Aurora Borealis!

First, a graph of the function should look like ( this ) so that it has a chance to have 3 distinct roots(ignore the fact that all are integers for now!).

So $\displaystyle f^{ ' }\left( x \right)$ must have 2 distinct roots.(And let ignore this fact for now as well!)

Since $\displaystyle f\left( { x }^{ 2 }+2x+2 \right)$ has no real roots, this means $\displaystyle { x }^{ 2 }+2x+2=m$ , where m is a root of the original polynomial, has no real roots. Then, $\displaystyle m<1$ .(that is, all roots must be less than 1)

Minimising c can be done by putting the leftmost root equal to 0(now you can use the two above facts to help figuring why this is the case!).

So, f=0

Now $\displaystyle f\left( x \right) =x\left( { x }^{ 2 }+ax+b \right)$ and, by noticing that b is the product of roots and -a is the sum of roots, we then know that the minimum values for a and b are 3 and 2, respectively.

So, $\displaystyle p\left( x \right) =x({ x }^{ 2 }+3x+2)$ and $\displaystyle p^{ ' }\left( x \right) =3{ x }^{ 2 }+6x+2)$

$\displaystyle p^{ ' }\left( x \right) =3{ x }^{ 2 }+6x+2)=k$ Now, the discriminant must be 0.

$\displaystyle 36=12(2-k)$ and $k=\boxed{-1}$

Here's some insight as to why C should have a minimum value of 0 Suppose for the sake of brevity, that all the roots (integers)are negative This would imply that the product of the roots can become as small as we want it to and thus a minimum or maximum value of C does not exist( C could tend to very large values as far as the roots are concerned) Same arguments hold for a and B Thus we arrive at a contradiction that there exists no minimum value of the coefficients Hence our assumption of all roots being negative was incorrect Thus we see that as 1 Is not a root of F(x) 0 can be a root which minimises C to 0 And takes care of A and B too Now as A and B are integers themselves We easily see that the other roots (-1,-2)minimise A and B Note that the roots are all distinct so repeated roots being 0 is not an option here :)