Question

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.

Answer 1

It is copied correctly; a 3 term sum.

Answer 2

Have you copies it correctly or is Y^4 + (1/2) (y^2) + (1/16)(6^4)?

Answer 3

Does the middle term have the variable y^2 in it?

Answer 4

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.:)

Answer 5

O.K. I was thinking of a perfect square, but now I see that I was on the wrong track.

Answer 6

there is a typo in your question somewhere for sure

Answer 7

\[ y^2+\frac{1}{4}y^2=\frac{5}{4}y^2\] once you combine like terms

Answer 8

Sorry, I just noticed it in the second part. The original question is correct, but the term should be squared.

Answer 9

I have since corrected it; thank you.

Answer 10

still a typo, since \((\frac{5}{4}y^2)^2=\frac{25}{16}y^4\)

Answer 11

Sorry again; I didn't see the missing parentheses. Now it is corrected(knock on wood);)

Answer 12

oooh now i get it
\[\left(y^2+\frac{1}{4y^2}\right)^2\]

Answer 13

happy holidaze! you done?

Answer 14

\[y^4+\frac{1}{2}+\frac{1}{16y^4} \]
\[(y^2)^2+2 y^2 \frac{1}{4y^2}+(\frac{1}{4y^2})^2\]
This is in the form \[a^2+2ab+b^2 \]
which can be factored into the form:
\[(a+b)^2 \]

Answer 15

probably easiest to see if you multiply out and see that it works

Answer 16

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!

Answer 17

yw (from all of us)

Answer 18

I hope you all have a great happy holiday! :)