If the quadratic $3x^2+bx+10$ can be written in the form $a(x+m)^2+n$, where $m$ and $n$ are integers, what is the largest integer that must be a divisor of $b$?
+9466 \(\ \phantom{=\qquad}3x^2+bx+10\\~\\ =\qquad3(x^2+\frac{b}{3}x)+10\\~\\ =\qquad3(x^2+\frac{b}{3}x+(\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3((x+\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3(x+\frac{b}{6})^2-3(\frac{b}{6})^2+10\)
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
