Factoring, why don't you distrubute instead?

Question

Example:
(y-x)²+2(x-y)=

work:
=-(x-y)²+2(x-y)
=-(x-y)(2x-2y)
=-2(x-y)(x-y)

What's even wrong with that?
Why is it wrong to distrubute?

Answer 1

You have a plus sign. If you were to pull out a common factor it would be

(y-x)((y-x) +2) which is pretty useless.

Your method

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

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

Answer 2

what's wrong with that is that you got the wrong answer.

Here's what you should have gotten:
(y-x)^2 + 2(x-y)
(y-x)(y-x) - 2(y-x)
(y-x-2)(y-x)

Assuming that x and y are both integers, this *is* factoring.

If your teacher doesn't want you to do this for some reason, it's not because the technique is wrong, but because he wants you to know how to do something else.

Sorta like walking across the park. Nothing wrong with that, except that if you are running a race, that shortcut probably constitutes cheating....

Answer 3

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

Answer 4

there is one problem:
(y - x)^2 = (-1(x-y))^ 2 = (x-y)^2

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

Answer 5

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

-(x-y)^2 = -x^2 +2xy -y^2 which is not = y^2-2xy +x^2

Answer 6

you have written (y-x)² = -(x-y)², which is wrong

(y-x)²=(x-y)²