Use the Rational Zero Theorem to list all possible rational zeros for the given function. \( f(x)=-3 x^{4}-15 x^{3}+19 x^{2}-16 x+6 \) Choose the answer below that lists all possible rational zeros. A. \( -1,1,-3,3,-\frac{1}{2}, \frac{1}{2},-\frac{1}{6}, \frac{1}{6},-\frac{1}{3}, \frac{1}{3},-\frac{3}{2}, \frac{3}{2} \) B. \( -1,1,-2,2,-3,3,-6,6,-\frac{1}{3}, \frac{1}{3},-\frac{2}{3}, \frac{2}{3} \) C. \( -1,1,-2,2,-3,3,-6,6,-\frac{1}{2}, \frac{1}{2},-\frac{1}{3}, \frac{1}{3},-\frac{1}{6}, \frac{1}{6} \) (D. \( -1,1,-2,2,-3,3,-6,6,-\frac{1}{2}, \frac{1}{2},-\frac{1}{6}, \frac{1}{6},-\frac{1}{3}, \frac{1}{3},-\frac{2}{3}, \frac{2}{3},-\frac{3}{2}, \frac{3}{2} \)

To apply the Rational Zero Theorem effectively, we'll consider the possible rational roots of the polynomial \( f(x) = -3x^4 - 15x^3 + 19x^2 - 16x + 6 \). The possible rational zeros are given by the formula \( \frac{p}{q} \), where \( p \) is a factor of the constant term (in this case, \( 6 \)), and \( q \) is a factor of the leading coefficient (which is \( -3 \)). The factors of \( 6 \) are \( \pm 1, \pm 2, \pm 3, \pm 6 \) and the factors of \( -3 \) are \( \pm 1, \pm 3 \). Thus, the possible rational zeros include combinations of these factors resulting in potential values like \( -1, 1, -2, 2, -3, 3, -6, 6, -\frac{1}{3}, \frac{1}{3}, -\frac{2}{3}, \frac{2}{3}, -\frac{1}{2}, \frac{1}{2}, -\frac{3}{2}, \frac{3}{2} \). The correct answer that encapsulates all these possibilities is option D: \( -1,1,-2,2,-3,3,-6,6,-\frac{1}{2}, \frac{1}{2},-\frac{1}{6}, \frac{1}{6},-\frac{1}{3}, \frac{1}{3},-\frac{2}{3}, \frac{2}{3},-\frac{3}{2}, \frac{3}{2} \). Happy polynomial hunting!