Solve the following inequalities:[tex]-5(x-1)(x-3)^{2} \ \textless \ 0[/tex]

Answer:[tex]\boxed{(1,\infty)}[/tex]Step-by-step explanation:By using the Test Intervals for Polynomial we can solve this problem. The following steps helps us to determine the intervals on which the values of a polynomial are entirely negative or entirely positive.1. You need to 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]5(x-1)(x-3)^2 > 0[/tex]So the real zeros are:[tex]x_{1}=1 \\ \\ x_{2}=3[/tex]2. Then you must use the real zeros of the polynomial to determine its test intervals.[tex](-\infty,1) \\ \\ (1,3) \\ \\ (3,\infty)[/tex]3. Take one representative x-value in each test interval so you must 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. On the other hand, 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,1): \ 5(0-1)(0-3)^2=-45 \ NEGATIVE \\ \\(1,3): \ 5(2-1)(2-3)^2=-5 \ POSITIVE \\ \\(3,\infty): \ 5(4-1)(4-3)^2=15\ POSITIVE[/tex]___________________So the solution is:[tex]\boxed{(1,\infty)}[/tex]So the interval at x = 1 is open because  because the inequality does not include the value at which the polynomial equals zero.