Question 27864
It can only be factored if the discriminant is positive and a perfect square. The discriminant is defined by:
{{{b^2-4ac}}} for {{{ax^2+bx+c}}} 
For your problem {{{x^2 + x - k}}} a=1, b=1 and c=-k
so the descriminant is {{{1+4k}}}
So any number that when multiplied by 4 is one less than a square number would work. Such as:
0 --> 1+4*0=1 --> {{{x^2+x=(x+1)(x)}}}
2 --> 1+4*2=9 --> {{{x^2+x-2=(x+2)(x-1)}}}
6 --> 1+4*6=25 --> {{{x^2+x-6=(x+3)(x-2)}}}
12 --> 1+4*12=49 --> {{{x^2+x-12=(x+4)(x-3)}}}
20 --> 1+4*20=81 --> {{{x^2+x-20=(x+5)(x-4)}}}
30 --> 1+4*30=121 --> {{{x^2+x-30=(x+6)(x-5)}}}
42 --> 1+4*42=169 --> {{{x^2+x-42=(x+7)(x-6)}}}
Starting to see some paterns, no doubt?...
So there's an infinite number of k's that'll work. To find the next value in the pattern, square each odd number, subtract one, and divide by 4.
{{{(1^2-1)/4=0}}}
{{{(3^2-1)/4=2}}}
{{{(5^2-1)/4=6}}}
{{{(7^2-1)/4=12}}}
{{{(9^2-1)/4=20}}}
{{{(11^2-1)/4=30}}}
{{{(13^2-1)/4=42}}}
{{{(15^2-1)/4=56}}}
{{{(17^2-1)/4=72}}}
Answer your question?