5x^2+8 &= Ax(x+1) + B(x+1) + Cx^2 \end{align*}

7. anonymous

where did the A/x come from?

8. anonymous

So the $$\frac{A}{x}$$ comes from the fact that the $$x$$ factor is taken to a power. As another example, if we had taken $$\frac{1}{(x+1)^2}$$, the decomposition would split into $$\frac{A}{x+1} + \frac{B}{(x+1)^2}$$.

9. anonymous

oh ok

10. anonymous

Can you solve the problem from there, then?

11. anonymous

im trying it

12. anonymous

shouldnt the next step be 5x^2 + 8= (Ax^2+Cx^2) + (Ax+Bx) +B? After combining terms?

13. anonymous

Exactly.

14. anonymous

15. anonymous

Is -7/x + 8/x^2 + 12/x+1 the right answer?

16. anonymous

I got a different answer from that. You can see that something is off because $$A+B$$ should equal $$0$$, as there is no $$x$$ term on the left hand side.

17. anonymous

im lost.

18. anonymous

So how did you go about solving once you combined terms?

19. anonymous

I'll show you

20. anonymous

5x^2+8=(A+C)x^2 + (A+B)x + B 5x^2 +0x+8=(A+C) + (A+B) + B then i put it on the matrix function on my calculator and it gave me -7, 8 and 12.

21. anonymous

thats how i got Is -7/x + 8/x^2 + 12/x+1.

22. anonymous

You probable plugged some numbers in wrong. I get -8, 8, and 13.

23. anonymous

probably*

24. anonymous

let me see if that works.

25. anonymous

Thats it. I dont understand what i did arong though?

26. anonymous

I had 1 0 1 5 1 1 0 1 0 1 0 8 as my matrix.

27. anonymous

Row 2, column 4 should be a 0, not a one, as there's no $$x$$ term.

28. anonymous

Ah i see the 0x term stays zero and not 1.

29. anonymous

Exactly.

30. anonymous

Alright I have now. Thank you so much for helping me. I really appreciate it!

31. anonymous

No problem, glad you understood it!