`f(x)=3x^(3)-5x^(2)-48x+80`
If the zeros of function f defined above are represented by r, s, and t, what is the value of the sum `r+s+t`?

Options

A. `(3)/(5)`
B. `(5)/(13)`
C. `(17)/(3)`
D. `8`

Answer

To find the sum of the zeros \( r + s + t \) of the polynomial function \( f(x) = 3x^3 - 5x^2 - 48x + 80 \), we can use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots.

### Step-by-Step Solution:

1. **Identify the Coefficients:**
The polynomial is given as \( f(x) = 3x^3 - 5x^2 - 48x + 80 \).
Here, the coefficients are:
- \( a = 3 \) (coefficient of \( x^3 \))
- \( b = -5 \) (coefficient of \( x^2 \))
- \( c = -48 \) (coefficient of \( x \))
- \( d = 80 \) (constant term)

2. **Apply Vieta's Formulas:**
According to Vieta's formulas, for a cubic polynomial of the form \( ax^3 + bx^2 + cx + d \), the sum of the roots \( r + s + t \) is given by:
\[ r + s + t = -\frac{b}{a} \]

3. **Calculate the Sum of the Roots:**
Substituting the values of \( b \) and \( a \):
\[ r + s + t = -\frac{-5}{3} = \frac{5}{3} \]

4. **Conclusion:**
Therefore, the sum of the zeros \( r + s + t \) is:
\[ \boxed{\frac{5}{3}} \]