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danielcb114
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.
It is copied correctly; a 3 term sum.
Have you copies it correctly or is Y^4 + (1/2) (y^2) + (1/16)(6^4)?
Does the middle term have the variable y^2 in it?
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.:)
O.K. I was thinking of a perfect square, but now I see that I was on the wrong track.
there is a typo in your question somewhere for sure
\[ y^2+\frac{1}{4}y^2=\frac{5}{4}y^2\] once you combine like terms
Sorry, I just noticed it in the second part. The original question is correct, but the term should be squared.
I have since corrected it; thank you.
still a typo, since \((\frac{5}{4}y^2)^2=\frac{25}{16}y^4\)
Sorry again; I didn't see the missing parentheses. Now it is corrected(knock on wood);)
oooh now i get it \[\left(y^2+\frac{1}{4y^2}\right)^2\]
happy holidaze! you done?
\[y^4+\frac{1}{2}+\frac{1}{16y^4} \] \[(y^2)^2+2 y^2 \frac{1}{4y^2}+(\frac{1}{4y^2})^2\] This is in the form \[a^2+2ab+b^2 \] which can be factored into the form: \[(a+b)^2 \]
probably easiest to see if you multiply out and see that it works
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!
I hope you all have a great happy holiday! :)