Respuesta :

irspow
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}

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