# How do you find the product of 9/(t-2)*((t+2)(t-2))/3?

Mar 19, 2017

The final answer is 3(t+2) or 3t + 6 if you prefer.

#### Explanation:

You can cancel full binomials just like you can numerical values. So, the $t - 2$ term in the denominator of the first factor cancels the one in the numerator of the second factor.

$\frac{9}{\cancel{t - 2}} \cdot \frac{\left(t + 2\right) \cancel{t - 2}}{3}$

Also, since 9 is 3 x 3, you can cancel the denominator of the second factor:

$\left(\cancel{9}\right) \textcolor{red}{3} \cdot \frac{t + 2}{\cancel{3}} = 3 \left(t + 2\right)$

Mar 19, 2017

$3 t + 6$

#### Explanation:

$\textcolor{b l u e}{\text{cancel common factors}}$ on the numerators/denominators of the fractions.

$\Rightarrow \frac{\textcolor{red}{9}}{\textcolor{b l u e}{\left(t - 2\right)}} \times \frac{\left(t + 2\right) \textcolor{b l u e}{\left(t - 2\right)}}{\textcolor{red}{3}}$

$= {\textcolor{red}{\cancel{9}}}^{3} / \left(\textcolor{b l u e}{\cancel{{\left(t - 2\right)}^{1}}}\right) \times \frac{\left(t + 2\right) \textcolor{b l u e}{\cancel{{\left(t - 2\right)}^{1}}}}{\textcolor{red}{\cancel{3}}} ^ 1$

$= \frac{3 \left(t + 2\right)}{1}$

$= 3 \left(t + 2\right)$

$= 3 t + 6$