Answer
$-\frac{c((3c+5)}{(5c+4)}$
$a\ne -\frac{4}{5}$ and $a \ne 2$
Work Step by Step
To simplify the expression, we need to arrange the numerator and denominator in descending powers and then factor the numerator and denominator.
$\frac{10c+c^2-3c^3}{5c^2 - 6c - 8}$ = $\frac{-3c^3+c^2+10c}{5c^2 - 6c - 8}$
Now we need to factor out $-c$ from the denominator and then factor:
$\frac{-3c^3+c^2+10c}{5c^2 - 6c - 8}$ = $\frac{(-c)(3c^2-c-10)}{5c^2 - 6c - 8}$ = $\frac{(-c)((3c+5)(c-2)}{(5c+4)(c-2)}$
Now we can divide out the common factor of $(c-2)$
$\frac{(-c)((3c+5)(c-2)}{(5c+4)(c-2)}$ = $\frac{(-c)((3c+5)}{(5c+4)}$ = $-\frac{c((3c+5)}{(5c+4)}$
To find the excluded values, we need to look at the factored expression before simplifying:
$\frac{(-c)((3c+5)(c-2)}{(5c+4)(c-2)}$
Setting each factor of the denominator equal to 0, we will find the excluded values:
$5c+4 = 0$
$5c=-4$
$c=\frac{-4}{5}$
$c-2= 0$
$c = 2$
Therefore, $a\ne -\frac{4}{5}$ and $a \ne 2$
