Answer:
watts
Step-by-step explanation:
Answer: C dont ask me why
Step-by-step explanation:
<h2>Steps:</h2>
So firstly, we need to factor out the GCF. Since all the terms share a GCF of x, factor that out as such:
![(x)(x^2y^2-x^4+y^3-yx^2)](https://tex.z-dn.net/?f=%28x%29%28x%5E2y%5E2-x%5E4%2By%5E3-yx%5E2%29)
Next, I will be factoring by grouping. For this, factor x²y² + y³ and -x⁴ - yx². Make sure that they have the same quantity on the inside:
![(x)(y^2(x^2+y)-x^2(x^2+y))](https://tex.z-dn.net/?f=%28x%29%28y%5E2%28x%5E2%2By%29-x%5E2%28x%5E2%2By%29%29)
Now rewrite the expression as:
![(x)(y^2-x^2)(x^2+y)](https://tex.z-dn.net/?f=%28x%29%28y%5E2-x%5E2%29%28x%5E2%2By%29)
However, we can factor this expression further. Next, we will be applying the difference of squares rule, which is
. In this case:
![y^2-x^2=(y+x)(y-x)\\\\(x)(y+x)(y-x)(x^2+y)](https://tex.z-dn.net/?f=y%5E2-x%5E2%3D%28y%2Bx%29%28y-x%29%5C%5C%5C%5C%28x%29%28y%2Bx%29%28y-x%29%28x%5E2%2By%29)
<h2>Answer:</h2>
In short, the factored form of this expression is:
![(x)(y+x)(y-x)(x^2+y)](https://tex.z-dn.net/?f=%28x%29%28y%2Bx%29%28y-x%29%28x%5E2%2By%29)
It is either negative 1968594
Or it is negative 2150814