PLEASE HELPPP ASAP It's time to show off your creativity and marketing skills!

You are going to design an advertisement for a new polynomial identity that you are going to invent. Your goal for this activity is to demonstrate the proof of your polynomial identity through an algebraic proof and a numerical proof in an engaging way! Make it so the whole world wants to purchase your polynomial identity and can't imagine living without it!

You may do this by making a flier, a newspaper or magazine advertisement, making an infomercial video or audio recording, or designing a visual presentation for investors through a flowchart or PowerPoint.

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

Warning! No identities used in the lesson may be submitted. 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 Column B

(x − y) (x2 + 2xy + y2)

(x + y) (x2 − 2xy + y2)

(y + x) (ax + b)

(y − x) (cy + d)

asdfghkl

step-by-step explanation:

asdfghkl

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.