Question

WILL FAN AND MEDAL!!!!!!!!!!!!!

a^4+4b^4

Answer 1

Instructions, please?

Answer 2

factor

Answer 3

thats all it says ;p

Answer 4

Suggestions: Let A=a^2 and B=b^2. Please make these substitution now.
Are you certain that that sign in the middle is + and not -?

Answer 5

you cannot factor the sum of two fourth powers

Answer 6

It's a positive

Answer 7

Actually I think you may...

Answer 8

at least not with real numbers

Answer 9

Do one thing.. add and subtract 4*a^2 * b^2 and then group +4a^2 b^2 with the a^4 + 4b^4

Answer 10

oh am i ever WRONG sorry

Answer 11

Then, unfortunately, you cannot find real factors; you could factor this, but your results will be complex (part imaginary). Are you familiar with complex numbers? imaginary numbers?

Answer 12

do what @vishweshshrimali5 is suggesting, it will work

Answer 13

Okay... let me give the solution.. part of it... you can work out the rest

Answer 14

\[a^4 + 4b^4\]
\[=a^4 + 4b^4 + 4a^2 b^2 - 4a^2 b^2\]
\[=(a^2)^2 + (2b^2)^2 + 2*(a^2)*(2b^2) - 4a^2b^2\]
\[=(a^2)^2 + (2b^2)^2 + 2*(a^2)*(2b^2) - (2ab)^2\]
\[=((a^2)^2 + (2b^2)^2 + 2*(a^2)*(2b^2)) - (2ab)^2\]

Answer 15

Now, first use this formula:

\[x^2 +2xy + y^2 = (x+y)^2\]

for the first bracket thing: \(a^2 + (2b)^2 + 2(a^2)(2b^2)\)

What will you get?

Answer 16

think this method of completing the square is called the "sophie germain" trick

Answer 17

Wow! Never knew that :D

Answer 18

she pretended to be a man during the french revolution, got instructions from legrange

Answer 19

Now, if only I knew that story -.- Maths is definitely interesting.. and I see proofs of that every single day :D

Answer 20

Let's not forget that this is Maddiem's question, and she deserves to be involved in this discussion.

Answer 21

i friend of mine wrote some notes for a class, included that as information
i assume he was right, did not check, but i imagine a quick google search would get it

Answer 22

It did ;)

Answer 23

a google search actually calls this question "germain's identity"!!

Answer 24

yes ^_^ and the identity is exactly this given question!

Answer 25

Please note: Maddiem, the person who posted this question, has left (gone offline). I admire your ability to do this particular math (I would not have thought about such an approach), but urge you not to forget that a student was waiting for your attention and help.

Answer 26

Sorry everyone for leaving my computer died ;P Thank you so much for all the help :*. I'll fan all of you!!

Answer 27

you got this?