Looking for something else?

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

## More answers

Looking for something else?

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

- anonymous

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

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Create 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

So 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

@Nixy so I'm literally just adding them together?
It's just \[(x+y)^{2} + (x^{2} + 2xy + y^{2})\] ?

Looking for something else?

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

- anonymous

Yes. 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

For example. \( \huge \frac{a}{2} = a × 0.5 \) is an identitiy and is always true

- anonymous

so for the example you gave me, am I supposed to use the distributive property then?

- anonymous

i mean, for the first example

- anonymous

You need to expand this first (x+y)^2 and then add all like terms

- anonymous

Expand \( \huge(x+y)^2 \) and then add all like terms

- anonymous

so, expanding (x+y)^2 \[(x+y) \times (x+y)\]
right?

- anonymous

\( \huge (x+y)(x+y) = ???\) is correct

- anonymous

Now times them using foil

- anonymous

x^2 + xy^2 + y^2 ?

- anonymous

So we have \( \huge x^2 + 2xy + y^2 \)

- anonymous

Now we have \( \huge x^2 + 2xy + y^2 + x^2 + 2xy + y^2 \) combine all like terms now.

- anonymous

Add all terms that can be added together.

- anonymous

x^4 + 4xy + y^4 ?

- anonymous

We should have \( \huge 2x^2+4xy+2y^2 \)
You don't add the exponents x^2 + x^2 = 2x^2

- anonymous

So 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

Thank you so much for helping me, I'm actually understanding it a lot better now.

- anonymous

You can combine them in all types of ways to make an identity

- anonymous

You can subtract, divide, times, add and square them. They can get pretty complex or they can be simple. Any questions?

- anonymous

I think I've got it, thank you c:

- anonymous

YW, time for me to get to bed :-) Almost 12 AM here :-)

- anonymous

Haha, 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.

Looking for something else?

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