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

- anonymous

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

- chestercat

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- anonymous

really? this is a large pain

- anonymous

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

- anonymous

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

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## More answers

- anonymous

first you have to divide

- anonymous

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

- anonymous

\[(x-2)-(7x+4/(x+1)^2(x-3)\]

- 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

- 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
\[\frac{-7-x}{(x+1)^2(x-3)}=\frac{a}{x-1}+\frac{b}{(x-1)^2}+\frac{c}{x-3}\]

- anonymous

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

- anonymous

\[-7x-4=a(x-1)^2+b(x-3)(x-1)+c(x-3)\]

- anonymous

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

- anonymous

*by

- anonymous

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

- anonymous

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

- 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

- anonymous

ah it is
\[-7x+4\]!

- anonymous

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

- anonymous

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

- 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

- anonymous

Yup, you did it all right

- anonymous

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

- anonymous

now maybe let
\[x=-1\] and find c

- anonymous

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

- anonymous

ok thanks.

- 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

- anonymous

i was just trying to get to it

- 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

- anonymous

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

- 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

- anonymous

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

- anonymous

yw

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