Posted by Raj on Tuesday, June 12, 2012 at 11:15pm.
let's look at the discriminant.
if b^2 - 4ac is a perfect square, then it can be factored over the rationals, so we start with that.
I will do the 2nd, since it has smaller numbers
for 3x^2 - 16x + k
b^2 - 4ac
= 256-12k
= 4(64-3k)
remember we have to take the square root of that
√(4(64-3k))
= 2√(64-3k)
For 64-3k to be a perfect square
we need 3k to be 0,15,28,39,48,55,60 or 63
of those only 0,15,39,48,60, and 63 are multiples of 3
So for rationals, k could be 0, 5, 13, 16, 20 or 21
testing:
let's try k = 13
3x^2 - 16x + 13
x = (-16 ± √100)/6
= -1 or 13/3
so 3x^2 - 16x + 13 = (x+1)(3x-13)
k = 0, 5, 13, 16, 20, or 21