Use the remainder theorem to determine if the given number \( c \) is a zero of the polynomial. \( f(x)=3 x^{3}-7 x^{2}-18 x+42 \) (a) \( c=-2 \) (b) \( c=-\sqrt{6} \)

To check if \( c \) is a zero of the polynomial \( f(x) = 3x^3 - 7x^2 - 18x + 42 \) using the remainder theorem, we need to evaluate \( f(c) \) for each value of \( c \). (a) For \( c = -2 \): \[ f(-2) = 3(-2)^3 - 7(-2)^2 - 18(-2) + 42 \] Calculating it step-by-step: \[ = 3(-8) - 7(4) + 36 + 42 \] \[ = -24 - 28 + 36 + 42 \] \[ = -52 + 78 \] \[ = 26 \]. Since \( f(-2) = 26 \neq 0 \), \( c = -2 \) is not a zero of the polynomial. (b) For \( c = -\sqrt{6} \): \[ f(-\sqrt{6}) = 3(-\sqrt{6})^3 - 7(-\sqrt{6})^2 - 18(-\sqrt{6}) + 42 \] Calculating it step-by-step: \[ = 3(-6\sqrt{6}) - 7(6) + 18\sqrt{6} + 42 \] \[ = -18\sqrt{6} - 42 + 18\sqrt{6} + 42 \] All terms involving \(\sqrt{6}\) cancel out, and the constant terms also cancel: \[ = 0 \]. Since \( f(-\sqrt{6}) = 0 \), \( c = -\sqrt{6} \) is a zero of the polynomial.