# How do you simplify (64qt)/(16q^2t^3) and find the excluded values?

Nov 5, 2017

See a solution process below:

#### Explanation:

To simplify this expression, first rewrite the expression as:

$\frac{64}{16} \left(\frac{q}{q} ^ 2\right) \left(\frac{t}{t} ^ 3\right) \implies$

$4 \left(\frac{q}{q} ^ 2\right) \left(\frac{t}{t} ^ 3\right)$

Next, use this rule of exponents to rewrite the expression:

$a = {a}^{\textcolor{red}{1}}$

$4 \left({q}^{\textcolor{red}{1}} / {q}^{2}\right) \left({t}^{\textcolor{red}{1}} / {t}^{3}\right)$

Now, use this rule for exponents to simplify the $q$ and $t$ terms:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$

$4 \left({q}^{\textcolor{red}{1}} / {q}^{\textcolor{b l u e}{2}}\right) \left({t}^{\textcolor{red}{1}} / {t}^{\textcolor{b l u e}{3}}\right) \implies$

$4 \left(\frac{1}{q} ^ \left(\textcolor{b l u e}{2} - \textcolor{red}{1}\right)\right) \left(\frac{1}{t} ^ \left(\textcolor{b l u e}{3} - \textcolor{red}{1}\right)\right) \implies$

$4 \left(\frac{1}{q} ^ 1\right) \left(\frac{1}{t} ^ 2\right) \implies$

$4 \left(\frac{1}{q}\right) \left(\frac{1}{t} ^ 2\right) \implies$

$\frac{4}{q {t}^{2}}$

To find the exclude values for this expression we need to equate the denominator of the original expression to $0$ and solve for each term equal to $0$:

$16 {q}^{2} {t}^{3} = 0$

First Excluded Value:

${q}^{2} = 0$

$q = 0$

Second Excluded Value:

${t}^{3} = 0$

$t = 0$

The Excluded Values Are: $q = 0$ and/or $t = 0$