Find all integers x for which $ |x^3 + 6x^2 + 2x − 6| $ is prime. I think that this means that  $|x^3 + 6x^2 + 2^x − 6| \equiv 0 \bmod p$. Ignoring the absolute value first, I get $
 x(x^2+2)+6(x^2−1) \equiv 0 \mod p \Rightarrow x(x^2+2) \equiv -6(x^2−1) \bmod p$. But I'm not sure how to proceed. 

Following your thoughts of grouping the terms into:
$$x(x^2+2)+6(x^2−1)$$
consider the following:


*

*The second term, $6(x^2-1)$ is always divisible by $6$ and consequently is divisible by $3$.

*For the first term $x(x^2+2)$, it is also divisible by 3. Why? $x$ is either $0,-1,1 \pmod 3$. Hence, $x^2 \equiv 0,1 \pmod 3$. If $x^2 \equiv 0 \pmod 3$, then $3|x^2 \implies 3|x$. On the other hand, if $x^2 \equiv 1 \pmod 3$, then $x^2 + 2 \equiv 0 \pmod 3$.


Now, what can you conclude? 
