The function below has at least one rational zero. Use this fact to find all zeros of the function. \[ h(x)=5 x^{3}+12 x^{2}+12 x+7 \] If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

To find the rational zeros of the polynomial \( h(x) = 5x^3 + 12x^2 + 12x + 7 \), we can use the Rational Root Theorem. This theorem states that any potential rational zero, in the form of \( \frac{p}{q} \), is derived from the factors of the constant term and the leading coefficient. Here, the factors of 7 (constant term) are \( \pm 1, \pm 7 \) and the factors of 5 (leading coefficient) are \( \pm 1, \pm 5 \). Therefore, the potential rational zeros to test are \( \pm 1, \pm 7, \pm \frac{1}{5}, \pm \frac{7}{5} \). Testing \( x = -1 \): \[ h(-1) = 5(-1)^3 + 12(-1)^2 + 12(-1) + 7 = -5 + 12 - 12 + 7 = 2 \quad (\text{not a zero}) \] Testing \( x = -\frac{1}{5} \): \[ h\left(-\frac{1}{5}\right) = 5\left(-\frac{1}{5}\right)^3 + 12\left(-\frac{1}{5}\right)^2 + 12\left(-\frac{1}{5}\right) + 7 = -\frac{1}{5} + \frac{12}{25} - \frac{12}{5} + 7 = 0 \quad (\text{this is a zero}) \] Since \( -\frac{1}{5} \) is a zero, we can factor \( h(x) \) using synthetic division to find the other roots. Using synthetic division: ``` -1/5 | 5 12 12 7 | -1 -11 -7 ------------------------- 5 11 1 0 ``` This gives us \( 5x^2 + 11x + 1 \). We can now solve \( 5x^2 + 11x + 1 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 5 \cdot 1}}{2 \cdot 5} = \frac{-11 \pm \sqrt{121 - 20}}{10} = \frac{-11 \pm \sqrt{101}}{10} \] Thus, the zeros of the function \( h(x) \) are: \[ -\frac{1}{5}, \frac{-11 + \sqrt{101}}{10}, \frac{-11 - \sqrt{101}}{10} \]