I need help solving this?

Question

[ x+y+2 ] [ x - (y+2) ]

Answer 1

First, get rid of the parentheses inside the second brackets, so the expression now reads [x+y+2] [x-y-2]. Now multiply each term in the first brackets by each term in the second brackets. x^2 - xy -2x + xy - y^2 - 2y + 2x - 2y -4. Now collect like terms. x^2 - y^2 - 4y - 4 (the xy and 2x terms cancel).

Answer 2

This fits into the standard (a+b)(a-b)=a^2-b^2 formula, where a=x and b=y+2. So without even doing all the multiplication, it's "obvious" that the first step to an answer is...
x^2- (y+2)^2.

If you expand the (y+2)^2 part, you get an answer of...
x^2 - y^2 - 4y - 4

(...which is pretty much what the One Gun Kid said above. I would just point out that you usually want the higher powers to the left, so all the x^2, y^2 and xy terms first; then the x and y terms; then the constants.)

Answer 3

[ x+y+2 ] [ x - (y+2) ]

ok, first get rid of the internal ( ) by distributing the -1 through
[ x+y+2 ] [ x - y - 2 ]

Now, distribute each term through (multiply everything in the second parethesis by x, then by y and then by 2
x^2 - xy - 2x + xy - y^2 - 2y + 2x - 2y - 4

Combine like terms
x^2 - y^2 - 4y - 4

Woolah :)

Answer 4

(a+b)(a-b) = a^2- b^2
So [x+(y+2)] [x-(y+2)] = x^2 - (y+2)^2
You can expand the (y+2)^2 and get the previous person's answer, but I think it is better to leave it in the above form.

Answer 5

Well, I got this answer: x²-y²-4y-4

This is how I did it;

1st step:
x²-xy-2x
xy-y²-2y
2x-2y-4


2nd Step:
x²-xy-2x+xy-y²-2y+2x-2y-4
= x²-y²-4y-4 <-----------------------

As I’ve done this on the computer, you may have to copy it out again on paper and try it for yourself. You must check my answer with yours.

Answer 6

x^2-4y-y^2-4

Answer 7

[x+y+2][x-(y+2)]
= 2xy[x-y-2]

=2xy+x-2y