The zeros appear to be {-2, 0, 0, +3}. These x-intercepts suggest the factors
F'(x) = (x+2)(x^2)(x-3)
When we evaluate F'(5), we get
F'(5) = (7)(25)(2) = 350
We know F(5) = -175, so there must be an additional factor (lead coefficient 'a') of -175/350 = -1/2.
The polynomial is
F(x) = -1/2(x+2)(x^2)(x-3)
F(x) = -1/2x^4 + 1/2x^3 + 3x^2
F'(5) = (7)(25)(2) = 350
We know F(5) = -175, so there must be an additional factor (lead coefficient 'a') of -175/350 = -1/2.
The polynomial is
F(x) = -1/2(x+2)(x^2)(x-3)
F(x) = -1/2x^4 + 1/2x^3 + 3x^2