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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 $b$?

 May 14, 2019
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\(\ \phantom{=\qquad}3x^2+bx+10\\~\\ =\qquad3(x^2+\frac{b}{3}x)+10\\~\\ =\qquad3(x^2+\frac{b}{3}x+(\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3((x+\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3(x+\frac{b}{6})^2-3(\frac{b}{6})^2+10\)

 

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

 May 14, 2019

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