# If p( a ) = 6a ^ { 3} - 20a - 10, what is p ( - 6) and p ( \frac { 1} { 6} )?

Dec 2, 2017

$p \left(- 6\right) = - 1196$

$p \left(\setminus \frac{1}{6}\right) = - \setminus \frac{479}{36}$ or $- 13 \setminus \frac{11}{36}$

#### Explanation:

To find the values of $p \left(- 6\right)$ and $p \left(\frac{1}{6}\right)$, we just need to plug those values into the function $6 {a}^{3} - 20 a - 10$ for $a$.

Let’s plug in $- 6$ for $p$ first:

$p \left(a\right) = 6 {a}^{3} - 20 a - 10$

$\setminus \implies p \left(- 6\right) = 6 {\left(- 6\right)}^{3} - 20 \left(- 6\right) - 10$

$\setminus \implies p \left(- 6\right) = 6 \left(- 216\right) - \left(- 120\right) - 10$

$\setminus \implies p \left(- 6\right) = - 1296 + 120 - 10$

$\setminus \implies p \left(- 6\right) = - 1176 - 10$

$\setminus \implies p \left(- 6\right) = - 1186$

Now we do the same for $p = \setminus \frac{1}{6}$:

$p \left(a\right) = 6 {a}^{3} - 20 a - 10$

$\setminus \implies p \left(\setminus \frac{1}{6}\right) = 6 {\left(\setminus \frac{1}{6}\right)}^{3} - 20 \left(\setminus \frac{1}{6}\right) - 10$

$\setminus \implies p \left(\setminus \frac{1}{6}\right) = 6 \left(\setminus \frac{1}{216}\right) - \setminus \frac{10}{3} - 10$

$\setminus \implies p \left(\setminus \frac{1}{6}\right) = \setminus \frac{1}{36} - \setminus \frac{10}{3} - 10$

$\setminus \implies p \left(\setminus \frac{1}{6}\right) = \setminus \frac{2}{72} - \setminus \frac{240}{72} - \setminus \frac{720}{72}$

$\setminus \implies p \left(\setminus \frac{1}{6}\right) = - \setminus \frac{958}{72}$

That can also be expressed as $- 13 \setminus \frac{11}{33}$