Question 1030147
Let {{{f(x) = ax^3 + bx^2 +cx + d}}}.
Since f(0)= 0, ==> d = 0

==> {{{f(x) = ax^3 + bx^2 +cx}}}
Now f(-1) = 15 ==> -a + b - c = 15, while
f(1) = -5 ==> a + b + c = -5.
Adding the corresponding sides of the two preceding equations, we get b = 5.
==> a + c = -10  <--------Equation (A)

f(2) = 12 ==> 8a + 4b + 2c = 12 ==> 8a+20+2c = 12, or 

4a+c = -4 <----------Equation (B)

after simplifying...
Solving for a and c from Equations A and B, we get a = 2 and c = -12.

==> {{{f(x) = 2x^3 + 5x^2 - 12x = x(2x-3)(x+4)}}}.

The roots correspond to the x-coordinates of the x-intercepts.  Thus the x-intercepts are (0,0), (3/2,0), and (-4,0).