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For how many value(s) of b does the equation $x^3+b^2x^2+2x+3=0$ have integral solutions?

This equation can either have one or three real (not necessarily integral) solutions. Using Descarte's rule of signs I found that it has either one or three negative roots. I also know this functions is increasing so I am assuming it only has one negative real root.

However, how do I relate this to b?

user7090
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I assume $b$ is supposed to be an integer?

The easiest approach to this problem is probably to use the usual method for finding the rational roots to an integer polynomial -- the rational root theorem. All of the rational roots are of the form $p/q$ where $p$ divides the constant term and $q$ divides the leading coefficient. (of course, you only need the integer roots... but all of the candidate roots the rational root theorem suggests are already integers)

  • That $b$ is an integer is never explicitly stated. With the assumption that b is an integer there is only one possible value being $b=0$. If $b$ needs not be an integer there are three values for b namely $b=0$,or $b=\pm\frac{\sqrt{30}}{3}$. Using the rational root theorem I found $x=-3$ and $x=-1$ to be the two possible negative roots. – user7090 Oct 04 '13 at 01:23
  • Note that if $b$ is not an integer, the rational root theorem doesn't apply. e.g. if you plug in $x=-4$, you can still solve for $b$. Maybe there's something interesting to be said if $b$ is a rational number: you'd still have to clear denominators in order to apply the rational root theorem, though. –  Oct 04 '13 at 01:35