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anonymous
 one year ago
will fan and medal~
polynomials and identities ~
Basically I have an assignment where I need to make my own polynomial identity
anonymous
 one year ago
will fan and medal~ polynomials and identities ~ Basically I have an assignment where I need to make my own polynomial identity

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Create your own using the columns below. See what happens when different binomials or trinomials are combined. Square one factor from column A and add it to one factor from column B to develop your own identity. Column A: (x − y) (x + y) (y + x) (y  x) Column B: (x2 + 2xy + y2) (x2 − 2xy + y2) (ax + b) (cy + d)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So do what they say. Square one factor from column A and add it to one factor from column B to develop your own identity. Column A: \( (x + y)^2 \) Column B: \( (x^2 + 2xy + y^2) \) add them \( (x + y)^2 + (x^2 + 2xy + y^2) = ?? \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Nixy so I'm literally just adding them together? It's just \[(x+y)^{2} + (x^{2} + 2xy + y^{2})\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. An identity is an equation that is always true. Once you solve by adding them together and put it on the other side of the = you will have an equation that is always true (identity)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For example. \( \huge \frac{a}{2} = a × 0.5 \) is an identitiy and is always true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for the example you gave me, am I supposed to use the distributive property then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean, for the first example

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You need to expand this first (x+y)^2 and then add all like terms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Expand \( \huge(x+y)^2 \) and then add all like terms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, expanding (x+y)^2 \[(x+y) \times (x+y)\] right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\( \huge (x+y)(x+y) = ???\) is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now times them using foil

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we have \( \huge x^2 + 2xy + y^2 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now we have \( \huge x^2 + 2xy + y^2 + x^2 + 2xy + y^2 \) combine all like terms now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Add all terms that can be added together.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We should have \( \huge 2x^2+4xy+2y^2 \) You don't add the exponents x^2 + x^2 = 2x^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we have an identity now and no matter what value we use for x or y is always true.. Below is our identity So we have an identity below now. \( \large (x+y)^{2} + (x^{2} + 2xy + y^{2}) = 2x^2+4xy+2y^2 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much for helping me, I'm actually understanding it a lot better now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can combine them in all types of ways to make an identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can subtract, divide, times, add and square them. They can get pretty complex or they can be simple. Any questions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I've got it, thank you c:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0YW, time for me to get to bed :) Almost 12 AM here :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha, same here. This is part of my last assignment, and I just wanted to get it done right! Thanks again for all your help. You're a lifesaver.
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