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. \( -1,1,-2,2,-3,3,-6,6,-\frac{1}{2}, \frac{1}{2},-\frac{1}{6}, \frac{1}{6},-\frac{1}{3}, \frac{1}{3},-\frac{2}{3}, \frac{2}{3},-\frac{3}{2}, \frac{3}{2} \)

We start with the polynomial \[ f(x)=-3x^4-15x^3+19x^2-16x+6. \] The Rational Zero Theorem tells us that any rational zero, written as \(\frac{p}{q}\) in lowest terms, must have \(p\) as a factor of the constant term and \(q\) as a factor of the leading coefficient. 1. The constant term is \(6\). Its factors are \[ \pm 1, \pm 2, \pm 3, \pm 6. \] 2. The leading coefficient is \(-3\). Its factors (ignoring the negative sign, since we consider \(\pm\)) are \[ \pm 1, \pm 3. \] 3. The possible values for \(\frac{p}{q}\) are all fractions where \(p\) comes from \(\{1,2,3,6\}\) and \(q\) comes from \(\{1,3\}\). Thus, the possible rational zeros are: - For \(q=1\): \[ \pm 1, \pm 2, \pm 3, \pm 6. \] - For \(q=3\): \[ \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{3}{3}, \pm \frac{6}{3}. \] Note that \(\frac{3}{3}=1\) and \(\frac{6}{3}=2\), so these do not add any new values. Thus, the complete list of distinct possible rational zeros is \[ \pm 1,\ \pm 2,\ \pm 3,\ \pm 6,\ \pm \frac{1}{3},\ \pm \frac{2}{3}. \] Among the answer choices, only option B lists exactly these values: \[ -1, \; 1, \; -2, \; 2, \; -3, \; 3, \; -6, \; 6, \; -\frac{1}{3}, \; \frac{1}{3}, \; -\frac{2}{3}, \; \frac{2}{3}. \] Therefore, the correct answer is B.