Solve the inequality. Write the solution set in interval notation if possible. Simplify your answer, if necessary. \( -5 d(d+3)^{2}(4-d)>0 \) The solution set is \( \square . \square \) \( (\square, \square)[\square, \square][\square, \square) \)

To solve the inequality \( -5 d(d+3)^{2}(4-d)>0 \), we first identify the critical points by setting the expression equal to zero: 1. \( d = 0 \) 2. \( d + 3 = 0 \) → \( d = -3 \) 3. \( 4 - d = 0 \) → \( d = 4 \) Thus, our critical points are \( d = -3, 0, 4 \). Next, we determine the intervals: 1. \( (-\infty, -3) \) 2. \( (-3, 0) \) 3. \( (0, 4) \) 4. \( (4, +\infty) \) Now we test a point from each interval in the original inequality. - For \( d = -4 \) (in \( (-\infty, -3) \)): \( -5(-4)((-4)+3)^{2}(4-(-4)) = -5(-4)(-1)^{2}(8) < 0 \) (False) - For \( d = -2 \) (in \( (-3, 0) \)): \( -5(-2)((-2)+3)^{2}(4-(-2)) = -5(-2)(1)^{2}(6) > 0 \) (True) - For \( d = 2 \) (in \( (0, 4) \)): \( -5(2)((2)+3)^{2}(4-2) = -5(2)(5)^{2}(2) < 0 \) (False) - For \( d = 5 \) (in \( (4, +\infty) \)): \( -5(5)((5)+3)^{2}(4-5) = -5(5)(8)^{2}(-1) > 0 \) (True) The intervals where the inequality holds true are \( (-3, 0) \) and \( (4, +\infty) \). Hence, the solution set in interval notation is: \((-3, 0) \cup (4, +\infty)\)