Yea, I like to factor, the question is How?

Question

Given:
x(2y-x)²+2x²(x-2y)=

I anwsered:
=-x(x-2y)²+2x²(x-2y)
=(x-2y)[2x²-x+(x-2y)]
=x(x-2y)[2x-1+x-2y]

Where did I go wrong?

Answer 1

expand: x(2y - x)² + 2x²(x - 2y)
= x(4y² - 4xy + x²) + 2x³ - 4x²y
= 4xy² - 4x²y + x³ + 2x³ - 4x²y
= 3x³ - 8x²y + 4xy²
= x(3x² - 8xy + 4y²)

factorise (3x² - 8xy + 4y)
3x.......-2y........-2xy
x.........-2y........-6xy
________________
3x².....+4y².......-8xy

x(2y - x)² + 2x²(x - 2y) = x(3x² - 8xy + 4y²) = x(3x - 2y)(x - 2y)

Answer 2

Step 1 to Step 2 x(2y-x)^2 + 2x^2(x-2y) =
x(x-2y)^2 + 2x^2 (x-2y) = x(x-2y) [1 + 2x(x-2y)]

(x-2y)^2 = (2y-x)^2 because you multiply by -1 on both and you get +1...(x-2y)(x-2y) = (-1)(2y-x)(-1)(2y-x) = (2y-x)(2y-x)

Step 2 to Step 3
your answer -x(x-2y)^2 + 2x^2(x-2y)
(x-2y)(2x^2 -x + (x - 2y))
what it should be
(x-2y)(2x^2(x-2y) - x) = x(x-2y)(2x(x-2y) - 1)

Answer 3

= x.(2y - x)² - 2x².(2y - x)
= (2y - x).(x).[(2y - x) - 2x ]
= (2y - x).(x).(2y - 3x)
= x.(2y - x).(2y - 3x)
Your error is in first line of working:-
- x.(x - 2y)² DOES NOT EQUAL x.(2y - x)²