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The number of prime values of the polynomial $n^3 − 10n^2 − 84n + 840 $ where $n$ is an integer is??

I do not get what they're asking us to do. Is there a specific method to solve this question? I factored the cubic, but idk what to do next.

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    HINT: Put $n=10$ – 2017-01-27
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    after factoring, if we want to find the prime values, why do we let each term = +/- 1????????? – 2017-01-27

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Note that $$n^3-10n^2-84n+840=(n-10)(n^2-84)$$ So, because of the definition of prime numbers, $n-10 = \pm 1$ or $n^2-84= \pm 1$. Since the latter is impossible we have that $n$ is $11$ or $9$. Since for $n=11$, it is $37$, and for $n=9$ the result is $3$, the answer to the question is ${2}$.

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    thank u, i factored and got (x-10)(x^2-84), however i did not get what u did after that. a prime number is a number that can be divided by 1 and itself only. why did u let (n-10( (x^2-84) equal to plus or minus 1?? Why? – 2017-01-27
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    and also why is it impossible to use sqrt 83 and sqrt 85????? – 2017-01-27
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    @exchangehelpforuni Because $\sqrt{83}$, $\sqrt{85}$ ain't integers. – 2017-01-27
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    @exchangehelpforuni In the positive numbers, a prime number $p$ can only be divided by $1$ and $p$. Over the integers, one may wish to include $-1$ and $-p$. If $n^2-84 \neq \pm 1 , n-10 \neq \pm 1$, then we have that there exists another integer that divides the prime number. – 2017-01-27