Apply Descartes' Rule of Signs. Determine the number of possible positive and negative real zeros for the given function. \[ (x)=0.93 x^{4}-0.91 x^{3}+0.28 x^{2}-0.96 x+0.74 \] Each part may have more than one possibility. Use commas to separate multiple responses. Number of possible positive real zeros: Part 2 of 2 Number of possible negative real zeros: Pat

To determine the number of possible positive real zeros using Descartes' Rule of Signs, we look at the coefficients of the polynomial function given: \(0.93, -0.91, 0.28, -0.96, 0.74\). The sign changes are from positive to negative (0.93 to -0.91), negative to positive (-0.91 to 0.28), positive to negative (0.28 to -0.96), and negative to positive (-0.96 to 0.74). There are four sign changes, which means there can be 4, 2, or 0 positive real zeros. For the negative real zeros, we evaluate \(f(-x)\), which gives us \((-x)\) substituted: \(f(-x)=0.93(-x)^{4}-0.91(-x)^{3}+0.28(-x)^{2}-0.96(-x)+0.74\). This simplifies to \(0.93x^{4}+0.91x^{3}+0.28x^{2}+0.96x+0.74\) which has no sign changes (all positive coefficients), indicating there are 0 negative real zeros. Thus, the answers are: Number of possible positive real zeros: 4, 2, 0 Number of possible negative real zeros: 0