Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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)

Answer 5

You're all set.

Answer 6

Are you sure?

Answer 7

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.!

Answer 8

My mistake, I meant the 128 has a y with an exponent of 1

Answer 9

I'll go with what you say because I don't know otherwise

Answer 10

Trust me on this one, that is simplified to it's fullest content.

Answer 11

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.

Answer 12

oh, i think you are wrong, what about 2x(8x^3 + y^3)^2

Answer 13

Yes, you could use that, but they are both simplified.

Answer 14

i need the completely simplified version though, as the question states

Answer 15

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.

Answer 16

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

Answer 17

The one with 2xy in the front?

Answer 18

yes

Answer 19

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 :/

Answer 20

its okay, i think 8x^3 + y^3 can be further factored

Answer 21

It can't, since there is an x and y. If they were both the same variable, then yes, but no here.

Answer 22

Oh wait!

Answer 23

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

Answer 24

I'm in it now

Answer 25

does factor theorem apply here?

Answer 26

It is a perfect cube trinomial so It can be simplified to
\[(2x+y)^3\]

Answer 27

it doesnt work that way, try multiply that back out and see if you get the same answer

Answer 28

Oh yeah I JUST realized that, god what a dumb mistake.

Answer 29

its okay. do you know synthetic division?

Answer 30

No, but look, there are no alternate forms:
http://www.wolframalpha.com/input/?i=%282x%2By%29%5E3

Answer 31

i typed in 8x^3 + y^3 and it gave me the fully factored form:
(2x + y)(4x^2 - 2xy + y^2)

Answer 32

I guess, I'm not as far as you though haha, besides I'm only in 8th grade :P

Answer 33

I'm in 8th grade algebra 2

Answer 34

probably different books, we have a book with a picture of another book on the front of it

Answer 35

Mine has a picture of some chick diving for a volley ball, lol. It's made by Glencoe

Answer 36

Well, i gotta go. It's eleven and im sleepy

Answer 37

Me too, even though I'm finishing up a report due tomorrow (I usually never do this)

Answer 38

Good bye