The zeros appear to be {

F'(5) = (7)(25)(2) = 350

We know F(5) = -175, so there must be an additional factor (

The polynomial is

F(x) = -1/2(x+2)(x^2)(x-3)

**-2, 0, 0, +3**}. These**x-intercepts**suggest the**factors**F'(x) =**(x+2)(x^2)(x-3)**When we evaluate F'(5), we getF'(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**