I won't do yours for you, but I'll do one exactly like yours 
that you can use as a model to do yours by:

 

So write x = 5 once, x = -4 twice

       x = 5;       x = -4;      x = -4

Get 0 on the right of each of the three equations:

   x - 5 = 0    x + 4 = 0    x + 4 = 0

Multiply all three left sides together and set it equal to what you get
when you multiply all three right sides together, 0∙0∙0 = 0

        (x - 5)(x + 4)(x + 4) = 0

Then you multiply that out

   (x² + 4x - 5x - 20)(x + 4) = 0

         (x² - x - 20)(x + 4) = 0

x³ + 4x² - x² - 4x - 20x - 80 = 0

          x³ - 3x² - 24x - 80 = 0

The polynomial that has the given zeros is the polynomial 
that when set equal to 0 has those solutions, so the 
polynomial that when set equal to zero is the polynomial 
that's set equal to 0 above, which is this polynomial, 
which we'll call P(x):

   P(x) = x³ - 3x² - 24x - 80    <---answer

Now do yours exactly the same way.

Edwin