<s>[INST] Integral values of $k$ for which polynomial has integral roots. How integral values of $k$ exists such that all roots of the polynomial:
$$f(x)=x^3-(k-3)x^2-11x+(4k-8)$$ are also integers.
Could someone please provide me some direction to proceed in this. First I thought first root would be obtained by hit and trial but it is not the case here. I also tried rewriting $x^3-(k-3)x^2-11x+(4k-8)=0$ as $x^3+3x^2-11x-8=k(x^2-4)$ and solve the question graphically but didn't succeed in that. Please give some direction 
 [/INST] Hint:
Since $$k ={x^3+3x^2-11x-8\over x^2-4}\implies x^2-4 \mid x^3+3x^2-11x-8$$
Since also $$x^2-4\mid (x+3)(x^2-4) = x^3+3x^2 -4x-12$$
we get
$$ x^2-4\mid ( x^3+3x^2-11x-8)-( x^3+3x^2-4x-12)= -7x+4$$
thus $$x-2\mid -7x+4\implies x-2 \mid -7x+4+7(x-2) = -10$$
Finally we have $x-2\in \{-10,-5,-2,-1,1,2,5,10\}$ so $x\in \{-8,-3,0,1,3,4,7,12\}$
Now, of course not all these are good...
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