Completely factor this expression (and show your work too): 128x^7y + 32x^4y^4 + 2xy^7

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Completely factor this expression (and show your work too): 128x^7y + 32x^4y^4 + 2xy^7

Mathematics
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\[128x ^{7}y + 32x^4y^4 + 2xy^7\]
\[2xy(64x ^{6}+16x ^{3}y ^{3}+y ^{6})\]
Yes i got there but it can be simplified further I'm pretty sure

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It can't since 1 is the lowest factor in there and it is already simplified at 1, the same goes for the lowest exponent of 1 (x,y)
You're all set.
Are you sure?
Yes because if you look at the remaining 2, it goes into 64 and 16, but not 1, which is the y to the sixth. You can't reduce any exponents either because the 128 has an x with an exponent of 1, and the 2 has a y with an exponent of 1. Hope that helps.!
My mistake, I meant the 128 has a y with an exponent of 1
I'll go with what you say because I don't know otherwise
Trust me on this one, that is simplified to it's fullest content.
http://www.wolframalpha.com/input/?i=128x%5E7y+%2B+32x%5E4y%5E4+%2B+2xy%5E7+ Look at alternate forms, and look at the middle one. You could use the other forms but they are just moved a little.
oh, i think you are wrong, what about 2x(8x^3 + y^3)^2
Yes, you could use that, but they are both simplified.
i need the completely simplified version though, as the question states
I know, I have taken tests with very similar questions and I wrote the middle one and it was correct. Whatever your preference may be, pick one.
i think what i wrote last can be simplified further, any help?
The one with 2xy in the front?
yes
Well, technically, the \[2xy(64x^6+16x^3y^3+y^6)\] can be simplified to \[2xy(8x^3+y^3)^2\] because if you square \[8x^3+y^3\] you will get 64x^6, then if you keep foiling you will get the original question. Sorry I'm really slow at the equation thing :/
its okay, i think 8x^3 + y^3 can be further factored
It can't, since there is an x and y. If they were both the same variable, then yes, but no here.
Oh wait!
I'm thinking you haven't taken algebra 2 yet or at least where I am, because you cannot automatically say that it is prime just because the variables are different
I'm in it now
does factor theorem apply here?
It is a perfect cube trinomial so It can be simplified to \[(2x+y)^3\]
it doesnt work that way, try multiply that back out and see if you get the same answer
Oh yeah I JUST realized that, god what a dumb mistake.
its okay. do you know synthetic division?
No, but look, there are no alternate forms: http://www.wolframalpha.com/input/?i=%282x%2By%29%5E3
i typed in 8x^3 + y^3 and it gave me the fully factored form: (2x + y)(4x^2 - 2xy + y^2)
I guess, I'm not as far as you though haha, besides I'm only in 8th grade :P
I'm in 8th grade algebra 2
probably different books, we have a book with a picture of another book on the front of it
Mine has a picture of some chick diving for a volley ball, lol. It's made by Glencoe
Well, i gotta go. It's eleven and im sleepy
Me too, even though I'm finishing up a report due tomorrow (I usually never do this)
Good bye

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