Solve the inequality. Write the solution set in interval notation if possible. Simplify your answer if necessary \( -3 d^{6}+20 d^{5}-25 d^{4} \leq 0 \) The solution set is \( \square . \square \) \( (\square, \square)[\square, \square][\square, \square) \)

To solve the inequality \(-3 d^{6} + 20 d^{5} - 25 d^{4} \leq 0\), we first factor out the common term, which is \(-d^{4}\): \[ -d^{4}(3d^{2} - 20d + 25) \leq 0 \] Next, we need to factor the quadratic expression \(3d^{2} - 20d + 25\). We can use the quadratic formula, \(d = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a=3\), \(b=-20\), \(c=25\): \[ d = \frac{20 \pm \sqrt{(-20)^{2} - 4 \cdot 3 \cdot 25}}{2 \cdot 3} = \frac{20 \pm \sqrt{400 - 300}}{6} = \frac{20 \pm \sqrt{100}}{6} = \frac{20 \pm 10}{6} \] This gives us two solutions: \[ d = \frac{30}{6} = 5 \quad \text{and} \quad d = \frac{10}{6} = \frac{5}{3} \] The critical points are \(d = 0\), \(d = \frac{5}{3}\), and \(d = 5\). We can test the sign of the product in each interval defined by these critical points: \((-∞, 0)\), \((0, \frac{5}{3})\), \((\frac{5}{3}, 5)\), and \((5, ∞)\). 1. For \(d < 0\): Choose \(d = -1\), \[ -(-1)^4 (3(-1)^2 - 20(-1) + 25) = -1(3 + 20 + 25) < 0 \] 2. For \(0 < d < \frac{5}{3}\): Choose \(d = 1\), \[ -(1^4)(3(1)^2 - 20(1) + 25) = -1(3 - 20 + 25) > 0 \] 3. For \(\frac{5}{3} < d < 5\): Choose \(d = 4\), \[ -(4^4)(3(4)^2 - 20(4) + 25) < 0 \] 4. For \(d > 5\): Choose \(d = 6\), \[ -(6^4)(3(6)^2 - 20(6) + 25) < 0 \] The sign analysis shows that the inequality is satisfied in the intervals \((-∞, 0)\) and \((\frac{5}{3}, 5)\). The endpoints \(\frac{5}{3}\) and \(5\) must be included as well since they make the product equal to zero. Thus, the solution set in interval notation is: \[ (-\infty, 0] \cup \left[\frac{5}{3}, 5\right] \] So, the final answer is: \((-∞, 0][\frac{5}{3}, 5)\)