All possible rational zeros are {±1,2,3,6,and all fractions formed by a combination of the integers)
By inspection using integers first...
1-7-6=-12 1 is not a root
8-14-6=-12 2 is not a root
27-21-6=0 so 3 is a root so (a-3) is a factor....
(a^3-7a-6)/(a-3)
a^2 rem 3a^2-7a-6
3a rem 2a-6
2 rem 0 so now we know that we have:
(a-3)(a^2+3a+2) and we can factor the second expression...
(a-3)(a^2+a+2a+2)
(a-3)(a(a+1)+2(a+1))
(a-3)(a+2)(a+1)
So there are three values of a that make the equation true....
a={-1, -2, 3}