Question 470160
we have to find polynomials {{{ax^3 + bx^2 + cx + d}}} to add to {{{x^3 - 3x^2 -12x +19}}} so that the sum is divisible by {{{x^2 + x - 6 = (x+3)(x - 2)}}}.
==> -3 and 2 are roots of the divisor {{{x^2 + x - 6}}}.

Now  {{{(ax^3 + bx^2 + cx + d) + (x^3 - 3x^2 -12x +19) = (a+1)x^3 + (b-3)x^2 + (c-12)x + (d + 19) = 0}}} .

By the factor theorem, 
after substituting -3 into the preceding polynomial and simplifying, we must have
27a - 5b + 5c - d = 1;
after substituting 2 into the preceding polynomial and simplifying, we must have
8a +4b + 2c + d = 9;

Adding both equations, we get 7a - b + c = 2, or b = 7a + c - 2.
Substituting this into 8a +4b + 2c + d = 9, we get d = -36a - 6c + 17.

The values for a and c are free to vary.
As an example, let a = c = 1. ==> b = 7*1 + 1 - 2 = 6, and 
d = -36*1 - 6*1 + 17 = -25
Then {{{x^3 + 6x^2 + x -25}}}, when added to {{{x^3 - 3x^2 -12x +19}}}, will be divisible by {{{x^2 + x -6}}}.