The function below has at least one rational zero. Use this fact to find all zeros of the function. \[ h(x)=6 x^{3}-43 x^{2}+6 x+7 \] If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

To find all zeros of the function \( h(x) = 6x^3 - 43x^2 + 6x + 7 \), we can follow these steps: 1. **Rational Root Theorem**: Possible rational zeros are factors of the constant term (7) divided by factors of the leading coefficient (6). Therefore, the possible rational zeros are: \[ \pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{3}, \pm \frac{7}{3}, \pm \frac{1}{6}, \pm \frac{7}{6} \] 2. **Testing Possible Zeros**: - **\( x = 7 \)**: \[ h(7) = 6(7)^3 - 43(7)^2 + 6(7) + 7 = 2058 - 2107 + 42 + 7 = 0 \] So, \( x = 7 \) is a zero. - **Synthetic Division**: Divide \( h(x) \) by \( (x - 7) \) to find the quadratic factor: \[ h(x) = (x - 7)(6x^2 - x - 1) \] 3. **Finding Remaining Zeros**: Solve the quadratic equation \( 6x^2 - x - 1 = 0 \) using the quadratic formula: \[ x = \frac{1 \pm \sqrt{1 + 24}}{12} = \frac{1 \pm 5}{12} \] This gives: \[ x = \frac{6}{12} = \frac{1}{2}, \quad x = \frac{-4}{12} = -\frac{1}{3} \] 4. **Conclusion**: All zeros of the function \( h(x) \) are: \[ 7, \frac{1}{2}, -\frac{1}{3} \] **Answer**: 7, $\dfrac{1}{2}$, $-\dfrac{1}{3}$