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danielcb114
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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.
 8 months ago
 8 months ago
danielcb114 Group Title
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.
 8 months ago
 8 months ago

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danielcb114 Group TitleBest ResponseYou've already chosen the best response.0
It is copied correctly; a 3 term sum.
 8 months ago

radar Group TitleBest ResponseYou've already chosen the best response.0
Have you copies it correctly or is Y^4 + (1/2) (y^2) + (1/16)(6^4)?
 8 months ago

radar Group TitleBest ResponseYou've already chosen the best response.0
Does the middle term have the variable y^2 in it?
 8 months ago

danielcb114 Group TitleBest ResponseYou've already chosen the best response.0
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.:)
 8 months ago

radar Group TitleBest ResponseYou've already chosen the best response.0
O.K. I was thinking of a perfect square, but now I see that I was on the wrong track.
 8 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
there is a typo in your question somewhere for sure
 8 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[ y^2+\frac{1}{4}y^2=\frac{5}{4}y^2\] once you combine like terms
 8 months ago

danielcb114 Group TitleBest ResponseYou've already chosen the best response.0
Sorry, I just noticed it in the second part. The original question is correct, but the term should be squared.
 8 months ago

danielcb114 Group TitleBest ResponseYou've already chosen the best response.0
I have since corrected it; thank you.
 8 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
still a typo, since \((\frac{5}{4}y^2)^2=\frac{25}{16}y^4\)
 8 months ago

danielcb114 Group TitleBest ResponseYou've already chosen the best response.0
Sorry again; I didn't see the missing parentheses. Now it is corrected(knock on wood);)
 8 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
oooh now i get it \[\left(y^2+\frac{1}{4y^2}\right)^2\]
 8 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
happy holidaze! you done?
 8 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[y^4+\frac{1}{2}+\frac{1}{16y^4} \] \[(y^2)^2+2 y^2 \frac{1}{4y^2}+(\frac{1}{4y^2})^2\] This is in the form \[a^2+2ab+b^2 \] which can be factored into the form: \[(a+b)^2 \]
 8 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
probably easiest to see if you multiply out and see that it works
 8 months ago

danielcb114 Group TitleBest ResponseYou've already chosen the best response.0
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!
 8 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
yw (from all of us)
 8 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
I hope you all have a great happy holiday! :)
 8 months ago
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