Have you learned the relationship between roots and coefficients?
[tex]S_n = x_1x_2...x_n = (-1)^n \dfrac{a_0}{a_n}[/tex]
That is, the sum of the roots taken n at a time (in all possible combinations) equals the constant term divided by the nth coefficient multiplied by negative one raised to the nth power. I encourage you to research why this is true, so you don't blindly use the theorem. Regardless, let
[tex]P(x) \textrm{ have roots } x_1, x_2, \textrm{ and } x_3[/tex]
Then [tex]S_1 = x_1 + x_2 + x_3 ; S_2 = x_1x_2 + x_1x_3 + x_2x_3 ; S_3 = x_1x_2x_3[/tex] If not, and you are given at least one root and there is one coefficient missing, you can do synthetic division with x = -3 and deduce what that value must be.
I can't really type synthetic division out here, but try doing it, because you know that the some value times a must equal 72.