Polynomial Sign 0A0502

1. **State the problem:** We want to find the sign of the function $$f(x) = (2x - 1)(3x + 5)(x + 1)$$ on the interval $$-\frac{5}{3} < x < \frac{1}{2}$$ given that the zeros are at $$x = -\frac{5}{3}, x = -1,$$ and $$x = \frac{1}{2}$$. 2. **Recall the rule for sign changes:** A polynomial changes sign only at its zeros. Between zeros, the sign is constant. We will test the sign of $$f(x)$$ in the intervals determined by the zeros: - $$(-\infty, -\frac{5}{3})$$ - $$(-\frac{5}{3}, -1)$$ - $$(-1, \frac{1}{2})$$ - $$(\frac{1}{2}, \infty)$$ 3. **Check the sign of each factor on the interval $$-\frac{5}{3} < x < \frac{1}{2}$$:** - For $$2x - 1$$: - At $$x = 0$$ (which is in the interval), $$2(0) - 1 = -1 < 0$$, so $$2x - 1$$ is negative on the interval. - For $$3x + 5$$: - At $$x = 0$$, $$3(0) + 5 = 5 > 0$$, so $$3x + 5$$ is positive on the interval. - For $$x + 1$$: - At $$x = 0$$, $$0 + 1 = 1 > 0$$, so $$x + 1$$ is positive on the interval. 4. **Determine the sign of $$f(x)$$ on the interval:** $$f(x) = (2x - 1)(3x + 5)(x + 1) = (\text{negative})(\text{positive})(\text{positive}) = \text{negative}$$ 5. **Conclusion:** On the interval $$-\frac{5}{3} < x < \frac{1}{2}$$, $$f(x)$$ is always negative. **Final answer:** B f is always negative on the interval.