# How do you find the product of t^2/((t-4)(t+4))*(t-4)/(6t)?

Mar 19, 2018

See a solution process below:

#### Explanation:

First, cancel common terms in the numerator and denominator:

${t}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} / \left(\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\left(y - 4\right)}}} \left(t + 4\right)\right) \cdot \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{t - 4}}}}{6 \textcolor{red}{\cancel{\textcolor{b l a c k}{t}}}} \implies$

$\frac{t}{t + 4} \cdot \frac{1}{6} \implies$

$\frac{t \cdot 1}{\left(t + 4\right) 6} \implies$

$\frac{t}{\left(6 \cdot t\right) + \left(6 \cdot 4\right)} \implies$

$\frac{t}{6 t + 24}$

However, because we cannot divide by $0$ in the original equation we must exclude some values for $t$:

• $t - 4 = 0$ means $t \ne 4$

• $t + 4 = 0$ means $t \ne - 4$

• $6 t = 0$ means $t \ne 0$

Therefore:

$\frac{t}{6 t + 24}$ where $t \ne 4$ and $t \ne - 4$ and $t \ne 0$