## anonymous 4 years ago Expand the expression in powers of x to x^3 1 -4 ---- - ----- 2x+1 4+x^2

1. anonymous

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

2. anonymous

I think... $(x^2-8)/(2x^3+x^2+8x+4)$

3. anonymous

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

4. anonymous

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

5. ash2326

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

6. 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 $\frac{1}{1+x}=1-x+x^2-x^3+...$

7. anonymous

How do you do it that way, satellite?

8. ash2326

order which grade question is this??

9. anonymous

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

10. ash2326

then we'll have to use satellite's method

11. 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

12. 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

13. 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

14. anonymous

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

15. anonymous

Thanks!

16. anonymous

yw

17. 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

18. anonymous

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