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 \(b\)?

Guest Jul 9, 2018

You could try developing the expression \(a(x+m)^2+n=ax^2+2amx+am^2+n\)  and, next, identifying coefficients.





If \(n\geqslant0\), as \(a=3\), the last equation would have limited solutions. You should choose the one with the largest \(m\) (that produces the largest \(b=2am\)), so that you can guess the largest divisor of \(b\) that will be equal to \(b\). I get \(m=1\) and \(b=2am=2·3·1=6\), so the largest divisor of \(b\) is \(6\).


If it's possible \(n<0\), there would not be limitation for the value of \(m\) and, in consequence, we could define any \(b\) (and its largest divisor) as large as we wanted just fixing any desired value of \(m\) (that would always have an associated value of \(n=10-3m^2\)). 

Guest Jul 9, 2018

15 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.