5x² + kx - 6 

5 can only be factored as 5*1, so every possible
factorization must be of this form:

(5x    )(x    )

Next all the ways to factor 6 are

6*1, 3*2, 2*3 and 1*6

so any possible factorization is in one of
these four forms:

(5x   6)(x   1)
(5x   3)(x   2)
(5x   2)(x   3)
(5x   1)(x   6)

Since the last term of 5x² + kx - 6 is negative 
the two signs must be opposite. So we have these
8 possible factorizations, which we multiply out 
to find k in each case:

(5x + 6)(x - 1)  =  5x² +   x - 6  so k =   1
(5x + 3)(x - 2)  =  5x² -  7x - 6  so k =  -7  
(5x + 2)(x - 3)  =  5x² - 13x - 6  so k = -13
(5x + 1)(x - 6)  =  5x² - 29x - 6  so k = -29
(5x - 6)(x + 1)  =  5x² -   x - 6  so k =  -1
(5x - 3)(x + 2)  =  5x² +  7x - 6  so k =   7
(5x - 2)(x + 3)  =  5x² + 13x - 6  so k =  13
(5x - 1)(x + 6)  =  5x² + 29x - 6  so k =  29

So all possible integer values of k are ±1, ±7, ±13, and ±29.

Edwin