Question

Expand the expression in powers of x to x^3

1 -4
---- - -----
2x+1 4+x^2

Answer 1

Sorry, it's mean to be 4 and not -4

Answer 2

I think... \[(x^2-8)/(2x^3+x^2+8x+4)\]

Answer 3

ahh, shoot, (x^2-8x), sry

Answer 4

Um no... it's expansion... not simplifying...

Answer 5

\(1/ (2x+1)-4/(4+x^2)\)
\((4+x^2-8x-4)/( 2x^3+x^2+8x+4)\)
\((x^2-8x)/( 2x^3+x^2+8x+4)\)

Answer 6

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
\[\frac{1}{1+x}=1-x+x^2-x^3+...\]

Answer 7

How do you do it that way, satellite?

Answer 8

order which grade question is this??

Answer 9

It's last year of highschool... so, pretty high (Uni first year)

Answer 10

then we'll have to use satellite's method

Answer 11

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

Answer 12

\[\frac{4}{4+x^2}\] 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

Answer 13

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

Answer 14

your last job is to combine like terms for
\[1-2x+4x^2-8x^3-\left(1-\frac{x^2}{4}\right )\]

Answer 15

Thanks!

Answer 16

yw

Answer 17

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

Answer 18

It is right :) I know how you got it now. Thank you