Math Help ASAP!

$\ \phantom{=\qquad}3x^2+bx+10\\~\\ =\qquad3(x^2+\frac{b}{3}x)+10\\~\\ =\qquad3(x^2+\frac{b}{3}x+(\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3((x+\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3(x+\frac{b}{6})^2-3(\frac{b}{6})^2+10$

Now it is in the form  a(x + m)² + n   where

$a=3\qquad \text{and}\qquad m=\frac{b}{6}\qquad \text{and}\qquad n=-3(\frac{b}{6})^2+10$

In order for  m  to be an integer,  b  must be a multiple of  6

In order for  n  to be an integer,  b  must be a multiple of  6

The largest integer that must be a divisor of  b  in order for both  m  and  n  to be integers is  6