# How do you divide (4p ^ { 3} q ^ { 5} ) ^ { 0} \div ( 6p ^ { 2} ) ^ { - 1}?

Mar 18, 2017

$6 {p}^{2}$

#### Explanation:

Any number to the 0 power is equal to 1

${\left(4 {p}^{3} {q}^{5}\right)}^{0}$ = 1

Distribute the -1 to 6 and ${p}^{2}$

$\frac{1}{\left({6}^{-} 1\right) \left({p}^{2 \left(- 1\right)}\right)}$ = $\frac{1}{\left({6}^{-} 1\right) \left({p}^{-} 2\right)}$

Any number to a negative exponent is equal to its reciprocal
For example: ${a}^{-} 2$ = $\frac{1}{{a}^{2}}$

$\frac{1}{\left({6}^{-} 1\right) \left({p}^{-} 2\right)}$= $6$${p}^{2}$

ANSWER: $6$${p}^{2}$

Mar 18, 2017

$6 {p}^{2}$

#### Explanation:

I have used substitutions a and b to demonstrate what is happening to groups. I do this in an attempt reduce confusion about the action of mathematical processes

Set $a = 4 {p}^{3} {q}^{5}$ so that we have ${a}^{0}$

But ${a}^{0} = 1$ so ${\left(4 {p}^{3} {q}^{5}\right)}^{0} = 1$

Now we have: $1 \div {\left(6 {p}^{2}\right)}^{- 1}$

Set $b = {\left(6 {p}^{2}\right)}^{- 1}$

$1 \div b \text{ "=" } 1 \times \frac{1}{b}$

=1xx1/((6p^2)^(-1)

$= 1 \times 6 {p}^{2}$

$= 6 {p}^{2}$