Question

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

Answer 1

`((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=p^2/q`

Explanation:

We start off with what's given:

`((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))`

Recall: `(a/b)/(c/d)=(a/b)/(c/d)*(d/c)/(d/c)=a/b*d/c`

So,

`((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=((2p^3q^2)/(8p^4q))*((16p^4)/(4pq^2))`

Recall: `a^5/a^3=a^5*a^-3=a^(5-3)=a^2`

So,

`((2p^3q^2)/(8p^4q))((16p^4)/(4pq^2))` `=` `((p^3q^2)/(4p^4q))((4p^4)/(pq^2))`

The `((p^3q^2)/(4p^4q))` part becomes:

`((p^3q^2)/(4p^4q))=(p^3q^2)(4^(-1)p^(-4)q^(-1))=4^(-1)p^(3-4)q^(2-1)=4^(-1)p^(-1)q^1`

`=q/(4p)`

The `((4p^4)/(pq^2))` part becomes:

`((4p^4)/(pq^2))=(4p^4)(p^(-1)q^(-2))=4p^(4-1)q^(-2)=4p^3q^(-2)`

`=(4p^3)/q^2`

So, our original setup was:

`((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=((2p^3q^2)/(8p^4q))*((16p^4)/(4pq^2))=((p^3q^2)/(4p^4q))((4p^4)/(pq^2))`

`=(q/(4p))((4p^3)/q^2)=(4p^3q)/(4pq^2)=p^2/q`

Answer 2

The answer is `p^2/q`.

Explanation:

`(2p^3q^2)/(8p^4q)` `-:` `(4pq^2)/(16p^4)`

Reduce the numerical part of each fraction.

`(cancel2^1p^3q^2)/(cancel8^4p^4q)` `-:` `(cancel4^1pq^2)/(cancel16^4p^4)` =

`(p^3q^2)/(4p^4q)` `-:` `(pq^2)/(4p^4)` =

Simplify each fraction.

Use the exponent rule `x^m/x^n=x^(m-n)` for each fraction.

`(p^3q^2)/(4p^4q)=(p^(3-4)q^(2-1))/(4)=(p^(-1)q)/4`

`(pq^2)/(4p^4)=(p^(1-4)q^2)/(4)=(p^-3q^2)/4`

Use exponent rule `x^(-m)=1/(x^m)` for each fraction.

`(q)/(4p)-:(q^2)/(4p^3)`

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

`(q)/(4p)xx(4p^3)/(q^2)` =

`(4p^3q)/(4pq^2)` =

Cancel the `4`.

`(cancel4^1p^3q)/(cancel4^1pq^2)`

Apply the exponent rules `x^m/x^n=x^(m-n)` and `x^(-m)=1/(x^m)`.

`(p^(3-1)q^(1-2))=p^2q^(-1)=(p^2)/q`