Oh, can someone finally help me solve this problem? Thank you!

Sep 24, 2017

frac((a + 2)(a - 3))(a + 1); $a \ne - 1 , 3 , 5$

Explanation:

We have: $\frac{a + 2}{a - 5} \div \frac{a + 1}{{a}^{2} - 8 a + 15}$

$= \frac{a + 2}{a - 5} \times \frac{{a}^{2} - 8 a + 15}{a + 1}$

$= \frac{a + 2}{a - 5} \times \frac{{a}^{2} - 3 a - 5 a + 15}{a + 1}$

$= \frac{a + 2}{a - 5} \times \frac{a \left(a - 3\right) - 5 \left(a - 3\right)}{a + 1}$

$= \frac{a + 2}{a - 5} \times \frac{\left(a - 3\right) \left(a - 5\right)}{a + 1}$

$= \frac{\left(a + 2\right) \left(a - 3\right) \left(a - 5\right)}{\left(a + 1\right) \left(a - 5\right)}$

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

Now, let's find the restricted values of $a$.

The denominators of the original expression contained $a - 5$ and ${a}^{2} - 8 a + 15$, which factorised into $\left(a - 3\right) \left(a - 5\right)$.

These values cannot be equal to zero, i.e. $a \ne 5$ and $a \ne 3 , 5$.

Combining these results we get $a \ne 3 , 5$.

Also, in the last line, the denominator is $a + 1 \ne 0$, or $a \ne - 1$.

Therefore, the restricted values of $a$ are $- 1 , 3 ,$ and $5$.