Find the zeros and their multiplicities. Consider using Descartes' rule of signs and the upper and lower bound theorem to limit your search for rational zeros. Write numbers as integers or simplified fractions. \[ h(x)=9 x^{4}-60 x^{3}+151 x^{2}-182 x+98 \]

To find the zeros of the polynomial \( h(x) = 9x^4 - 60x^3 + 151x^2 - 182x + 98 \), we can first apply Descartes' rule of signs. Observing the coefficients, we note that there are two sign changes, indicating there could be 2 or 0 positive real roots. For negative roots, consider \( h(-x) = 9x^4 + 60x^3 + 151x^2 + 182x + 98 \), which has no sign changes. Hence, there are no negative roots. Now, we apply the Rational Root Theorem. The potential rational roots are the factors of the constant term (98) divided by the factors of the leading coefficient (9). This gives us candidates like \( \pm 1, \pm 2, \pm 7, \pm 14, \pm 49, \pm 98, \pm \frac{1}{9}, \pm \frac{2}{9}, \ldots \). Testing these candidates, we discover that \( x = 2 \) is a root. We can then perform synthetic division of \( h(x) \) by \( x - 2 \) to simplify the polynomial, leading us to \( 9x^3 - 34x^2 + 83x - 49 \). Repeating the same process with Descartes’ rule and the Rational Root Theorem on the reduced polynomial ultimately leads us to find the other roots, yielding \( x = 2 \) with multiplicity 1, and a pair of conjugate complex roots with multiplicity 2. Thus, the complete factorization leads to zeros at \( x = 2 \) and a complex conjugate pair derived from the cubic nature of the factor remaining after division. Each corresponding multiplicity can now be determined through further exploration or polynomial long division. Enjoy the adventure of discovering roots!