## Algebra 1

$-\frac{c((3c+5)}{(5c+4)}$ $a\ne -\frac{4}{5}$ and $a \ne 2$
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 $a \ne 2$