please do it completely plzzz

Peacewanters007 · #1 $Old$ January 3rd, 2008, 04:15 AM

(2x-2y-2z)^2
solve above!I mean expand using sonme sort of formula
plz dont joke me 4 such a questone I am a student of low level I mean low class oh junior class

DivideBy0 · #2 $Old$ January 3rd, 2008, 04:26 AM

Hi Peacewanters2007!

What exactly do you want to do with that? Do you need to expand it? Or is it part of an equation you need to solve?

wingless · #3 $Old$ January 3rd, 2008, 05:31 AM

$(a + b + c)^2 = (a + b + c)(a + b + c) =$ $a.a + a.b + a.c + b.a + b.b + b.c + c.a + c.b + c.c = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$

So, take $a=2x$ , $b=-2y$ , $c=-2z$ ,
$4x^2 + 4y^2 + 4z^2 - 8xy - 8xz + 8yz$

Or you could do this without using a formula,

$(2x - 2y - 2z)^2 = (2x - 2y - 2z)(2x - 2y - 2z) =$ $4x^2 - 4xy - 4xz -4xy + 4y^2 + 4yz - 4xz + 4yz + 4z^2 = 4x^2 + 4y^2 + 4z^2 - 8xy - 8xz + 8yz$

topsquark · #4 $Old$ January 3rd, 2008, 02:25 PM

Quote:

Originally Posted by Peacewanters007 $View Post$

(2x-2y-2z)^2
solve above!I mean expand using sonme sort of formula
plz dont joke me 4 such a questone I am a student of low level I mean low class oh junior class

Perhaps you should look at it as
$(2x - 2y - 2z) = ((2x - 2y) - 2z)^2 = (2x - 2y)^2 - 2(2x - 2y)(2z) + (2z)^2$
which you should be able to expand on your own from here.

-Dan

Jhevon · #5 $Old$ January 3rd, 2008, 08:28 PM

or you could use the multinomial theorem

(topsquark's method is perhaps easier, but you'll get a messy expression anyway)