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anonymous
 3 years ago
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.
anonymous
 3 years ago
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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is copied correctly; a 3 term sum.

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

radar
 3 years ago
Best ResponseYou've already chosen the best response.0Does the middle term have the variable y^2 in it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Unfortunately, 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
 3 years ago
Best ResponseYou've already chosen the best response.0O.K. I was thinking of a perfect square, but now I see that I was on the wrong track.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there is a typo in your question somewhere for sure

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ y^2+\frac{1}{4}y^2=\frac{5}{4}y^2\] once you combine like terms

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, I just noticed it in the second part. The original question is correct, but the term should be squared.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have since corrected it; thank you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0still a typo, since \((\frac{5}{4}y^2)^2=\frac{25}{16}y^4\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry again; I didn't see the missing parentheses. Now it is corrected(knock on wood);)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oooh now i get it \[\left(y^2+\frac{1}{4y^2}\right)^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0happy holidaze! you done?

myininaya
 3 years ago
Best 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: \[(a+b)^2 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0probably easiest to see if you multiply out and see that it works

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ahh, 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!

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2I hope you all have a great happy holiday! :)
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