# (x+y)2 and the answer i got was x2+y2?

As you just read, i figured out the question.. but i know the answer is wrong! I cant find another way to get another answer!.Could you explain to me what error i did wrong and show me the proper way to do it?

### 9 Answers

- 1 decade agoFavorite Answer
If that "2" meant "²", then the answer is wrong. In this case, the answer is "x² + 2xy + y²". But if it really meant 2, your answer is correct. Oh well, good luck :D

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- 1 decade ago
Well, one thing is that generally you put the number before the letter, therefore 2x + 2y

Another thing that may have gone wrong is that you have misread the question. If the '2' was after the (x+y), (x+y) may actually be squared (the number is smaller and slightly raised next to the bracket).

If that is the case, then just do (x+y)(x+y) which is x^2 + 2xy + y^2

In case you didn't know, in general notation '^' means, to the power of; so x^2 means x squared

Hope this helps!

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- BieberLv 61 decade ago
Squared means that something is multiplied by itself.. meaning:

(x+y)^2 = (x+y) * (x+y)

Now just distribute everything using the FOIL method if you like:

x^2 + xy + xy + y^2, or

x^2 + 2xy + y^2

Have a great day! :-)

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- Ron JLv 51 decade ago
you have to multiply the x in one bracket by the y in the other. and the y by the other x

so x^2 +2xy +y^2

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- RU MattLv 61 decade ago
(x + y)^2, unfortunately you cannot just square both terms. Expand it, by writing:

(x + y) * (x + y), which is equal to (x + y)^2

Now, use the FOIL method, to multiply the First, Outer, Inner, and Last:

First: x^2

Outer: xy

Inner: yx (or xy)

Last: y^2

Now, add them up:

x^2 + xy + xy + y^2

So, (x + y)^2 = x^2 + 2xy + y^2

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- cidyahLv 71 decade ago
(x+y)^2 =(x+y)(x+y) = x^2+xy+xy+y^2 = x^2+2xy+y^2

2xy is missing in your answer,

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- Amar SoniLv 71 decade ago
(x+y)2 = (x+y)(x+y) = x*x+x*y+y*x+y*y =x^2+x*y+x*y+y^2 =x^2+2xy+y^2................Ans

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- ComoLv 71 decade ago
(x + y) ² = (x + y) (x + y) = x ² + xy + xy + y ² = x ² + 2xy + y ²

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- Anonymous1 decade ago
Distributive Property, your answer is correct.

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