Question

I have absolutely no clue how to solve this PLEASE HELP.

I'm suppose to factor this... but how??
x(2y-x)^2 + 2x^2 (x-2y)

Answer 1

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

Explanation:

We want to factor the following expression:

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

First, notice that both of the terms have an x. Thus, let's factor out the x first:

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

Now, notice the similarity between (2y-x) and (x-2y). If we multiply either of them by a negative, they will be the same.

Therefore, we can factor out a negative from (x-2y). This will give us:

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

Since multiplication is commutative:

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

Now, we can factor out the (2y-x):

`=x(2y-x)((2y-x)-2x)`

And we can simplify this to get:

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

Alternate Method:

Starting from here:

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

We can factor out a negative from the (2y-x)² term. This yields:

`=x((-1(x-2y))^2+2x(x-2y))`

Since -1 squared is positive, we can simplify:

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

Now, we can factor out the (x-2y):

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

Simplifying gives:

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

While this looks different, they are exactly the same. If we factor out a negative from both the second and third term, we get:

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

The negatives will cancel, leaving us with:

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

This is exactly the same as what we acquired previously.

Therefore, both answers are correct.