For which integral values of can 4x^2 + kx + 3 be factored over the integers?

If you have a shape ax^2 + bx + c, it can be rewritten as (px + q)(kx + m) with pk = a, qm = c, and pm + qk = b

Since our shape is 4x^2 + bx + 3, we have pk = 4 and qm = 3. Without accounting for any permutations since they will be equivalent, we have (p, k, q, m) = (2, 2, 3, 1) as a solution or even (4, 1, 3, 1)

Let's look at (p, k, q, m) = (2, 2, 3, 1)

pm + qk = b
2 * 3 + 2 * 1 = b, 8 = b is a solution and your factorization is
4x^2 + 8x + 3 = (2x + 3)(2x + 1)

Let's look at (p, k, q, m) = (4, 1, 3, 1)

pm + qk = b
4 * 1 + 3 * 1 = b
7 = b is a solution

and your factorization is (4x + 3)(x + 1)

So your solution set is b E {7, 8}

In order for the function to be factored in integers its discriminant must be a perfect square.  That is:
k^2 - 4(4)(3) = k^2 - 48 = n^2
where n is an integer.
Or you could rearrange and say (k + n)(k - n) = 48
This is true for
k = ± 7, ± 8, and ± 13
Checking:
 4x^2 + 7x + 3 = (x + 1) (4x + 3)
 4x^2 - 7x + 3 = (x - 1) (4x - 3)
 4x^2 + 8x + 3 = (2x + 1) (2x + 3)
 4x^2 - 8x + 3 = (2x - 1) (2x - 3)
 4x^2 + 13x + 3 = (x + 3) (4x + 1)
 4x^2 - 13x + 3 = (x - 3) (4x - 1)