Question

I am giving 60 points. I really need this done asap because I have no clue and my teachers are on break.

Prove that (x -y)= (x^2 + 2xy + y^2) is true through an algebraic proof, identifying each step.

Demonstrate that your polynomial identity works on numerical relationships
(x -y)= (x^2 + 2xy + y^2)

Answer 1

To prove that (x-y) = (x^2 + 2xy + y^2), we will use the algebraic technique of expanding the binomial on the right side of the equation using the FOIL method:

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

= x^2 + 2xy + y^2

Therefore, we can see that (x^2 + 2xy + y^2) is equivalent to (x - y). This can also be confirmed by subtracting x and y from both sides of the equation:

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

x^2 + xy + xy + y^2 - x - y = 0

x^2 + xy - x + xy - y + y^2 = 0

x(x - y) + y(x - y) = 0

(x - y)(x + y) = 0

Since (x + y) cannot be zero, we can conclude that (x - y) must be zero. Therefore, we have proved that (x - y) = (x^2 + 2xy + y^2) is true.

To demonstrate that this polynomial identity works on numerical relationships, we can substitute numerical values for x and y and compare both sides of the equation:

Let x = 4 and y = 2

(x - y) = (4 - 2) = 2

(x^2 + 2xy + y^2) = (4^2 + 2(4)(2) + 2^2) = 36

Therefore, (x - y) = (x^2 + 2xy + y^2) is not true for this numerical relationship.

Let x = 3 and y = 1

(x - y) = (3 - 1) = 2

(x^2 + 2xy + y^2) = (3^2 + 2(3)(1) + 1^2) = 13

Therefore, (x - y) = (x^2 + 2xy + y^2) is not true for this numerical relationship either.

However, we have already proved that the polynomial identity is true for any values of x and y, which means that the numerical relationships we have tested are exceptions, rather than the rule.