# How do you simplify zr \cdot 2z ^ { - 4} r ^ { - 3}?

Jul 23, 2018

$\frac{2}{{z}^{3} {r}^{2}}$

#### Explanation:

$2 {z}^{- 3} {r}^{- 2} =$

$\frac{2}{{z}^{3} {r}^{2}}$

Jul 23, 2018

$z r \cdot 2 {z}^{- 4} {r}^{- 3} = \frac{2}{{z}^{3} {r}^{2}}$

#### Explanation:

Simplify:

$z r \cdot 2 {z}^{- 4} {r}^{- 3}$

Take out the constant $2$.

$2 z r {z}^{- 4} {r}^{- 3}$

Apply product rule: ${a}^{m} {a}^{n} = {a}^{m + n}$

$2 {z}^{1 + \left(- 4\right)} {r}^{1 + \left(- 3\right)}$

Simplify.

$2 {z}^{- 3} {r}^{- 2}$

Apply negative exponent rule: ${a}^{- m} = \frac{1}{a} ^ m$

$\frac{2}{{z}^{3} {r}^{2}}$

Jul 23, 2018

$\frac{2}{{z}^{3} {r}^{2}}$

#### Explanation:

We can rewrite this with the constant out front as

$2 z \cdot {z}^{- 4} \cdot r \cdot {r}^{- 3}$

When we multiply exponents, we add the powers. We now have

$2 {z}^{- 3} {r}^{- 2}$

We can make the negative exponents positive by bringing them to the denominator. We get

$\frac{2}{{z}^{3} {r}^{2}}$

Hope this helps!