At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

\[128x ^{7}y + 32x^4y^4 + 2xy^7\]

\[2xy(64x ^{6}+16x ^{3}y ^{3}+y ^{6})\]

Yes i got there but it can be simplified further I'm pretty sure

You're all set.

Are you sure?

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.

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 think what i wrote last can be simplified further, any help?

The one with 2xy in the front?

yes

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 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