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blurbendy
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I know about Pascal's triangle, but I don't know if that's what you're referring to.
blurbendy
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Sure, do need an explanation of what it is, or do you have a problem you want me to look at?
blurbendy
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you*
blurbendy
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Sure, do you know how to expand a binomial (something)^2?
blurbendy
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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?
blurbendy
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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|
blurbendy
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Make sense?
precal
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pascal's triangle is use to expand
precal
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|dw:1358369183479:dw|
blurbendy
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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
precal
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|dw:1358369198853:dw|do you see the pattern?
blurbendy
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@precal, that's what im attempting to show. thanks for the visual though :)
blurbendy
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perfect, so we have x^2 + xy + xy + y^2 which equals
x^2 + 2xy + y^2, which is what precal's picture shows
precal
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|dw:1358369366989:dw|
precal
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|dw:1358369387693:dw|then notice what is happening to the x powers and then the y powers
precal
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x powers start at 2 while y start at 0
precal
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this is the pattern you will follow for you problem
blurbendy
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Now, if we wanted to take that further like (x +y)^3 (like your problem) we can do two things.
precal
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|dw:1358369548166:dw|you will use the 3rd row
precal
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|dw:1358369588020:dw|
blurbendy
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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.
blurbendy
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That would give us the correct answer, OR we could use the pattern precal is showing us.
precal
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|dw:1358369600649:dw|
precal
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be careful with your second term because of the negative, just simplify and you are done
precal
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yw