# How do you simplify (2x^2y^4*4x^2y^4*3x)/(3x^-3y^2) and write it using only positive exponents?

Feb 17, 2017

See the entire simplification process below:

#### Explanation:

First, to simplify the numerator we will rewrite this expression as:

$\frac{\left(2 \cdot 3 \cdot 4\right) \left({x}^{2} \cdot {x}^{2} \cdot x\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 \left({x}^{2} \cdot {x}^{2} \cdot x\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}}$

We can now use these two rules for exponents to simplify the numerator:

$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$\frac{24 \left({x}^{2} \cdot {x}^{2} \cdot {x}^{\textcolor{red}{1}}\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 {x}^{2 + 2 + 1} {y}^{4 + 4}}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 {x}^{5} {y}^{8}}{3 {x}^{-} 3 {y}^{2}} \to$

$\frac{8 {x}^{5} {y}^{8}}{{x}^{-} 3 {y}^{2}}$

We can now use this rule of exponents to complete the simplification:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$\frac{8 {x}^{\textcolor{red}{5}} {y}^{\textcolor{red}{8}}}{{x}^{\textcolor{b l u e}{- 3}} {y}^{\textcolor{b l u e}{2}}} \to 8 {x}^{\textcolor{red}{5} - \textcolor{b l u e}{- 3}} {y}^{\textcolor{red}{8} - \textcolor{b l u e}{2}} \to 8 {x}^{\textcolor{red}{5} + \textcolor{b l u e}{3}} {y}^{\textcolor{red}{8} - \textcolor{b l u e}{2}} \to$

$8 {x}^{8} {y}^{6}$