[tex](x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2[/tex]
So the trinomial [tex]x^2+kx-35[/tex] can be factored as long as
[tex]\begin{cases}r_1+r_2=-k\\r_1r_2=-35\end{cases}[/tex]
has integer solutions for [tex]r_1,r_2[/tex]. Clearly, both have to be factors of -35, which leaves only a handful of cases:
[tex](r_1,r_2)=(1,-35)\implies r_1+r_2=-34[/tex]
[tex](r_1,r_2)=(-1,35)\implies r_1+r_2=34[/tex]
[tex](r_1,r_2)=(5,-7)\implies r_1+r_2=-2[/tex]
[tex](r_1,r_2)=(-5,7)\implies r_1+r_2=2[/tex]
So the possible values of [tex]k[/tex] are [tex]\pm34[/tex] and [tex]\pm2[/tex].
