Multiplying and simplifying expressions The expression is:
$$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2}$$
What I did was subtract the exponents of the numerator to the exponents of the denominator. I did a cross multiplication too. 
What confuses me is the whole numbers and how i should deal with the negatives.
Please clear it for me.

First, we'll express the product as a single fraction, and use the commutativity of multiplication to "rearrange" terms:
$$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2} = \frac{(3\cdot \color{red}{\bf 8})\cdot( 3\cdot \color{blue}{\bf 5})\;a^4b^2c^3\; x^3y^3 z^3}{(5\cdot \color{blue}{\bf 5})\cdot(2\cdot \color{red}{\bf 8})\;a^2b^2c^2\;xy^2z^5}$$
Now, we can cancel common integer factors that appear in both numerator and denominator, and multiply the remaining integer factors:
$$\frac{(3\cdot \color{red}{\bf 8})\cdot( 3\cdot \color{blue}{\bf 5})\;a^4b^2c^3\; x^3y^3 z^3}{(5\cdot \color{blue}{\bf 5})\cdot(2\cdot \color{red}{\bf 8})\;a^2b^2c^2\;xy^2z^5}= \frac{9a^4b^2c^3x^3y^3z^3}{10\;a^2 b^2 c^2 x^2z^5}$$
Now, using the fact that $\dfrac {k^n}{k^m} = k^{n-m}$ for $n \geq m$, and that $\dfrac {k^n}{k^m} = \dfrac 1{k^{m-n}}$ for $n \lt m$, then simplifying, we have:
$$\begin{align} \frac{9a^4b^2c^3x^3y^3z^3}{10\;a^2 b^2 c^2 x^2z^5} & = \frac{9 a^{4-2}b^{2-2}c^{3-2}x^{3-1}y^{3-2}}{10z^{5-3}}\\ \\ & = \frac{9a^2b^0cx^2y}{10z^2} \\ \\ & = \frac{9a^2cx^2y}{10z^2}\end{align}$$

Putting all steps together, we have $$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2} = \frac{9a^2cx^2y}{10z^2}$$
