By the fundamental theorem of algebra, we can write
[tex]x^2+bx+10=(x-r_1)(x-r_2)[/tex]
Expanding the right hand side, we get
[tex]x^2+bx+10=x^2-(r_1+r_2)x+r_1r_2\implies\begin{cases}r_1+r_2=-b\\r_1r_2=10\end{cases}[/tex]
We want [tex]b[/tex] to be an integer, which means [tex]r_1,r_2[/tex] must also be integers. This means [tex]r_1,r_2[/tex] must be factors of 10. There are several possibilities:
[tex]r_1=\pm10,r_2=\pm1\implies b=-(10+1)=-11\text{ or }b=-(-10-1)=11[/tex]
[tex]r_1=\pm5,r_2=\pm2\implies b=-(5+2)=-7\text{ or }b=-(-5-2)=7[/tex]
So there are 4 possible values for [tex]b[/tex]: -11, -7, 7, and 11.
