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 and their multiplicities for the polynomial \( h(x) = 9x^4 - 60x^3 + 151x^2 - 182x + 98 \), we will follow these steps: 1. **Use Descartes' Rule of Signs** to determine the number of positive and negative real roots. 2. **Apply the Upper and Lower Bound Theorem** to find bounds for the rational roots. 3. **Use the Rational Root Theorem** to list possible rational roots. 4. **Test the possible rational roots** to find actual zeros. ### Step 1: Descartes' Rule of Signs **Positive Roots:** We count the sign changes in \( h(x) \): - \( 9 \) (positive) to \( -60 \) (negative) → 1 sign change - \( -60 \) (negative) to \( 151 \) (positive) → 2 sign changes - \( 151 \) (positive) to \( -182 \) (negative) → 3 sign changes - \( -182 \) (negative) to \( 98 \) (positive) → 4 sign changes Thus, there are 4, 2, or 0 positive roots. **Negative Roots:** Now we evaluate \( h(-x) \): \[ h(-x) = 9(-x)^4 - 60(-x)^3 + 151(-x)^2 - 182(-x) + 98 = 9x^4 + 60x^3 + 151x^2 + 182x + 98 \] Counting the sign changes in \( h(-x) \): - \( 9 \) (positive) to \( 60 \) (positive) → 0 sign changes - \( 60 \) (positive) to \( 151 \) (positive) → 0 sign changes - \( 151 \) (positive) to \( 182 \) (positive) → 0 sign changes - \( 182 \) (positive) to \( 98 \) (positive) → 0 sign changes Thus, there are 0 negative roots. ### Step 2: Upper and Lower Bound Theorem To apply the Upper and Lower Bound Theorem, we can use synthetic division with potential bounds. We will test \( x = 1 \) and \( x = 2 \) to see if they are upper bounds. **Testing \( x = 1 \):** \[ \begin{array}{r|rrrrr} 1 & 9 & -60 & 151 & -182 & 98 \\ & & 9 & -51 & 100 & -82 \\ \hline & 9 & -51 & 100 & -82 & 16 \\ \end{array} \] The last value is positive, so \( x = 1 \) is not an upper bound. **Testing \( x = 2 \):** \[ \begin{array}{r|rrrrr} 2 & 9 & -60 & 151 & -182 & 98 \\ & & 18 & -84 & 134 & -96 \\ \hline & 9 & -42 & 67 & -48 & 2 \\ \end{array} \] The last value is positive, so \( x = 2 \) is not an upper bound. **Testing \( x = 3 \):** \[ \begin{array}{r|rrrrr} 3 & 9 & -60 & 151 & -182 & 98 \\ & & 27 & -99 & 156 & -78 \\ \hline & 9 & -33 & 52 & -26 & 20 \\ \end{array} \] The last value is positive, so \( x = 3 \) is not an upper bound. **Testing \( x = 4 \):** \[ \begin{array}{r|rrrrr} 4 & 9 & -60 & 151 & -182 & 98 \\ & & 36 & -96 & 220 & -248 \\ \hline & 9 & -24 & 55 & 38 & -150 \\ \end{array} \] The last value is negative, so \( x = 4 \) is an upper bound. ### Step 3: Rational Root Theorem The possible rational roots are the factors of the constant term (98) over the factors of the leading coefficient (9). The factors of 98 are \( \pm 1, \pm 2, \pm 7, \pm 14, \pm 49, \pm 98 \) and the factors of 9 are \( \pm 1, \pm 3, \pm 9 \). Thus, the possible rational roots are: \[ \pm 1, \pm \frac{1}{3}, \pm \frac{1}{9}, \pm 2, \pm \frac{2}{3}, \pm \frac{2}{9}, \pm 7, \pm \frac{7}{3}, \pm \frac{7}{9}, \pm 14, \pm \frac{14}{3}, \pm \frac{14}{9}, \pm 49, \pm \frac{49}{3}, \pm \frac{49}{9}, \pm 98, \pm \frac{98}{3}, \pm \frac{98}{9} \] ### Step 4: Testing Possible Rational Roots We will test the rational roots starting from the simplest ones. **Testing \( x = 2 \):** \[ h(2) = 9(2)^4 - 60(2)^3 + 151(2)^2 - 182(2) + 98 = 144 - 480 + 604 - 364 + 98 = 2 \] Not a root. **Testing \( x = 1 \):** \[ h(1) = 9(1)^4 - 60(1)^3 + 151(1)^2 - 182(1) + 98 = 9 - 60 + 151 - 182 + 98 = 16 \] Not a root. **Testing \( x = 7 \):** \[ h(7) = 9(7)^4 - 60(7)^3 + 151(7)^2 - 182(7) + 98 \] Calculating: \[ = 9(2401) - 60(343) + 151(49) - 182(7) + 98 \] \[ = 21609 - 20580 + 7399 - 1274 + 98 = 0 \] Thus, \( x = 7 \) is a root. Now we can perform synthetic division of \( h(x) \) by \( (x - 7) \) to find