# Are there any restrictions on the domain of \frac { b - 3} { - b ^ { 2} + 18b - 80} \cdot \frac { b ^ { 2} - 18b + 80} { 9b }?

Apr 7, 2017

$b \ne 0 , 8 , 10$
$\therefore b \in \mathbb{R} - \left\{0 , 8 , 10\right\}$

#### Explanation:

Since the denominator cannot have $0$, we can find the restrictions on the domain in the following way.

$9 b \ne 0$
$\implies b \ne 0$

$- {b}^{2} + 18 b - 80 \ne 0$
$\implies - {b}^{2} + 10 b + 8 b - 80 \ne 0$
$\implies - b \left(b - 10\right) + 8 \left(b - 10\right) 1 = 0$
$\implies \left(b - 10\right) \left(- b + 8\right) \ne 0$
$\implies b \ne 10 , 8$

Hence $b$ cannot be either $0$ or $8$ or $10$.
$\therefore b \ne 0 , 8 , 10$ .
$\therefore b \in \mathbb{R} - \left\{0 , 8 , 10\right\}$