anonymous
  • anonymous
green
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
blurbendy
  • blurbendy
I know about Pascal's triangle, but I don't know if that's what you're referring to.
blurbendy
  • blurbendy
Sure, do need an explanation of what it is, or do you have a problem you want me to look at?
blurbendy
  • blurbendy
you*

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

hba
  • hba
http://mathworld.wolfram.com/PascalsTheorem.html
blurbendy
  • blurbendy
Sure, do you know how to expand a binomial (something)^2?
blurbendy
  • blurbendy
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
  • blurbendy
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
  • blurbendy
Make sense?
precal
  • precal
pascal's triangle is use to expand
precal
  • precal
|dw:1358369183479:dw|
blurbendy
  • blurbendy
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
  • precal
|dw:1358369198853:dw|do you see the pattern?
blurbendy
  • blurbendy
@precal, that's what im attempting to show. thanks for the visual though :)
blurbendy
  • blurbendy
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
  • precal
|dw:1358369366989:dw|
precal
  • precal
|dw:1358369387693:dw|then notice what is happening to the x powers and then the y powers
precal
  • precal
x powers start at 2 while y start at 0
precal
  • precal
this is the pattern you will follow for you problem
blurbendy
  • blurbendy
Now, if we wanted to take that further like (x +y)^3 (like your problem) we can do two things.
precal
  • precal
|dw:1358369548166:dw|you will use the 3rd row
precal
  • precal
|dw:1358369588020:dw|
blurbendy
  • blurbendy
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
  • blurbendy
That would give us the correct answer, OR we could use the pattern precal is showing us.
precal
  • precal
|dw:1358369600649:dw|
precal
  • precal
be careful with your second term because of the negative, just simplify and you are done
precal
  • precal
yw

Looking for something else?

Not the answer you are looking for? Search for more explanations.