$\displaystyle \frac{(-4)^2x^2(yz)^4}{-16xy^{-4}z}$

So:

$\displaystyle \frac{-16x^2y^4z^4}{-16xy{-4}z}$

Then:

$\displaystyle xy^8z^3$

However, I'm told the correct answer is $\displaystyle -xy^8z^3$

Why $\displaystyle -x$ instead of $\displaystyle x$ ?