Answer:
![(u/v)(x)=-x^(3)+x^(2)-1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ah4axfwsg0wwdoxuvj7wx5qwwjqyttf7jd.png)
Explanation:
The arithmetic operation of the two equations presented in this case can be solved as a division of polynomials.
We can write the problem as:
![(u/v)(x)=(u(x))/(v(x)) \\\\(u/v)(x)=(x^5-x^4+x^2)/(-x^2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/njpp6hmyt85usqbbd0qgvrrs5xv0tlyh2u.png)
We need not to forget the negative in the denominator. Then we can factor the numerator as follows:
![(u/v)(x)=((-x^2)(-x^3+x^2-1))/((-x^2))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wb5s36ifexwuchcgk250i7uk37qzngey7a.png)
Now we can easily spot the solution, and get there with the following steps:
![(u/v)(x)=((-x^2)/(-x^2)) (-x^3+x^2-1)\\\\(u/v)(x)=(1)(-x^3+x^2-1)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/eu245x8jcwj95s70s7aiokz70zlv0xk6rs.png)
As I said earlier we need to remember that the negative will change the symbols in the equation once we factor the polynomial.
And like that we get to the answer:
![(u/v)(x)=-x^(3)+x^(2)-1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ah4axfwsg0wwdoxuvj7wx5qwwjqyttf7jd.png)