Consider:
(x+y)^2
= (x+y)(x+y)
= x^2 + xy + xy + y^2
= x^2 +2xy + y^2
Question:
(-2-5)^-3
= 1/(-2-5)^3
= 1/(-2-5)(-2-5)(-2-5)
= 1/?
How to foil such an expression?
(x+y)^3
= (x+y)(x+y)(x+y)
= ?
As again, thanks for the help.
Consider:
(x+y)^2
= (x+y)(x+y)
= x^2 + xy + xy + y^2
= x^2 +2xy + y^2
Question:
(-2-5)^-3
= 1/(-2-5)^3
= 1/(-2-5)(-2-5)(-2-5)
= 1/?
How to foil such an expression?
(x+y)^3
= (x+y)(x+y)(x+y)
= ?
As again, thanks for the help.
You can either look it up as a formula or do it in steps. As you said:
(x + y)^2 = x^2 + 2xy + y^2. Thus:
(x + y)^3 = (x + y)(x^2 + 2xy + y^2) = x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)
= x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3
So
(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
Similarly:
(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
-Dan
think of the distributive property:
x(a+b)=ax+bx
now try: (a+b)(x+y)
use distributive property: x(a+b)+y(a+b)
use distributive property again: ax+bx+ay+by
now try: (x+y)^3
rewrite: (x+y)(x+y)(x+y)
use foil: (x^2+2xy+y^2)(x+y)
use distributive property: x^2(x+y)+2xy(x+y)+y^2(x+y)
can you solve from there?
With a little work you can find the following:
(x+y)^2 = x^2 + 2xy + y^2
(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5
etc.
Note that in the expansion for (x+y)^n each term takes the form cx^a*y^b with a+b = n, and all possible terms are listed.
As far as the constants are concerned there are various ways to produce them, the simplest being the combinitorial function, usually written as
nCr = (n!)/[(r!)(n - r)!]
You can also use Pascal's Triangle. (Which, if I may be allowed to brag, I discovered on my own. )
So, to write out, say, (x+y)^4 we do the following:
(x+y)^4 = (4C0)*x^4*y^0 + (4C1)x^(4-1)*y^1 + (4C2)x^(4-2)*y^2 + (4C3)x^(4-3)*y^3 + (4C4)x^(4-4)*y^4
(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
-Dan