Solve the following inequalities:[tex]-(x+2)^{2} (x-3)(x+3)(x-4)[/tex] ≥ 0

Answer:[tex]\boxed{(-\infty,-3] \ U \ [3,4] \ U \ \{-2\}}[/tex]Step-by-step explanation:To solve this problem we need to use the Test Intervals for Polynomial. The following steps helps us to determine the intervals on which the values of a polynomial are entirely  negative or entirely positive.1. Find all real zeros of the polynomial arranging them in increasing order from smallest to largest. We call these zeros the critical numbers of the polynomial.Before we start applying this steps, let's multiply the entire inequality by -1 changing the direction of the inequality, so the result is:[tex](x+2)^2(x-3)(x+3)(x-4)\leq 0[/tex]So the real zeros are:[tex]x_{1}=-3 \\ \\ x_{2}=-2 \\ \\ x_{3}=3 \\ \\ x_{4}=4[/tex]2. Use the real zeros of the polynomial to determine its test intervals.[tex](-\infty,-3) \\ \\ (-3,-2) \\ \\ (-2,3) \\ \\ (3,4)[/tex]3. Take one representative x-value in each test interval, then evaluate the polynomial at this value. If the value of the polynomial is negative, the polynomial will have negative values for every x-value in the interval. If the value of the polynomial is positive, the polynomial will have positive values for every x-value in the interval.Polynomial values___________________[tex](-\infty,-3): \ (-4+2)^2(-4-3)(-4+3)(-4-4)=-224 \ NEGATIVE \\ \\(-3,-2): \ (-2.5+2)^2(-2.5-3)(-2.5+3)(-2.5-4)=4.46 \ POSITIVE \\ \\ (-2,3): \ (-1+2)^2(-1-3)(-1+3)(-1-4)=40 \ POSITIVE \\ \\ (3,4): \ (3.5+2)^2(3.5-3)(3.5+3)(3.5-4)=-49.15 \ NEGATIVE \\ \\ (4,\infty): \ (5+2)^2(5-3)(5+3)(5-4)=784 \ POSITIVE[/tex]___________________At x = -3, x = 3, and x = 4 we use brackets because the inequality includes the value at which the polynomial equals zero.Finally, if you set [tex]x=-2[/tex]:[tex](x+2)^2(x-3)(x+3)(x-4)\leq 0 \\ \\ (-2+2)^2(-2-3)(-2+3)(-2-4)\leq 0 \\ \\ 0\leq 0[/tex]So [tex]x=-2[/tex] is also a solution to the system. Finally, the solution is:[tex]\boxed{(-\infty,-3] \ U \ [3,4] \ U \ \{-2\}}[/tex]