Answer:
[tex]=x(2y-x)(2y-3x)[/tex]
Step-by-step explanation:
We want to factor the following expression:
[tex]x(2y-x)^2+2x^2(x-2y)[/tex]
First, notice that both of the terms have an x. Thus, let's factor out the x first:
[tex]=x((2y-x)^2+2x(x-2y))[/tex]
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:
[tex]=x((2y-x)^2+2x(-(2y-x))[/tex]
Since multiplication is commutative:
[tex]=x((2y-x)^2-2x(2y-x))[/tex]
Now, we can factor out the (2y-x):
[tex]=x(2y-x)((2y-x)-2x)[/tex]
And we can simplify this to get:
[tex]=x(2y-x)(2y-3x)[/tex]
Alternate Method:
Starting from here:
[tex]x((2y-x)^2+2x(x-2y))[/tex]
We can factor out a negative from the (2y-x)² term. This yields:
[tex]=x((-1(x-2y))^2+2x(x-2y))[/tex]
Since -1 squared is positive, we can simplify:
[tex]=x((x-2y)^2+2x(x-2y))[/tex]
Now, we can factor out the (x-2y):
[tex]=x(x-2y)((x-2y)+2x)[/tex]
Simplifying gives:
[tex]=x(x-2y)(3x-2y)[/tex]
While this looks different, they are exactly the same. If we factor out a negative from both the second and third term, we get:
[tex]=x(-(2y-x))(-(2y-3x))[/tex]
The negatives will cancel, leaving us with:
[tex]=x(2y-x)(2y-3x)[/tex]
This is exactly the same as what we acquired previously.
Therefore, both answers are correct.
