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I know about Pascal's triangle, but I don't know if that's what you're referring to.
Sure, do need an explanation of what it is, or do you have a problem you want me to look at?
you*

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Other answers:

  • hba
http://mathworld.wolfram.com/PascalsTheorem.html
Sure, do you know how to expand a binomial (something)^2?
Okay, let's start there then. Say you have something like (x + y)^2 and we want to expand that. You can rewrite (x + y)^2 so that it's like: (x + y)(x + y) Make sense so far?
Good, now we're going to multiply the first x in (x + y) to BOTH the (x + y) in the other term to get: x^2 + xy |dw:1358369073609:dw|
Make sense?
pascal's triangle is use to expand
|dw:1358369183479:dw|
Great, then we move on and do the same thing with the first y. and get xy + y^2, so in total we have x^2 + xy + xy + y^2
|dw:1358369198853:dw|do you see the pattern?
@precal, that's what im attempting to show. thanks for the visual though :)
perfect, so we have x^2 + xy + xy + y^2 which equals x^2 + 2xy + y^2, which is what precal's picture shows
|dw:1358369366989:dw|
|dw:1358369387693:dw|then notice what is happening to the x powers and then the y powers
x powers start at 2 while y start at 0
this is the pattern you will follow for you problem
Now, if we wanted to take that further like (x +y)^3 (like your problem) we can do two things.
|dw:1358369548166:dw|you will use the 3rd row
|dw:1358369588020:dw|
We could multiply (x + y)(x^2 + 2xy + y^2) And do the EXACT same thing like how we were multiplying the x to the second term and then the y to the second term.
That would give us the correct answer, OR we could use the pattern precal is showing us.
|dw:1358369600649:dw|
be careful with your second term because of the negative, just simplify and you are done
yw

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