The critical information we need are the relative extrema of P(x)
P'(x) = 12x^3-12x^2+24x = 12x*(x^2-x+2). The quadratic factor of x^2-x+2 can be expressed as (x-1/2)^2 + 7/4 which is clearly strictly positive. So there is only one real root of P'(x), which is x=0.
Then since x=0 is the only extremum it must be the global minimum. Then any choice of c<0 will have the minimum below the x-axis and then P(x) will have two real roots.
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A slight change in P(x) makes for a more interesting problem, lets call it Q(x) = 3x^4-4x^3-12x^2+c. For Q(x) we have Q'(0)=12x^3-12x^2-24x=12x*(x-2)*(x+1) which then has three local extrema: x=2, x=0, and x=-1.
If c=0 then the three extrema are two local minimums at (2,-32) and (-1,-5) and a local maximum at (0,0). Any choice of c<0 will have Q(x) with exactly two real roots but we also have the interval between the local minimums corresponding to 5<c<32 where again Q(x) will have two real roots.
Solution