anonymous
  • anonymous
Help!!! Decompose into partial fractions. (x^4-3x^3-3x^2+10)/(x+1)^2(x-3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
really? this is a large pain
anonymous
  • anonymous
you need all the work, or just the answer? because it is annoying as hell
anonymous
  • anonymous
I can help. I have it all except for a few parts. let me post it h/o

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anonymous
  • anonymous
first you have to divide
anonymous
  • anonymous
because degree of numerator is bigger than degree of denominator, so that is step one
anonymous
  • anonymous
\[(x-2)-(7x+4/(x+1)^2(x-3)\]
anonymous
  • 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
  • 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
  • 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
  • anonymous
\[-7x-4=a(x-1)^2+b(x-3)(x-1)+c(x-3)\]
anonymous
  • anonymous
i would not use the "equate like coefficients" method yet. i would find a right away, but letting x = 3
anonymous
  • anonymous
*by
anonymous
  • anonymous
its "(x+1)" and "(x+1)^2" I believe, not "(x-1)"
anonymous
  • anonymous
so it is... \[-7x-4=a(x+1)^2+b(x-3)(x+1)+c(x-3)\]
anonymous
  • 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
  • anonymous
ah it is \[-7x+4\]!
anonymous
  • anonymous
Im almost positive I got -7x+4 for the remainder when I divided earlier
anonymous
  • 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
  • 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
  • anonymous
Yup, you did it all right
anonymous
  • anonymous
you had a minus and then parentheses so i thought you meant minus the whole thing
anonymous
  • anonymous
now maybe let \[x=-1\] and find c
anonymous
  • anonymous
ohhh, my fault. The -17/16 is right for sure
anonymous
  • anonymous
ok thanks.
anonymous
  • 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
  • anonymous
i was just trying to get to it
anonymous
  • 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
  • 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
  • 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
  • anonymous
This really helped a lot.. I know it involved a lot so thanks a bunch:)
anonymous
  • anonymous
yw

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