anonymous
  • anonymous
How to factor y^4+1/2+1/(16y^4)? I know that it factors into (y^2+1/(4y^2))^2, but I can't figure out why. I need to simplify this for an arc length integral. Any help is greatly appreciated.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
It is copied correctly; a 3 term sum.
radar
  • radar
Have you copies it correctly or is Y^4 + (1/2) (y^2) + (1/16)(6^4)?
radar
  • radar
Does the middle term have the variable y^2 in it?

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anonymous
  • anonymous
Unfortunately, no. I have checked my algebra up to that point and I'm also looking at a solution manual; which skips the factoring step. I haven't seen anything like it before. I'm thinking something similar to completing the square is the way to go, but I'm lost at this point. Thanks for your help, by the way.:)
radar
  • radar
O.K. I was thinking of a perfect square, but now I see that I was on the wrong track.
anonymous
  • anonymous
there is a typo in your question somewhere for sure
anonymous
  • anonymous
\[ y^2+\frac{1}{4}y^2=\frac{5}{4}y^2\] once you combine like terms
anonymous
  • anonymous
Sorry, I just noticed it in the second part. The original question is correct, but the term should be squared.
anonymous
  • anonymous
I have since corrected it; thank you.
anonymous
  • anonymous
still a typo, since \((\frac{5}{4}y^2)^2=\frac{25}{16}y^4\)
anonymous
  • anonymous
Sorry again; I didn't see the missing parentheses. Now it is corrected(knock on wood);)
anonymous
  • anonymous
oooh now i get it \[\left(y^2+\frac{1}{4y^2}\right)^2\]
anonymous
  • anonymous
happy holidaze! you done?
myininaya
  • myininaya
\[y^4+\frac{1}{2}+\frac{1}{16y^4} \] \[(y^2)^2+2 y^2 \frac{1}{4y^2}+(\frac{1}{4y^2})^2\] This is in the form \[a^2+2ab+b^2 \] which can be factored into the form: \[(a+b)^2 \]
anonymous
  • anonymous
probably easiest to see if you multiply out and see that it works
anonymous
  • anonymous
Ahh, yes; I don't know why I didn't see that before. I wanted to express it that way, but for some reason my mind turned off the possibility. Thank you, myininaya, satellite73 and radar. You've made me very happy!
anonymous
  • anonymous
yw (from all of us)
myininaya
  • myininaya
I hope you all have a great happy holiday! :)

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