The function below has at least one rational zero. Use this fact to find all zeros of the function. \[ f(x)=2 x^{3}+9 x^{2}+5 x-7 \] If there is more than one zero, separate them with commas. Winite evact values, not decimal approximations. \( \square \)

To find the rational zeros of the function \( f(x) = 2x^3 + 9x^2 + 5x - 7 \), we can use the Rational Root Theorem, which suggests that possible rational zeros are the factors of the constant term divided by the factors of the leading coefficient. Here, the factors of \(-7\) (the constant term) are \(\pm 1, \pm 7\), and the factors of \(2\) (the leading coefficient) are \(\pm 1, \pm 2\). This gives us potential rational zeros of \(\pm 1, \pm \frac{1}{2}, \pm 7, \pm \frac{7}{2}\). By testing these possible values, we find that \(x = 1\) is a zero. We can now use synthetic division to divide \(f(x)\) by \(x - 1\): \[ \begin{array}{r|rrrr} 1 & 2 & 9 & 5 & -7 \\ & & 2 & 11 & 16 \\ \hline & 2 & 11 & 16 & 9 \\ \end{array} \] This gives us a quotient of \(2x^2 + 11x + 9\). Now, we need to find the zeros of \(2x^2 + 11x + 9\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 2 \cdot 9}}{2 \cdot 2} = \frac{-11 \pm \sqrt{121 - 72}}{4} = \frac{-11 \pm \sqrt{49}}{4} = \frac{-11 \pm 7}{4}. \] This gives us two more zeros: 1. \(x = \frac{-4}{4} = -1\) 2. \(x = \frac{-18}{4} = -\frac{9}{2}\) Thus, the zeros of the function \(f(x)\) are: \(x = 1, -1, -\frac{9}{2}\)