anonymous
  • anonymous
Expand the expression in powers of x to x^3 1 -4 ---- - ----- 2x+1 4+x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Sorry, it's mean to be 4 and not -4
anonymous
  • anonymous
I think... \[(x^2-8)/(2x^3+x^2+8x+4)\]
anonymous
  • anonymous
ahh, shoot, (x^2-8x), sry

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anonymous
  • anonymous
Um no... it's expansion... not simplifying...
ash2326
  • ash2326
\(1/ (2x+1)-4/(4+x^2)\) \((4+x^2-8x-4)/( 2x^3+x^2+8x+4)\) \((x^2-8x)/( 2x^3+x^2+8x+4)\)
anonymous
  • anonymous
i think you have to do this piece by piece, using \[\frac{1}{1-r}=1+r+r^2+r^3 + ...\] or in this case \[\frac{1}{1+x}=1-x+x^2-x^3+...\]
anonymous
  • anonymous
How do you do it that way, satellite?
ash2326
  • ash2326
order which grade question is this??
anonymous
  • anonymous
It's last year of highschool... so, pretty high (Uni first year)
ash2326
  • ash2326
then we'll have to use satellite's method
anonymous
  • anonymous
ok that was wrong, it is just the way you want it. \[\frac{1}{1+2x}=1-(2x)+(2x)^2-(2x)^3+...\] but we can stop there because you only need first four terms
anonymous
  • anonymous
\[\frac{4}{4+x^2}\] is more of a pain because you have to have a one in the denominator, so divide top and bottom by 4 to get \[\frac{1}{1+\frac{x^2}{4}}\] and repeat the process
anonymous
  • anonymous
with judicious use of parentheses you get \[\frac{1}{1+\frac{x^2}{4}}=1-(\frac{x^2}{4})+(\frac{x^2}{4})^2\] but really we can stop here because you only need up to \[x^3\] for your problem
anonymous
  • anonymous
your last job is to combine like terms for \[1-2x+4x^2-8x^3-\left(1-\frac{x^2}{4}\right )\]
anonymous
  • anonymous
Thanks!
anonymous
  • anonymous
yw
anonymous
  • anonymous
oh look, we can even check that it is right, by looking at the first 3 terms here http://www.wolframalpha.com/input/?i=1%2F%282x%2B1%29-4%2F%284%2Bx^2%29
anonymous
  • anonymous
It is right :) I know how you got it now. Thank you

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