Question

Demonstrate the proof of a polynomial identity through an algebraic proof and a numerical proof.
(Some Examples)
(x – y)
(x + y)
(y + x)
(y – x)
(x + a)
(y + b)
(x2 + 2xy + y2)
(x2 – 2xy + y2)
(ax + b)
(cy + d)
I am unclear on what I am supposed to do.

Answer 1

I believe you have to create your own proof. I did something like this once

Answer 2

By just plugging number in or is their a better way than plug 'n chug?

Answer 3

*numbers

Answer 4

Well, basically I just took one of those, like (x+y), then cubed it, \((x+y)^3\), found what that equaled, then plugged in a number for x and y, then confirmed the identity

Answer 5

So say \[(x+y)^3\] where x=3 and y=6 I would add x+y and then cube it correct?
So \[9^3\] Which = 729

Answer 6

Well, actually,\((x+y)^3 = (x+y)(x+y)(x+y)\). But I believe that would come out to be the same once you substitute, but first we have to find what \((x+y)^3\) is in terms of x and y

Answer 7

What?

Answer 8

(x+y)^3 = (x+y)(x+y)(x+y)

Answer 9

To make it a smaller number I'm going to use \[(x-y)^3\]
So \[(6-3)^3=(6-3)(6-3)(6-3)\]
\[6-3=3\]
Soo
\[(3)(3)(3)=27\]

Answer 10

Do you know FOIL

Answer 11

First Outside Inside Last

Answer 12

Ahhh your joking

Answer 13

Yes, we need to use that with \((x+y)(x+y)(x+y)\) without substituting in any numbers

Answer 14

You know sometimes I despise FOIL

Answer 15

\[x^2-xy-xy+y^2(x-y)\]

Answer 16

Then rinse and repeat

Answer 17

\[x^3-x^2y-x^2y+xy^2\]

Answer 18

Why -xy-xy?

Answer 19

\[(x-y)(x-y)\]
First:x*x=x^2
Outside:x*-y=-xy

Answer 20

Inside and last are the same thing

Answer 21

Oh, I thought we were doing x+y :P

Answer 22

Haha Yea sorry bout didn't like the 729, 27 seemed more amiable

Answer 23

So then \[x^3−x^2y−x^2y+xy^2\] is correct?

Answer 24

From here I plug in 6 for x and 3 for y yes?

Answer 25

Remember, we need to multiply each term by x and -y
\(x^2 * x\\-xy*x\\-xy*x\\+y^2 * x\\+x^2*y\\-xy*-y\\-xy*-y\\+y^2*-y\)

Answer 26

\[3^2*3\] ?

Answer 27

\(\color{blue}{\text{Originally Posted by}}\) @Prometheus777
So then \[x^3−x^2y−x^2y+xy^2\] is correct?
\(\color{blue}{\text{End of Quote}}\)
All you did here was multiply it all by x

Answer 28

What I did was FOIL \[(x-y)^3\]
which gave me
\[(x-y)(x-y)(x-y)\]
from there I foiled \[(x-y)(x-y)\]
which gave me \[x^2-xy-xy+y^2\]
So far so good?

Answer 29

Yes, that is correct so far

Answer 30

Alright, with \[x^2-xy-xy+y^2\]
I proceeded to FOIL that against \[(x-y)\]
\[x^2-xy-xy+y^2*(x-y)\]
Which gave me \[x^3-x^2y-x^2y+xy^3\]

Answer 31

So do I plug in the numbers or is there another step?

Answer 32

When I plug 6 in for x and 3 in for y I get 162...

Answer 33

I got
\(x^3+2xy^2-y^3\)

Answer 34

How the what?

Answer 35

I understand all but the last part the \[-y^3\]

Answer 36

\(x^2 * x=x^3\\-xy*x=-x^2y\\-xy*x=-x^2y\\+y^2 * x=xy^2\\+x^2*y=+x^2y\\-xy*-y=xy^2\\-xy*-y=xy^2\\+y^2*-y=-y^3\)

\(x^3-x^2y-x^2y+xy^2+x^2y+xy^2+xy^2-y^3\)
combine like terms
\(x^3-3x^2y+3xy^2-y^3\)

I feel like I did something wrong now...

Answer 37

You did it correct Jess :)

Answer 38

I plug in 6 for x and 3 for y and it works :-)
Now can you explain how you got that?

Answer 39

You could even use pascal's triangle: