# How do you simplify (2p^3q^2)/(8p^4q) div (4pq^2)/(16p^4)?

Jul 16, 2015

$\frac{\frac{2 {p}^{3} {q}^{2}}{8 {p}^{4} q}}{\frac{4 p {q}^{2}}{16 {p}^{4}}} = {p}^{2} / q$

#### Explanation:

We start off with what's given:

$\frac{\frac{2 {p}^{3} {q}^{2}}{8 {p}^{4} q}}{\frac{4 p {q}^{2}}{16 {p}^{4}}}$

Recall: $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{\frac{a}{b}}{\frac{c}{d}} \cdot \frac{\frac{d}{c}}{\frac{d}{c}} = \frac{a}{b} \cdot \frac{d}{c}$

So,

$\frac{\frac{2 {p}^{3} {q}^{2}}{8 {p}^{4} q}}{\frac{4 p {q}^{2}}{16 {p}^{4}}} = \left(\frac{2 {p}^{3} {q}^{2}}{8 {p}^{4} q}\right) \cdot \left(\frac{16 {p}^{4}}{4 p {q}^{2}}\right)$

Recall: ${a}^{5} / {a}^{3} = {a}^{5} \cdot {a}^{-} 3 = {a}^{5 - 3} = {a}^{2}$

So,

$\left(\frac{2 {p}^{3} {q}^{2}}{8 {p}^{4} q}\right) \left(\frac{16 {p}^{4}}{4 p {q}^{2}}\right)$$=$$\left(\frac{{p}^{3} {q}^{2}}{4 {p}^{4} q}\right) \left(\frac{4 {p}^{4}}{p {q}^{2}}\right)$

The $\left(\frac{{p}^{3} {q}^{2}}{4 {p}^{4} q}\right)$ part becomes:

$\left(\frac{{p}^{3} {q}^{2}}{4 {p}^{4} q}\right) = \left({p}^{3} {q}^{2}\right) \left({4}^{- 1} {p}^{- 4} {q}^{- 1}\right) = {4}^{- 1} {p}^{3 - 4} {q}^{2 - 1} = {4}^{- 1} {p}^{- 1} {q}^{1}$

$= \frac{q}{4 p}$

The $\left(\frac{4 {p}^{4}}{p {q}^{2}}\right)$ part becomes:

$\left(\frac{4 {p}^{4}}{p {q}^{2}}\right) = \left(4 {p}^{4}\right) \left({p}^{- 1} {q}^{- 2}\right) = 4 {p}^{4 - 1} {q}^{- 2} = 4 {p}^{3} {q}^{- 2}$

$= \frac{4 {p}^{3}}{q} ^ 2$

So, our original setup was:

$\frac{\frac{2 {p}^{3} {q}^{2}}{8 {p}^{4} q}}{\frac{4 p {q}^{2}}{16 {p}^{4}}} = \left(\frac{2 {p}^{3} {q}^{2}}{8 {p}^{4} q}\right) \cdot \left(\frac{16 {p}^{4}}{4 p {q}^{2}}\right) = \left(\frac{{p}^{3} {q}^{2}}{4 {p}^{4} q}\right) \left(\frac{4 {p}^{4}}{p {q}^{2}}\right)$

$= \left(\frac{q}{4 p}\right) \left(\frac{4 {p}^{3}}{q} ^ 2\right) = \frac{4 {p}^{3} q}{4 p {q}^{2}} = {p}^{2} / q$

Jul 17, 2015

The answer is ${p}^{2} / q$.

#### Explanation:

$\frac{2 {p}^{3} {q}^{2}}{8 {p}^{4} q}$$\div$$\frac{4 p {q}^{2}}{16 {p}^{4}}$

Reduce the numerical part of each fraction.

$\frac{{\cancel{2}}^{1} {p}^{3} {q}^{2}}{{\cancel{8}}^{4} {p}^{4} q}$$\div$$\frac{{\cancel{4}}^{1} p {q}^{2}}{{\cancel{16}}^{4} {p}^{4}}$ =

$\frac{{p}^{3} {q}^{2}}{4 {p}^{4} q}$$\div$$\frac{p {q}^{2}}{4 {p}^{4}}$ =

Simplify each fraction.

Use the exponent rule ${x}^{m} / {x}^{n} = {x}^{m - n}$ for each fraction.

$\frac{{p}^{3} {q}^{2}}{4 {p}^{4} q} = \frac{{p}^{3 - 4} {q}^{2 - 1}}{4} = \frac{{p}^{- 1} q}{4}$

$\frac{p {q}^{2}}{4 {p}^{4}} = \frac{{p}^{1 - 4} {q}^{2}}{4} = \frac{{p}^{-} 3 {q}^{2}}{4}$

Use exponent rule ${x}^{- m} = \frac{1}{{x}^{m}}$ for each fraction.

$\frac{q}{4 p} \div \frac{{q}^{2}}{4 {p}^{3}}$

In order to divide fractions, invert the second fraction to get its reciprocal, and then multiply.

$\frac{q}{4 p} \times \frac{4 {p}^{3}}{{q}^{2}}$ =

$\frac{4 {p}^{3} q}{4 p {q}^{2}}$ =

Cancel the $4$.

$\frac{{\cancel{4}}^{1} {p}^{3} q}{{\cancel{4}}^{1} p {q}^{2}}$

Apply the exponent rules ${x}^{m} / {x}^{n} = {x}^{m - n}$ and ${x}^{- m} = \frac{1}{{x}^{m}}$.

$\left({p}^{3 - 1} {q}^{1 - 2}\right) = {p}^{2} {q}^{- 1} = \frac{{p}^{2}}{q}$