Question

I really need help! It won't take long.

You must:

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


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

Answer 1

@welshfella do you know where to begin?

Answer 2

I guess it wants you to pick out an identity from some of the above expressions

one would be

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

if you expand the left side using the distributive law you will get the right side

Answer 3

thats the proof .
Then you can plug in some values of x and y to demonstrate the last part.

Answer 4

@welshfella dude you totally stole that.. smh you could have gotten me in trouble!!! I REALLY DO NEED HELP. if i change those 2's into 5's do you know how carry on with the problem??

Answer 5

(x + y)^5 = x^5 + 5xy + y^5

Answer 6

@welshfella

Answer 7

no that is not correct

Answer 8

i really dont know how to to do this. can you give an original example

Answer 9

the right side is much longet
YOu need to know the Binomial Theorem to get that one.

Answer 10

???

Answer 11

OK
Prove the identity
( a - 2b)^2 = a^2 - 4ab + b^2

applying the distributive law

LHS = (a - 2b)(a - 2b) = a(a - 2b) - 2b(a - 2b)
= a^2 - 2ab - 2ab + b^2
= a^2 - 4ab + b^2
= RHS

that is the algebraic proof

Answer 12

now plug in values like a = 2 and b = 3 and find if LHS = RHS for 3rd part

Answer 13

LHS = left hand side
RHS = right hand side

Answer 14

oh inade one error the last term is 4b^2 not b^2

Answer 15

so what would be the final answer

Answer 16

for the proof where i have b^2 replace it with 4b^2

you can do the last part just put a=2 and b = 3 into
(a - 2b)^2 = a^2 - 4ab + 4b^2

to show that left side = right side

left side = (2 -2*3)^2 = -4^2 = 16

so see if you can get right side = 16 It should do.

Answer 17

ok thank you so much

Answer 18

yw