Question 799653
Many details to the process, but you would want to do synthetic division using "divisors" of x+2, and of x-2, each separately.  The roots correspondingly being checked are -2 and +2.  


The dividend would be one of the factors of {{{ax^3+bx^2+x+6}}}, once a has been factored.  This dividend is {{{x^3+(b/a)x^2+x/a+6/a}}}.  During the division, the intermediary expressions become more complicated...


The root, -2 gives remainder {{{(4+4b-8a)/a=4}}}, equated according to the description of the problem; and the root +2 gives remainder {{{(8+4b+8a)/a=4}}}.
These seem to be two equations in the unknowns, a and b.  


Finishing the solution for a and b from here should be no trouble, seems to appear straightforward.