Question

Question:
-Label and display your new polynomial identity

-Prove that it is true through an algebraic proof, identifying each step

-Demonstrate that your polynomial identity works on numerical relationships

Below is a list of some sample factors you may use to help develop your own identity.

(x – y)
(x + y)
(y + x)
(y – x)
(x + a)
(y + b)
(x2 + 2xy + y2)
(x2 – 2xy + y2)
(ax + b)
(cy + d)

*** I do not need a straight answer, I need guidance. For example, I do not know which is the best set of factors to chose.

Answer 1

@freckles

Answer 2

@Michele_Laino

Answer 3

@Mehek14

Answer 4

haven't learned polynomial yet

Answer 5

@jim_thompson5910

Answer 6

so it sounds like you have to use those factors and multiply them out

you don't have to choose all of them, just 2 or 3 maybe

Answer 7

(x – y)*(x+y) = ???

Answer 8

I'm not allowed to use that order because it was used in the lesson

Answer 9

Would (y+x)(y-x) work?

Answer 10

sure that works

Answer 11

what does that multiply out to?

Answer 12

FOIL right?

Answer 13

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

Answer 14

very good

Answer 15

can you simplify that at all?

Answer 16

Yep! do the -xy and +xy cancel out or do they become -2xy? If so it would be
y^2-2xy-x^2

Answer 17

nope, more like 0xy since the -1 and the +1 add to 0

Answer 18

it would be -2xy if you had -xy-xy

Answer 19

Thats what I was thinking

Answer 20

0xy is the same as 0, so it goes away

Answer 21

Okay then so it would be y^2-x^2?

Answer 22

yes it is

Answer 23

(y+x)(y-x) = y^2 - x^2 is an identity

Answer 24

Awesome!

Answer 25

an identity is any equation that is always true for the allowed input variables

Answer 26

What's next, how would I prove it

Answer 27

well you effectively did really

when you used the distributive property, that is an algebraic property

Answer 28

when you FOIL, you're really using the distributive property twice

Answer 29

Is there a name to this identity?

Answer 30

Okay I see

Answer 31

(y+x)(y-x) = y^2 - x^2

difference of squares

Answer 32

Ohhh okay cool! So when the question states: Demonstrate that your polynomial identity works on numerical relationships. What does that mean?

Answer 33

you just plug in various x,y pairs

Answer 34

and prove the equation holds true

Answer 35

And the numbers that I plug in can be random?

Answer 36

yes

Answer 37

And I would plug in those numbers to the identity that we just made? to both sides?

Answer 38

For example

(x,y) = (7,2)

-------------

(y+x)(y-x) = y^2 - x^2

(2+7)(2-7) = 2^2 - 7^2

(9)(-5) = 4 - 49

-45 = -45

Answer 39

so (y+x)(y-x) = y^2 - x^2 has been proven true for the specific case when x = 7, y = 2

Answer 40

Wow great example! I understand now

Answer 41

what's another example?

Answer 42

Okay here is another one:

(x,y) = (2,4)

(y+x)(y-x) = y^2 - x^2

(4+2)(4-2) = 4^2 - 2^2

(6)(2)=16-4

12=12

Answer 43

Is that correct or did I make a mistake?

Answer 44

you nailed it, nice work

Answer 45

so if you managed to do EVERY possible x,y pair allowed, then you'd find that they all work. Of course, there are infinitely many pairs, so it's not possible to do this. This is why the algebraic proof is much more effective

Answer 46

Yay!!! So the whole thing is done?

Answer 47

I really appreciate you helping me

Answer 48

yes it is and I'm glad I could help