Find all zeros of the function. \( f(x)=6 x^{3}-25 x^{2}-24 x-5 \) Entry tip: Enter the zeros (exact values, not decimal approximations) separated by commas.

To find the zeros of the function \( f(x) = 6x^3 - 25x^2 - 24x - 5 \), we can use the Rational Root Theorem to test potential rational roots. After testing, we find that \( x = -1 \) is a root. We can then use synthetic division to factor the polynomial. Dividing \( f(x) \) by \( x + 1 \) gives us \( 6x^2 - 31x - 5 \). Next, we can use the quadratic formula on \( 6x^2 - 31x - 5 = 0 \): \[ x = \frac{-(-31) \pm \sqrt{(-31)^2 - 4 \cdot 6 \cdot (-5)}}{2 \cdot 6} \] Calculating this gives us: \[ x = \frac{31 \pm \sqrt{961}}{12} = \frac{31 \pm 31}{12} \] So, we get \( x = \frac{62}{12} = \frac{31}{6} \) and \( x = 0 \). Putting it all together, the zeros of the function are \( x = -1, \frac{31}{6}, 0 \).