Hi Hope,

I am going to try a similar problem.

Simplify

\[\frac{3 z^2 \times 4 z^3}{2 z^4}\]

What I see in the numerator is 3 times z times z, times 4 times z times z times z. Since multiplication is commutative I can rearrange the order to get $3 \times 4$ times a string of 5 z's multiplied together. Written in exponential form

\[3 z^2 \times 4 z^3 = 12 \times z^5\]

It is important to stop here and see what I did. The exponent of 5 in $z^5$ is the sum of the exponents in $z^2$ and $z^3.$ Hence the expression I started with can be written

\[\frac{3 z^2 \times 4 z^3}{2 z^4} = \frac{12 z^5}{2 z^4}\]

12 divided by 2 is 6 but what about the z's? Again I have a string of 5 z's multiplied together in the numerator and 4 z's multiplied together in the denominator. The 4 z's in the denominator cancel 4 of the 5 z's in the numerator so I end up with

\[\frac{3 z^2 \times 4 z^3}{2 z^4} = \frac{12 z^5}{2 z^4} = 6z.\]

If you look at the exponents you see that I subtracted the exponent of 4 from the exponent of 5. Thus rewriting my solution using the laws of exponents I get

\[\frac{3 z^2 \times 4 z^3}{2 z^4} = \frac{12 z^{3 + 2}}{2 z^4} = 6z^{5 - 4} = 6z.\]

Now try $\large \frac{2x^3 \times 2x^2}{3x^2} .$

Penny