Let the roots be a, b, c then F(x)=2(x-a)(x-b)(x-c)=0, and a+b+c=6.
F(x)=2(x2-(a+b)x+ab)(x-c)=2(x3-(a+b)x2+abx-cx2+c(a+b)-abc)=
2(x3-(a+b+c)x2+(ab+ac+bc)x-abc)=2(x3-6x2+(ab+ac+bc)x-abc)=
2x3-12x2+2(ab+ac+bc)x-2abc=2x3-3kx2+...
Therefore -12=-3k, 3k=12, k=4.