## anonymous 4 years ago Help!!! Decompose into partial fractions. (x^4-3x^3-3x^2+10)/(x+1)^2(x-3)

1. anonymous

really? this is a large pain

2. anonymous

you need all the work, or just the answer? because it is annoying as hell

3. anonymous

I can help. I have it all except for a few parts. let me post it h/o

4. anonymous

first you have to divide

5. anonymous

because degree of numerator is bigger than degree of denominator, so that is step one

6. anonymous

$(x-2)-(7x+4/(x+1)^2(x-3)$

7. anonymous

Im not getting the next step right.. I get $Ax^2+2Ax+A+Bx^3-Bx^2-5Bx-3B+Cx-3x=x^3-x^2-5x-3$ This is from A/(x-3)+B/(x+1)+C(x+1)^2

8. anonymous

ok so we have $x-2$ we can ignore, and now you have $\frac{-7x-4}{(x+1)^2(x-3)}$ i will take your word for it. i know the $x-2$ is right now you need $\frac{-7-x}{(x+1)^2(x-3)}=\frac{a}{x-1}+\frac{b}{(x-1)^2}+\frac{c}{x-3}$

9. anonymous

ok i will use your notation $\frac{-7-x}{(x+1)^2(x-3)}=\frac{a}{x-3}+\frac{b}{x-1}+\frac{c}{(x-1)^2}$

10. anonymous

$-7x-4=a(x-1)^2+b(x-3)(x-1)+c(x-3)$

11. anonymous

i would not use the "equate like coefficients" method yet. i would find a right away, but letting x = 3

12. anonymous

*by

13. anonymous

its "(x+1)" and "(x+1)^2" I believe, not "(x-1)"

14. anonymous

so it is... $-7x-4=a(x+1)^2+b(x-3)(x+1)+c(x-3)$

15. anonymous

ok i still let x = 3, get $-21-4=16a$ $a=-\frac{25}{16}$ but that is wrong, so maybe the -7x-4 is wrong

16. anonymous

ah it is $-7x+4$!

17. anonymous

Im almost positive I got -7x+4 for the remainder when I divided earlier

18. anonymous

ok good, so we get $-7x+4=a(x+1)^2+b(x-3)(x+1)+c(x-3)$ $x=3$ $-21+4=16a$ $a=-\frac{17}{16}$

19. anonymous

yeah ok, you had a minus sign in front of the whole thing and i didn't check it, but you are right , it is -7x+4

20. anonymous

Yup, you did it all right

21. anonymous

you had a minus and then parentheses so i thought you meant minus the whole thing

22. anonymous

now maybe let $x=-1$ and find c

23. anonymous

ohhh, my fault. The -17/16 is right for sure

24. anonymous

ok thanks.

25. anonymous

oh yeah, i have the answer for sure \ http://www.wolframalpha.com/input/?i=partial+fractions+%28x^4-3x^3-3x^2%2B10%29%2F%28%28x%2B1%29^2%28x-3%29%29

26. anonymous

i was just trying to get to it

27. anonymous

you can also use the "equate like coefficients" method but it requires solving a 3 by 3 system. i think substitution is much easier usually

28. anonymous

for example, if i replace x by -1 i get $c=-\frac{11}{4}$ pretty much in my head $7+4=-4c$ $c=-\frac{11}{4}$

29. anonymous

now that you have two out of the three the third one should be easy enough to find. plug in the numbers and see what is missing

30. anonymous

This really helped a lot.. I know it involved a lot so thanks a bunch:)

31. anonymous

yw